Chapter 7 Area (Class 8 - Latest Maths NCERT (Ganita Prakash II) Solutions)

Solution for Part (i)

Given:

1. Area of top-left rectangle = $28\text{ in}^2$ and its height = $4\text{ in}$.

2. Area of bottom-right rectangle = $14\text{ in}^2$ and its height = $2\text{ in}$.

To Find:

The missing width ($?$) of the $14\text{ in}^2$ rectangle.

Solution:

We know that for any rectangle:

In the bottom-right rectangle, the height is given as $2\text{ in}$ and the area is $14\text{ in}^2$.

To find the missing width ($?$):

Hence, the missing side length in figure (i) is $7\text{ in}$.

Solution for Part (ii)

Given:

1. Top-left rectangle: Area = $29\text{ m}^2$, Height = $4\text{ m}$.

2. Top-right rectangle: Area = $11\text{ m}^2$, Height = $4\text{ m}$.

3. Bottom-left rectangle: Area = $50\text{ m}^2$.

To Find:

1. Width of the top-left rectangle.

2. Width of the top-right rectangle.

3. Height of the bottom rectangle section.

Solution:

Step 1: Find the width of the top-left section.

Step 2: Find the width of the top-right section.

Step 3: Find the missing height of the bottom section ($?$).

The bottom-left rectangle has an area of $50\text{ m}^2$ and shares the same width ($7.25\text{ m}$) as the rectangle above it.

Thus, the missing side lengths for figure (ii) are $7.25\text{ m}$, $2.75\text{ m}$, and approximately $6.9\text{ m}$.

(i) Measurements needed and Formula:

To find the area of the shaded path, we need the following measurements:

• The length and width of the outer rectangle $ABCD$.
• The length and width of the inner rectangular park $EFGH$.

Let us assign possible values:

Let the outer rectangle $ABCD$ have length $L = 20\text{ m}$ and width $W = 15\text{ m}$.

Let the inner rectangle $EFGH$ have length $l = 16\text{ m}$ and width $w = 11\text{ m}$.

Calculation:

Formula:

(ii) Area with Path Width:

If only the width of the path along each side is given, we cannot find the area because the absolute size of the park is unknown. We also need the measurements of either the inner park or the outer rectangle.

Let us assign values:

Let the length of the inner park $EFGH$ be $l = 10\text{ m}$, width $w = 8\text{ m}$, and the width of the path be $x = 2\text{ m}$ all around.

We can break the path into four rectangles to find the area:

• Two horizontal rectangles: Each with length $l + 2x$ and width $x$.
• Two vertical rectangles: Each with length $w$ and width $x$.

Calculation:

General Formula:

(iii) Variation in Position:

No, the area of the path does not change when the outer rectangle is moved, provided the dimensions of both the outer rectangle ($ABCD$) and the inner rectangle ($EFGH$) remain the same.

The area of the path is strictly the difference between the outer area and the inner area:

Since the individual areas of the two rectangles do not change when they are shifted relative to each other, their difference (the path area) remains constant regardless of the position of the park within the outer boundary.

Measurements Needed

To find the area of the crosspath, we need the following measurements in addition to the plot dimensions:

• The width of the horizontal path (running parallel to the $14\text{ m}$ side).
• The width of the vertical path (running parallel to the $12\text{ m}$ side).

Usually, in such plots, the width of both paths is kept the same for uniformity.

Assigning Values and Finding Area

Given:

Length of the rectangular plot ($L$) $= 14\text{ m}$

Width of the rectangular plot ($B$) $= 12\text{ m}$

Assumed Values:

Let the width of both the horizontal and vertical paths be $x = 2\text{ m}$.

To Find:

Area of the crosspath.

Solution:

The crosspath consists of two rectangular paths that intersect each other at the center. When we calculate the area of both paths, the central square (intersection) is counted twice. Therefore, we must subtract it once.

Step 1: Calculate the area of the horizontal path.

Step 2: Calculate the area of the vertical path.

Step 3: Calculate the area of the common intersection (square).

Step 4: Total area of the crosspath.

General Formula

If $L$ is the length of the plot, $B$ is the width of the plot, and $x$ is the uniform width of the crosspath, the formula is:

Or, in a simplified form:

I. Area of the Spiral Tube

Given:

The spiral tube has a uniform width of $1$ unit. The labeled outer lengths of the segments forming the spiral are:

Horizontal segments: $20$, $20$, $15$, $10$, and $5$.

Vertical segments: $20$, $15$, $10$, and $5$.

To Find:

The total area of the spiral tube.

Solution:

To find the area, we can sum the lengths of all the segments. However, we must account for the fact that each corner where two segments meet is a shared square area. There are 9 segments in total, which means there are 8 corners (turns).

Step 1: Calculate the sum of all the labeled segment lengths.

Step 2: Subtract the overlapping corner areas. Since the width is $1$, each corner is a $1 \times 1$ square ($1\text{ unit}^2$).

II. Length of the Straight Tube

Given:

A bent "L-shaped" tube with outer lengths of $5$ and $5$ units. The width is $1$ unit.

To Find:

The length ($?$) of a straight tube with the same area.

Solution:

First, calculate the area of the bent tube on the left. It consists of two segments of length $5$ with one shared corner.

For the straight tube to have the same area with a width of $1$ unit:

Therefore, the length of the straight tube should be $9\text{ units}$.

Given:

A square is divided into three regions: 1, 2, and 3. Let the original side length of the square be $s$.

To Find:

The increase in the areas of regions 1, 2, and 3 when the side length of the square is doubled.

Solution:

First, let us understand the original areas of the regions:

1. Region 3 is a triangle formed by the diagonal, so its area is half of the square.

2. Regions 1 and 2 are smaller triangles. The tick marks on the diagonal indicate that the diagonal is bisected. Thus, regions 1 and 2 are equal in area, each being one-fourth of the square's total area.

Let the original total area be $A$.

When the side length of the square is doubled, the new side length becomes $2s$.

The new total area ($A'$) becomes:

Reason:

When the dimensions (side lengths) of a two-dimensional figure are scaled by a factor of $k$, its area increases by a factor of $k^2$. Here, the side is doubled ($k=2$), so the area becomes $2^2 = 4$ times the original.

This scaling applies to every part of the figure. Therefore:

• New Area of Region 1 $= 4 \times (\text{Original Area of Region 1})$
• New Area of Region 2 $= 4 \times (\text{Original Area of Region 2})$
• New Area of Region 3 $= 4 \times (\text{Original Area of Region 3})$

Conclusion:

The area of each region becomes 4 times its original area. The increase in area for each region is 3 times the original area (since New Area $-$ Original Area $= 4x - x = 3x$).

Activity Solution

This is a classic geometric puzzle. To solve this, follow these steps:

1. Construction: Draw a square on a piece of cardboard. Draw two perpendicular lines that intersect at a point inside the square (not necessarily the center). This divides the square into four quadrilaterals.

2. Cutting: Carefully cut along these two lines to separate the four pieces.

3. Rearrangement: To form a larger square with a hole in the middle, rotate each piece by $90^\circ$ and move them outwards. Specifically:

• Arrange the four pieces such that their original outer corners (the $90^\circ$ corners of the original square) now face inwards to form the corners of a central empty square (the hole).
• The cut edges will now form the outer boundary of the new, larger square.

Mathematical Reason:

The total area of the four pieces remains constant. When they are rearranged with a hole in the center, the outer boundary forms a larger square whose area is equal to the (Area of the original square + Area of the hole).

This activity demonstrates that the same set of shapes can occupy a larger bounding box if the internal arrangement is changed to include empty space.

Formula Used:

(i) For Triangle $ABC$:

Base ($BC$) $= 4\text{ cm}$ and Height ($AE$) $= 3\text{ cm}$.

(ii) For Triangle $DEF$:

Base ($EF$) $= 5\text{ cm}$ and Height ($DN$) $= 3.2\text{ cm}$.

(iii) For Triangle $NAT$:

It is a right-angled triangle where Base ($AT$) $= 3\text{ cm}$ and Height ($NA$) $= 4\text{ cm}$.

Given:

In $\Delta ABC$:

Case 1: Base ($BC$) $= 6\text{ units}$ and corresponding Height ($AX$) $= 4\text{ units}$.

Case 2: Base ($AC$) $= 8\text{ units}$ and corresponding Height ($BY$) $= ?$.

To Find:

The length of the altitude $BY$.

Solution:

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

First, calculate the area using base $BC$ and altitude $AX$:

Now, we use the same area to find altitude $BY$ using base $AC$:

Given:

1. $\Delta SUB$ is an isosceles triangle with $SU = SB$.

2. $SE \perp UB$ (Altitude from the vertex to the base).

3. Area of $\Delta SEB = 24\text{ sq. units}$.

To Find:

Area of $\Delta SUB$.

Solution:

In an isosceles triangle, the altitude drawn from the vertex connecting the equal sides to the base ($SE$) acts as a line of symmetry. This means it bisects the base $UB$ and divides the triangle into two congruent triangles of equal area.

The total area of $\Delta SUB$ is the sum of the areas of these two smaller triangles:

Given:

A rectangle with base $b$ and height $h$. The area of the rectangle is $b \times h$.

To Find:

A method to construct a triangle with the same area.

Solution:

According to the Baudhayana Śulba-Sūtra, to transform a quadrilateral (like a rectangle) into a triangle of equal area, we can manipulate the dimensions such that the triangle's formula $\frac{1}{2} \times \text{base} \times \text{height}$ yields the same result.

Method:

1. Keep the height of the triangle the same as the height of the rectangle ($h$).

2. Make the base of the triangle double the base of the rectangle ($2b$).

Alternatively, one can keep the same base ($b$) and double the height ($2h$) for the triangle to maintain the same area.

Given:

A triangle with base $b$ and height $h$. The area of the triangle is $\frac{1}{2} \times b \times h$.

To Find:

A method to construct a rectangle with the same area.

Solution:

In Indian geometry (Śulba-Sūtras), a common method to convert a triangle into a rectangle of equal area is through a process of cutting and rearranging (geometry by "cutting").

Method:

1. Identify the base ($b$) and the altitude (height $h$) of the triangle.

2. Construct a rectangle with the same base ($b$) but half the height ($\frac{h}{2}$) of the triangle.

Geometrically, this is done by drawing a line parallel to the base at half the height of the triangle. The top part of the triangle can be cut and rearranged to fill the sides of the bottom part, forming a rectangle.

Given:

Three identical squares $ABCD, BCEF,$ and $BFGH$. Let the side of each square be $x$.

From the figure, we observe the coordinates (assuming $D$ is at $(0,0)$):

• $D = (0,0), C = (x,0), E = (2x,0)$
• $A = (0,x), B = (x,x), F = (2x,x)$
• $H = (x,2x), G = (2x,2x)$

Analysis of Regions:

1. Red Region: This is triangle $DHE$. The base is $DE = 2x$. The altitude (height) from $H$ to $DE$ is the vertical distance $CH = 2x$.

2. Blue Region: The line $DH$ passes from $(0,0)$ to $(x,2x)$. Its equation is $y = 2x_{coord}$. It intersects the line $AB$ ($y=x$) at $x_{coord} = 0.5x$. The blue region is a triangle with base $AD=x$ and height $0.5x$.

Comparing the two areas:

(i) Find Blue Area if Red Area $= 49$:

(ii) Find Area of each Square if Total Area $= 180$:

Total Area $= \text{Red Area} + \text{Blue Area}$

Hence, the area of each square is $80\text{ sq. units}$.

Given:

In $\Delta XYZ$, $M$ is the midpoint of $XY$ and $N$ is the midpoint of $XZ$.

To Find:

The fraction of the area of $\Delta XYZ$ that is the area of $\Delta XMN$.

Construction Required:

Join the points $N$ and $Y$ to form a line segment $NY$.

Solution:

We know that a median of a triangle divides it into two triangles of equal area.

Step 1: Consider $\Delta XYZ$. Since $N$ is the midpoint of $XZ$, $YN$ is the median to the side $XZ$.

Step 2: Now consider $\Delta XNY$. Since $M$ is the midpoint of $XY$, $NM$ is the median to the side $XY$.

Step 3: Substitute the value of $\text{Area}(\Delta XNY)$ from equation (i) into equation (ii):

Therefore, the area of $\Delta XMN$ is $\frac{1}{4}$ (one-fourth) of the area of $\Delta XYZ$.

Shortest Path Solution

To find the shortest path from the House to the River and then to the Water Tank, we use the principle of reflection.

Steps to find the shortest path:

1. Imagine the river bank as a straight mirror line.

2. Find the reflection (image) of the "House" across the river line. Let's call this point House'.

3. Draw a straight line connecting House' directly to the Water Tank.

4. The point where this straight line crosses the River bank is the exact spot where Gopal should collect water. Let's call this point P.

5. The shortest path is to go from the House to point P, and then from point P to the Water Tank.

Reason:

The distance from House to $P$ is equal to the distance from House' to $P$ due to symmetry. Since the shortest distance between two points (House' and Water Tank) is a straight line, the path $House \rightarrow P \rightarrow Water\ Tank$ represents the minimum possible total distance.

Geometric Trace:

• Let $H$ be the House and $W$ be the Water Tank.
• Let $H'$ be the reflection of $H$ across the river.
• The path length is $HP + PW = H'P + PW$.
• This sum is minimized when $H', P,$ and $W$ are collinear (forming a straight line).

Given:

1. Length of diagonal $AC = 22\text{ cm}$.

2. Length of perpendicular (altitude) from $B$ to $AC$, $BM = 3\text{ cm}$.

3. Length of perpendicular (altitude) from $D$ to $AC$, $DN = 3\text{ cm}$.

To Find:

Area of quadrilateral $ABCD$.

Solution:

A quadrilateral can be divided into two triangles by its diagonal. Here, diagonal $AC$ divides quadrilateral $ABCD$ into $\Delta ABC$ and $\Delta ADC$.

We use the formula for the area of a triangle, which is $\frac{1}{2} \times \text{base} \times \text{height}$.

Now, adding the areas of the two triangles:

Alternate Formula:

Thus, the area of quadrilateral $ABCD$ is $66\text{ cm}^2$.

Given:

1. Rectangle $ABCD$ with length $DC = 18\text{ cm}$ and breadth $BC = 10\text{ cm}$.

2. Points $E$ and $F$ on sides $AB$ and $AD$ respectively, with dimensions:

• $AE = 10\text{ cm}$ and $EB = 8\text{ cm}$ (Total $AB = 18\text{ cm}$).
• $AF = 6\text{ cm}$ and $FD = 4\text{ cm}$ (Total $AD = 10\text{ cm}$).

To Find:

Area of the shaded region (Quadrilateral $FECD$).

Solution:

The area of the shaded region can be found by subtracting the areas of the unshaded triangles from the total area of the rectangle $ABCD$.

Step 1: Calculate the area of rectangle $ABCD$.

Step 2: Calculate the area of unshaded $\Delta AFE$.

Step 3: Calculate the area of unshaded $\Delta EBC$.

Step 4: Find the area of the shaded region.

The area of the shaded region is $110\text{ cm}^2$.

To Find:

The necessary measurements to calculate the area of a regular hexagon.

Solution:

A regular hexagon is a polygon with six equal sides and six equal interior angles. To find its area, you primarily need only one measurement:

1. The side length ($s$): If you know the length of one side, you can calculate the area because a regular hexagon can be divided into six identical equilateral triangles.

Alternative measurements that could be used:

2. The Apothem ($a$): The perpendicular distance from the center to the midpoint of any side. If the apothem is known, the area is given by:

3. The Radius ($r$): The distance from the center to any vertex. In a regular hexagon, the radius is equal to the side length ($r = s$).

In summary, knowing the side length is the most direct measurement required.

Given:

A rectangle with a blue region consisting of two triangles that share a common vertex at a point inside the rectangle. The bases of these triangles are the two opposite horizontal sides of the rectangle.

To Find:

The fraction of the total area of the rectangle that is covered by the blue region.

Solution:

Let the length of the rectangle be $l$ and the width (height) be $w$.

The blue region consists of two triangles: a top triangle and a bottom triangle.

1. Both triangles have a base equal to the length of the rectangle, $l$.

2. Let the height of the top triangle be $h_1$ and the height of the bottom triangle be $h_2$.

3. Since the point where they meet is inside the rectangle, the sum of their heights must equal the total width of the rectangle:

Now, let's calculate the total area of the blue region:

The total area of the rectangle is $l \times w$.

The area of the blue region is exactly $\frac{1}{2}$ (half) of the total area of the rectangle.

To Find:

A geometric method to construct a smaller quadrilateral with exactly half the area of a given original quadrilateral.

Solution:

The most effective method to achieve this is by using Varignon's Theorem. This theorem states that the figure formed by joining the midpoints of the sides of any quadrilateral is a parallelogram, and its area is always half of the original quadrilateral.

Step-by-Step Method:

1. Take any quadrilateral (let's call it $ABCD$).

2. Locate the midpoints of all four sides: $AB$, $BC$, $CD$, and $DA$. Let these midpoints be $P, Q, R,$ and $S$ respectively.

3. Join these midpoints in order ($P$ to $Q$, $Q$ to $R$, $R$ to $S$, and $S$ to $P$).

4. The resulting quadrilateral $PQRS$ is a parallelogram.

Reasoning:

If you draw a diagonal $AC$ in the original quadrilateral, the triangles $\Delta ABC$ and $\Delta ADC$ are formed. By the Midpoint Theorem, $PQ$ is parallel to $AC$ and $PQ = \frac{1}{2}AC$. Similarly, $RS$ is parallel to $AC$ and $RS = \frac{1}{2}AC$. This construction consistently leads to an interior figure whose area is exactly half of the total.

Alternate Method:

Divide the quadrilateral into two triangles by drawing a diagonal. Construct a new triangle for each half with the same base but half the altitude. Joining these results in a new quadrilateral with half the total area.

Solution for Part (i)

Given:

A set of parallelograms $(a)$ to $(g)$ drawn on a grid. All parallelograms share the same horizontal base length and are positioned between the same two horizontal parallel lines.

Observation:

• By counting the grid units, the base of each parallelogram is $3$ units.
• The perpendicular height (the distance between the top and bottom parallel lines) for each parallelogram is $2$ units.

To Find:

The relationship between the areas of all these parallelograms.

Solution:

The area of a parallelogram is calculated using the formula:

Since all the parallelograms have the same base ($3$ units) and the same height ($2$ units):

Reason:

According to the geometric property, "Parallelograms on the same base and between the same parallels are equal in area." Therefore, the areas of all these parallelograms are equal.

Solution for Part (ii)

To Find:

The relationship between their perimeters and identifying the maximum and minimum perimeter.

Solution:

The perimeter of a parallelogram is the sum of all its four sides. Since the top and bottom bases are equal for all figures ($3$ units), the perimeter depends on the length of the slant sides.

As the "slant" or "lean" of the parallelogram increases, the length of the slant side increases, even though the vertical height remains the same. This is because the slant side is the hypotenuse of a right-angled triangle formed by the height and the horizontal displacement.

Observations from the grid:

• Minimum Perimeter: Figure $(d)$ appears to have the minimum perimeter. Its side is almost vertical (it is a rectangle), meaning the slant side is nearly equal to the height ($2$ units), which is the shortest possible distance between the parallels.
• Maximum Perimeter: Figure $(g)$ has the maximum perimeter. It is the most "stretched" or slanted parallelogram. The horizontal distance between its top and bottom vertices is the largest, making its slant side the longest among all the figures.

Conclusion:

The perimeters of the parallelograms are not equal. The perimeter is minimum for figure $(d)$ and maximum for figure $(g)$.

Formula Used:

(i) Given: Base $= 7\text{ cm}$, Height $= 4\text{ cm}$

(ii) Given: Base $= 5\text{ cm}$, Height $= 3\text{ cm}$

(iii) Given: Base $= 5\text{ cm}$, Height $= 4.8\text{ cm}$

(iv) Given: Base $= 2\text{ cm}$, Height $= 4.4\text{ cm}$

Given:

In parallelogram $PQRS$:

• Base $SR = 12\text{ cm}$
• Corresponding height $QM = 6\text{ cm}$
• Other side $PS = 7.6\text{ cm}$ (acts as base for height $QN$)

To Find:

The length of altitude $QN$.

Solution:

We know that the area of a parallelogram is the same regardless of which base and height are chosen.

Step 1: Calculate the area using base $SR$ and height $QM$.

Step 2: Use the calculated area to find $QN$ using base $PS$.

The length of $QN$ is $9.47\text{ cm}$.

Given:

A rectangle with side lengths $5\text{ cm}$ and $4\text{ cm}$.

A parallelogram with the same side lengths $5\text{ cm}$ and $4\text{ cm}$.

To Find:

Which figure has the greater area?

Solution:

1. Area of the Rectangle: Since the sides are perpendicular, the height is equal to the breadth.

2. Area of the Parallelogram: Let the base be $5\text{ cm}$. The slant side is $4\text{ cm}$.

The height ($h$) of a parallelogram is the perpendicular distance between the bases. In any right-angled triangle formed inside the parallelogram, the slant side ($4\text{ cm}$) acts as the hypotenuse, and the height ($h$) acts as a leg. Since the hypotenuse is always the longest side:

Therefore, for the parallelogram:

Conclusion:

The rectangle has a greater area than the parallelogram with the same side lengths.

Given:

A triangle with base $b$ and height $h$.

To Find:

A method to obtain a rectangle with area twice that of the triangle ($Area = b \times h$).

Solution:

Since the area of the rectangle needs to be $b \times h$, we can consider the following different methods:

Method 1: Same Base and Same Height

Construct a rectangle with the same base ($b$) and the same height ($h$) as the triangle. The area of this rectangle will be $b \times h$, which is exactly twice the area of the triangle.

Method 2: Double Base and Half Height

Construct a rectangle with twice the base ($2b$) and half the height ($\frac{h}{2}$) of the triangle.

Method 3: Half Base and Double Height

Construct a rectangle with half the base ($\frac{b}{2}$) and twice the height ($2h$) of the triangle.

Given:

A triangle with base $b$ and height $h$. The area is $\frac{1}{2}bh$.

Solution:

According to the Śulba-Sūtras (ancient Indian geometric texts), a triangle can be converted into a rectangle of the same area using the following method:

Method:

1. Take the base of the triangle as the base of the rectangle.

2. Take half the altitude (height) of the triangle as the height of the rectangle.

By keeping the base the same and halving the height, the rectangle occupies the same total area as the original triangle.

Given:

An isosceles triangle $ABC$ with altitude $AD$ drawn to the base $BC$.

In an isosceles triangle, the altitude $AD$ bisects the base $BC$. Therefore:

Solution:

1. Dissection:

Cut the isosceles triangle $ABC$ along the altitude line $AD$. This results in two congruent right-angled triangles, $\Delta ABD$ and $\Delta ACD$.

2. Rearrangement:

Each right triangle is exactly half of a rectangle with sides equal to the legs of the triangle ($AD$ and $BD$).

• $\Delta ABD$ has legs $AD$ and $BD$.
• $\Delta ACD$ has legs $AD$ and $DC$.

Since $BD = DC$, these two right triangles can be assembled to form a rectangle of sides $AD$ and $BD$.

3. Assembly:

Keep $\Delta ABD$ as it is. Rotate $\Delta ACD$ and place its hypotenuse $AC$ against the hypotenuse $AB$ of the first triangle (this works if the triangles are flipped appropriately). More simply, if we construct a rectangle with length $AD$ and width $BD$, its area is:

Since $BD = \frac{1}{2} BC$, we have:

Thus, by cutting the isosceles triangle into two right-angled triangles and rearranging them, we obtain a rectangle with the same area.

Given:

A rectangle with length $L$ and width $W$. The area of the rectangle is $L \times W$.

To Find:

A method to convert this rectangle into an isosceles triangle of the same area using dissection (cutting and rearranging).

Solution:

In the Śulba-Sūtras, the process is the reverse of converting an isosceles triangle into a rectangle. To convert a rectangle into an isosceles triangle of equal area:

Step 1: Identify the Midpoint

Take the rectangle and find the midpoint of one of its longer sides (let's say the top side).

Step 2: Dissection (Cutting)

Draw two diagonal lines starting from this midpoint to the two opposite bottom corners of the rectangle. This divide the rectangle into three parts:

• One large isosceles triangle in the middle.
• Two identical right-angled triangles on the sides.

Step 3: Rearrangement

1. Cut out the two right-angled triangles from the sides.

2. Flip them and attach them to the top of the middle isosceles triangle such that their original vertical edges meet at the center.

3. This forms a larger isosceles triangle whose base is the same as the rectangle's length, but its height is doubled.

Verification:

Let the side length of both figures be $s$.

Case 1: Equilateral Triangle vs. Square

The formula for the area of a square is:

The formula for the area of an equilateral triangle is:

Using the value of $\sqrt{3} \approx 1.732$:

Conclusion: Since $1.0 s^2 > 0.433 s^2$, the square has a much greater area than an equilateral triangle of the same side length.

Case 2: Two Equilateral Triangles vs. Square

The area of two identical equilateral triangles is:

Using the value of $\sqrt{3} \approx 1.732$:

Comparing this to the area of the square ($1.0 s^2$):

Conclusion: Even two identical equilateral triangles combined have less area than a square of the same side length. The square still has a greater area.

Reason: In a square, the "height" is equal to the side length, whereas in an equilateral triangle, the height ($\frac{\sqrt{3}}{2}s \approx 0.866s$) is always less than the side length. This leads to a smaller overall area for the triangle.

Given:

Length of the first diagonal ($d_1$) $= 20\text{ cm}$

Length of the second diagonal ($d_2$) $= 15\text{ cm}$

To Find:

Area of the rhombus.

Solution:

We know that the area of a rhombus is half the product of its diagonals.

Substituting the given values:

Thus, the area of the rhombus is $150\text{ cm}^2$.

Method:

To convert a rectangle of area $L \times W$ into a rhombus of the same area, we can use the following dissection steps:

1. Identify Dimensions: A rhombus with diagonals $d_1$ and $d_2$ has an area of $\frac{1}{2} d_1 d_2$. To match a rectangle of area $L \times W$, we can set $d_1 = L$ and $d_2 = 2W$.

2. Dissection Steps:

• Divide the rectangle into four equal smaller rectangles by drawing its horizontal and vertical midlines.
• In each of the four smaller rectangles, draw a diagonal that connects the center of the original rectangle to the outer corners.
• Cut along these diagonals. You will have four right-angled triangles and a central diamond shape.
• Rearrangement: Move the four outer triangles and attach them to the opposite sides of the central shape to form a rhombus.

Alternative Simple Method:

Take a rectangle with length $L$ and width $W$. Mark the midpoints of all four sides. If you join these midpoints, you get a rhombus. However, this rhombus has exactly half the area of the rectangle. To get a rhombus of equal area, you would need to start with a rectangle that has double the width ($2W$) or double the length ($2L$) before performing the midpoint construction.

General Formula for Area of a Trapezium:

(i) Solution:

Given: Parallel sides $a = 10\text{ ft}$ and $b = 7\text{ ft}$, Height $h = 16\text{ ft}$.

(ii) Solution:

Given: Parallel sides $a = 24\text{ m}$ and $b = 36\text{ m}$, Height $h = 14\text{ m}$.

(iii) Solution:

Given: Parallel sides $a = 14\text{ in}$ and $b = 6\text{ in}$, Height $h = 10\text{ in}$.

(iv) Solution:

Given: Parallel sides $a = 12\text{ ft}$ and $b = 18\text{ ft}$, Height $h = 8\text{ ft}$.

Given:

An isosceles trapezium with parallel sides $a$ and $b$, and height $h$.

Solution:

According to the Śulba-Sūtras, an isosceles trapezium (where non-parallel sides are equal) can be converted into a rectangle of equal area by the following dissection method:

Steps of Dissection:

1. Drop a perpendicular from one of the top vertices to the base. Let's say from vertex $A$ to the base $CD$ at point $P$. This cuts off a small right-angled triangle from one side of the trapezium.

2. Cut this triangle out from the trapezium.

3. Flip the triangle and attach it to the other non-parallel side of the trapezium.

4. Since the trapezium is isosceles, the angles and the slant side of the triangle will perfectly match the other side, resulting in a rectangle.

Result:

The resulting rectangle will have a height equal to the original height ($h$) of the trapezium, and its base will be the average of the two parallel sides.

Solution:

Based on the figure and the hint provided, the vertices of the rectangle $EFGH$ are found using the midpoints of the non-parallel sides of the trapezium.

Steps to find the vertices:

1. Identify Midpoints: Locate the midpoints of the non-parallel sides $AD$ and $BC$. Let these midpoints be $I$ and $J$ respectively.

2. Draw Perpendiculars: Draw a vertical line (perpendicular to the parallel bases $AB$ and $CD$) passing through midpoint $I$. This line intersects the top parallel line at $H$ and the bottom parallel line at $G$.

3. Repeat for other side: Similarly, draw a vertical line passing through midpoint $J$. This line intersects the top parallel line at $E$ and the bottom parallel line at $F$.

Reasoning for equal area:

In the figure, since $I$ is the midpoint of $AD$:

Similarly, since $J$ is the midpoint of $BC$:

By cutting triangles $\Delta DGI$ and $\Delta CFJ$ from the bottom and rotating them to positions $\Delta AHI$ and $\Delta BEJ$ at the top, the trapezium $ABCD$ is transformed into the rectangle $EFGH$. Thus, the Area of rectangle $EFGH$ = Area of trapezium $ABCD$.

To Find:

Dimensions of a trapezium with an area of $144\text{ cm}^2$.

Solution:

We can use the idea that the Area of a rectangle = Area of a trapezium if the rectangle's base is the average of the trapezium's parallel sides and their heights are the same.

Step 1: Choose a rectangle with area $144\text{ cm}^2$.

Let's choose a rectangle with base $L = 18\text{ cm}$ and height $H = 8\text{ cm}$.

Step 2: Convert this rectangle into a trapezium.

For the trapezium to have the same area, the average of its parallel sides ($a$ and $b$) must equal the rectangle's base ($18\text{ cm}$), and the height must be $8\text{ cm}$.

We can choose any two numbers that add up to $36$. For example, let $a = 16\text{ cm}$ and $b = 20\text{ cm}$.

Conclusion:

A trapezium with parallel sides $16\text{ cm}$ and $20\text{ cm}$ and a height of $8\text{ cm}$ will have an area of $144\text{ cm}^2$.

Verification:

Given:

A regular hexagon is divided into three regions: an equilateral triangle, a rhombus, and a trapezium.

Solution:

A regular hexagon can be divided into 6 identical equilateral triangles by joining its center to each of its six vertices. Let the area of one such small equilateral triangle be $x$.

Based on the geometric properties of a regular hexagon, we can observe the following:

1. Equilateral Triangle: The smallest triangular part shown in the figure is exactly one of the 6 small triangles.

2. Rhombus: A rhombus in a regular hexagon (formed by joining two adjacent triangles) consists of 2 small equilateral triangles.

3. Trapezium: The remaining part is a large trapezium. Since the total area of the hexagon is $6x$, the area of the trapezium is:

To Find:

The ratio of their areas (Trapezium : Triangle : Rhombus).

Therefore, the ratio of the areas of the trapezium, equilateral triangle, and rhombus is $3 : 1 : 2$.

Given:

1. Trapezium $ZYXW$ with $ZY \parallel WX$.

2. $A$ is the midpoint of $XY$, so $YA = AX$.

3. $\Delta ZWB$ is formed by extending base $WX$ to $B$ and connecting $Z$ to $B$ through $A$.

To Prove:

Proof:

Consider $\Delta ZYA$ and $\Delta BXA$. In these two triangles:

Therefore, by ASA (Angle-Side-Angle) congruence criterion:

Since congruent triangles have equal areas:

Now, let's look at the areas of the figures:

Substituting equation (i) into this:

From the figure, the sum of Area(Quad $ZWXA$) and Area($\Delta BXA$) is exactly the area of $\Delta ZWB$:

Hence Proved.