Chapter 9 Perimeter & Area (Class 7 - Maths NCERT Exemplar Solutions)

Given:

Length of the rectangle $= 12$ cm

Breadth of the rectangle $= 6$ cm

Number of congruent parts $= 8$

To Find:

Area of each congruent part.

Solution:

First, we calculate the total area of the rectangle:

$\text{Area of rectangle} = \text{Length} \times \text{Breadth}$

$\text{Area of rectangle} = 12 \times 6$

$\text{Area of rectangle} = 72$ cm$^2$

Since the rectangle is composed of 8 congruent parts, the area of each part is equal.

$\text{Area of each part} = \frac{\text{Total Area}}{\text{Number of parts}}$

$\text{Area of each part} = \frac{\cancel{72}^{9}}{\cancel{8}_{1}}$

$\text{Area of each part} = 9$ cm$^2$

Hence, the correct option is (d).

Given:

Area of right triangle $= 54$ cm$^2$

Length of one leg (say, altitude AB) $= 12$ cm

To Find:

Perimeter of the triangle.

Solution:

Let the other leg be the base (BC).

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

$54 = 6 \times BC$

$BC = \frac{54}{6} = 9$ cm

Now, we find the hypotenuse (AC) using Pythagoras theorem:

$AC^2 = AB^2 + BC^2$

$AC^2 = 12^2 + 9^2$

$AC^2 = 144 + 81$

$AC^2 = 225$

$AC = \sqrt{225} = 15$ cm

Finally, we calculate the perimeter:

$\text{Perimeter} = AB + BC + AC$

$\text{Perimeter} = 12 + 9 + 15$

$\text{Perimeter} = 36$ cm

Hence, the correct option is (c).

Given:

In parallelogram QPON:

Base $QP = 6$ cm

Height (altitude) $QM = 8$ cm

Solution:

The area of a parallelogram is given by the product of its base and corresponding height.

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

$\text{Area} = 6 \times 8$

$\text{Area} = 48$ cm$^2$

Therefore, the area of parallelogram QPON is 48 cm$^2$.

Solution:

We know that $1$ hectare $= 10,000$ m$^2$.

Also, $1$ m $= 100$ cm, therefore $1$ m$^2 = (100 \times 100)$ cm$^2 = 10,000$ cm$^2$.

To convert hectare to cm$^2$:

$1 \text{ hectare} = 10,000 \times 10,000$ cm$^2$

$1 \text{ hectare} = 10,00,00,000$ cm$^2$

Using exponential notation, $1$ hectare $= 10^8$ cm$^2$.

Given:

Side of small square $= 1$ m

Side of large square $= 5$ km $= 5,000$ m

Solution:

$\text{Area of large square} = \text{side} \times \text{side}$

$\text{Area of large square} = 5,000 \times 5,000 = 2,50,00,000$ m$^2$

$\text{Area of small square} = 1 \times 1 = 1$ m$^2$

$\text{Number of small squares} = \frac{\text{Area of large square}}{\text{Area of small square}}$

$\text{Number of small squares} = \frac{2,50,00,000}{1} = 2,50,00,000$

Therefore, 2,50,00,000 (or 2.5 crore) squares are required.

Solution:

Congruent triangles are identical in shape and size. Since their corresponding bases and altitudes are equal, their calculated areas must be the same.

Therefore, all the congruent triangles have equal (or same) area.

Solution:

The given statement is False.

Triangles can have the same area if the product of their base and height is equal, even if their sides and angles (shape) are completely different.

For example, a triangle with base 4 cm and height 3 cm has an area of 6 cm$^2$. Another triangle with base 6 cm and height 2 cm also has an area of 6 cm$^2$, but they are clearly not congruent.

Statement: The area of any parallelogram $ABCD$, is $AB \times BC$.

Solution:

The given statement is False.

The area of a parallelogram is calculated by the formula:

In a parallelogram $ABCD$, $AB$ and $BC$ are adjacent sides. The product $AB \times BC$ would only represent the area if the angle between them is $90^\circ$ (making it a rectangle). In general parallelograms, the height is not equal to the adjacent side.

Statement: Ratio of the circumference and the diameter of a circle is more than $3$.

Solution:

The given statement is True.

We know that the circumference ($C$) of a circle is given by $C = \pi d$, where $d$ is the diameter.

The ratio of the circumference to the diameter is:

The value of $\pi$ is approximately $3.14$ (or $\frac{22}{7}$), which is clearly greater than $3$.

Given:

Total Length of playground $= 160$ m

Total Width of playground $= 80$ m

Area for swings $= 80 \text{ m} \times 80 \text{ m}$

Width of the path $= 1.5$ m

To Find:

Area covered by grass.

Solution:

The total area of the playground is $160 \times 80 = 12,800$ m$^2$.

The area remaining after the swing portion is:

Remaining Length $= 160 - 80 = 80$ m

Remaining Width $= 80$ m

Remaining Area $= 80 \times 80 = 6,400$ m$^2$

Now, within this $80 \times 80$ portion, there are two paths of width $1.5$ m:

1. Area of path parallel to width $= 80 \times 1.5 = 120$ m$^2$

2. Area of path parallel to length $= 80 \times 1.5 = 120$ m$^2$

3. Area of common square at intersection $= 1.5 \times 1.5 = 2.25$ m$^2$

Total Area of Path $= 120 + 120 - 2.25 = 237.75$ m$^2$

Area covered by grass $=$ Remaining Area $-$ Total Area of Path

Area of grass $= 6,400 - 237.75 = 6,162.25$ m$^2$

Thus, the area covered by grass is $6,162.25$ m$^2$.

Given:

Base $AB = 8$ cm, Altitude $AE = 4$ cm

Base $AD = 6$ cm

To Find:

Altitude corresponding to side $AD$ (Let it be $h$).

Solution:

The area of a parallelogram remains constant regardless of which side is taken as the base.

$\text{Area of parallelogram } ABCD = \text{Base } AB \times \text{Altitude } AE$

$\text{Area} = 8 \times 4 = 32$ cm$^2$

Now, using $AD$ as the base:

$\text{Area} = AD \times h$

$32 = 6 \times h$

$h = \frac{32}{6} = \frac{16}{3}$ cm

$h = 5.33$ cm (approx.)

The altitude corresponding to side $AD$ is $5.33$ cm.

Given:

Inner Length ($l$) $= 30$ m, Inner Width ($w$) $= 20$ m

Path width along length $= 5$ m, Path width along width $= 8$ m

Rate $= \textsf{₹} 200$ per m$^2$

To Find:

Total cost of cementing the path.

Solution:

The outer dimensions of the plot including the path are:

Outer Length $= 30 + 8 + 8 = 46$ m

Outer Width $= 20 + 5 + 5 = 30$ m

$\text{Area of path} = (\text{Outer Area}) - (\text{Inner Area})$

$\text{Area of path} = (46 \times 30) - (30 \times 20)$

$\text{Area of path} = 1,380 - 600 = 780$ m$^2$

$\text{Total Cost} = \text{Area} \times \text{Rate}$

$\text{Total Cost} = 780 \times \textsf{₹} 200$

$\text{Total Cost} = \textsf{₹} 1,56,000$

The cost of cementing the path is $\textsf{₹} 1,56,000$.

Given:

Circumference ($C$) $= 33$ cm

To Find:

Area of the circle.

Solution:

We first find the radius ($r$) using the circumference formula:

$2\pi r = 33$

$2 \times \frac{22}{7} \times r = 33$

$r = \frac{33 \times 7}{44} = \frac{3 \times 7}{4} = \frac{21}{4}$ cm

Now, we find the area:

$\text{Area} = \pi r^2$

$\text{Area} = \frac{22}{7} \times \left(\frac{21}{4}\right) \times \left(\frac{21}{4}\right)$

$\text{Area} = \frac{22 \times 3 \times 21}{16} = \frac{1,386}{16}$

$\text{Area} = 86.625$ cm$^2$

The area of the circle is $86.625$ cm$^2$.

Given:

In the given figure (Fig. 9.12):

1. Point $D$ is the center of the circle.

2. $DE$ is the radius of the circle.

3. $AD = 5 \text{ cm}$ and $AE = 8 \text{ cm}$.

4. $ABCD$ is a rectangle where $D$ is a vertex and $B$ lies on the circumference of the circle.

To Find:

The perimeter of the rectangle $ABCD$.

Solution:

First, we find the radius of the circle. Since $A$ lies on the line segment $DE$:

$\text{Radius } (r) = DE = AD + AE$

$r = 5 \text{ cm} + 8 \text{ cm}$

In a rectangle, the diagonals are equal. Since $D$ is the center and $B$ is a point on the circle, the segment $DB$ is also a radius of the circle.

Now, consider the right-angled triangle $\triangle DAB$ (since $\angle DAB = 90^\circ$ in a rectangle). By Pythagoras Theorem:

$AB^2 + AD^2 = DB^2$

$AB^2 + 5^2 = 13^2$

$AB^2 + 25 = 169$

$AB^2 = 169 - 25$

$AB^2 = 144$

$AB = \sqrt{144}$

Now, we calculate the perimeter of the rectangle $ABCD$:

$\text{Perimeter} = 2 \times (\text{Length} + \text{Breadth})$

$\text{Perimeter} = 2 \times (AB + AD)$

$\text{Perimeter} = 2 \times (12 \text{ cm} + 5 \text{ cm})$

$\text{Perimeter} = 2 \times 17 \text{ cm}$

Therefore, the perimeter of the rectangle $ABCD$ is $34 \text{ cm}$.

Given:

From the figure (Fig. 9.13), we have a rectangle $ABCD$ where:

The shaded region $DEBF$ is a parallelogram.

To Find:

1. Area of the shaded parallelogram $DEBF$.

2. Area of each triangle ($\triangle DAF$ and $\triangle BCE$).

3. Ratio of the area of the shaded portion to the remaining area of the rectangle.

Solution:

Step 1: Determine the unknown lengths.

In rectangle $ABCD$, the opposite sides are equal ($AB = CD = 10 \text{ cm}$ and $AD = BC = 6 \text{ cm}$).

We can find the length of the base of the shaded parallelogram ($FB$):

$FB = 10 \text{ cm} - 4 \text{ cm} = 6 \text{ cm}$

Since $DEBF$ is a parallelogram, opposite sides are equal, so $DE = FB = 6 \text{ cm}$.

Now find $EC$:

$EC = DC - DE = 10 \text{ cm} - 6 \text{ cm} = 4 \text{ cm}$.

Step 2: Calculate the area of the shaded parallelogram $DEBF$.

$\text{Area of parallelogram} = \text{Base} \times \text{Height}$

Here, Base ($FB$) = $6 \text{ cm}$ and Height ($AD$) = $6 \text{ cm}$.

$\text{Area of shaded region} = 6 \text{ cm} \times 6 \text{ cm}$

Step 3: Calculate the area of each triangle.

For $\triangle DAF$ (Right-angled at $A$):

$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$

$\text{Area of } \triangle DAF = \frac{1}{2} \times 4 \text{ cm} \times 6 \text{ cm} = 12 \text{ cm}^2$

For $\triangle BCE$ (Right-angled at $C$):

$\text{Area of } \triangle BCE = \frac{1}{2} \times 4 \text{ cm} \times 6 \text{ cm} = 12 \text{ cm}^2$

Step 4: Calculate the ratio.

The remaining area of the rectangle consists of the two triangles:

$\text{Remaining area} = \text{Area of } \triangle DAF + \text{Area of } \triangle BCE$

$\text{Remaining area} = 12 \text{ cm}^2 + 12 \text{ cm}^2 = 24 \text{ cm}^2$

Now, the ratio of shaded area to remaining area is:

$\text{Ratio} = \frac{\text{Shaded Area}}{\text{Remaining Area}}$

$\text{Ratio} = \frac{36}{24}$

Dividing both by their highest common factor (12):

Final Answers:

1. Area of the shaded parallelogram is $36 \text{ cm}^2$.

2. Area of each triangle is $12 \text{ cm}^2$.

3. The required ratio is $3 : 2$.

Given:

Four shapes (1, 2, 3, and 4) are drawn on a squared grid. Let each small square side be $1 \text{ unit}$ and the area of each small square be $1 \text{ sq. unit}$.

To Find:

Identify the statement that is not correct among the given options.

Solution:

To evaluate the statements, we first calculate the Area and Perimeter of each shape by counting the grid units.

1. For Shape 1:

2. For Shape 2:

3. For Shape 3:

4. For Shape 4:

Verification of Statements:

(a) Shapes 1, 3 and 4 have different areas and different perimeters: Let's check. Shapes 1 and 4 have area $18$, while Shape 3 has area $16$ (Areas are different). However, Shapes 1, 3, and 4 all have the same perimeter of $22 \text{ units}$. Therefore, the claim that they have "different perimeters" is incorrect.

(b) Shapes 1 and 4 have the same area as well as the same perimeter: This is correct as both have Area $= 18$ and Perimeter $= 22$.

(c) Shapes 1, 2 and 4 have the same area: This is correct as all three have an Area of $18 \text{ sq. units}$.

(d) Shapes 1, 3 and 4 have the same perimeter: This is correct as all three have a Perimeter of $22 \text{ units}$.

Based on the analysis, statement (a) is the one that is not correct.

Correct Option: (a)

Given:

Dimensions of original sheet $= 6 \text{ cm} \times 5 \text{ cm}$

Dimensions of cut piece $= 3 \text{ cm} \times 2 \text{ cm}$

Solution:

$\text{Area of original sheet} = 6 \times 5 = 30 \text{ cm}^2$

$\text{Area of cut piece} = 3 \times 2 = 6 \text{ cm}^2$

$\text{Area of remaining sheet} = 30 - 6 = 24 \text{ cm}^2$

Hence, the correct option is (c).

Given:

Total number of unit squares $= 36$.

Area of the rectangle $= 36 \text{ sq. units}$.

To Find:

The least perimeter of a rectangle (where length $\neq$ breadth) formed by these squares.

Solution:

A rectangle is formed by joining $36$ unit squares. Let the length and breadth of the rectangle be $l$ and $b$ respectively. The area is given by the product of its dimensions:

To find the possible dimensions, we look for the factor pairs of $36$. For each pair, we calculate the perimeter using the formula:

The possible integer dimensions and their respective perimeters are shown in the table below:

Conclusion:

Mathematically, a square is a special type of rectangle with the least perimeter ($24 \text{ units}$). However, if the question distinguishes a rectangle from a square (implying length and breadth must be different), we look for the next smallest perimeter.

From the table, the perimeter values are $74, 40, 30, 26,$ and $24$.

Excluding the square ($6 \times 6$), the rectangle with the least perimeter has dimensions $9 \text{ units} \times 4 \text{ units}$.

Calculating the perimeter for these dimensions:

Thus, considering a rectangle with distinct sides, the least perimeter is $26 \text{ units}$.

Hence, the correct option is (b).

Given:

Side of square $= 22 \text{ cm}$.

Solution:

The length of the wire is equal to the perimeter of the square.

$\text{Length of wire} = 4 \times 22 = 88 \text{ cm}$.

When the wire is rebent into a circle, its circumference ($C$) will be $88 \text{ cm}$.

$2\pi r = 88$

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

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

$r = \frac{88 \times 7}{44} = 2 \times 7 = 14 \text{ cm}$.

Hence, the correct option is (b).

Given:

Radius of circle ($r$) $= 14 \text{ cm}$ (from Question 4).

Solution:

$\text{Area of circle} = \pi r^2$

$\text{Area} = \frac{22}{7} \times 14 \times 14$

$\text{Area} = 22 \times 2 \times 14$

$\text{Area} = 44 \times 14 = 616 \text{ cm}^2$.

Hence, the correct option is (c).

Given:

Area of rectangle $= 14 \text{ cm} \times 11 \text{ cm} = 154 \text{ cm}^2$.

$\text{Area of circle} = \text{Area of rectangle}$.

Solution:

$\pi r^2 = 154$

$\frac{22}{7} \times r^2 = 154$

$r^2 = \frac{154 \times 7}{22}$

$r^2 = 7 \times 7 = 49$

$r = \sqrt{49} = 7 \text{ cm}$.

Hence, the correct option is (d).

Given:

The shaded portion is a trapezium (or can be seen as rectangle minus a triangle).

Solution:

Base of rectangle $= 5 \text{ cm}$; Height $= 3 \text{ cm} + 1 \text{ cm} = 4 \text{ cm}$.

$\text{Area of shaded portion} = \text{Area of Trapezium}$

Parallel sides are $1 \text{ cm}$ and $3 \text{ cm}$. Distance between them (height) is $5 \text{ cm}$.

$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

$\text{Area} = \frac{1}{2} \times (1 + 3) \times 5 = \frac{1}{2} \times 4 \times 5 = 10 \text{ cm}^2$.

Hence, the correct option is (d).

Solution:

Area of a parallelogram is $\text{Base} \times \text{Height}$.

1. Using base $AD$, height is $BE$. So, $\text{Area} = AD \times BE$. (Option b is correct)

2. Using base $BC$, height is $BE$. So, $\text{Area} = BC \times BE$. (Option d is correct)

3. Using base $CD$ (or $DC$), height is $BF$. So, $\text{Area} = DC \times BF$. (Option c is correct)

Option (a) says $DE \times DC$. Here $DE$ is only a segment of the side $AD$ and not an altitude to $DC$. Therefore, this does not represent the area.

Hence, the correct option is (a).

Given:

In Fig. 9.17, $MNOP$ is a parallelogram. $OQ$ is the altitude dropped from vertex $O$ to the side $MP$.

Solution:

In the given figure, for $\triangle MNO$, let us consider $NO$ as the base.

Since $MNOP$ is a parallelogram, $MP$ is parallel to $NO$ ($MP \parallel NO$). The perpendicular distance between these parallel lines is given by the altitude $OQ$.

The height of $\triangle MNO$ corresponding to the base $NO$ is equal to the distance between the parallel lines $MP$ and $NO$, which is $OQ$.

We know that the area of a triangle is given by the formula:

Substituting base $= NO$ and height $= OQ$:

Therefore, the correct option is (d).

Given:

$\triangle MNO$ and parallelogram $MNOP$ share the same base $NO$ and lie between the same parallel lines $MP \parallel NO$.

Solution:

Let the base $NO = b$ and the height (distance between parallels) be $h$.

The area of $\triangle MNO$ is:

The area of parallelogram $MNOP$ is:

Now, finding the ratio:

This can be written as $1 : 2$.

Therefore, the correct option is (c).

Given:

From Fig. 9.18, we have three triangles $\triangle MNO$, $\triangle MOP$, and $\triangle MPQ$.

All three triangles have a common vertex $M$ and their bases $NO$, $OP$, and $PQ$ lie on the same straight line. Thus, they all have the same height, $MO = 5 \text{ cm}$.

The lengths of the bases are: $NO = 4 \text{ cm}$, $OP = 2 \text{ cm}$, and $PQ = 6 \text{ cm}$.

Solution:

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

1. $\text{Area}(\triangle MNO) = \frac{1}{2} \times 4 \times 5 = 10 \text{ cm}^2$

2. $\text{Area}(\triangle MOP) = \frac{1}{2} \times 2 \times 5 = 5 \text{ cm}^2$

3. $\text{Area}(\triangle MPQ) = \frac{1}{2} \times 6 \times 5 = 15 \text{ cm}^2$

The ratio of their areas is:

Dividing throughout by $5$:

Therefore, the correct option is (a).

Given:

Parallelogram $EFGH$ with base $EF = 10 \text{ cm}$.

Altitude corresponding to base $EF$ is $FI = 4 \text{ cm}$ (since $FI \perp HG$ and $HG \parallel EF$).

Solution:

The area of a parallelogram is calculated as:

Taking $EF$ as the base and $FI$ as the height:

Therefore, the correct option is (c).

Solution:

By definition, $\pi$ (pi) is the ratio of the circumference of a circle to its diameter.

We know the formula for circumference ($C$) is:

Rearranging the formula to find $\pi$:

Therefore, the correct option is (c).

Solution:

The circumference ($C$) of a circle is given by the formula:

The value of $\pi$ is approximately $3.14$ or $\frac{22}{7}$.

Since $3.14 > 3$, the value of $\pi \times d$ will always be greater than $3 \times d$.

Hence, the circumference is more than three times of its diameter.

Therefore, the correct option is (a).

Given:

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

$\text{Altitude (height) } QT = 10 \text{ cm}$

To Find:

Base $PR$ of the triangle.

Solution:

We use the area formula for a triangle:

Substituting the given values:

Dividing both sides by $5$:

Therefore, the correct option is (a).

Given:

In $\triangle PQR$:

$PR = 12 \text{ cm}$, $QR = 6 \text{ cm}$

$PL \perp QR$ extended, so $PL = 8 \text{ cm}$ (height for base $QR$)

$QM \perp PR$ (height for base $PR$)

Solution:

The area of the triangle remains the same regardless of which side is chosen as the base.

Case 1: Taking $QR$ as base and $PL$ as height.

Case 2: Taking $PR$ as base and $QM$ as height.

Since the area is $24 \text{ cm}^2$:

Therefore, the correct option is (c).

Given:

In right-angled triangle $\triangle MNO$, legs $MN = 6 \text{ cm}$ and $NO = 8 \text{ cm}$. $NP \perp MO$.

To Find:

Length of perpendicular $NP$.

Solution:

First, we find the hypotenuse $MO$ using the Pythagoras theorem:

Now, we calculate the area of $\triangle MNO$ using legs $MN$ and $NO$:

We can also calculate the area using $MO$ as the base and $NP$ as the height:

Therefore, the correct option is (a).

Given:

$\text{Area of right-angled triangle} = 30 \text{ cm}^2$.

Smallest side (let it be the base) $= 5 \text{ cm}$.

To Find:

Length of the hypotenuse.

Solution:

In a right-angled triangle, the two sides containing the right angle are the base and height.

Now, using the Pythagoras theorem to find the hypotenuse ($h$):

Therefore, the correct option is (b).

Given:

$\text{Diameter (} d \text{)} = 5 \text{ cm}$.

Solution:

We know that the formula for the circumference ($C$) of a circle is:

Taking $\pi = 3.14$:

Therefore, the correct option is (c).

Given:

$\text{Circumference (} C \text{)} = 88 \text{ cm}$.

Solution:

The formula for circumference is $C = 2 \pi r$.

Solving for $r$:

Therefore, the correct option is (c).

Given:

$\text{Radius (} r \text{)} = 10 \text{ cm}$.

Solution:

The edge of a semicircular disc consists of the curved boundary (arc) and the straight boundary (diameter).

Length of curved boundary $= \pi r$

Length of straight boundary (diameter) $= 2r$

Using $\pi = 3.14$ and $r = 10$:

Therefore, the correct option is (b).

Given:

$\text{Diameter (} d \text{)} = 8 \text{ m}$.

Solution:

First, find the radius ($r$):

The formula for the area of a circle is:

Using $\pi = 3.14$:

Therefore, the correct option is (c).

Given:

Diameter $= m$, Radius $= n$, and Circumference $= p$.

Solution:

The area ($A$) of any circle is given by the formula:

Substituting the given radius $n$:

Therefore, the correct option is (d).

Given:

$\text{Diameter (} d \text{)} = 2.8 \text{ m}$.

Solution:

First, find the radius ($r$):

The formula for the area of a semicircle is:

Using $\pi = \frac{22}{7}$:

Therefore, the correct option is (a).

Given:

$1 \text{ m}^2 = x \text{ mm}^2$

Solution:

We know the relationship between meters and millimeters is:

Therefore,

To find the area in square millimeters, we square both sides:

Comparing this with $1 \text{ m}^2 = x \text{ mm}^2$, we get $x = 10,00,000$.

Therefore, the correct option is (d).

Solution:

Area of one small square with side $1 \text{ mm}$ is:

Area of the large square with side $1 \text{ cm}$ (where $1 \text{ cm} = 10 \text{ mm}$) is:

The number of small squares $p$ is calculated as:

Therefore, the correct option is (b).

Evaluation of Options:

(a) Square with side 12 m: $\text{Area} = 12 \times 12 = 144 \text{ m}^2$. (Incorrect)

(b) 12 squares with side 1 m each: $\text{Area of 1 square} = 1 \times 1 = 1 \text{ m}^2$. Total area for 12 such squares $= 12 \times 1 \text{ m}^2 = 12 \text{ m}^2$. (Correct)

(c) 3 squares with side 4 m each: $\text{Area of 1 square} = 4 \times 4 = 16 \text{ m}^2$. Total area $= 3 \times 16 = 48 \text{ m}^2$. (Incorrect)

(d) 4 squares with side 3 m each: $\text{Area of 1 square} = 3 \times 3 = 9 \text{ m}^2$. Total area $= 4 \times 9 = 36 \text{ m}^2$. (Incorrect)

Therefore, the correct option is (b).

Solution:

Let the side of the original rhombus be $s$ and its height be $h$.

When each side is doubled, the new side $s' = 2s$. Since the shape remains a rhombus, the altitude (height) $h'$ also doubles, $h' = 2h$.

The new area is 4 times the original area.

Therefore, the correct option is (d).

Solution:

Let the original sides of the parallelogram be $a$ and $b$.

The new sides are $2a$ and $2b$.

The perimeter of the new parallelogram will be 2 times the original perimeter.

Therefore, the correct option is (b).

Solution:

Let the original radius be $r$.

The new radius is $2r$.

The area increases to 4 times the original area.

Therefore, the correct option is (d).

Given:

$\text{Radius of the circle (} r \text{)} = 10 \text{ cm}$.

Solution:

For the largest square to be cut out of a circle, the diagonal of the square must be equal to the diameter of the circle.

The area of a square can be calculated using its diagonal as:

Therefore, the correct option is (b).

Given:

$\text{Length of rectangle} = 10 \text{ cm}$.

$\text{Breadth of rectangle} = 8 \text{ cm}$.

Solution:

The largest circle that can fit inside a rectangle must have its diameter equal to the shorter side of the rectangle. If the diameter were larger, the circle would extend beyond the boundaries of the rectangle.

To find the radius ($r$):

Therefore, the correct option is (a).

Given:

From the figure, the lengths of the vertical segments are: $CD = 4 \text{ cm}$, $AB = 6 \text{ cm}$, $JI = 3 \text{ cm}$, $HG = 5 \text{ cm}$, and $FE = 2 \text{ cm}$.

The total horizontal base $DE = 20 \text{ cm}$.

Solution:

The perimeter of the figure is the sum of all its boundary segments.

Step 1: Sum of Vertical Segments

Step 2: Sum of Horizontal Segments

The segments $CB$, $AJ$, $IH$, and $GF$ are horizontal. By observing the figure, the sum of these segments corresponds to the total horizontal span, which is equal to the base $DE$.

Total horizontal distance is the sum of the top segments and the bottom segment:

Step 3: Total Perimeter

Therefore, the correct option is (a).

Given:

$\text{Area of circle} = 81 \pi r^2$

Solution:

Let the radius of the circle be $R$. We know the formula for area is:

Canceling $\pi$ from both sides:

Now, calculate the circumference ($C$):

Therefore, the correct option is (b).

Given:

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

Solution:

First, find the side of the square ($s$):

The largest circle that can be cut out of a square has a diameter equal to the side of the square.

The circumference ($C$) is given by $\pi D$:

Therefore, the correct option is (b).

Solution:

Let the original radius be $r$. Then the original area is $A_1 = \pi r^2$.

When the radius is tripled, the new radius becomes $3r$.

The area becomes 9 times the original area.

Therefore, the correct option is (a).

Given:

$\text{Radius of semicircle (} R \text{)} = 4r$

Solution:

The formula for the area of a semicircle is:

Substituting $R = 4r$:

Therefore, the correct option is (a).

Solution:

For a regular polygon, all sides are of equal length. The total distance around the polygon is the product of the number of sides and the length of one side.

Blank: number of sides

Solution:

When a wire is rebent, its total length does not change. The length of the wire forms the boundary of the shape.

However, the amount of region enclosed by the boundary can change depending on the dimensions.

Blanks: perimeter, area

Given:

$\text{Area of square MNOP} = 144 \text{ cm}^2$

Solution:

In Fig. 9.24, the square $MNOP$ is divided by its two diagonals and two lines passing through the center (connecting the midpoints of opposite sides).

This construction divides the square into $8$ congruent (equal) triangles.

Blank: 18 cm²

Given:

$ACDF$ is a rectangle with length $FD = 10 \text{ cm}$ and breadth $CD = 5 \text{ cm}$.

In the figure, $AB = 3 \text{ cm}$ and $BC$ is a part of the side $AC$.

Solution:

Since $ACDF$ is a rectangle, opposite sides are equal. Therefore:

Now, we find the length of the base $BC$ of the parallelogram $BCEF$:

The height of parallelogram $BCEF$ is the perpendicular distance between the parallel lines $AC$ and $FD$, which is equal to the side $CD$ of the rectangle.

Area of a parallelogram is calculated as:

Blank: 35

Solution:

In a parallelogram, any of the four sides can be considered as the base for the purpose of calculating the area, provided the corresponding altitude (height) is used.

Blank: base

Solution:

The vertical or perpendicular distance from an opposite vertex to the chosen base is called the altitude or height.

Blank: altitude (or height)

Solution:

Just as perimeter refers to the boundary of a polygon, the total length of the boundary of a circle is specifically called its circumference.

Blank: circumference

Solution:

The mathematical constant representing the ratio $\frac{\text{Circumference}}{\text{Diameter}}$ is denoted by the Greek letter $\pi$ (pi).

Blank: $\pi$

Given:

$\text{Area} = 90 \text{ cm}^2$

$\text{Base} = 20 \text{ cm}$

To Find:

Length of the altitude ($h$).

Solution:

Using the area formula for a triangle:

Blank: 9

Solution:

In common calculations, $\pi$ is an irrational number approximated for ease of use.

Blank: $\frac{22}{7}$ (or $3.14$)

Solution:

From the definition of $\pi$, we have $C = \pi \times d$.

Blank: $\pi$

Solution:

The standard formula for the circumference ($C$) of a circle in terms of its radius ($r$) is given by:

Blank: radius (or r)

Solution:

We know that:

Squaring both sides to find the area conversion:

Blank: 10,000

Solution:

We know that:

Squaring both sides:

Blank: 100

Solution:

A hectare is a metric unit of area defined as the area of a square with sides of $100 \text{ metres}$.

Blank: 10,000

Solution:

The general formula to calculate the area of any triangle is half the product of its base and its corresponding vertical height.

Blank: height (or altitude)

Solution:

We know that:

Squaring both sides:

In the Indian numbering system, this is written as $10,00,000$.

Blank: 10,00,000 (or 1,000,000)

Solution:

First, calculate the area of the large square in square centimeters ($cm^2$).

Area of one small square with side $1 \text{ cm}$:

Number of small squares needed:

Blank: 3,60,000

Solution:

We know that:

Therefore,

For $10 \text{ cm}^2$:

Blank: 0.001 (or $\frac{1}{1000}$)

Solution:

1. Area: Figure (ii) is formed by removing a small rectangular portion from the interior edge of Figure (i). Since part of the region is removed, the area of (ii) is definitely smaller than the area of (i).

2. Perimeter: The perimeter is the length of the outer boundary. In Figure (ii), the flat bottom edge of (i) is replaced by three segments (two vertical and one horizontal) that "go inside" the shape. The sum of these three segments is greater than the single straight segment they replaced. Therefore, the perimeter of (ii) is greater than that of (i).

The statement is True.

Solution:

Let us count the unit squares and boundary segments on the grid for both shapes.

For Shape (i):

For Shape (ii):

This shape is created by taking the $4 \times 3$ rectangle, removing $1$ square from the bottom and adding $1$ square to the top.

Perimeter count for (ii): 3 (Left) + 1 (bottom segment) + 1 (up) + 1 (across) + 1 (down) + 1 (bottom segment) + 3 (Right) + 1 (top segment) + 1 (up) + 1 (across) + 1 (down) + 1 (top segment).

Verification:

(a) True (Both have area $12$).

(b) False (Perimeter of (i) is $14$, while (ii) is $18$).

(c) False (Area is $12$ unit squares, not $13$).

(d) True (By counting boundary segments, perimeter is $18$).

Solution:

Two parallelograms can have the same perimeter but different shapes. For example, a square (special parallelogram) and a very thin, slanted parallelogram can have equal perimeters, but the square will enclose a much larger area.

The statement is False.

Solution:

Congruent triangles are identical in all respects (sides and angles). Since their dimensions are exactly the same, the region they enclose (area) must also be identical.

The statement is True.

Solution:

The area of a parallelogram depends on the product of its base and height, whereas the perimeter depends on the sum of the lengths of its adjacent sides.

Consider two rectangles (which are special types of parallelograms) with the same area of $36 \text{ cm}^2$:

1. Rectangle A: sides are $6 \text{ cm}$ and $6 \text{ cm}$.

2. Rectangle B: sides are $9 \text{ cm}$ and $4 \text{ cm}$.

Since the areas are equal but the perimeters are different, the given statement is incorrect.

Therefore, the statement is False.

Solution:

In Fig. 9.28, all triangles ($\triangle FAB, \triangle EAB, \triangle DAB, \text{ and } \triangle CAB$) share the common base AB. Furthermore, the vertices $F, E, D, \text{ and } C$ all lie on the same horizontal line which is parallel to the base line. Thus, the perpendicular distance (altitude) from these vertices to the base $AB$ is the same for all.

The statement is True.

Solution:

While the triangles have the same base and height, their side lengths and interior angles are clearly different (e.g., $\triangle FAB$ is a right-angled triangle, while others are not). Shapes must be identical to be congruent.

The statement is False.

Solution:

The area of a triangle is given by the formula:

As established in Question 62, all these triangles have the same base and the same altitude. Therefore, their areas must be equal.

The statement is True.

Solution:

Referring to Fig. 9.28, all the triangles ($\triangle FAB, \triangle EAB, \triangle DAB, \text{ and } \triangle CAB$) are constructed on the same base AB and between the same parallel lines.

While these triangles have the same base and the same altitude (perpendicular height), the lengths of their other two sides (slanted sides) are different because their third vertices are at different positions on the top parallel line.

For example, in $\triangle FAB$, the side $AF$ is vertical, making it a right-angled triangle, whereas in $\triangle EAB$, the sides $AE$ and $EB$ have different lengths compared to $AF$ and $FB$.

Since perimeter is the sum of all three sides, and at least two sides vary in length for each triangle, their perimeters will not be the same.

Therefore, the statement is True.

Solution:

In Fig. 9.29, both triangles $\triangle ABC$ and $\triangle ACD$ share the same vertex $A$ and their bases $BC$ and $CD$ lie on the same straight line $BD$.

The perpendicular $AC$ is the common altitude (height) for both triangles.

Now, finding the ratio of their areas:

This shows the ratio of the areas is indeed the same as the ratio of their bases.

Therefore, the statement is True.

Solution:

The area of a triangle is given by the formula:

For two triangles to have equal area, they must have both the same base and the same altitude (height). If they only have the same base but different heights, their areas will be different.

Therefore, the statement is False.

Solution:

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

To find the ratio of circumference to radius, we divide $C$ by $r$:

This can be expressed as the ratio $2 \pi : 1$.

Therefore, the statement is True.

Solution:

In the metric system of measurement commonly used in India, $1 \text{ hectare}$ is defined as follows:

To find the value of $5 \text{ hectares}$:

Since $50,000 \neq 500$, the statement is incorrect.

Therefore, the statement is False.

Solution:

There is no direct proportional relationship between perimeter and area when comparing different shapes. It is possible for a figure to have a larger perimeter but a smaller area than another figure.

Example:

1. Consider a square with side $10 \text{ cm}$:

2. Consider a very thin rectangle with length $20 \text{ cm}$ and breadth $1 \text{ cm}$:

In this case, the perimeter increased from $40$ to $42$, but the area decreased from $100$ to $20$.

Therefore, the statement is False.

Solution:

This is a standard geometric property. Different shapes can enclose the same amount of space (area) using boundaries of different lengths (perimeter).

Example:

1. A rectangle of $9 \text{ cm} \times 4 \text{ cm}$ has $\text{Area} = 36 \text{ cm}^2$ and $\text{Perimeter} = 26 \text{ cm}$.

2. A rectangle (square) of $6 \text{ cm} \times 6 \text{ cm}$ has $\text{Area} = 36 \text{ cm}^2$ and $\text{Perimeter} = 24 \text{ cm}$.

Both figures have the same area of $36 \text{ cm}^2$ but different perimeters.

Therefore, the statement is True.

Solution:

A figure with a larger area does not automatically have a larger perimeter. The shape of the figure significantly influences these values.

Example:

1. Figure A: A square with side $4 \text{ cm}$.

2. Figure B: A rectangle with length $8 \text{ cm}$ and breadth $1 \text{ cm}$.

Here, Figure A has a larger area ($16 > 8$) but a smaller perimeter ($16 < 18$) than Figure B.

Therefore, the statement is True.

Given:

Length of the rectangular lawn ($l$) $= 72 \text{ m}$

Breadth of the rectangular lawn ($b$) $= 18 \text{ m}$

Number of shrubs per metre $= 3$

To Find:

Total number of shrubs to be planted along the boundary.

Solution:

The shrubs are planted along the boundary, which corresponds to the perimeter of the rectangle.

We know that:

Substituting the given values:

Now, to find the total number of shrubs:

Therefore, 540 shrubs will be planted in all.

Given:

Total land dimensions $= 15 \text{ m} \times 10 \text{ m}$

Camel area (rectangle) $= 5 \text{ m} \times 3 \text{ m}$

Ox area (circle) with diameter $= 2.8 \text{ m}$

Plants area (rectangle) $= 9 \text{ m} \times 1 \text{ m}$

Solution:

First, we calculate the area of each section:

For the Ox area, radius $r = \frac{2.8}{2} = 1.4 \text{ m}$:

Now, find the combined area for animals and plants:

Calculate the remaining living area:

Find the required ratio:

Dividing both by $8$:

Therefore, the ratio is 377 : 1498.

Given:

Perimeter ($P$) $= 40 \text{ m}$

Length ($l$) $= 5 \times \text{breadth} - 4$

Solution:

Let the breadth of the rectangle be $b$ metres.

Then, the length $l = 5b - 4$.

The formula for perimeter is:

Substituting the values:

Now, calculate the length:

Finally, find the area:

Therefore, the area of the rectangle is 64 m².

Given:

Wall dimensions $= 5 \text{ m} \times 4 \text{ m}$

Window dimensions $= 1.5 \text{ m} \times 1 \text{ m}$

Door dimensions $= 2.25 \text{ m} \times 1 \text{ m}$

To Find:

The area of the wall to be painted (excluding window and door).

Solution:

Calculate the areas of individual components:

The area to be painted is:

Therefore, the area to be painted is 16.25 m².

Given:

Area of one small rectangle $= 8 \text{ m}^2$

Breadth of one small rectangle $= 2 \text{ m}$

MNOP is composed of four such congruent rectangles.

Solution:

First, find the length ($l$) of one small rectangle:

From Fig. 9.31, the large rectangle $MNOP$ is arranged as follows:

1. The bottom part consists of one horizontal rectangle (Width $= 4 \text{ m}$, Height $= 2 \text{ m}$).

2. The top part consists of one horizontal rectangle (Width $= 4 \text{ m}$, Height $= 2 \text{ m}$).

3. The middle part consists of two vertical rectangles placed side-by-side. Each has height $= 4 \text{ m}$ and width $= 2 \text{ m}$. Together, their total width is $2 + 2 = 4 \text{ m}$.

Now, calculate the total dimensions of $MNOP$:

Find the perimeter of $MNOP$:

Therefore, the perimeter of MNOP is 24 m.

Given:

In Fig. 9.32, $ABCD$ is a parallelogram and $\triangle AFB$ is a triangle on the same base $AB$.

Base $AB = 10 \text{ cm}$.

Altitude of $\triangle AFB$ (EF) $= 16 \text{ cm}$.

Area of $\triangle AFB = \text{Area of parallelogram } ABCD$.

$O$ is the midpoint of $DC$.

To Find:

(i) Altitude of the parallelogram $ABCD$ to the base $AB$.

(ii) Area of $\triangle DAO$.

Solution:

First, we calculate the area of $\triangle AFB$:

According to the question, Area of parallelogram $ABCD = \text{Area of } \triangle AFB$.

Let $h$ be the altitude of the parallelogram to the base $AB$.

Now, for the second part, $O$ is the midpoint of $DC$. Since $ABCD$ is a parallelogram, $DC = AB = 10 \text{ cm}$.

In $\triangle DAO$, the base is $DO$ and the altitude corresponds to the distance between the parallel lines $AB$ and $DC$, which we found to be $8 \text{ cm}$.

Results:

Altitude of parallelogram $= 8 \text{ cm}$.

Area of $\triangle DAO = 20 \text{ cm}^2$.

Given:

Ratio of Area($\triangle WXY$) : Area($\triangle WZY$) $= 3 : 4$.

Total Area($\triangle WXZ$) $= 56 \text{ cm}^2$.

Altitude $WY = 8 \text{ cm}$.

To Find:

Lengths of $XY$ and $YZ$.

Solution:

Let the area of $\triangle WXY$ be $3k$ and the area of $\triangle WZY$ be $4k$.

Now, we find the individual areas:

Using the area formula for $\triangle WXY$ with base $XY$ and height $WY$:

Similarly, for $\triangle WZY$ with base $YZ$ and height $WY$:

Results:

$XY = 6 \text{ cm}$ and $YZ = 8 \text{ cm}$.

Given:

Side of the square field $= 70 \text{ m}$.

A semicircular lawn is made with the maximum area inside this field.

Solution:

For a semicircular lawn to have maximum area within a square of side $70 \text{ m}$, its diameter must be equal to the side of the square.

(i) Perimeter of the lawn:

The perimeter of a semicircular lawn consists of the curved arc and the straight diameter.

(ii) Area of square field excluding the lawn:

The excluded area is:

Results:

Perimeter of lawn $= 180 \text{ m}$.

Excluded area $= 2975 \text{ m}^2$.

Given:

In Fig. 9.35, $ABCD$ is a parallelogram.

Shaded triangle $\triangle ABE$ has Area $= 9 \text{ cm}^2$.

Base of triangle $BE = 3 \text{ cm}$.

Remaining part of parallelogram base $EC = 4 \text{ cm}$.

To Find:

Area of parallelogram $ABCD$.

Solution:

Let $h$ be the height of the triangle $ABE$ (perpendicular $AE$). This $h$ is also the height of the parallelogram $ABCD$.

The base of the parallelogram $ABCD$ is $BC$.

Now, calculate the area of the parallelogram:

Therefore, the area of the parallelogram is 42 cm².

Given:

Square Pizza: side $= 45 \text{ cm}$, Cost $= \textsf{₹} 150$.

Circular Pizza: diameter $= 50 \text{ cm}$, Cost $= \textsf{₹} 160$.

Solution:

To determine the better deal, we should find out which pizza gives more area per unit of money spent (or has a lower cost per unit area).

1. For Square Pizza:

2. For Circular Pizza:

Radius $r = \frac{50}{2} = 25 \text{ cm}$. Taking $\pi \approx 3.14$:

Comparing the two results:

Square pizza gives $13.5 \text{ cm}^2$ per rupee, while the circular pizza gives approximately $12.26 \text{ cm}^2$ per rupee.

Since the square pizza provides a larger area for every rupee spent, it is the better deal.

Given:

Three squares are attached such that each square to the left is attached at the midpoint of the side of the square to its right. From the figure:

Side of the largest square (left) $= 6 \text{ m}$

Side of the middle square $= 3 \text{ m}$ (Since it is attached to the midpoint of the $6 \text{ m}$ side)

Side of the smallest square (right) $= 1.5 \text{ m}$ (Since it is attached to the midpoint of the $3 \text{ m}$ side)

Solution:

The perimeter of the figure is the sum of all its external boundary segments.

1. Horizontal Boundary:

The bottom horizontal boundary is the sum of the sides of the three squares:

The top horizontal boundary consists of the top sides of each square:

2. Vertical Boundary:

The leftmost vertical side is $6 \text{ m}$.

The rightmost vertical side is $1.5 \text{ m}$.

Additionally, there are vertical segments where the squares meet. Since they are attached at midpoints:

Exposed vertical parts of the $6 \text{ m}$ square $= (6 - 3) = 3 \text{ m}$ ($1.5 \text{ m}$ above and $1.5 \text{ m}$ below the junction).

Exposed vertical parts of the $3 \text{ m}$ square $= (3 - 1.5) = 1.5 \text{ m}$ ($0.75 \text{ m}$ above and $0.75 \text{ m}$ below the junction).

3. Total Perimeter Calculation:

Therefore, the perimeter of the complete figure is 33 m.

Given:

$ABCD$ is a square with side $s = 15 \text{ cm}$.

$BDFE$ is a square formed on the diagonal $BD$ of square $ABCD$.

Solution:

First, we find the length of the diagonal $BD$ using the Pythagoras theorem in $\triangle BCD$:

Since $BDFE$ is a square with side $BD$, its area is given by the square of its side:

Therefore, the area of the square BDFE is 450 cm².

Given:

In $\triangle PQR$: sides are $6 \text{ cm}$, $10 \text{ cm}$, and $14 \text{ cm}$.

In $\triangle ABC$: $\angle B = 90^\circ$, $BC = 5 \text{ cm}$, and hypotenuse $AC = 13 \text{ cm}$.

$\text{Perimeter}(\triangle ABC) = \text{Perimeter}(\triangle PQR)$.

Solution:

First, calculate the perimeter of $\triangle PQR$:

So, $\text{Perimeter}(\triangle ABC) = 30 \text{ cm}$.

In $\triangle ABC$, let the altitude $AB$ be $x$. The perimeter is:

Now, we find the area of the right-angled $\triangle ABC$:

Therefore, the area of $\triangle ABC$ is 30 cm².

Given:

In parallelogram $MGHK$:

Side $GH = 6 \text{ cm}$. Altitude corresponding to side $GH$ is $MN = 8 \text{ cm}$.

Altitude corresponding to side $MG$ is $MO = 4 \text{ cm}$.

Solution:

The area of the parallelogram remains constant regardless of which base is chosen.

Step 1: Calculate Area using base GH

Step 2: Find side MG using Area

Step 3: Calculate Perimeter

Since $MGHK$ is a parallelogram, $MK = GH = 6 \text{ cm}$ and $KH = MG = 12 \text{ cm}$.

Therefore, the perimeter of MGHK is 36 cm.

Given:

$\text{Area}(\triangle PQR) = 20 \text{ cm}^2$.

$\text{Area}(\triangle PQS) = 44 \text{ cm}^2$.

$PQ \perp QS$ and base $QR = 5 \text{ cm}$.

To Find:

Length of the segment $RS$.

Solution:

Since $PQ \perp QS$, $PQ$ is the common altitude for both $\triangle PQR$ and $\triangle PQS$.

Step 1: Find the length of PQ

Step 2: Find the length of QS

Step 3: Find RS

From the figure, $S$, $R$, and $Q$ lie on the same straight line, so:

Therefore, the length of RS is 6 cm.

Given:

Area of an isosceles triangle $= 48 \text{ cm}^2$

Altitude (height) corresponding to the base $= 8 \text{ cm}$

To Find:

Perimeter of the triangle.

Solution:

Let the base of the isosceles triangle be $b$.

In an isosceles triangle, the altitude to the base bisects the base into two equal parts.

Let $a$ be the length of the equal sides. Using Pythagoras theorem in the right-angled triangle formed by the altitude and half of the base:

Now, we find the perimeter:

Therefore, the perimeter of the triangle is 32 cm.

Given:

Perimeter of parallelogram $= 96 \text{ m}$

Area of parallelogram $= 270 \text{ m}^2$

One side $= 18 \text{ m}$

To Find:

Other side, and altitudes $l$ and $m$.

Solution:

Let the other side of the parallelogram be $b$.

Finding Altitude $m$ (corresponding to side 18 m):

Finding Altitude $l$ (corresponding to side 30 m):

Therefore, the other side is 30 m, and the altitudes are $l = 9 \text{ m}$ and $m = 15 \text{ m}$.

Given:

In $\triangle PQR$, $\angle Q = 90^\circ$.

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

Smallest side (let $PQ$) $= 8 \text{ cm}$.

To Find:

The other two sides ($QR$ and hypotenuse $PR$).

Solution:

Since the triangle is right-angled at $Q$, sides $PQ$ and $QR$ are the base and height.

Now, find the hypotenuse $PR$ using Pythagoras theorem:

Therefore, the lengths of the other two sides are 15 cm and 17 cm.

Given:

A large rectangle is composed of 5 congruent (identical) small rectangles.

Perimeter of the large rectangle $= 264 \text{ cm}$.

Solution:

Let the length of one small rectangle be $l$ and its breadth be $b$.

From Fig. 9.44, we see that the top row has 3 rectangles placed vertically (width $b$), and the bottom row has 2 rectangles placed horizontally (width $l$). Since they form a single large rectangle, their total widths must be equal:

Now, consider the dimensions of the large rectangle:

Using the perimeter formula for the large rectangle:

Substitute $l = 1.5b$:

Finding $l$:

Now, calculate the perimeter of one small rectangle:

Therefore, the perimeter of one of the rectangles is 120 cm.

Given:

Radius of the circle ($r$) $= 7 \text{ cm}$.

To Find:

Area of the inscribed square.

Solution:

When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.

The area of a square can be found using the length of its diagonal:

Therefore, the area of the square is 98 cm².

Given:

Radius of the circle ($r$) $= 7 \text{ cm}$.

A square is inscribed in the circle as shown in Fig. 9.45. From Question 92, the area of this square is $98 \text{ cm}^2$.

To Find:

Area of the shaded portion (the region between the circle and the square).

Solution:

First, we calculate the area of the circle:

Now, to find the shaded portion, we subtract the area of the square from the area of the circle:

Therefore, the area of the shaded portion is 56 cm².

Given:

The figure (Fig. 9.46) consists of a rectangle and a semicircle.

Total base of the figure $= 10.2 \text{ cm}$.

Height of the rectangular part $= 1.5 \text{ cm}$.

Horizontal segment before the semicircle $= 3.9 \text{ cm}$.

Solution:

Step 1: Find the dimensions of the semicircle.

The diameter ($d$) of the semicircle is the remaining part of the total base:

Step 2: Calculate individual areas.

Step 3: Calculate total area.

Therefore, the area enclosed is 30.8925 cm².

Solution:

The figure (Fig. 9.47) can be divided into a rectangle and a right-angled triangle by drawing a horizontal line from the $4 \text{ cm}$ height.

1. Dimensions of the Rectangle:

Length $= 13 \text{ cm}$, Breadth $= 4 \text{ cm}$.

2. Dimensions of the Triangle:

Total Area:

Therefore, the total area enclosed is 82 cm².

Solution:

The figure (Fig. 9.48) is composed of a rectangle and a triangle.

1. Area of the Rectangle:

Length $= 15 \text{ cm}$, Breadth $= 3 \text{ cm}$.

2. Area of the Triangle:

Height $= 4 \text{ cm}$.

Base of the triangle $= \text{Total length} - \text{labeled segment} $$ = 15 - 10 = 5 \text{ cm}$.

Total Area:

Therefore, the area enclosed is 55 cm².

Solution:

The figure (Fig. 9.49) is a combination of a semicircle and a triangle.

1. Area of the Semicircle:

From the figure, the radius $r = 10 \text{ cm}$.

Using $\pi = 3.14$:

2. Area of the Triangle:

The base of the triangle is the diameter of the semicircle $= 2 \times 10 = 20 \text{ cm}$.

The altitude (height) of the triangle $= 17 - 10 = 7 \text{ cm}$.

Total Area:

Therefore, the area enclosed is 227 cm².

Given:

Diameter of the inner circle ($d$) $= 7 \text{ cm}$.

Width of the shaded ring $= 7 \text{ cm}$.

Solution:

First, we find the radius of the inner circle ($r$):

The radius of the outer circle ($R$) is the inner radius plus the width of the ring:

The area of the shaded region is the difference between the areas of the outer and inner circles:

Using the identity $a^2 - b^2 = (a+b)(a-b)$:

Therefore, the area of the shaded region is 308 cm².

Given:

Diameter of the large circle ($D$) $= 14 \text{ cm}$.

Diameter of each small circle ($d$) $= \frac{7}{4} \text{ cm}$.

Solution:

Step 1: Calculate the area of the large circle.

Radius of large circle $R = \frac{14}{2} = 7 \text{ cm}$.

Step 2: Calculate the area of the two small circles.

Radius of small circle $r = \frac{d}{2} = \frac{7}{4 \times 2} = \frac{7}{8} \text{ cm}$.

Step 3: Calculate shaded area.

Therefore, the area of the shaded region is 149.1875 cm².

Given:

Radius of the circle ($r$) $= 16 \text{ cm}$.

Solution:

1. Original Perimeter:

The perimeter of the original circle is its circumference ($C$).

2. Perimeter of the new shape:

The circle is cut into four quadrants. When rearranged, the boundary of the new shape consists of:

i. The four arcs of the quadrants, which together equal the original circumference ($2 \pi r$).

ii. Two straight radial segments (radii) that were internal in the original circle but are now part of the external boundary.

3. Change in Perimeter:

Therefore, the perimeter increases by 32 cm.

Given:

Perimeter of each congruent rectangle $= 16 \text{ cm}$.

Solution:

Let the length and breadth of one rectangle be $l$ and $b$ respectively.

By observing Fig. 9.53, we can see that the side of the large square is formed by the sum of the length of one rectangle and the breadth of another rectangle.

Now, calculate the area of the large square:

Therefore, the area of the large square is 64 cm².

Given:

$AE \perp CD$

$AC = 5 \text{ cm}$, $DE = 4 \text{ cm}$

$\text{Area}(\triangle AED) = 6 \text{ cm}^2$

Solution:

Step 1: Find the height AE.

In right-angled triangle $AED$:

Step 2: Find sides AD and EC.

In right $\triangle AED$, using Pythagoras theorem:

In right $\triangle AEC$, using Pythagoras theorem ($AC$ is hypotenuse):

Step 3: Calculate Area and Perimeter of ABCD.

Base $CD = DE + EC = 4 + 4 = 8 \text{ cm}$.

Therefore, the area is 24 cm² and the perimeter is 26 cm.

Given:

From the design shown in Fig. 9.55:

Length of each straight part of the track $= 52 \text{ cm}$.

The radius of the semicircular ends is indicated by the arrow from the center line to the boundary.

To Find:

Total distance around the track (rounded to the nearest whole cm).

Solution:

The total distance around the track is the sum of the lengths of the two straight parallel segments and the length of the two semicircular ends.

1. Length of straight parts:

There are two parallel straight segments of $52 \text{ cm}$ each.

2. Length of curved parts:

There are two semicircular ends with the same radius. Together, they form one complete circle.

Using $\pi \approx 3.14$ and $r = 16 \text{ cm}$:

3. Total Distance:

Rounding to the nearest whole number:

Therefore, the total distance around the track is 204 cm.

Given:

The cover hangs $30 \text{ cm}$ (or $0.3 \text{ m}$) all around the table.

To Find:

Area of hanging portion and the cost of polishing the table top.

Solution:

The dimensions of the table top are smaller than the cover by twice the hanging length (once for each side).

1. Area of hanging part:

2. Cost of polishing:

Area of hanging cover is 2.97 m² and cost of polishing is $\textsf{₹}$ 72.08.

Given:

Dimensions of plot $= 200 \text{ m} \times 150 \text{ m}$.

Three roads, each $3 \text{ m}$ wide, run along the length ($200 \text{ m}$).

Solution:

Since the roads are along the length, the area of each road is a rectangle of $200 \text{ m} \times 3 \text{ m}$.

The area available for building houses is the total area minus the road area:

Therefore, the area for building houses is 28,200 m².

Given:

Room dimensions $= 4.5 \text{ m} \times 4 \text{ m}$.

Tile dimensions $= 15 \text{ cm} \times 10 \text{ cm}$. Rate $= \textsf{₹ } 4.50 \text{ per tile}$.

Solution:

Convert room dimensions to cm ($1 \text{ m} = 100 \text{ cm}$):

Calculate the number of tiles required:

Therefore, the total cost of covering the floor is $\textsf{₹}$ 5,400.

Given:

Diameter of circular garden ($d$) $= 28 \text{ m}$.

Cost of fencing per metre $= \textsf{₹ } 300$.

Solution:

Fencing is done along the boundary of the garden, which is its circumference.

Taking $\pi = \frac{22}{7}$:

Now, calculate the total cost of fencing:

Therefore, the total cost of the wooden fencing is $\textsf{₹}$ 26,400.

Given:

Radius of the circle ($r$) $= 14 \text{ cm}$.

One side of the rectangle ($l$) $= 24 \text{ cm}$.

To Find:

Length of the wire and comparison of areas of the circle and the rectangle.

Solution:

The length of the wire is equal to the circumference of the circle.

The same wire is used to form a rectangle, so the perimeter of the rectangle is also $88 \text{ cm}$. Let the breadth be $b$.

Now, let us calculate the areas of both shapes:

Comparing the areas, $616 \text{ cm}^2 > 480 \text{ cm}^2$.

Therefore, the length of the wire is 88 cm and the circle encloses more area.

Given:

Radius of the wheel ($r$) $= 25 \text{ cm}$.

Number of rotations $= 350$.

Solution:

The distance covered in one rotation is equal to the circumference of the wheel.

Now, calculate total distance for 350 rotations:

Convert distance to metres ($1 \text{ m} = 100 \text{ cm}$):

Therefore, the wheel will cover a distance of 550 metres.

Given:

Outer circumference of the path $= 44 \text{ m}$.

Width of the circular path $= 2 \text{ m}$.

Solution:

Let the outer radius be $R$.

Inner radius $r = R - \text{width} = 7 - 2 = 5 \text{ m}$.

1. Inner Circumference:

2. Area of Path:

The inner circumference is approx 31.43 m and area of the path is approx 75.43 m².

Given:

Outer dimensions of carpet: $L = 5 \text{ m}$, $B = 2 \text{ m}$.

Width of red border $= 25 \text{ cm} = 0.25 \text{ m}$.

Solution:

1. Area of blue portion:

The blue portion is the inner rectangle.

2. Area of red portion:

3. Ratio:

Area of blue portion is 6.75 m² and the ratio of red to blue area is 13 : 27.

Solution:

(a) Area for Hay:

From the figure, the hay region is a rectangle of $17.8 \text{ m}$ by $10.6 \text{ m}$.

(b) Cost for Vegetable Garden:

The vegetable garden dimensions are $49 \text{ m}$ by $15.2 \text{ m}$.

(c) Fencing for the house:

Fencing length is the perimeter of the house.

(d) Banana Orchard:

The orchard dimensions are $15.7 \text{ m}$ by $20 \text{ m}$.

Since the number of trees must be a whole number, there can be 251 trees.

Solution:

From the layout (Fig. 9.58), the dimensions are as follows:

Total Width $= 15 \text{ m}$

Bedroom 1 & 2 Width $= x$, Height $= 5 \text{ m}$

Bathroom Width $= 2 \text{ m}$, Height $= 5 \text{ m}$

Kitchen Width $= 15 - (x + 2) = (13 - x) \text{ m}$, Height $= 5 \text{ m}$

(a) Expression for total area of both bedrooms and kitchen:

(b) Expression for the perimeter of the living room:

The living room is L-shaped. Its boundary consists of segments of lengths $15, (5+2), (15-x), 2, x,$ and $5$ metres.

(c) Cost of carpeting bedrooms and living room:

(d) Cost of tiling bathroom and kitchen:

(e) Find x if bedroom area is 35 m²:

Given:

Lawn Dimensions $= 10 \text{ m} \times 4 \text{ m}$.

Width of flower bed $= 50 \text{ cm} = 0.5 \text{ m}$.

Solution:

The flower bed is along three sides of the lawn. By looking at Fig. 9.59, the bed is along the two widths ($4 \text{ m}$ sides) and one length ($10 \text{ m}$ side).

The dimensions of the inner remaining rectangular portion are:

Therefore, the area of the remaining portion is 31.5 m².

Given:

Playground Dimensions $= 120 \text{ m} \times 80 \text{ m}$.

Path 1 Width (along length) $= 3 \text{ m}$.

Path 2 Width (along width) $= 2 \text{ m}$.

Solution:

Step 1: Calculate total area of the paths.

The paths intersect at the center. The area of the intersection ($3 \text{ m} \times 2 \text{ m}$) is counted twice, so we subtract it once.

Step 2: Calculate remaining area.

Therefore, the area of the remaining playground is 9086 m².

Given:

Park Dimensions $= 20 \text{ m} \times 15 \text{ m}$.

Flower bed width $= 1 \text{ m}$. Rate for manuring $= \textsf{₹ } 45/\text{m}^2$.

Solution:

The L-shaped bed consists of two rectangular strips.

Total Cost:

Therefore, the total cost of manuring is $\textsf{₹}$ 1,530.

Given:

Painting Dimensions $= 60 \text{ cm} \times 38 \text{ cm}$.

Width of frame $= 6 \text{ cm}$.

Solution:

The frame is "around" the painting, so it increases the overall dimensions of the object.

Area of the Frame:

Therefore, the area of the wooden frame is 1320 cm².

Given:

The figure shows an outer square containing four congruent right-angled triangles. The dimensions provided are the legs of these triangles.

Solution:

The shaded portion is an inner square. Its area can be found by subtracting the area of the four congruent triangles from the area of the outer square.

1. Area of the four triangles:

2. Area of the outer square:

The side of the outer square is the sum of the base and height of the triangles.

3. Area of the shaded portion:

Therefore, the area of the shaded portion is 1000 cm².

Given:

Side of the square tile $= 20 \text{ cm}$.

Solution:

Case (i): The tile has four quarter circles at the corners. From the figure, these quarter circles meet, meaning the radius of each is half the side of the square.

The area of 4 quarter circles is equal to the area of 1 full circle.

The shaded area is the region left in the center:

Case (ii): A central circle touches all sides. Its diameter is equal to the side of the square ($20 \text{ cm}$), so its radius is $10 \text{ cm}$.

The shaded area is the region at the corners:

Conclusion:

The area of the shaded portion in both tiles is 86 cm². The areas are equal.

Given:

Field Dimensions $= 48 \text{ m} \times 12 \text{ m}$.

Triangle legs $= 2 \text{ m}$ and $4 \text{ m}$.

Solution:

Number of beds that can be laid:

Therefore, 144 flower beds can be laid in the field.

Given:

Original Length $= 32 \text{ m}$, Original Width $= 26 \text{ m}$.

Increase in Area $= 650 \text{ m}^2$. Width remains constant ($26 \text{ m}$).

Solution:

Let the new length be $L$. Since width is still $26 \text{ m}$:

Therefore, the length of the expanded wheat field is 57 metres.

Given:

In $\triangle AEC$, $\angle E = 90^\circ$. $AC = 25 \text{ cm}, AE = 15 \text{ cm}, BC = 7 \text{ cm}$. $BD \perp AC$.

Solution:

Step 1: Find the length of EC.

Using Pythagoras theorem in $\triangle AEC$:

Step 2: Find the area of $\triangle ABC$.

In $\triangle ABC$, considering $BC$ as the base, the corresponding altitude is $AE$ (since $AE \perp EC$).

Step 3: Find the length of DB.

The area of $\triangle ABC$ can also be expressed using base $AC$ and altitude $DB$.

The area of $\triangle ABC$ is 52.5 cm² and the length of $DB$ is 4.2 cm.

Given:

Dimensions of the large chocolate sheet $= 18 \text{ cm} \times 18 \text{ cm}$

Dimensions of one small chocolate piece $= 1.5 \text{ cm} \times 2 \text{ cm}$

Solution:

To find the total number of pieces, we divide the total area of the sheet by the area of one piece.

Number of pieces:

Alternative Check:

Along the length ($18 \text{ cm}$), we can cut $18 / 1.5 = 12$ pieces.

Along the breadth ($18 \text{ cm}$), we can cut $18 / 2 = 9$ pieces.

Total pieces $= 12 \times 9 = 108$.

Therefore, 108 pieces of chocolate can be cut.

Given:

The figure (Fig. 9.66) consists of a shaded square with a unshaded central region. Every small step segment is $1 \text{ cm}$ long.

Solution:

By observing the steps, we can determine the dimensions of the outer square. There are 8 horizontal segments of $1 \text{ cm}$ each across the span.

The shaded region is composed of four identical corner shapes. Let's count the number of unit squares ($1 \text{ cm} \times 1 \text{ cm}$) in one corner:

Row 1: 4 squares

Row 2: 3 squares

Row 3: 2 squares

Row 4: 1 square

Since there are four such corners:

Therefore, the area of the shaded region is 40 cm².

Solution:

Step 1: House Dimensions

(i) Cost of paving the path:

The path is $1 \text{ m}$ wide around the house. The outer dimensions including the path are:

(ii) Cost of wooden flooring (except bathroom):

(iii) Area of Living Room:

The total area of the house is $63 \text{ m}^2$. Subtract areas of other rooms:

Bedroom: $4 \times 3 = 12 \text{ m}^2$ | Kids Room: $4 \times 3 = 12 \text{ m}^2$ | Kitchen: $4 \times 3 = 12 \text{ m}^2$ | Bath: $5 \text{ m}^2$

Solution:

Moulding is installed around the ceiling, which is equal to the perimeter of the room. Carpeting is done on the floor, equal to the area.

(c) Total Cost of Moulding:

Given:

Rectangle $ABCD$: $L = 80 \text{ cm}, B = 60 \text{ cm}$.

A rhombus $PQRS$ is formed by connecting midpoints. A central circle has radius $r = 10 \text{ cm}$.

Solution:

The area of the shaded portion is the area of the rhombus minus the area of the circle.

Area of circular rangoli (using $\pi \approx 3.14$):

Area of shaded portion:

Therefore, the area of the shaded portion is 2086 cm².

Given:

Dimensions of rectangular sheet $= 100 \text{ cm} \times 80 \text{ cm}$.

Side of each of the 4 squares $= 10 \text{ cm}$.

Equal sides of the isosceles right triangle $= 10 \text{ cm}$.

Solution:

We first calculate the area of the original sheet and the areas of the parts removed.

1. Area of the original rectangular sheet:

2. Area of the 4 squares removed:

3. Area of the isosceles right triangle removed:

Since it is an isosceles right triangle, the two equal sides are the base and the height.

4. Area of the remaining part:

Therefore, the area of the remaining part of the paper is 7550 cm².

Given:

Inner circumference ($C_1$) $= 352 \text{ mm}$

Outer circumference ($C_2$) $= 396 \text{ mm}$

To Find:

The width of the design region ($R - r$).

Solution:

Let $r$ be the inner radius and $R$ be the outer radius.

1. Find the inner radius (r):

2. Find the outer radius (R):

3. Find the width:

Therefore, the width of the circular design is 7 mm.

Given:

Distance of the moon from the earth (radius of circular path, $r$) $= 384000 \text{ km}$.

$\pi = 3.14$.

Solution:

The length of the path described in one complete revolution is equal to the circumference of the circular path.

Therefore, the length of the path is 24,11,520 km.

Given:

Scale of photograph $= 1:10$.

Dimensions in photograph: $L = 25 \text{ cm}, B = 10 \text{ cm}, \text{hole diameter} = 0.5 \text{ cm}$.

Rate of polishing $= \textsf{₹ } 200/\text{m}^2$.

Solution:

Step 1: Find actual dimensions of the table and holes.

Step 2: Calculate areas in square metres.

Step 3: Area to be polished and total cost.

Therefore, the cost of polishing is approximately $\textsf{₹}$ 497.65.