Chapter 11 Perimeter and Area (Additional Questions)

Given:

Side length of the square = $5$ cm

To Find:

The perimeter of the square.

Solution:

The formula for the perimeter of a square is given by:

Perimeter $= 4 \times \text{side length}$

Substitute the given side length into the formula:

Perimeter $= 4 \times 5$ cm

Perimeter $= 20$ cm

The perimeter of the square is $20$ cm.

Comparing this with the given options, we find that option (B) is correct.

The final answer is $\boxed{\text{20 cm}}$.

Given:

Length of the rectangle ($l$) $= 8$ cm

Width of the rectangle ($w$) $= 4$ cm

To Find:

The area of the rectangle.

Solution:

The formula for the area of a rectangle is given by:

Area $= \text{Length} \times \text{Width}$

Area $= l \times w$

Substitute the given values into the formula:

Area $= 8 \text{ cm} \times 4 \text{ cm}$

Area $= 32 \text{ cm}^2$

The area of the rectangle is $32 \text{ cm}^2$.

Comparing this with the given options, we find that option (D) is correct.

The final answer is $\boxed{\text{32 cm}^2}$.

Given:

Length of the rectangle $= l$

Width of the rectangle $= w$} $

To Find:

The formula for the perimeter of the rectangle.

Solution:

The perimeter of a rectangle is the total distance around its boundary.

A rectangle has four sides, with opposite sides being equal in length.

The sides are of lengths $l, w, l, w$.

To find the perimeter, we add the lengths of all four sides:

Perimeter $= l + w + l + w$

Combining like terms:

Perimeter $= (l + l) + (w + w)$

Perimeter $= 2l + 2w$

We can factor out the common factor of 2:

Perimeter $= 2(l + w)$

The formula for the perimeter of a rectangle with length $l$ and width $w$ is $2(l+w)$.

Comparing this with the given options, we find that option (B) is correct.

Option (A) $l \times w$ is the formula for the area of a rectangle.

Option (C) $l+w$ is half the perimeter.

Option (D) $4l$ is the formula for the perimeter of a square with side length $l$.

The final answer is $\boxed{2(l+w)}$.

Given:

Area of the square $= 64\text{ cm}^2$

To Find:

The length of the side of the square.

Solution:

Let the side length of the square be $s$ cm.

The formula for the area of a square is given by:

Area $= \text{side} \times \text{side} = s^2$

We are given that the area is $64\text{ cm}^2$. So, we have:

$s^2 = 64$

To find the value of $s$, we take the square root of both sides of the equation:

$s = \sqrt{64}$

$s = 8$

Since the side length must be a positive value, we take the positive square root.

Therefore, the length of the side of the square is $8$ cm.

The length of the side of the square is $8$ cm.

Comparing this with the given options, we find that option (B) is correct.

The final answer is $\boxed{\text{8 cm}}$.

Given:

Base of the parallelogram $= b$

Height of the parallelogram $= h$

To Find:

The formula for the area of the parallelogram.

Solution:

The area of a parallelogram is found by multiplying its base by its corresponding height.

The formula for the area of a parallelogram with base $b$ and height $h$ is:

Area $= \text{base} \times \text{height}$

Area $= b \times h$

The formula for the area of a parallelogram is $b \times h$.

Comparing this with the given options, we find that option (B) is correct.

Option (A) is the formula for the area of a triangle.

Option (C) is the sum of the base and height.

Option (D) is related to the perimeter if $b$ and $h$ were adjacent sides (which they are not necessarily for perimeter calculation).

The final answer is $\boxed{b \times h}$.

Given:

Base of the parallelogram ($b$) $= 10$ cm

Height of the parallelogram ($h$) $= 6$ cm

To Find:

The area of the parallelogram.

Solution:

The formula for the area of a parallelogram is given by:

Area $= \text{base} \times \text{height}$

Area $= b \times h$

Substitute the given values into the formula:

Area $= 10 \text{ cm} \times 6 \text{ cm}$

Area $= 60 \text{ cm}^2$

The area of the parallelogram is $60 \text{ cm}^2$.

Comparing this with the given options, we find that option (C) is correct.

The final answer is $\boxed{\text{60 cm}^2}$.

Given:

Base of the triangle ($b$) $= 12$ cm

Height of the triangle ($h$) $= 5$ cm

To Find:

The area of the triangle.

Solution:

The formula for the area of a triangle is given by:

Area $= \frac{1}{2} \times \text{base} \times \text{height}$

Area $= \frac{1}{2} \times b \times h$

Substitute the given values into the formula:

Area $= \frac{1}{2} \times 12 \text{ cm} \times 5 \text{ cm}$

Area $= \frac{1}{2} \times 60 \text{ cm}^2$

Area $= 30 \text{ cm}^2$

The area of the triangle is $30 \text{ cm}^2$.

Comparing this with the given options, we find that option (B) is correct.

The final answer is $\boxed{\text{30 cm}^2}$.

Given:

The question asks for the name of the distance around a circle.

To Find:

The correct term for the distance around a circle from the given options.

Solution:

Let's define each term given in the options:

(A) Area: The area of a circle is the measure of the space enclosed within the boundary of the circle. It is calculated using the formula $\text{Area} = \pi r^2$, where $r$ is the radius.

(B) Diameter: The diameter of a circle is a line segment that passes through the center of the circle and has its endpoints on the circle. It is the longest chord of the circle. The diameter is equal to twice the radius ($d = 2r$).

(C) Radius: The radius of a circle is a line segment from the center of the circle to any point on the circle. It is half the length of the diameter ($r = d/2$).

(D) Circumference: The circumference of a circle is the distance around its outer boundary. It is equivalent to the perimeter of other shapes. The formula for the circumference is $\text{Circumference} = 2\pi r$ or $\text{Circumference} = \pi d$, where $r$ is the radius and $d$ is the diameter.

Based on these definitions, the distance around a circle is called its circumference.

Comparing this with the given options, we find that option (D) is correct.

The final answer is $\boxed{\text{Circumference}}$.

Given:

Radius of the circle = $r$

Relationship between diameter and radius: $d = 2r$

To Find:

The formula for the circumference of the circle.

Solution:

The circumference of a circle is the distance around its boundary.

The formula for the circumference of a circle in terms of its radius $r$ is:

Circumference $= 2\pi r$

The diameter $d$ of a circle is related to its radius $r$ by the equation $d = 2r$.

We can substitute $2r$ with $d$ in the circumference formula:

Circumference $= \pi \times (2r)$

Since $2r = d$, we have:

Circumference $= \pi \times d = \pi d$

Therefore, the circumference of a circle can be expressed as either $2\pi r$ or $\pi d$, where $d$ is the diameter and $d=2r$.} $

Let's examine the given options:

(A) $\pi r^2$: This is the formula for the area of a circle, not the circumference.

(B) $2\pi r$: This is a valid formula for the circumference in terms of the radius.

(C) $\pi d$: This is a valid formula for the circumference in terms of the diameter.

(D) Both (B) and (C) (where $d=2r$): Since $2\pi r = \pi(2r) = \pi d$, both formulas are equivalent ways to calculate the circumference. This option correctly identifies that both expressions represent the circumference given the relationship between $d$ and $r$.

The final answer is $\boxed{\text{Both (B) and (C) (where } d=2r \text{)}}$.

Given:

Radius of the circle ($r$) $= 7$ cm

To Find:

The area of the circle.

Solution:

The formula for the area of a circle is given by:

Area $= \pi r^2$

We use the value of $\pi = \frac{22}{7}$ as the radius is a multiple of 7.

Substitute the given radius into the formula:

Area $= \frac{22}{7} \times (7 \text{ cm})^2$

Area $= \frac{22}{7} \times (7 \text{ cm} \times 7 \text{ cm})$

Area $= \frac{22}{\cancel{7}} \times \cancel{7} \text{ cm} \times 7 \text{ cm}$

Area $= 22 \times 7 \text{ cm}^2$

Area $= 154 \text{ cm}^2$

The area of the circle is $154 \text{ cm}^2$.

Comparing this with the given options, we find that option (B) is correct.

The final answer is $\boxed{\text{154 cm}^2}$.

Given:

Perimeter of the rectangle ($P$) $= 40$ cm

Length of the rectangle ($l$) $= 12$ cm

To Find:

The width of the rectangle ($w$).

Solution:

The formula for the perimeter of a rectangle is given by:

$P = 2(l + w)$

Substitute the given values of perimeter ($P$) and length ($l$) into the formula:

$40 = 2(12 + w)$

Divide both sides of the equation by 2:

$\frac{40}{2} = 12 + w$

$20 = 12 + w$}

To find $w$, subtract 12 from both sides of the equation:

$w = 20 - 12$

$w = 8$}

The width of the rectangle is $8$ cm.

The width of the rectangle is $8$ cm.

Comparing this with the given options, we find that option (A) is correct.

The final answer is $\boxed{\text{8 cm}}$.

Given:

The area formula $\frac{1}{2} \times (\text{product of diagonals})$

To Find:

The shape among the given options that has this area formula.

Solution:

Let's consider the area formulas for each of the given shapes:

(A) Square: A square is a special type of rhombus where all angles are $90^\circ$. The area of a square with side $s$ is $s^2$. The diagonals of a square are equal in length, say $d$, and are perpendicular bisectors of each other. The area can also be calculated as $\frac{1}{2} \times d \times d = \frac{1}{2} d^2$. Since it's a type of rhombus, this formula applies.

(B) Rectangle: The area of a rectangle with length $l$ and width $w$ is $l \times w$. The diagonals are equal but are not necessarily perpendicular unless it's a square.

(C) Rhombus: A rhombus is a quadrilateral where all four sides are equal in length. The diagonals of a rhombus are perpendicular bisectors of each other. Let the lengths of the diagonals be $d_1$ and $d_2$. The area of a rhombus is given by the formula:

Area $= \frac{1}{2} \times d_1 \times d_2$

This matches the given formula $\frac{1}{2} \times (\text{product of diagonals})$.

(D) Parallelogram: The area of a parallelogram with base $b$ and corresponding height $h$ is $b \times h$. The diagonals of a parallelogram are not necessarily perpendicular or equal in length.

The formula $\frac{1}{2} \times (\text{product of diagonals})$ is the standard formula for the area of a rhombus. While it also applies to a square (which is a specific type of rhombus), the rhombus is the general shape described by this formula among the options.

Comparing this with the given options, we find that option (C) is correct.

The final answer is $\boxed{\text{Rhombus}}$.

Given:

Circumference of the circle ($C$) $= 88$ cm

Value of $\pi = \frac{22}{7}$

To Find:

The radius of the circle ($r$).

Solution:

The formula for the circumference of a circle is given by:

$C = 2\pi r$

Substitute the given values into the formula:

$88 = 2 \times \frac{22}{7} \times r$

Simplify the right side of the equation:

$88 = \frac{44}{7} \times r$

To find $r$, multiply both sides by $\frac{7}{44}$:

$r = 88 \times \frac{7}{44}$} $

Cancel out the common factor:

$r = \cancel{88}^{2} \times \frac{7}{\cancel{44}_{1}}$

$r = 2 \times 7$

$r = 14$}

The radius of the circle is $14$ cm.

The radius of the circle is $14$ cm.

Comparing this with the given options, we find that option (B) is correct.

The final answer is $\boxed{\text{14 cm}}$.

Given:

Area of the triangle ($A$) $= 40\text{ cm}^2$

Base of the triangle ($b$) $= 10$ cm

To Find:

The corresponding height of the triangle ($h$).

Solution:

The formula for the area of a triangle is given by:

$A = \frac{1}{2} \times \text{base} \times \text{height}$

$A = \frac{1}{2} \times b \times h$

Substitute the given values of area ($A$) and base ($b$) into the formula:

$40 = \frac{1}{2} \times 10 \times h$

Simplify the right side of the equation:

$40 = 5 \times h$

To find $h$, divide both sides by 5:

$h = \frac{40}{5}$

$h = 8$}

The corresponding height of the triangle is $8$ cm.

The corresponding height of the triangle is $8$ cm.

Comparing this with the given options, we find that option (B) is correct.

The final answer is $\boxed{\text{8 cm}}$.

We need to match each geometric shape from the first column with its corresponding area formula from the second column.

(i) Square: The area of a square is found by multiplying the length of one side by itself. The formula is side $\times$ side. This matches option (c).

Match: (i) - (c)

(ii) Rectangle: The area of a rectangle is found by multiplying its length by its width. The formula is length $\times$ width. This matches option (d).

Match: (ii) - (d)

(iii) Parallelogram: The area of a parallelogram is found by multiplying the length of its base by its corresponding height. The formula is base $\times$ height. This matches option (b).

Match: (iii) - (b)

(iv) Circle: The area of a circle is found using the formula $\pi$ times the square of its radius ($r$). The formula is $\pi r^2$. This matches option (a).

Match: (iv) - (a)

Putting the matches together, we get:

(i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)

Now, we compare this result with the given options.

Option (A) is (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a).

Option (B) is (i)-(c), (ii)-(b), (iii)-(d), (iv)-(a).

Option (C) is (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a).

Option (D) is (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

The correct matching is given in option (A).

The correct option is (A) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a).

Let the side length of a square be $s$.

The formula for the perimeter of a square is given by:

$P = 4 \times s$

Let's evaluate the Assertion (A): Doubling the side length of a square doubles its perimeter.

If the side length is $s$, the perimeter is $P = 4s$.

If we double the side length, the new side length becomes $s' = 2s$.

The new perimeter, $P'$, is given by:

$P' = 4 \times s'$

$P' = 4 \times (2s)$

$P' = 8s$

Since $P = 4s$, we can write $P' = 2 \times (4s) = 2P$.

Thus, doubling the side length results in doubling the perimeter.

So, Assertion (A) is true.

Now, let's evaluate the Reason (R): Perimeter of a square is $4 \times$ side length, so it is directly proportional to the side length.

The formula for the perimeter of a square is indeed $P = 4 \times s$.

In a relationship of the form $y = kx$, where $k$ is a constant, $y$ is directly proportional to $x$.

In the formula $P = 4s$, the perimeter $P$ is related to the side length $s$ by a constant factor of 4. This indicates that $P$ is directly proportional to $s$.

So, Reason (R) is true.

Finally, let's determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that doubling the side doubles the perimeter. This fact arises directly from the property of direct proportionality stated in Reason (R).

If $P \propto s$, then for any constant $k$, $P = ks$. If $s$ is replaced by $2s$, the new perimeter $P'$ is $k(2s) = 2(ks) = 2P$.

Since the perimeter of a square is directly proportional to its side length ($P=4s$), doubling the side length will indeed double the perimeter.

Therefore, Reason (R) correctly explains Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

This matches option (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

Let the original radius of the circle be $r$.

The original area of the circle, denoted by $A$, is given by the formula:

$A = \pi r^2$

Now, let's consider the Assertion (A): If the radius of a circle is doubled, its area becomes four times the original area.

If the radius is doubled, the new radius, $r'$, is $r' = 2r$.

The new area of the circle, denoted by $A'$, is given by:

$A' = \pi (r')^2$

Substitute $r' = 2r$ into the formula:

$A' = \pi (2r)^2$

$A' = \pi (4r^2)$

$A' = 4 \pi r^2$

Since the original area $A = \pi r^2$, we can write the new area as:

$A' = 4A$

This shows that when the radius is doubled, the area becomes four times the original area.

Thus, Assertion (A) is true.

Next, let's consider the Reason (R): Area of a circle is $\pi r^2$, so it is proportional to the square of the radius.

The formula for the area of a circle is $A = \pi r^2$.

This formula shows that the area $A$ is directly proportional to the square of the radius $r$ (i.e., $r^2$). The constant of proportionality is $\pi$. This means that $A \propto r^2$.

So, Reason (R) is true.

Finally, we need to determine if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that doubling the radius quadruples the area. Reason (R) states that the area is proportional to the square of the radius ($A \propto r^2$).

If a quantity is proportional to the square of another quantity (say, $y \propto x^2$), then if $x$ is multiplied by a factor $k$, $y$ will be multiplied by $k^2$.

In this case, the area ($A$) is proportional to the square of the radius ($r^2$). If the radius ($r$) is multiplied by a factor of 2 (doubled), the area ($A$) will be multiplied by a factor of $2^2 = 4$.

Therefore, the fact that the area of a circle is proportional to the square of its radius directly explains why doubling the radius results in four times the area.

Reason (R) is the correct explanation for Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

This matches option (A).

The correct option is (A) Both A and R are true, and R is the correct explanation of A.

The question provides the diameter of the circular park.

Given: Diameter of the circular park = 28 meters.

We need to find the radius of the circular park.

The relationship between the diameter ($D$) and the radius ($r$) of a circle is that the radius is half of the diameter.

$r = \frac{D}{2}$

Substitute the given diameter value into the formula:

$r = \frac{28\text{ m}}{2}$

$r = 14\text{ m}$

The radius of the circular park is 14 meters.

Now we compare this result with the given options:

(A) 28 m

(B) 14 m

(C) 56 m

(D) 7 m

The calculated radius matches option (B).

The correct option is (B) 14 m.

From the case study in Question 18:

Diameter of the circular park = 28 meters.

The radius of the circular park (inner radius) is half of the diameter.

Inner radius = $\frac{28\text{ m}}{2} = 14\text{ m}$.

A path of width 1.4 m is built around the park on the outside.

This means the path adds 1.4 m to the radius of the park.

The radius of the circular park including the path is the sum of the inner radius (radius of the park) and the width of the path.

Radius including path = Inner radius + Width of path

Radius including path = $14\text{ m} + 1.4\text{ m}$

Radius including path = $15.4\text{ m}$

The radius of the circular park including the path is 15.4 meters.

Now we compare this result with the given options:

(A) 14 m

(B) 15.4 m

(C) 28 m

(D) 29.4 m

The calculated radius matches option (B).

The correct option is (B) 15.4 m.

We are given the diameter of a circular park and the width of a path built around it on the outside.

Given: Diameter of the circular park = 28 meters.

Given: Width of the path = 1.4 meters.

We need to find the area of the path using $\pi = \frac{22}{7}$.

First, find the radius of the circular park (this is the inner radius of the circular area including the path).

Let the inner radius be $r_1$.

$\begin{array}{rcl} \text{Inner radius } (r_1) & = & \frac{\text{Diameter}}{2} \\ & = & \frac{28\text{ m}}{2} \\ & = & 14\text{ m} \end{array}$

Next, find the radius of the circular park including the path (this is the outer radius).

Let the outer radius be $r_2$.

The path is built on the outside, so the outer radius is the inner radius plus the width of the path.

$\begin{array}{rcl} \text{Outer radius } (r_2) & = & r_1 + \text{Width of path} \\ & = & 14\text{ m} + 1.4\text{ m} \\ & = & 15.4\text{ m} \end{array}$

The path is the region between the outer circle and the inner circle (the park). This region is called an annulus.

The area of the path is the difference between the area of the outer circle and the area of the inner circle.

Area of inner circle = $\pi r_1^2$

Area of outer circle = $\pi r_2^2$

Area of path = Area of outer circle - Area of inner circle

$A_{path} = \pi r_2^2 - \pi r_1^2$

$A_{path} = \pi (r_2^2 - r_1^2)$

We can calculate $r_1^2$ and $r_2^2$ first.

$r_1^2 = (14)^2 = 196$

$r_2^2 = (15.4)^2 = 237.16$

$r_2^2 - r_1^2 = 237.16 - 196 = 41.16$

Now, substitute the values into the formula for the area of the path, using $\pi = \frac{22}{7}$.

$A_{path} = \frac{22}{7} \times (r_2^2 - r_1^2)$

$A_{path} = \frac{22}{7} \times 41.16$

To simplify the calculation, divide 41.16 by 7:

$\frac{41.16}{7} = 5.88$

Now, multiply the result by 22:

$A_{path} = 22 \times 5.88$

$\begin{array}{c} & & 5 & . & 8 & 8 \\ \times & & & 2 & 2 \\ \hline & 1 & 1 & . & 7 & 6 \\ 1 & 1 & 7 & . & 6 & \times \\ \hline 1 & 2 & 9 & . & 3 & 6 \\ \hline \end{array}$

$A_{path} = 129.36 \text{ m}^2$

The calculated area of the path is $129.36\text{ m}^2$. Now, we compare this result with the given options:

(A) $616\text{ m}^2$

(B) $748\text{ m}^2$

(C) $132\text{ m}^2$

(D) $722.24\text{ m}^2$

The calculated value $129.36\text{ m}^2$ is closest to option (C) $132\text{ m}^2$.

The correct option is (C) $132\text{ m}^2$ (This is the closest option to the calculated value based on the given numbers).

Area is a measure of the two-dimensional space a surface occupies.

The unit of area is derived from the unit of length. Since area is a two-dimensional measurement (like length $\times$ width or side $\times$ side), its unit is the square of a unit of length.

Let's examine the given options:

(A) meters: This is a unit of length or distance in the International System of Units (SI).

(B) kilograms: This is a unit of mass in the SI system.

(C) square meters: This is a unit formed by squaring the unit of length (meter). It represents the area of a square with sides 1 meter long. This is a standard unit for measuring area.

(D) cubic centimeters: This is a unit formed by cubing the unit of length (centimeter). It represents the volume of a cube with sides 1 centimeter long. This is a standard unit for measuring volume.

Based on the units, square meters is the unit used to measure area.

The correct option is (C) square meters.

Given:

Area of the parallelogram = $100\text{ cm}^2$

Base of the parallelogram = 20 cm

To Find:

The corresponding height of the parallelogram.

Solution:

The formula for the area of a parallelogram is given by:

Area = base $\times$ height

We are given the Area ($100\text{ cm}^2$) and the base ($b = 20\text{ cm}$). We need to find the height ($h$).

Substitute the given values into the formula:

$100 = 20 \times h$

To find $h$, divide both sides of the equation by 20:

$\frac{100}{20} = h$

$h = \frac{100}{20}$

$h = 5$

The height is in the same unit of length as the base, which is centimeters.

So, the corresponding height of the parallelogram is 5 cm.

Now, compare this result with the given options:

(A) 5 cm

(B) 10 cm

(C) 20 cm

(D) 50 cm

The calculated height matches option (A).

The correct option is (A) 5 cm.

Given:

Perimeter of the square field = 60 meters.

To Find:

The area of the square field.

Solution:

Let the side length of the square field be $s$ meters.

The formula for the perimeter of a square is:

Perimeter = $4 \times$ side length

We are given that the perimeter is 60 meters. Substitute this value into the formula:

$60 = 4s$

To find the side length $s$, divide both sides of the equation by 4:

$s = \frac{60}{4}$

$s = 15$

So, the side length of the square field is 15 meters.

Now, we need to find the area of the square field.

The formula for the area of a square is:

Area = side length $\times$ side length

Substitute the side length $s = 15$ meters into the formula:

$A = (15\text{ m})^2$

$A = 15 \times 15 \text{ m}^2$

$A = 225 \text{ m}^2$

The area of the square field is $225\text{ m}^2$.

Now, compare this result with the given options:

(A) $15\text{ m}^2$

(B) $30\text{ m}^2$

(C) $225\text{ m}^2$

(D) $900\text{ m}^2$

The calculated area matches option (C).

The correct option is (C) $225\text{ m}^2$.

Let the original radius of the circle be $r_{orig}$.

The formula for the circumference of a circle is given by:

$C = 2\pi r$

So, the original circumference is:

$C_{orig} = 2\pi r_{orig}$

Now, the radius of the circle is halved.

The new radius, $r_{new}$, is:

$r_{new} = \frac{r_{orig}}{2}$

Let the new circumference be $C_{new}$. Using the formula for circumference with the new radius:

$C_{new} = 2\pi r_{new}$

Substitute the expression for $r_{new}$:

$C_{new} = 2\pi \left(\frac{r_{orig}}{2}\right)$

Simplify the expression for $C_{new}$:

$C_{new} = \cancel{2}\pi \frac{r_{orig}}{\cancel{2}}$

$C_{new} = \pi r_{orig}$

Now, compare $C_{new}$ with $C_{orig}$.

We know $C_{orig} = 2\pi r_{orig}$.

We can rewrite $C_{new}$ in terms of $C_{orig}$:

$C_{new} = \pi r_{orig} = \frac{1}{2} (2\pi r_{orig})$

$C_{new} = \frac{1}{2} C_{orig}$

This shows that the new circumference is half of the original circumference.

Therefore, if the radius of a circle is halved, its circumference will be halved.

Now, compare this result with the given options:

(A) Halved

(B) Doubled

(C) Quartered

(D) The same

The result matches option (A).

The correct option is (A) Halved.

Given:

Original shape: Rectangle with length $l = 10\text{ cm}$ and width $w = 8\text{ cm}$.

New shape: Square.

To Find:

The side length of the square.

Solution:

When a wire is bent from one shape to another, its total length remains the same. The total length of the wire is equal to the perimeter of the shape it forms.

So, the perimeter of the rectangle is equal to the perimeter of the square.

First, calculate the perimeter of the rectangle.

The formula for the perimeter of a rectangle is:

Perimeter = $2 \times (\text{length} + \text{width})$

$P_{rectangle} = 2(l + w)$

Substitute the given values for length and width:

$P_{rectangle} = 2(10\text{ cm} + 8\text{ cm})$

$P_{rectangle} = 2(18\text{ cm})$

$P_{rectangle} = 36\text{ cm}$

The length of the wire is 36 cm.

Now, the wire is bent to form a square. Let the side length of the square be $s$.

The formula for the perimeter of a square is:

Perimeter = $4 \times$ side length

$P_{square} = 4s$

Since the perimeter of the square is equal to the length of the wire (perimeter of the rectangle):

$P_{square} = P_{rectangle}$

$4s = 36\text{ cm}$

To find the side length $s$, divide both sides of the equation by 4:

$s = \frac{36\text{ cm}}{4}$

$s = 9\text{ cm}$

The side length of the square is 9 cm.

Now, compare this result with the given options:

(A) 9 cm

(B) 18 cm

(C) 36 cm

(D) 80 cm

The calculated side length matches option (A).

The correct option is (A) 9 cm.

The question asks about the relationship between the area of a triangle and the area of a parallelogram under specific conditions.

The given conditions are:

1. The triangle and the parallelogram share the same base.

2. The triangle and the parallelogram are between the same parallel lines.

Let the shared base be denoted by $b$.

Let the distance between the parallel lines be denoted by $h$. This distance is the perpendicular height corresponding to the base $b$ for both the parallelogram and the triangle because they are between the same parallel lines and share the base.

The formula for the area of a parallelogram is:

Area of parallelogram = base $\times$ height

$A_{parallelogram} = b \times h$

The formula for the area of a triangle is:

Area of triangle = $\frac{1}{2} \times$ base $\times$ height

$A_{triangle} = \frac{1}{2} \times b \times h$

Comparing the two area formulas:

$A_{triangle} = \frac{1}{2} \times (b \times h)$

Since $A_{parallelogram} = b \times h$, we can substitute this into the equation for the area of the triangle:

$A_{triangle} = \frac{1}{2} \times A_{parallelogram}$

This relationship, that the area of the triangle is exactly half the area of the parallelogram when they share the same base and are between the same parallel lines, is a fundamental theorem in geometry.

This relationship holds true whenever the specified conditions are met, regardless of the specific dimensions of the shapes (as long as they exist with that shared base and height).

Therefore, the statement is always true.

Now, compare this conclusion with the given options:

(A) Always true

(B) Always false

(C) Sometimes true

(D) Not related to area

Our conclusion matches option (A).

The correct option is (A) Always true.

Given:

Shape of the garden: Rectangle.

Length of the garden ($l$) = 20 m.

Width of the garden ($w$) = 15 m.

Rate of fencing = $\textsf{₹ }50$ per meter.

To Find:

The cost of fencing the rectangular garden.

Solution:

Fencing is done along the boundary of the garden. The length of the boundary is the perimeter of the rectangular garden.

The formula for the perimeter of a rectangle is:

Perimeter = $2 \times (\text{length} + \text{width})$

Substitute the given values of length and width into the formula:

$P = 2(20\text{ m} + 15\text{ m})$

$P = 2(35\text{ m})$

$P = 70\text{ m}$

The perimeter of the garden is 70 meters. This is the total length of the fence required.

Now, we need to find the total cost of fencing at the rate of $\textsf{₹ }50$ per meter.

Total cost = Perimeter $\times$ Rate per meter

Total cost = $70\text{ m} \times \textsf{₹ }50/\text{m}$

Total cost = $70 \times 50 \textsf{₹}$

$70 \times 50 = 3500$

Total cost = $\textsf{₹ }3500$

The cost of fencing the rectangular garden is $\textsf{₹ }3500$.

Now, compare this result with the given options:

(A) $\textsf{₹ }3500$

(B) $\textsf{₹ }1750$

(C) $\textsf{₹ }1000$

(D) $\textsf{₹ }7500$

The calculated cost matches option (A).

The correct option is (A) $\textsf{₹ }3500$.

Given:

Diameter of the circle ($D$) = 14 cm.

To Find:

The area of the circle.

Solution:

The formula for the area of a circle is $A = \pi r^2$, where $r$ is the radius of the circle.

The radius is half of the diameter:

$r = \frac{D}{2}$

Substitute the given diameter into the formula to find the radius:

$r = \frac{14\text{ cm}}{2}$

$r = 7\text{ cm}$

Now, use the formula for the area of the circle. We can use the value $\pi = \frac{22}{7}$ as is common for problems involving multiples of 7.

$A = \pi r^2$

$A = \frac{22}{7} \times (7\text{ cm})^2$

$A = \frac{22}{7} \times (7 \times 7)\text{ cm}^2$

$A = \frac{22}{\cancel{7}} \times (\cancel{7} \times 7)\text{ cm}^2$

$A = 22 \times 7\text{ cm}^2$

$A = 154\text{ cm}^2$

The area of the circle is $154\text{ cm}^2$.

Now, compare this result with the given options:

(A) $22\text{ cm}^2$

(B) $44\text{ cm}^2$

(C) $154\text{ cm}^2$

(D) $616\text{ cm}^2$

The calculated area matches option (C).

The correct option is (C) $154\text{ cm}^2$.

The perimeter of a two-dimensional shape is defined as the total length of its boundary.

Let's consider each of the given shapes:

(A) Square: A square is a polygon with four equal sides. Its boundary consists of these four sides. The perimeter of a square is the sum of the lengths of its four sides ($s+s+s+s = 4s$). This is equal to the sum of the lengths of its boundaries.

(B) Rectangle: A rectangle is a polygon with four sides, where opposite sides are equal in length. Its boundary consists of these four sides. The perimeter of a rectangle is the sum of the lengths of its four sides ($l+w+l+w = 2(l+w)$). This is equal to the sum of the lengths of its boundaries.

(C) Triangle: A triangle is a polygon with three sides. Its boundary consists of these three sides. The perimeter of a triangle is the sum of the lengths of its three sides ($a+b+c$). This is equal to the sum of the lengths of its boundaries.

In all the cases (Square, Rectangle, Triangle), which are polygons, the perimeter is indeed the sum of the lengths of their sides, which form their boundaries.

Therefore, the statement "Which of the following has a perimeter equal to the sum of the lengths of its boundaries?" is true for Square, Rectangle, and Triangle.

Since the property holds for all the listed shapes (A), (B), and (C), the correct option is (D).

The correct option is (D) All of the above.

Given:

Side of the square ($s$) = 10 cm.

Length of the rectangle ($l$) = 20 cm.

Area of the square = Area of the rectangle.

To Find:

The width of the rectangle ($w$).

Solution:

First, calculate the area of the square.

The formula for the area of a square is:

Area = side $\times$ side

Area of square = $s^2$

Area of square = $(10\text{ cm})^2$

Area of square = $10 \times 10\text{ cm}^2$

Area of square = $100\text{ cm}^2$

Next, consider the rectangle.

The formula for the area of a rectangle is:

Area = length $\times$ width

Area of rectangle = $l \times w$

Area of rectangle = $20\text{ cm} \times w$

We are given that the area of the square is equal to the area of the rectangle.

Area of rectangle = Area of square

$20\text{ cm} \times w = 100\text{ cm}^2$

To find the width ($w$), divide the area by the length:

$w = \frac{100\text{ cm}^2}{20\text{ cm}}$

$w = \frac{\cancel{100}^{5}}{\cancel{20}_{1}}\text{ cm}$

$w = 5\text{ cm}$

The width of the rectangle is 5 cm.

Now, compare this result with the given options:

(A) 5 cm

(B) 10 cm

(C) 20 cm

(D) 50 cm

The calculated width matches option (A).

The correct option is (A) 5 cm.

Let the original base of the parallelogram be $b$ and the original height be $h$.

The formula for the area of a parallelogram is:

Area = base $\times$ height

$A_{original} = b \times h$

According to the question, the height of the parallelogram is doubled, while the base remains the same.

New height ($h_{new}$) = $2h$

New base ($b_{new}$) = $b$

Now, calculate the new area of the parallelogram using the new height and base:

Area$_{new}$ = New base $\times$ New height

$A_{new} = b_{new} \times h_{new}$

$A_{new} = b \times (2h)$

$A_{new} = 2 \times (b \times h)$

Since $A_{original} = b \times h$, we can substitute this into the expression for $A_{new}$:

$A_{new} = 2 \times A_{original}$

This shows that the new area of the parallelogram is twice the original area.

Therefore, if the height of a parallelogram is doubled while the base remains the same, the area of the parallelogram will be doubled.

Now, compare this result with the given options:

(A) Halved

(B) Doubled

(C) Quartered

(D) The same

The result matches option (B).

The correct option is (B) Doubled.

Given:

Inner radius of the circular track ($r_1$) = 7 m.

Outer radius of the circular track ($r_2$) = 14 m.

Use $\pi = \frac{22}{7}$.

To Find:

The area of the circular track.

Solution:

A circular track is the region between two concentric circles. This region is called an annulus.

The area of the track is found by subtracting the area of the inner circle from the area of the outer circle.

The formula for the area of a circle with radius $r$ is given by $A = \pi r^2$.

Area of the inner circle ($A_1$) with radius $r_1 = 7\text{ m}$:

$A_1 = \pi r_1^2$

$A_1 = \frac{22}{7} \times (7\text{ m})^2$

$A_1 = \frac{22}{7} \times (7 \times 7)\text{ m}^2$

$A_1 = \frac{22}{\cancel{7}} \times (\cancel{7} \times 7)\text{ m}^2$

$A_1 = 22 \times 7\text{ m}^2$

$A_1 = 154\text{ m}^2$

Area of the outer circle ($A_2$) with radius $r_2 = 14\text{ m}$:

$A_2 = \pi r_2^2$

$A_2 = \frac{22}{7} \times (14\text{ m})^2$

$A_2 = \frac{22}{7} \times (14 \times 14)\text{ m}^2$

$A_2 = \frac{22}{\cancel{7}} \times (\cancel{14}^2 \times 14)\text{ m}^2$

$A_2 = 22 \times (2 \times 14)\text{ m}^2$

$A_2 = 22 \times 28\text{ m}^2$

$A_2 = 616\text{ m}^2$

Area of the track = Area of outer circle - Area of inner circle

$A_{track} = A_2 - A_1$

$A_{track} = 616\text{ m}^2 - 154\text{ m}^2$

$A_{track} = 462\text{ m}^2$

Alternate Method:

The area of an annulus (circular track) can also be calculated directly using the formula:

$A_{track} = \pi (r_2^2 - r_1^2)$

Substitute the values of $r_1$ and $r_2$:

$A_{track} = \frac{22}{7} \times (14^2 - 7^2)\text{ m}^2$

$A_{track} = \frac{22}{7} \times (196 - 49)\text{ m}^2$

$A_{track} = \frac{22}{7} \times 147\text{ m}^2$

$A_{track} = 22 \times \frac{147}{7}\text{ m}^2$

$A_{track} = 22 \times 21\text{ m}^2$

$A_{track} = 462\text{ m}^2$

The area of the track is $462\text{ m}^2$.

Now, compare this result with the given options:

(A) $154\text{ m}^2$

(B) $616\text{ m}^2$

(C) $462\text{ m}^2$

(D) $308\text{ m}^2$

The calculated area matches option (C).

The correct option is (C) $462\text{ m}^2$.

Given:

The relationship between hectares and square meters is provided in the question.

$1\text{ hectare} = 10000\text{ m}^2$

To Find:

The number of square meters in 1 hectare.

Solution:

According to the information given in the question, 1 hectare is equal to 10000 square meters.

Therefore, there are 10000 square meters in 1 hectare.

Now, compare this result with the given options:

(A) 100

(B) 1000

(C) 10000

(D) 100000

The number matches option (C).

The correct option is (C) 10000.

Let the original base of the triangle be $b_{orig}$ and the original height be $h_{orig}$.

The formula for the area of a triangle is:

Area = $\frac{1}{2} \times \text{base} \times \text{height}$

The original area of the triangle, $A_{orig}$, is:

$A_{orig} = \frac{1}{2} \times b_{orig} \times h_{orig}$

According to the question, the base of the triangle is tripled and the height is halved.

New base ($b_{new}$) = $3 \times b_{orig}$

New height ($h_{new}$) = $\frac{1}{2} \times h_{orig}$

Now, calculate the area of the new triangle, $A_{new}$, using the new base and height:

$A_{new} = \frac{1}{2} \times b_{new} \times h_{new}$

Substitute the expressions for $b_{new}$ and $h_{new}$:

$A_{new} = \frac{1}{2} \times (3 \times b_{orig}) \times (\frac{1}{2} \times h_{orig})$

$A_{new} = \frac{1}{2} \times 3 \times \frac{1}{2} \times b_{orig} \times h_{orig}$

$A_{new} = \left(\frac{1}{2} \times 3 \times \frac{1}{2}\right) \times (b_{orig} \times h_{orig})$

$A_{new} = \frac{3}{4} \times (b_{orig} \times h_{orig})$

To compare $A_{new}$ with $A_{orig}$, we can rewrite the expression for $A_{new}$ by incorporating the term $\frac{1}{2}$ to match the formula for $A_{orig}$:

$A_{new} = \frac{3}{2} \times \left(\frac{1}{2} \times b_{orig} \times h_{orig}\right)$

Since $A_{orig} = \frac{1}{2} \times b_{orig} \times h_{orig}$, we have:

$A_{new} = \frac{3}{2} \times A_{orig}$

The factor $\frac{3}{2}$ is equal to 1.5.

So, the area of the new triangle is 1.5 times the original area.

Now, compare this result with the given options:

(A) The same as the original area

(B) 1.5 times the original area

(C) 3 times the original area

(D) Half the original area

The result matches option (B).

The correct option is (B) 1.5 times the original area.

Given:

Length of the first diagonal ($d_1$) = 6 cm.

Length of the second diagonal ($d_2$) = 8 cm.

To Find:

The area of the rhombus.

Solution:

The area of a rhombus can be calculated using the lengths of its diagonals. The formula is:

Area = $\frac{1}{2} \times \text{length of diagonal 1} \times \text{length of diagonal 2}$

Substitute the given values of the diagonals into the formula:

$A = \frac{1}{2} \times 6\text{ cm} \times 8\text{ cm}$

Multiply the lengths of the diagonals:

$6 \times 8 = 48$

So, $d_1 \times d_2 = 48\text{ cm}^2$.

Now, substitute this product back into the area formula:

$A = \frac{1}{2} \times 48\text{ cm}^2$

$A = \frac{48}{2}\text{ cm}^2$

$A = 24\text{ cm}^2$

The area of the rhombus is $24\text{ cm}^2$.

Now, compare this result with the given options:

(A) $24\text{ cm}^2$

(B) $48\text{ cm}^2$

(C) $14\text{ cm}^2$

(D) $28\text{ cm}^2$

The calculated area matches option (A).

The correct option is (A) $24\text{ cm}^2$.

Given:

Shape of the plot: Rectangle.

Length of the plot ($l$) = 50 m.

Width of the plot ($w$) = 30 m.

Rate of levelling = $\textsf{₹ }10$ per square meter.

To Find:

The total cost of levelling the rectangular plot.

Solution:

Levelling covers the entire surface area of the plot. So, we first need to calculate the area of the rectangular plot.

The formula for the area of a rectangle is:

Area = length $\times$ width

Substitute the given values of length and width into the formula:

$A = 50\text{ m} \times 30\text{ m}$

$A = (50 \times 30)\text{ m}^2$

$A = 1500\text{ m}^2$

The area of the rectangular plot is 1500 square meters.

Now, we need to find the total cost of levelling at the rate of $\textsf{₹ }10$ per square meter.

Total cost = Area $\times$ Rate per square meter

Total cost = $1500\text{ m}^2 \times \textsf{₹ }10/\text{m}^2$

Total cost = $(1500 \times 10) \textsf{₹}$

$1500 \times 10 = 15000$

Total cost = $\textsf{₹ }15000$

The cost of levelling the rectangular plot is $\textsf{₹ }15000$.

Now, compare this result with the given options:

(A) $\textsf{₹ }800$

(B) $\textsf{₹ }1500$

(C) $\textsf{₹ }8000$

(D) $\textsf{₹ }15000$

The calculated cost matches option (D).

The correct option is (D) $\textsf{₹ }15000$.

Given:

The numerical value of the circumference of a circle is equal to the numerical value of its area.

To Find:

The radius of the circle.

Solution:

Let the radius of the circle be $r$ units.

The formula for the circumference of a circle with radius $r$ is:

Circumference ($C$) = $2\pi r$

The formula for the area of a circle with radius $r$ is:

Area ($A$) = $\pi r^2$

According to the problem statement, the numerical value of the circumference is equal to the numerical value of the area.

$C = A$

$2\pi r = \pi r^2$

We need to solve this equation for $r$. Since $r$ is a radius, $r \ge 0$. For a circle to exist with non-zero area/circumference, $r > 0$. We can divide both sides by $\pi r$ (assuming $r \neq 0$).

Divide both sides by $\pi$: $\frac{2\pi r}{\pi} = \frac{\pi r^2}{\pi}$

$2r = r^2$

Now, rearrange the equation to solve for $r$:

$r^2 - 2r = 0$

Factor out the common term $r$:

$r(r - 2) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1: $r = 0$

Case 2: $r - 2 = 0 \implies r = 2$

If $r = 0$, the circumference is $2\pi(0) = 0$ and the area is $\pi(0)^2 = 0$. While numerically equal, this represents a point, not a standard circle with a radius. In the context of such problems, a non-zero radius is usually implied.

Thus, the relevant solution is $r = 2$.

When the radius is 2 units, the circumference is $C = 2\pi(2) = 4\pi$ and the area is $A = \pi(2)^2 = 4\pi$. The numerical values are equal.

So, the radius of the circle is 2 units.

Now, compare this result with the given options:

(A) 1 unit

(B) 2 units

(C) $\pi$ units

(D) $2\pi$ units

The calculated radius matches option (B).

The correct option is (B) 2 units.

Given:

The side of a square park is $45$ meters.

Let the side of the square park be $s$. So, $s = 45$ m.

To Find:

The perimeter and the area of the park.

Solution:

The perimeter of a square is calculated by the formula:

Substituting the given side length into the formula (i):

$P = 4 \times 45 \text{ m}$

$P = 180 \text{ m}$

The area of a square is calculated by the formula:

Substituting the given side length into the formula (ii):

$A = (45 \text{ m})^2$

$A = 45 \times 45 \text{ m}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 4 & 5 \\ \times & & 4 & 5 \\ \hline && 2 & 2 & 5 \\ & 1 & 8 & 0 & \times \\ \hline 2 & 0 & 2 & 5 \\ \hline \end{array}$

$A = 2025 \text{ m}^2$

Thus, the perimeter of the square park is $180$ meters and the area of the square park is $2025$ square meters.

Given:

Length of the rectangular field, $l = 60$ meters.

Breadth of the rectangular field, $b = 40$ meters.

To Find:

The perimeter and the area of the rectangular field.

Solution:

The perimeter of a rectangle is calculated by the formula:

Substituting the given values of length and breadth into formula (i):

$P = 2 \times (60 \text{ m} + 40 \text{ m})$

$P = 2 \times (100 \text{ m})$

$P = 200 \text{ m}$

The area of a rectangle is calculated by the formula:

Substituting the given values of length and breadth into formula (ii):

$A = 60 \text{ m} \times 40 \text{ m}$

$A = (60 \times 40) \text{ m}^2$

$A = 2400 \text{ m}^2$

Thus, the perimeter of the rectangular field is $200$ meters and the area of the rectangular field is $2400$ square meters.

Given:

Perimeter of the square, $P = 64$ cm.

To Find:

The length of its side and its area.

Solution:

Let the side length of the square be $s$ cm.

The formula for the perimeter of a square is:

Substitute the given perimeter into formula (i):

$64 = 4s$

To find the side length, divide both sides by 4:

$s = \frac{64}{4}$

$s = 16$ cm

The formula for the area of a square is:

Substitute the calculated side length into formula (ii):

$A = (16 \text{ cm})^2$

$A = 16 \times 16 \text{ cm}^2$

$A = 256 \text{ cm}^2$

Thus, the length of the side of the square is $16$ cm and its area is $256$ cm$^2$.

Given:

Area of the rectangular room, $A = 120$ sq meters.

Length of the rectangular room, $l = 15$ meters.

To Find:

The breadth and perimeter of the rectangular room.

Solution:

Let the breadth of the rectangular room be $b$ meters.

The formula for the area of a rectangle is given by:

Substitute the given values into formula (i):

$120 \text{ m}^2 = 15 \text{ m} \times b$

To find the breadth $b$, divide the area by the length:

$b = \frac{120 \text{ m}^2}{15 \text{ m}}$

$b = 8 \text{ m}$

Now that we have the length ($l = 15$ m) and the breadth ($b = 8$ m), we can find the perimeter.

The formula for the perimeter of a rectangle is given by:

Substitute the values of length and breadth into formula (ii):

$P = 2 \times (15 \text{ m} + 8 \text{ m})$

$P = 2 \times (23 \text{ m})$

$P = 46 \text{ m}$

Thus, the breadth of the rectangular room is $8$ meters and its perimeter is $46$ meters.

The formula for the area of a parallelogram is:

In mathematical notation, the formula can be written as:

$A = b \times h$

Explanation of the terms involved:

• Base ($b$): The base of a parallelogram is the length of any one of its sides. You can choose any side to be the base.

• Height ($h$): The height of a parallelogram is the perpendicular distance from the chosen base to the opposite side. It is important that the height is perpendicular to the base.

For example, if you have a parallelogram with a base of length $b$ and the perpendicular distance from that base to the opposite side is $h$, then its area is given by the product of $b$ and $h$.

Given:

Base of the parallelogram, $b = 12$ cm.

Corresponding height of the parallelogram, $h = 7$ cm.

To Find:

The area of the parallelogram.

Solution:

The formula for the area of a parallelogram is:

Substitute the given values into formula (i):

$A = 12 \text{ cm} \times 7 \text{ cm}$

$A = (12 \times 7) \text{ cm}^2$

$A = 84 \text{ cm}^2$

Thus, the area of the parallelogram is $84$ cm$^2$.

Given:

Area of the parallelogram, $A = 240$ sq cm.

Base of the parallelogram, $b = 16$ cm.

To Find:

The corresponding height, $h$, of the parallelogram.

Solution:

The formula for the area of a parallelogram is:

Substitute the given values into formula (i):

$240 \text{ cm}^2 = 16 \text{ cm} \times h$

To find the height $h$, divide the area by the base:

$h = \frac{240 \text{ cm}^2}{16 \text{ cm}}$

$h = \frac{240}{16} \text{ cm}$

Now, we calculate the value:

$h = 15 \text{ cm}$

Thus, the corresponding height of the parallelogram is $15$ cm.

The formula for the area of a triangle is:

In mathematical notation, the formula can be written as:

$A = \frac{1}{2} b h$

Explanation of the terms involved:

• Base ($b$): The base of a triangle is the length of any one of its sides. You can choose any side to be the base.

• Height ($h$): The height (or altitude) of a triangle is the perpendicular distance from the vertex opposite the chosen base to the base (or the extension of the base). The height forms a right angle with the base.

The area of a triangle is always half the area of a parallelogram with the same base and height.

Given:

Base of the triangle, $b = 10$ cm.

Height of the triangle, $h = 6$ cm.

To Find:

The area of the triangle.

Solution:

The formula for the area of a triangle is:

Substitute the given values into formula (i):

$A = \frac{1}{2} \times 10 \text{ cm} \times 6 \text{ cm}$

$A = \frac{1}{2} \times (10 \times 6) \text{ cm}^2$

$A = \frac{1}{2} \times 60 \text{ cm}^2$

$A = 30 \text{ cm}^2$

Thus, the area of the triangle is $30$ cm$^2$.

Given:

Area of the triangle, $A = 90$ sq cm.

Base of the triangle, $b = 15$ cm.

To Find:

The corresponding height, $h$, of the triangle.

Solution:

The formula for the area of a triangle is:

Substitute the given values into formula (i):

$90 \text{ cm}^2 = \frac{1}{2} \times 15 \text{ cm} \times h$

Multiply both sides by 2:

$2 \times 90 \text{ cm}^2 = 15 \text{ cm} \times h$

$180 \text{ cm}^2 = 15 \text{ cm} \times h$

To find the height $h$, divide both sides by $15$ cm:

$h = \frac{180 \text{ cm}^2}{15 \text{ cm}}$

$h = \frac{180}{15} \text{ cm}$

$h = 12 \text{ cm}$

Thus, the corresponding height of the triangle is $12$ cm.

The circumference of a circle is the total distance around the outside edge of the circle. It is essentially the perimeter of the circle.

The formula for the circumference of a circle with radius $r$ is:

Alternatively, if the diameter of the circle is $d$, since $d = 2r$, the formula can also be written as:

$C = \pi d$

In these formulas:

• $C$ represents the Circumference.

• $r$ represents the radius of the circle (distance from the center to any point on the edge).

• $d$ represents the diameter of the circle (distance across the circle through the center, $d = 2r$).

• $\pi$ (pi) is a mathematical constant, approximately equal to $3.14159$ or $\frac{22}{7}$. It represents the ratio of a circle's circumference to its diameter.

Given:

Radius of the circle, $r = 7$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

The circumference of the circle.

Solution:

The formula for the circumference of a circle is:

Substitute the given values into formula (i):

$C = 2 \times \frac{22}{7} \times 7 \text{ cm}$

$C = 2 \times 22 \times \frac{\cancel{7}}{\cancel{7}} \text{ cm}$

$C = 2 \times 22 \text{ cm}$

$C = 44 \text{ cm}$

Thus, the circumference of the circle is $44$ cm.

Given:

Diameter of the circle, $d = 21$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

The circumference of the circle.

Solution:

The formula for the circumference of a circle using its diameter is:

Substitute the given values into formula (i):

$C = \frac{22}{7} \times 21 \text{ cm}$

$C = \frac{22}{\cancel{7}_{1}} \times \cancel{21}^{3} \text{ cm}$

$C = 22 \times 3 \text{ cm}$

$C = 66 \text{ cm}$

Thus, the circumference of the circle is $66$ cm.

The formula for the area of a circle with radius $r$ is:

In this formula:

• $A$ represents the Area of the circle.

• $r$ represents the radius of the circle (the distance from the center of the circle to any point on its boundary).

• $\pi$ (pi) is a mathematical constant, approximately equal to $3.14159$ or $\frac{22}{7}$. It is the ratio of a circle's circumference to its diameter.

Given:

Radius of the circle, $r = 14$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

The area of the circle.

Solution:

The formula for the area of a circle is:

Substitute the given values into formula (i):

$A = \frac{22}{7} \times (14 \text{ cm})^2$

$A = \frac{22}{7} \times (14 \times 14) \text{ cm}^2$

$A = \frac{22}{\cancel{7}_{1}} \times (\cancel{14}^{2} \times 14) \text{ cm}^2$

$A = 22 \times 2 \times 14 \text{ cm}^2$

$A = 44 \times 14 \text{ cm}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 4 & 4 \\ \times & & 1 & 4 \\ \hline && 1 & 7 & 6 \\ & 4 & 4 & \times \\ \hline 6 & 1 & 6 \\ \hline \end{array}$

$A = 616 \text{ cm}^2$

Thus, the area of the circle is $616$ cm$^2$.

Given:

Diameter of the circle, $d = 28$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

The area of the circle.

Solution:

First, find the radius of the circle. The radius ($r$) is half of the diameter ($d$).

Substitute the given diameter into formula (i):

$r = \frac{28 \text{ cm}}{2}$

$r = 14 \text{ cm}$

The formula for the area of a circle is:

Substitute the value of $\pi$ and the calculated radius into formula (ii):

$A = \frac{22}{7} \times (14 \text{ cm})^2$

$A = \frac{22}{7} \times (14 \times 14) \text{ cm}^2$

$A = \frac{22}{\cancel{7}_{1}} \times (\cancel{14}^{2} \times 14) \text{ cm}^2$

$A = 22 \times 2 \times 14 \text{ cm}^2$

$A = 44 \times 14 \text{ cm}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 4 & 4 \\ \times & & 1 & 4 \\ \hline && 1 & 7 & 6 \\ & 4 & 4 & \times \\ \hline 6 & 1 & 6 \\ \hline \end{array}$

$A = 616 \text{ cm}^2$

Thus, the area of the circle is $616$ cm$^2$.

Given:

Circumference of the circle, $C = 88$ cm.

Value of $\pi = \frac{22}{7}$.

To Find:

The radius, $r$, of the circle.

Solution:

The formula for the circumference of a circle is:

Substitute the given circumference and the value of $\pi$ into formula (i):

$88 \text{ cm} = 2 \times \frac{22}{7} \times r$

$88 = \frac{44}{7} r$

To find the radius $r$, multiply both sides by $\frac{7}{44}$:

$r = 88 \times \frac{7}{44} \text{ cm}$

$r = \cancel{88}^{2} \times \frac{7}{\cancel{44}_{1}} \text{ cm}$

$r = 2 \times 7 \text{ cm}$

$r = 14 \text{ cm}$

Thus, the radius of the circle is $14$ cm.

Given:

Area of the square = $144$ sq cm.

To Find:

The perimeter of the square.

Solution:

Let the side length of the square be $s$ cm.

The formula for the area of a square is:

Substitute the given area into formula (i):

$144 \text{ cm}^2 = s^2$

To find the side length, take the square root of the area:

$s = \sqrt{144} \text{ cm}$

$s = 12 \text{ cm}$

Now that we have the side length ($s = 12$ cm), we can find the perimeter.

The formula for the perimeter of a square is:

Substitute the calculated side length into formula (ii):

$P = 4 \times 12 \text{ cm}$

$P = 48 \text{ cm}$

Thus, the perimeter of the square is $48$ cm.

Given:

Length of the rectangle, $l = 15$ cm.

Breadth of the rectangle, $b = 10$ cm.

The wire is first bent into a rectangle and then rebent into a square.

To Find:

The side length of the square.

Solution:

When a wire is bent into a shape, its length is equal to the perimeter of that shape.

First, we find the perimeter of the rectangular shape. The formula for the perimeter of a rectangle is:

Substitute the given length and breadth into formula (i):

$P_{rect} = 2 \times (15 \text{ cm} + 10 \text{ cm})$

$P_{rect} = 2 \times (25 \text{ cm})$

$P_{rect} = 50 \text{ cm}$

Since the same wire is rebent into a square, the perimeter of the square is equal to the perimeter of the rectangle.

So, $P_{sq} = 50 \text{ cm}$.

Let the side length of the square be $s$ cm.

The formula for the perimeter of a square is:

Substitute the perimeter of the square into formula (ii):

$50 \text{ cm} = 4s$

To find the side length $s$, divide both sides by 4:

$s = \frac{50 \text{ cm}}{4}$

$s = 12.5 \text{ cm}$

Thus, the side length of the square is $12.5$ cm.

Given:

Side of the square garden, $s = 30$ meters.

Rate of fencing = $\textsf{₹}20$ per meter.

To Find:

The cost of fencing the square garden.

Solution:

Fencing is done along the boundary of the garden. Therefore, the length of the fence required is equal to the perimeter of the square garden.

The formula for the perimeter of a square is:

Substitute the given side length into formula (i):

$P = 4 \times 30 \text{ m}$

$P = 120 \text{ m}$

The cost of fencing is calculated by multiplying the total length of the fence (perimeter) by the rate per meter.

Substitute the calculated perimeter and the given rate into formula (ii):

Total Cost = $120 \text{ m} \times \textsf{₹}20/\text{m}$

Total Cost = $120 \times 20 \textsf{₹}$

Total Cost = $2400 \textsf{₹}$

Thus, the cost of fencing the square garden is $\textsf{₹}2400$.

Given:

Radius of the circular track, $r = 105$ meters.

Number of rounds taken by the runner = $2$.

Value of $\pi = \frac{22}{7}$.

To Find:

The total distance covered by the runner.

Solution:

The distance covered by the runner in one round is equal to the circumference of the circular track.

The formula for the circumference of a circle is:

Substitute the given radius and the value of $\pi$ into formula (i):

$C = 2 \times \frac{22}{7} \times 105 \text{ m}$

$C = 2 \times 22 \times \frac{\cancel{105}^{15}}{\cancel{7}_{1}} \text{ m}$

$C = 2 \times 22 \times 15 \text{ m}$

$C = 44 \times 15 \text{ m}$

Let's perform the multiplication:

$\begin{array}{cc}& & 4 & 4 \\ \times & & 1 & 5 \\ \hline && 2 & 2 & 0 \\ & 4 & 4 & \times \\ \hline 6 & 6 & 0 \\ \hline \end{array}$

$C = 660 \text{ m}$

The distance covered in one round is $660$ meters.

The runner takes $2$ rounds of the track.

Total distance covered = Distance in one round $\times$ Number of rounds

Substitute the calculated circumference into formula (ii):

Total distance = $660 \text{ m} \times 2$

Total distance = $1320 \text{ m}$

Thus, the total distance covered by the runner is $1320$ meters.

Given:

Base of the parallelogram, $b = 1.5$ meters.

Height of the parallelogram, $h = 0.8$ meters.

To Find:

The area of the parallelogram in square centimeters.

Solution:

First, we convert the given dimensions from meters to centimeters.

We know that $1 \text{ meter} = 100 \text{ centimeters}$.

Base $b = 1.5 \text{ m} = 1.5 \times 100 \text{ cm} = 150 \text{ cm}$.

Height $h = 0.8 \text{ m} = 0.8 \times 100 \text{ cm} = 80 \text{ cm}$.

The formula for the area of a parallelogram is:

Substitute the values of the base and height in centimeters into formula (i):

$A = 150 \text{ cm} \times 80 \text{ cm}$

$A = (150 \times 80) \text{ cm}^2$

$A = 12000 \text{ cm}^2$

Thus, the area of the parallelogram is $12000$ square centimeters.

Given:

Area of the triangular plot, $A = 150$ sq meters.

Height of the triangular plot, $h = 12$ meters.

To Find:

The length of its base, $b>.

Solution:

The formula for the area of a triangle is:

Substitute the given values into formula (i):

$150 \text{ m}^2 = \frac{1}{2} \times b \times 12 \text{ m}$

$150 = \frac{12}{2} \times b$

$150 = 6 \times b$

To find the base $b$, divide the area by $6$:

$b = \frac{150}{6} \text{ m}$

$b = 25 \text{ m}$

Thus, the length of the base of the triangular plot is $25$ meters.

Given:

The ratio of the radii of two circles is $2:3$.

To Find:

The ratio of their circumferences.

Solution:

Let the radii of the two circles be $r_1$ and $r_2$.

Given that the ratio of their radii is $2:3$, we can write:

The formula for the circumference of a circle with radius $r$ is $C = 2 \pi r$.

Let the circumferences of the two circles be $C_1$ and $C_2$.

$C_1 = 2 \pi r_1$

$C_2 = 2 \pi r_2$

Now, we find the ratio of their circumferences:

We can cancel out the common term $2 \pi$ from the numerator and the denominator in formula (ii):

$\frac{C_1}{C_2} = \frac{\cancel{2 \pi} r_1}{\cancel{2 \pi} r_2}$

$\frac{C_1}{C_2} = \frac{r_1}{r_2}$

From equation (i), we know that $\frac{r_1}{r_2} = \frac{2}{3}$.

Therefore, the ratio of the circumferences is:

$\frac{C_1}{C_2} = \frac{2}{3}$

The ratio of the circumferences of the two circles is $2:3$.

Conclusion: The ratio of the circumferences of two circles is equal to the ratio of their radii.

Given:

The ratio of the radii of two circles is $2:3$.

To Find:

The ratio of their areas.

Solution:

Let the radii of the two circles be $r_1$ and $r_2$.

Given that the ratio of their radii is $2:3$, we can write:

The formula for the area of a circle with radius $r$ is $A = \pi r^2$.

Let the areas of the two circles be $A_1$ and $A_2$.

Now, we find the ratio of their areas by dividing equation (ii) by equation (iii):

We can cancel out the common term $\pi$ from the numerator and the denominator:

$\frac{A_1}{A_2} = \frac{\cancel{\pi} r_1^2}{\cancel{\pi} r_2^2}$

$\frac{A_1}{A_2} = \frac{r_1^2}{r_2^2}$

From equation (i), we know that $\frac{r_1}{r_2} = \frac{2}{3}$. Substitute this into equation (iv):

$\frac{A_1}{A_2} = \left(\frac{2}{3}\right)^2$

$\frac{A_1}{A_2} = \frac{2^2}{3^2}$

$\frac{A_1}{A_2} = \frac{4}{9}$

The ratio of the areas of the two circles is $4:9$.

Conclusion: The ratio of the areas of two circles is equal to the square of the ratio of their radii.

Given:

Length of the rectangular garden = $100$ meters.

Breadth of the rectangular garden = $80$ meters.

Width of the path built inside the garden = $5$ meters.

To Find:

The area of the path.

Solution:

Let the outer rectangle represent the garden including the path, and the inner rectangle represent the garden excluding the path.

The dimensions of the outer rectangle are:

Length of outer rectangle ($L$) = $100$ m.

Breadth of outer rectangle ($B$) = $80$ m.

The path is $5$ meters wide and is built inside the garden along its boundary. This means the path reduces the length and breadth of the inner rectangular area.

The length of the inner rectangle will be the length of the outer rectangle minus twice the width of the path (once from each end).

Substitute the values into formula (i):

$l = 100 \text{ m} - 2 \times 5 \text{ m}$

$l = 100 \text{ m} - 10 \text{ m}$

$l = 90 \text{ m}$

Similarly, the breadth of the inner rectangle will be the breadth of the outer rectangle minus twice the width of the path.

Substitute the values into formula (ii):

$b = 80 \text{ m} - 2 \times 5 \text{ m}$

$b = 80 \text{ m} - 10 \text{ m}$

$b = 70 \text{ m}$

Now, calculate the area of the outer rectangle (garden including the path).

Substitute the dimensions into formula (iii):

$A_{outer} = 100 \text{ m} \times 80 \text{ m}$

$A_{outer} = 8000 \text{ m}^2$

Next, calculate the area of the inner rectangle (garden excluding the path).

Substitute the calculated dimensions into formula (iv):

$A_{inner} = 90 \text{ m} \times 70 \text{ m}$

$A_{inner} = 6300 \text{ m}^2$

The area of the path is the difference between the area of the outer rectangle and the area of the inner rectangle.

Substitute the calculated areas into formula (v):

$A_{path} = 8000 \text{ m}^2 - 6300 \text{ m}^2$

$A_{path} = 1700 \text{ m}^2$

Thus, the area of the path is $1700$ square meters.

Given:

Length of the door, $l_{door} = 2$ m.

Breadth of the door, $b_{door} = 1$ m.

Length of the wall, $l_{wall} = 4.5$ m.

Breadth of the wall, $b_{wall} = 3.6$ m.

Rate of whitewashing = $\textsf{₹}20$ per sq meter.

Area of the door is not to be whitewashed.

To Find:

The cost of whitewashing the wall.

Solution:

First, we find the area of the door. The door is rectangular in shape.

The formula for the area of a rectangle is:

Using formula (i), the area of the door is:

$A_{door} = l_{door} \times b_{door}$

$A_{door} = 2 \text{ m} \times 1 \text{ m}$

$A_{door} = 2 \text{ m}^2$

Next, we find the area of the entire wall, assuming it is rectangular.

Using formula (i), the area of the wall is:

$A_{wall} = l_{wall} \times b_{wall}$

$A_{wall} = 4.5 \text{ m} \times 3.6 \text{ m}$

$A_{wall} = 16.20 \text{ m}^2$

The area to be whitewashed is the area of the wall minus the area of the door.

Substitute the calculated areas into formula (ii):

$A_{white} = 16.20 \text{ m}^2 - 2 \text{ m}^2$

$A_{white} = 14.20 \text{ m}^2$

The cost of whitewashing is calculated by multiplying the area to be whitewashed by the rate per square meter.

Substitute the calculated area to be whitewashed and the given rate into formula (iii):

Total Cost = $14.20 \text{ m}^2 \times \textsf{₹}20/\text{m}^2$

Total Cost = $14.20 \times 20 \textsf{₹}$

Total Cost = $284 \textsf{₹}$

Thus, the cost of whitewashing the wall (excluding the door) is $\textsf{₹}284$.

Given:

Perimeter of the rectangular sheet, $P = 100$ cm.

Length of the rectangular sheet, $l = 35$ cm.

The sheet is cut into $10$ equal square pieces.

To Find:

The breadth of the sheet, the area of the sheet, and the area of each square piece.

Solution:

First, we find the breadth of the rectangular sheet using the perimeter formula.

The formula for the perimeter of a rectangle is:

Substitute the given values into formula (i):

$100 \text{ cm} = 2 \times (35 \text{ cm} + b)$

Divide both sides by 2:

$\frac{100}{2} \text{ cm} = 35 \text{ cm} + b$

$50 \text{ cm} = 35 \text{ cm} + b$

Subtract 35 cm from both sides to find $b$:

$b = 50 \text{ cm} - 35 \text{ cm}$

$b = 15 \text{ cm}$

Next, we find the area of the rectangular sheet.

The formula for the area of a rectangle is:

Substitute the length and the calculated breadth into formula (ii):

$A_{rect} = 35 \text{ cm} \times 15 \text{ cm}$

$A_{rect} = (35 \times 15) \text{ cm}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 3 & 5 \\ \times & & 1 & 5 \\ \hline && 1 & 7 & 5 \\ & 3 & 5 & \times \\ \hline 5 & 2 & 5 \\ \hline \end{array}$

$A_{rect} = 525 \text{ cm}^2$

The rectangular sheet is cut into $10$ equal square pieces.

The area of each square piece is the total area divided by the number of pieces.

Substitute the calculated area and the number of pieces into formula (iii):

$A_{sq\_piece} = \frac{525 \text{ cm}^2}{10}$

$A_{sq\_piece} = 52.5 \text{ cm}^2$

Thus, the breadth of the rectangular sheet is $15$ cm, the area of the sheet is $525$ sq cm, and the area of each square piece is $52.5$ sq cm.

Given:

Side of the square plot, $s_{sq} = 60$ meters.

Length of the rectangular plot, $l_{rect} = 90$ meters.

Area of the square plot = Area of the rectangular plot.

To Find:

The breadth of the rectangular plot ($b_{rect}$) and compare the perimeters of the two plots.

Solution:

First, calculate the area of the square plot.

The formula for the area of a square is:

Substitute the side length into formula (i):

$A_{sq} = (60 \text{ m})^2$

$A_{sq} = 60 \times 60 \text{ m}^2$

$A_{sq} = 3600 \text{ m}^2$

Given that the area of the rectangular plot is the same as the area of the square plot:

So, $A_{rect} = 3600 \text{ m}^2$.

The formula for the area of a rectangle is:

Substitute the known area and length into formula (ii):

$3600 \text{ m}^2 = 90 \text{ m} \times b_{rect}$

To find the breadth $b_{rect}$, divide the area by the length:

$b_{rect} = \frac{3600 \text{ m}^2}{90 \text{ m}}$

$b_{rect} = \frac{3600}{90} \text{ m}$

$b_{rect} = 40 \text{ m}$

Now, calculate the perimeter of the square plot.

The formula for the perimeter of a square is:

Substitute the side length into formula (iii):

$P_{sq} = 4 \times 60 \text{ m}$

$P_{sq} = 240 \text{ m}$

Next, calculate the perimeter of the rectangular plot.

The formula for the perimeter of a rectangle is:

Substitute the given length and the calculated breadth into formula (iv):

$P_{rect} = 2 \times (90 \text{ m} + 40 \text{ m})$

$P_{rect} = 2 \times (130 \text{ m})$

$P_{rect} = 260 \text{ m}$

Finally, compare the perimeters of the two plots:

$P_{sq} = 240$ m

$P_{rect} = 260$ m

Comparing the values, we see that $260 \text{ m} > 240 \text{ m}$.

So, the perimeter of the rectangular plot is greater than the perimeter of the square plot.

Thus, the breadth of the rectangular plot is $40$ meters, and the perimeter of the rectangular plot ($260$ m) is greater than the perimeter of the square plot ($240$ m).

Given:

Diameter of the circular garden = $42$ meters.

Width of the path around the garden = $3.5$ meters.

The path runs around the garden on the outside.

Value of $\pi = \frac{22}{7}$.

To Find:

The area of the path.

Solution:

Let the radius of the circular garden be $r_{garden}$ and the radius of the outer circle (garden plus path) be $r_{outer}$.

First, find the radius of the circular garden.

Using formula (i):

$r_{garden} = \frac{42 \text{ m}}{2}$

$r_{garden} = 21 \text{ m}$

Since the path of width $3.5$ m runs around the garden on the outside, the radius of the outer circle will be the radius of the garden plus the width of the path.

Using formula (ii):

$r_{outer} = 21 \text{ m} + 3.5 \text{ m}$

$r_{outer} = 24.5 \text{ m}$

Now, find the area of the circular garden (inner circle).

The formula for the area of a circle is:

Using formula (iii), the area of the garden is:

$A_{garden} = \pi \times (r_{garden})^2$

$A_{garden} = \frac{22}{7} \times (21 \text{ m})^2$

$A_{garden} = \frac{22}{7} \times 21 \times 21 \text{ m}^2$

$A_{garden} = 22 \times \frac{\cancel{21}^{3}}{\cancel{7}_{1}} \times 21 \text{ m}^2$

$A_{garden} = 22 \times 3 \times 21 \text{ m}^2$

$A_{garden} = 66 \times 21 \text{ m}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 6 & 6 \\ \times & & 2 & 1 \\ \hline && 6 & 6 \\ & 1 & 3 & 2 & \times \\ \hline 1 & 3 & 8 & 6 \\ \hline \end{array}$

$A_{garden} = 1386 \text{ m}^2$

Next, find the area of the outer circle (garden plus path).

Using formula (iii), the area of the outer circle is:

$A_{outer} = \pi \times (r_{outer})^2$

$A_{outer} = \frac{22}{7} \times (24.5 \text{ m})^2$

$A_{outer} = \frac{22}{7} \times 24.5 \times 24.5 \text{ m}^2$

$A_{outer} = 22 \times \frac{\cancel{24.5}^{3.5}}{\cancel{7}_{1}} \times 24.5 \text{ m}^2$

$A_{outer} = 22 \times 3.5 \times 24.5 \text{ m}^2$

$A_{outer} = 77 \times 24.5 \text{ m}^2$

Let's perform the multiplication:

$\begin{array}{cc}& & 7 & 7 \\ \times & & 2 & 4 & . & 5 \\ \hline && 3 & 8 & 5 \\ & 3 & 0 & 8 & \times \\ 1 & 5 & 4 & \times & \times \\ \hline 1 & 8 & 8 & 6 & . & 5 \\ \hline \end{array}$

$A_{outer} = 1886.5 \text{ m}^2$

The area of the path is the difference between the area of the outer circle and the area of the inner circle (garden).

Substitute the calculated areas into formula (iv):

$A_{path} = 1886.5 \text{ m}^2 - 1386 \text{ m}^2$

Let's perform the subtraction:

$\begin{array}{cccccc} & 1 & 8 & 8 & 6 & . & 5 \\ - & 1 & 3 & 8 & 6 & . & 0 \\ \hline & & 5 & 0 & 0 & . & 5 \\ \hline \end{array}$

$A_{path} = 500.5 \text{ m}^2$

Thus, the area of the path is $500.5$ square meters.

Given:

Length of the rectangle, $l = 40$ cm.

Breadth of the rectangle, $b = 22$ cm.

The wire is rebent from the shape of a rectangle into a circle.

Value of $\pi = \frac{22}{7}$.

To Find:

The radius and area of the circle formed, and which shape encloses more area.

Solution:

When a wire is rebent from one shape to another, the length of the wire remains constant. The length of the wire is equal to the perimeter of the shape it forms.

First, calculate the perimeter of the rectangular shape. This is the total length of the wire.

The formula for the perimeter of a rectangle is:

Substitute the given length and breadth into formula (i):

$P_{rect} = 2 \times (40 \text{ cm} + 22 \text{ cm})$

$P_{rect} = 2 \times (62 \text{ cm})$

$P_{rect} = 124 \text{ cm}$

So, the length of the wire is $124$ cm.

When the wire is rebent into a circle, its length becomes the circumference of the circle.

So, $C = 124$ cm.

The formula for the circumference of a circle with radius $r$ is:

Substitute the circumference and the value of $\pi$ into formula (iii) to find the radius $r$:

$124 \text{ cm} = 2 \times \frac{22}{7} \times r$

$124 = \frac{44}{7} r$

To find $r$, multiply both sides by $\frac{7}{44}$:

$r = 124 \times \frac{7}{44} \text{ cm}$

$r = \frac{\cancel{124}^{31}}{\cancel{44}_{11}} \times 7 \text{ cm}$

$r = \frac{31 \times 7}{11} \text{ cm}$

$r = \frac{217}{11} \text{ cm}$

$r \approx 19.73 \text{ cm}$

Next, calculate the area of the circle formed.

The formula for the area of a circle is:

Substitute the value of $\pi$ and the calculated radius into formula (iv):

$A_{circle} = \frac{22}{7} \times \left(\frac{217}{11} \text{ cm}\right)^2$

$A_{circle} = \frac{22}{7} \times \frac{217^2}{11^2} \text{ cm}^2$

$A_{circle} = \frac{22}{7} \times \frac{217 \times 217}{11 \times 11} \text{ cm}^2$

$A_{circle} = \frac{2 \times \cancel{11}}{\cancel{7}} \times \frac{(\cancel{7} \times 31) \times 217}{\cancel{11} \times 11} \text{ cm}^2$

$A_{circle} = \frac{2 \times 31 \times 217}{11} \text{ cm}^2$

$A_{circle} = \frac{62 \times 217}{11} \text{ cm}^2$

Calculate $62 \times 217$:

$\begin{array}{cc}& & 2 & 1 & 7 \\ \times & & & 6 & 2 \\ \hline && 4 & 3 & 4 \\ & 1 & 3 & 0 & 2 & \times \\ \hline 1 & 3 & 4 & 5 & 4 \\ \hline \end{array}$

$A_{circle} = \frac{13454}{11} \text{ cm}^2$

$A_{circle} = 1223.09... \text{ cm}^2$

Now, calculate the area of the original rectangular shape.

The formula for the area of a rectangle is:

Substitute the given length and breadth into formula (v):

$A_{rect} = 40 \text{ cm} \times 22 \text{ cm}$

$A_{rect} = 880 \text{ cm}^2$

Compare the areas of the circle and the rectangle:

$A_{circle} \approx 1223.09 \text{ cm}^2$

$A_{rect} = 880 \text{ cm}^2$

Since $1223.09 > 880$, the area of the circle is greater than the area of the rectangle.

Thus, the radius of the circle is $\frac{217}{11}$ cm (approximately $19.73$ cm), the area of the circle is $\frac{13454}{11}$ sq cm (approximately $1223.09$ sq cm), and the circle encloses more area than the rectangle.

Given:

Area of the parallelogram, $A = 180$ sq cm.

First base of the parallelogram, $b_1 = 20$ cm.

Height corresponding to another base, $h_2 = 15$ cm.

To Find:

The height corresponding to the first base ($h_1$) and the length of the other base ($b_2$).

Solution:

The formula for the area of a parallelogram is:

Using formula (i) with the first base ($b_1$) and its corresponding height ($h_1$), we have:

Substitute the given area and the first base into formula (ii):

$180 \text{ cm}^2 = 20 \text{ cm} \times h_1$

To find the height $h_1$, divide the area by the base $b_1$:

$h_1 = \frac{180 \text{ cm}^2}{20 \text{ cm}}$

$h_1 = \frac{180}{20} \text{ cm}$

$h_1 = 9 \text{ cm}$

Now, we use the area formula (i) again with the other base ($b_2$) and its corresponding height ($h_2$). The area of the parallelogram remains the same.

Substitute the given area and the second height into formula (iii):

$180 \text{ cm}^2 = b_2 \times 15 \text{ cm}$

To find the length of the other base $b_2$, divide the area by the height $h_2$:

$b_2 = \frac{180 \text{ cm}^2}{15 \text{ cm}}$

$b_2 = \frac{180}{15} \text{ cm}$

$b_2 = 12 \text{ cm}$

Thus, the height corresponding to the base of $20$ cm is $9$ cm, and the length of the other base corresponding to the height of $15$ cm is $12$ cm.

Given:

Base of the triangular field, $b = 24$ meters.

Height of the triangular field, $h = 15$ meters.

Rate of leveling = $\textsf{₹}15$ per square meter.

Cost of fencing rate = $\textsf{₹}25$ per meter.

To Find:

The cost of leveling the field and the additional information needed to find the cost of fencing.

Solution (Cost of Leveling):

The cost of leveling depends on the area of the field.

The formula for the area of a triangle is:

Substitute the given base and height into formula (i):

$A = \frac{1}{2} \times 24 \text{ m} \times 15 \text{ m}$

$A = \frac{1}{\cancel{2}_{1}} \times \cancel{24}^{12} \times 15 \text{ m}^2$

$A = 12 \times 15 \text{ m}^2$

$A = 180 \text{ m}^2$

The cost of leveling is calculated by multiplying the area by the rate per square meter.

Substitute the calculated area and the given rate into formula (ii):

Cost of leveling = $180 \text{ m}^2 \times \textsf{₹}15/\text{m}^2$

Cost of leveling = $180 \times 15 \textsf{₹}$

Let's perform the multiplication:

$\begin{array}{cc}& & 1 & 8 & 0 \\ \times & & & 1 & 5 \\ \hline && 9 & 0 & 0 \\ & 1 & 8 & 0 & \times \\ \hline 2 & 7 & 0 & 0 \\ \hline \end{array}$

Cost of leveling = $2700 \textsf{₹}$

Additional Information Needed for Cost of Fencing:

Fencing is done along the boundary of the field. The length of the fence required is equal to the perimeter of the triangular field.

The perimeter of a triangle is the sum of the lengths of its three sides.

We are given the length of one side (the base, $24$ m) and the height ($15$ m). The height is the perpendicular distance from a vertex to the base, not necessarily the length of another side.

To calculate the perimeter, we need the lengths of the other two sides of the triangle.

Without knowing the lengths of the other two sides, or other properties of the triangle (like whether it's a right-angled triangle, isosceles, or equilateral), we cannot calculate its perimeter.

The cost of fencing would be calculated by multiplying the perimeter by the rate per meter ($\textsf{₹}25$), but we cannot find the perimeter with the given information.

Thus, the cost of leveling the triangular field is $\textsf{₹}2700$. To find the cost of fencing, we need the lengths of the other two sides of the triangular field.

Given:

Length of the rectangular park, $L = 80$ meters.

Width of the rectangular park, $B = 60$ meters.

Width of each cross road = $5$ meters.

The roads run at right angles through the center and parallel to the sides.

To Find:

The area of the roads and the area of the park excluding the area of the roads.

Solution (Area of the roads):

There are two cross roads:

1. One road parallel to the length of the park.

2. One road parallel to the width of the park.

The road parallel to the length has a length equal to the length of the park and a width of $5$ meters.

Substitute the values into formula (i):

$A_{road\_L} = 80 \text{ m} \times 5 \text{ m}$

$A_{road\_L} = 400 \text{ m}^2$

The road parallel to the width has a length equal to the width of the park and a width of $5$ meters.

Substitute the values into formula (ii):

$A_{road\_B} = 60 \text{ m} \times 5 \text{ m}$

$A_{road\_B} = 300 \text{ m}^2$

The two roads cross each other at the center. The area where they cross is a square with side length equal to the width of the roads.

Substitute the width of the road into formula (iii):

$A_{overlap} = 5 \text{ m} \times 5 \text{ m}$

$A_{overlap} = 25 \text{ m}^2$

When calculating the area of the two roads by adding $A_{road\_L}$ and $A_{road\_B}$, the area of the central square overlap is counted twice. To find the total area of the roads, we add the areas of the two individual roads and subtract the area of the overlap once.

Substitute the calculated areas into formula (iv):

$A_{roads} = 400 \text{ m}^2 + 300 \text{ m}^2 - 25 \text{ m}^2$

$A_{roads} = 700 \text{ m}^2 - 25 \text{ m}^2$

$A_{roads} = 675 \text{ m}^2$

Solution (Area of the park excluding roads):

First, calculate the total area of the rectangular park.

Substitute the given dimensions into formula (v):

$A_{park} = 80 \text{ m} \times 60 \text{ m}$

$A_{park} = 4800 \text{ m}^2$

The area of the park excluding the area of the roads is the total area of the park minus the total area of the roads.

Substitute the calculated areas into formula (vi):

Area of park excluding roads = $4800 \text{ m}^2 - 675 \text{ m}^2$

Let's perform the subtraction:

$\begin{array}{ccccc} & 4 & 8 & 0 & 0 \\ - & & 6 & 7 & 5 \\ \hline & 4 & 1 & 2 & 5 \\ \hline \end{array}$

Area of park excluding roads = $4125 \text{ m}^2$

Thus, the area of the roads is $675$ square meters, and the area of the park excluding the area of the roads is $4125$ square meters.

Given:

Length of the rectangle, $l = 14$ cm.

Breadth of the rectangle, $b = 7$ cm.

The circle is inscribed in the rectangle such that its diameter is equal to the breadth of the rectangle.

Diameter of the circle, $d = 7$ cm.

Assume $\pi = \frac{22}{7}$.

To Find:

The area of the shaded region (Area of the rectangle excluding the area of the inscribed circle).

Solution:

First, we find the area of the rectangle.

The formula for the area of a rectangle is:

Substitute the given dimensions into formula (i):

$A_{rect} = 14 \text{ cm} \times 7 \text{ cm}$

$A_{rect} = 98 \text{ cm}^2$

Next, we find the area of the inscribed circle.

The radius of the circle is half of its diameter.

Substitute the given diameter into formula (ii):

$r = \frac{7 \text{ cm}}{2}$

$r = 3.5 \text{ cm}$ or $\frac{7}{2}$ cm.

The formula for the area of a circle is:

Substitute the value of $\pi$ and the calculated radius into formula (iii):

$A_{circle} = \frac{22}{7} \times \left(\frac{7}{2} \text{ cm}\right)^2$

$A_{circle} = \frac{22}{7} \times \frac{7^2}{2^2} \text{ cm}^2$

$A_{circle} = \frac{22}{7} \times \frac{49}{4} \text{ cm}^2$

$A_{circle} = \frac{\cancel{22}^{11}}{\cancel{7}_{1}} \times \frac{\cancel{49}^{7}}{\cancel{4}_{2}} \text{ cm}^2$

$A_{circle} = \frac{11 \times 7}{2} \text{ cm}^2$

$A_{circle} = \frac{77}{2} \text{ cm}^2$

$A_{circle} = 38.5 \text{ cm}^2$

The area of the shaded region is the area of the rectangle minus the area of the circle.

Substitute the calculated areas into formula (iv):

$A_{shaded} = 98 \text{ cm}^2 - 38.5 \text{ cm}^2$

Let's perform the subtraction:

$\begin{array}{ccccc} & 9 & 8 & . & 0 \\ - & 3 & 8 & . & 5 \\ \hline & 5 & 9 & . & 5 \\ \hline \end{array}$

$A_{shaded} = 59.5 \text{ cm}^2$

Thus, the area of the shaded region is $59.5$ square centimeters.

Given:

A triangular field with base length = $30$ meters.

A line segment is drawn from the opposite vertex to the base, dividing the field into two smaller triangles with equal area.

To Find:

Where the line segment should meet the base.

Explanation of the reasoning.

Solution and Reasoning:

Let the triangular field be represented by triangle ABC, where BC is the base of length $30$ meters, and A is the opposite vertex.

Let the line segment from vertex A meet the base BC at point D.

This divides the original triangle ABC into two smaller triangles: $\triangle ABD$ and $\triangle ADC$.

The base of $\triangle ABD$ is BD, and the base of $\triangle ADC$ is DC.

The sum of the lengths of these bases is the length of the original base:

BD + DC = $30$ meters.

Let $h$ be the height of the original triangle ABC from vertex A to the base BC. This height is the perpendicular distance from A to the line containing BC.

The area of $\triangle ABC$ is given by:

$A_{ABC} = \frac{1}{2} \times \text{BC} \times h$

Now consider the two smaller triangles $\triangle ABD$ and $\triangle ADC$. Both triangles share the same vertex A. Their bases (BD and DC) lie on the same straight line (the base BC of the original triangle).

Therefore, the height of $\triangle ABD$ from vertex A to its base BD is also $h$.

Similarly, the height of $\triangle ADC$ from vertex A to its base DC is also $h$.

The area of $\triangle ABD$ is:

The area of $\triangle ADC$ is:

The problem states that the farmer wants to divide the field into two smaller triangles with equal area.

Substitute the expressions for the areas from (ii) and (iii):

$\frac{1}{2} \times \text{BD} \times h = \frac{1}{2} \times \text{DC} \times h$

Assuming the height $h$ is not zero (which would mean it's not a triangle), we can cancel $\frac{1}{2} \times h$ from both sides of the equation:

$\text{BD} = \text{DC}$

This result shows that the point D must divide the base BC into two equal segments. In other words, the line segment from vertex A must meet the base BC at its midpoint.

Since the total length of the base BC is $30$ meters, the midpoint D will be located halfway along the base.

Length of BD = Length of DC = $\frac{1}{2} \times \text{BC}$

Length of BD = Length of DC = $\frac{1}{2} \times 30 \text{ m}$

Length of BD = Length of DC = $15 \text{ m}$

Reasoning Summary:

When a triangle is divided by a line segment from a vertex to the opposite base, the two resulting smaller triangles share the same height (the perpendicular distance from the common vertex to the line containing the base). For these two triangles to have equal areas, their bases (the segments on the original base) must also be equal in length. Therefore, the line segment must bisect the base.

Thus, the line segment should meet the base exactly at its midpoint, which is $15$ meters from either end of the base.

Given:

Square with side $s = 10$ cm.

Parallelogram with base $b = 10$ cm and height $h = 10$ cm.

To Compare:

The area of the square and the area of the parallelogram.

Solution:

Calculate the area of the square.

The formula for the area of a square is:

Substitute the given side length into formula (i):

$A_{sq} = (10 \text{ cm})^2$

$A_{sq} = 10 \times 10 \text{ cm}^2$

$A_{sq} = 100 \text{ cm}^2$

Calculate the area of the parallelogram.

The formula for the area of a parallelogram is:

Substitute the given base and height into formula (ii):

$A_{para} = 10 \text{ cm} \times 10 \text{ cm}$

$A_{para} = 100 \text{ cm}^2$

Comparison of Areas:

Area of square ($A_{sq}$) = $100 \text{ cm}^2$

Area of parallelogram ($A_{para}$) = $100 \text{ cm}^2$

Comparing the areas, we find that $A_{sq} = A_{para}$.

Conclusion:

The area of the square with side $10$ cm is equal to the area of the parallelogram with base $10$ cm and height $10$ cm.

This shows that a parallelogram with the same base length as the side of a square and a height equal to the side of the square will have the same area as the square, even if its shape is different (i.e., it is not a square). The area of a parallelogram depends on the base and the perpendicular height, not the length of the slanted side.

Rough Diagrams:

Imagine a Square with all four sides of length $10$ cm and all interior angles as $90^\circ$. Its area calculation is straightforward as side $\times$ side.

Imagine a Parallelogram with a horizontal base of length $10$ cm. From one vertex on the top side, draw a perpendicular line straight down to the base. This perpendicular line represents the height, and its length is given as $10$ cm. The sides of the parallelogram that are not the base will be slanted, and their length will be greater than or equal to the height (it equals the height only if it's a rectangle, which is a specific type of parallelogram). The area calculation involves multiplying the base by this perpendicular height.