ANSWERs for figure it out Class 8 Mathematics – NCERT (Ganita Prakash) Part 2 CH7 AREA CLASS 8

ANSWERs for figure it out Class 8 Mathematics – NCERT (Ganita Prakash) Part 2 CH7  AREA CLASS 8 
Figure it Out Page number 150
1. Identify the missing side lengths

(i) 

(ii) Given:

(iii) Given:

• One side = 7 in 

(ii) (a) Given:

• 
  

(i) Given:

• Area = 21 in²

• One side = 7 in

• Answer: Missing side = 3 in

• Area = 28 in²

• One side = 7 in 

Answer: Missing side = 4 in

• Area = 35 in²

Answer: Missing side = 5 in

(iv) Given

• Area = 14 in²

• One side = 7 in 

Answer: Missing side = 2 in

• Area = 29 m²

• One side = 4 m

• 
  

Answer: Missing side = 7.25 m

(b) Given:

• Area = 11 m² & Height =4m

• One side = ?

(c) Given:

Area = 50 m²
one side = ?
The total area of the shape is given as 50m²

   Area(c)=Total Area−(Area(a)+Area(b))

   Area (c)=50m² − ( 29m² + 11m² )  

• Width = same as 29 m² rectangle = 7.25 m

• Area = 10 m²

​

2. The figure shows a path (the shaded portion) laid around a rectangular park EFGH. (i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. An example of a formula — Area of a rectangle = length × width. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.]

Path around a rectangular park

A rectangular park EFGH is surrounded by a path (shaded region).

(i) Measurements needed to find the area of the path

To find the area of the path, we need:

• Length of the outer rectangle (ABCD)

• Breadth of the outer rectangle

• Length of the inner rectangle (EFGH)

• Breadth of the inner rectangle

Let:

• Outer rectangle = 20 m × 14 m

• Inner park = 16 m × 10 m

• 
  

  Formula

​

(ii) If the width of the path along each side is given, can you find 

its area? If not, what other measurements do you need? Assign 

values of your choice to these measurements and find the area of 

the path. Give a formula for the area using these measurements. 

[Hint: Break the path into rectangles.]

(ii) If width of path is given

Suppose:

• Width of path = x m

• Inner park dimensions = l × b

Then:

• Outer length =
  

• Outer breadth =
  

(iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?

Answer: ❌ No

Reason:
Area depends only on dimensions, not on position.
As long as the inner park and width of the path remain the same, area remains unchanged.

3. The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.

Cross path in a rectangular plot
Plot dimensions = 14 m × 12 m
To find the area of the cross path, we need:
Width of the horizontal path
Width of the vertical path

Let:

Width of each path = 2 m

Area of horizontal path:

Area of vertical path:

Overlapping square counted twice:

​General Formula

If width of each path = w:

4. Find the area of the spiral tube shown in the figure. The tube has the same width throughout [Hint: There are different ways of finding the area. Here is one method.] What should be the length of the straight tube if it is to have the same area as the bent tube on the left?

Area of spiral tube

The tube has uniform width.

Break the spiral into rectangular strips of width 5 units.

Add the areas of all straight rectangular parts.

Length of straight tube

To have the same area:

The total length ( L ) of the spiral tube is the sum of the lengths of all its straight segments. 

  L=20+20+20+15+15+10+10+5+5 = 120 units

The tube has a uniform width of 1 unit The area can be found by multiplying the total length by the width.   

A = L x W

A =  120 x 1 = 120

The area of the spiral tube is 120 Sq.units

The area of the bent tube is calculated above as 120 square units. The width of the straight tube is also 1 unit.

A = L x W

120 =  L x 1 

L= 120

The length of the straight tube should be 120 units

​

5. In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

Doubling the side of a square

If original side = s
New side = 2s

Areas

• Original area = s²

• New area = (2s)² = 4s²

• Increase in regions 1, 2, and 3

The increase in the total Area is A2-A1 =  4s² - s² = 3s²

Each region’s area becomes 3 times its original area.

Reason:
Area is proportional to the square of the side.

 The total area increases by a factor of 4.
Therefore, the area of each region (1, 2, and 3) will also increase by a factor of 4, or increase by three times their original area.

6. Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside. You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials.

Rearranging square pieces

• Square divided into 4 equal parts.

• Rearranged to form a larger square with a hole.

• Draw two perpendicular lines inside a square as shown in the figure. This divides the square into four parts, which are all right-angled trapezoids.

• To form a larger square with a hole in the center, you can arrange the four trapezoidal pieces around a central square hole. Orient the pieces so that the shorter parallel side of each trapezoid faces inwards towards the hole, and the longer parallel side faces outwards. The perpendicular sides of adjacent trapezoids will align to form the sides of the larger square.The area of the larger square minus the hole is equal to the area of the original square.Area depends on the surface covered, not on the shape or arrangement.

Answer:

The four right-angled trapezoidal pieces can be rearranged to form a larger square with a square hole in the center. The area of the central hole will be equal to the area of the smallest right-angled triangle that can be formed by rearranging the pieces.

• Conclusion:

✔️ Area remains the same

✔️ Shape changes, area does not

Figure it Out Page number 157-158

1. Find the areas of the following triangles:

(i) From the figure:

• Base = 4 cm
  Height = 3 cm

(ii) From the figure:

• Base = 5 cm
  Height = 3.2 cm

(iii)

Base = 4 cm

Height = 3 cm

2. Find the length of the altitude BY

Given from the figure

•  units

•  units (altitude corresponding to base BC)

•  units

•  is the altitude corresponding to base
  

area of

• Base  units

• Height  units

• Base  units

• Corresponding altitude BY

Since the area is already known to be 12 square units:

Answer : The length of the altitude BY=3 units

3. Find the area of ∆SUB, given that it is isosceles, SE is perpendicular to UB, and the area of ∆SEB is 24 sq. units.

• Given

  •  is isosceles

  • 
    

  • Area of  sq. units

  In an isosceles triangle:

  • The perpendicular drawn from the vertex to the base is also the median.

  • Therefore, point E is the midpoint of UB.

  • Hence,UE=EB

  So, the two triangles:△USE and △BSE are congruent right-angled triangles.

  Since the two triangles are congruent:

  The whole triangle  is made up of these two equal triangles.

In the Śulba-Sūtras, which are ancient Indian geometric texts that deal with the construction of altars, we can find many interesting problems on the topic of areas. When altars are built, they must have the exact prescribed shape and area. This gives rise to problems of the kind where one has to transform a given shape into another of the same area. The Śulba-Sūtras give solutions to many such problems. Such problems are also posed and solved in Euclid’s Elements. Here are two problems of this kind.

4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area. 

Method

• Take the given rectangle.

• Draw a diagonal of the rectangle.

• The diagonal divides the rectangle into two congruent triangles.

Explanation

Each triangle has half the area of the rectangle.
So, one of the triangles obtained has an area equal to half the rectangle.

 Thus, a triangle of equal area is obtained.

or

5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area. 

Method

• Take the given triangle.

• Draw a line through the vertex parallel to the base to form a second, congruent triangle.

• Join the two congruent triangles to form a rectangle.

Explanation

The rectangle formed has:

• The same base

• Twice the height,
  so its area equals the area of the original triangle.

 Thus, a rectangle of equal area is obtained.

or

6. ABCD, BCEF, and BFGH are identical squares. 

(i) If the area of the red region is 49 sq. units, then what is the area of the blue region? 

(ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?

Let the side of each square = s units.

(i) If the area of the red region is 49 sq. units, find the area of the blue region

Look at square ABCD

• The diagonal line goes from D to B

• This diagonal divides square ABCD into two equal triangles

• Each triangle has area =
• The blue/purple region

• The blue region is half of one of those triangles

• Therefore its area is:

So visually and mathematically, it is one–fourth of the square.

• The red triangular piece from HB

• Can be rearranged 4 times to exactly fill one full square

The ratio is: 

(i) If red area = 49 sq units

(ii) If the total area of the red and blue regions is 180 sq. units, find the area of each square

(ii) Another version of the figure

Total area of red + blue = 180 sq units
(This is a fresh case — not using 49)

7. If M and N are the midpoints of XY and XZ, what fraction of the area of ∆XYZ is the area of ∆XMN? [Hint: Join NY]

Given:

•  is the midpoint of
  

•  is the midpoint of
  

•  is the given triangle

Find what fraction of the area of  is the area of .

Step 1: Join

Since  is the midpoint of , the line  is a median of .

A median divides a triangle into two triangles of equal area.

Step 2: Use the midpoint  in

In ,  is the midpoint of .
So,  is a median of .

A median again divides a triangle into two equal-area triangles.

Answer

📌 Key idea to remember

Each median divides a triangle into two equal-area triangles.
Using two medians successively divides the area into four equal parts.

8. Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

Explanation

The shortest distance between two points is a straight line.
But here, Gopal must touch the river line in between.

To find the shortest path, we use the reflection method.

1. Draw the river as two parallel horizontal lines and label it River.

2. Mark a point below the river and label it House.

3. Mark another point below the river (but on the other side) and label it Water Tank.

4. Reflect the Water Tank across the river line to get T′ (reflected tank) above the river.

5. Join House to T′ with a straight line.

6. Mark the point where this line meets the river as Touch point.

7. Join the Touch point to the Water Tank.

 Explanation

The shortest path is obtained by reflecting the water tank across the river and drawing a straight line from the house to the reflected point.
The point where this line meets the river gives the shortest route:

House → River → Water Tank.

📌 Key idea to remember

Reflection helps convert a broken path into a straight line, which gives the shortest distance.

Shortest version of Answer:

 The shortest path is a straight line from the house to the river, and then another straight line from that point on the river to the water tank.

To find this path, one can use a geometric principle: The shortest distance between two points is a straight line. The path consists of two straight-line segments.

The shortest path is found using reflection.

Reflect the water tank across the river and join the house to this reflected point by a straight line.

The point where this line meets the river gives the shortest path: House → River → Water Tank.

Figure it Out page number 160

 1. Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC

Given:
 cm,  cm,  cm
 and

Explanation:

Draw diagonal
The quadrilateral  is divided into two triangles: △ABC and , both having the same base .

Area of :

Area of :

Total Area of quadrilateral :

2. Find the area of the shaded region given that ABCD is a rectangle

Given

• ABCD is a rectangle

• AB = 10 cm + 8 cm = 18 cm

• BC = 10 cm

• Point E on AB such that AE = 10 cm

• Point F on AD such that AF = 6 cm and FD = 4 cm

• The shaded region is D–F–E–C

Triangle AFE

• Base = AE = 10 cm

• Height = AF = 6 cm

Triangle EBC

• Base = EB = 8 cm

• Height = BC = 10 cm

​3. What measurements would you need to find the area of a regular hexagon?

Answer:

The length of one side is sufficient.

Explanation:

A regular hexagon has six equal sides. If the side length is s, its area is:

(Alternatively, the apothem can also be used, but side length is enough.)

What is an apothem?

Apothem is the perpendicular distance from the centre of a regular polygon to the midpoint of any side.

👉 In simple words:
It is the shortest distance from the centre to a side, not to a corner.

Why is apothem mentioned for a regular hexagon?

For a regular hexagon:

• All sides are equal

• The centre is exactly in the middle

• The apothem helps find the area using another formula

Area formula using apothem

So what does the sentence mean?

“Alternatively, the apothem can also be used, but side length is enough.”

It means:

• You can find the area using only the side length (simpler method), or

• You can also find it using the apothem and perimeter

For exams, side length alone is usually sufficient and easier, so apothem is optional.

        ●

       / \

      /   \

     /     \

    ●-------●

    |   ⟂   |

    |   a   |   a = apothem

    |       |

    ●-------●

     \     /

      \   /

       \ /

        ●

Labels to remember

• ● = vertices of the regular hexagon

• Centre = point where the ⟂ (right angle) meets

• a (apothem) = perpendicular distance from centre to the midpoint of a side

Memory tip 

Apothem = centre → side (perpendicular)

4. What fraction of the total area of the rectangle is the area of the blue region?

Explanation:

The rectangle is divided into two coloured regions: blue and white.
Each region is made up of two triangles.

The two blue triangles together form one large triangle, and the two white triangles together form another triangle.

Both triangles have:

• the same base (a side of the rectangle)

• the same height (the other side of the rectangle)

Since the area of a triangle is

both combined regions have equal area.

Answer:

​​

The blue region occupies half of the rectangle’s area.

5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral. One can derive special formulae to find the areas of a parallelogram, rhombus and trapezium.

Answer: 

Draw one of the diagonals of the given quadrilateral.

A diagonal divides any quadrilateral into two triangles. The area of each triangle is not necessarily half the area of the quadrilateral unless the diagonal bisects the other diagonal. However, a different method is to join the midpoints of two adjacent sides of the quadrilateral to one of the opposite vertices.

Alternatively, a more general method involves finding the midpoint of any two adjacent sides of the given quadrilateral and connecting them to form a smaller triangle. The area of this triangle will be a specific fraction of the total area, but the most direct method to get a new quadrilateral with half the area is to use a diagonal.

A simpler, more general method that creates a new quadrilateral is to connect the midpoints of the four sides of the original quadrilateral. The resulting inner quadrilateral (a parallelogram) will have exactly half the area of the original quadrilateral.

Explanation:

Join the midpoints of all four sides of the given quadrilateral.

The figure formed inside is a parallelogram.
This inner parallelogram always has exactly half the area of the original quadrilateral (this is a known geometric result).

Answer:

To get a quadrilateral of half the area, join the midpoints of the four sides of the given quadrilateral.
The inner quadrilateral formed has half the area of the original one.

Figure it Out Page number 162 - 163

1. Observe the parallelograms in the figure below. 

(i) What can we say about the areas of all these parallelograms?

 (ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?

i) What can we say about the areas of all these parallelograms?

Answer:
All the given parallelograms have the same area.

Explanation:
The area of a parallelogram is given by

From the grid:

• Each parallelogram has the same base length (4 units).

• Each has the same vertical height (3 units).

Since both base and height are the same for all figures, their areas are equal.

(ii) What can we say about their perimeters? Which has the maximum and minimum perimeter?

Answer:
The perimeters are different for the parallelograms.

• Minimum perimeter: Figure (a)

• Maximum perimeter: Figures (c) and (g)

Explanation:
The perimeter of a parallelogram is the sum of all its sides.

• The base is the same for all figures.

• The slanted sides are different because the horizontal shift varies.

• As the horizontal displacement increases, the slanted side becomes longer (by the Pythagorean theorem), increasing the perimeter.

Figure (a) has vertical sides (shortest slanted sides), so it has the smallest perimeter.
Figures (c) and (g) have the greatest slant, so they have the largest perimeter.

Final Summary 

• All areas: Same

• Perimeters: Different

• Minimum perimeter: (a)

• Maximum perimeter: (c) and (g)

2. Find the areas of the following parallelograms:

The area of a parallelogram is given by the formula 

   Area=𝐛×𝐡

. (i) Area = 7 x 4 = 28 cm²

(ii) Area = 5 x 3 = 15 cm²

(iii) Area = 5 x 4.8 = 24cm²

(iv) Area = 4.4 x 2 = 8.8 cm²

   

3. Find QN.

Area of the parallelogram

Using the other base PS=7.6 cm,

Answer:  cm

 4. Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]

4. Compare areas of rectangle and parallelogram 

Answer: The rectangle 

A rectangle with side lengths of 5 cm and 4 cm has an area of 

   

5 cm×4 cm=20cm². 

A parallelogram with the same side lengths (5 cm and 4 cm) will have a smaller area unless it is also a rectangle (i.e., has 90-degree angles). 

The area of a parallelogram is calculated by multiplying its base by its perpendicular height. 

The height of a parallelogram with a 4 cm side length will be less than or equal to 4 cm. 

The maximum area for a parallelogram with these side lengths occurs when it is a rectangle.

Therefore, the rectangle has the greater area.

5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

Area of a triangle

Area of a rectangle

Method 1 

• Length = base of the triangle

• Width = height of the triangle

So the rectangle has twice the area of the triangle.

Method 2 

Choose any convenient length  for the rectangle.
Calculate width using:

This rectangle will also have twice the area of the triangle.

Method 3 (Fix the width, calculate the length)

Choose any convenient width
Calculate length using:

This again gives a rectangle with double the area.

Method 4 (Using duplication of the triangle)

Make a congruent copy of the given triangle and place it beside the original triangle along the base.
The combined figure forms a parallelogram, which can be rearranged into a rectangle of the same area.

Answer

A rectangle whose length × width = base × height of the triangle will have twice the area of the given triangle.
This can be obtained by taking the triangle’s base and height as the rectangle’s dimensions or by suitably choosing one dimension and calculating the other.

6. [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.

 [Śulba-Sūtras] Method to obtain a rectangle of the same area as a given triangle

Answer with explanation:

Consider a triangle with base  and height .

• The area of the triangle is

• Take a rectangle whose:

  • length = ​ (half of the triangle’s base)

  • breadth =  (same as the triangle’s height)

Then, the area of the rectangle is

which is equal to the area of the triangle.

Hence, a rectangle with half the base and the same height has the same area as the given triangle.

Key Point (for memory):

Triangle area = Rectangle area when rectangle length = b ⁄ 2 and height = h

 7. [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it? 

[Hint: Show that triangles ∆ADB and ∆ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]

Śulba-Sūtras] Converting an isosceles triangle into a rectangle

Answer with explanation:

Take an isosceles triangle ABC and draw the altitude AD to the base BC.

• Since the triangle is isosceles, AD bisects BC, so BD = DC.

• Cutting along AD divides the triangle into two congruent right-angled triangles:
   and .

• Rearrange these two triangles by placing their hypotenuses (AB and AC) next to each other.

• The legs AD and BD + DC form the sides.

Thus, the two triangles together form a rectangle with the same area as the original triangle.

Isosceles triangle → rectangle: Cut along altitude and rearrange

8. [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection. 

[Śulba-Sūtras] Convert a rectangle into an isosceles triangle

Answer with explanation:

• Draw a diagonal of the rectangle.

• This divides the rectangle into two congruent right-angled triangles.

• Rearrange the two triangles by joining their hypotenuses together.

• The joined figure forms an isosceles triangle.

Hence, a rectangle can be converted into an isosceles triangle by cutting and rearranging without changing the area.

Rectangle → isosceles triangle: Cut along diagonal and rearrange

9. Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.

Area comparison

(a) Which has greater area:

An equilateral triangle or a square of the same side length?

Answer:
The square has the greater area.

Explanation:
Let the side length be .

• Area of an equilateral triangle

• Area of a square

Since , the square has a greater area.

(b) Which has greater area:

Two identical equilateral triangles together or a square of the same side length?

Answer: The square still has the greater area.

Explanation:

• Area of two equilateral triangles

• Area of a square

Since , the square has a greater area.

Square has greater area than:

one equilateral triangle
two equilateral triangles of the same side length

Figure it Out Page number 169-170

 1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm. 

Given: Diagonal₁ = 20 cm & Diagonal₂ = 15 cm

Area:

Answer: 150 cm²

2. Give a method to convert a rectangle into a rhombus of equal area using dissection. 

Method:

1. Take a rectangle.

2. Draw both diagonals.

3. Cut along one diagonal to get two congruent triangles.

4. Rotate one triangle and join it to the other along equal sides.

5. The new figure formed is a rhombus.

Reason:

• Area is preserved because only cutting and rearranging is done.

• All sides of the new figure are equal ⇒ rhombus.

 3. Find the areas of the following figures:

(i) Given:

• Parallel sides = 10 ft, 7 ft

• Height = 16 ft (perpendicular shown)

1/2 x 16 x (10+7)

8 x 17 

Answer: 136 ft²

(ii) Given:

• Parallel sides = 24 m, 36 m

• Height = 14 m

Formula:

Answer: 420 m²

(iii) Given:

• Height = 10 in

• Parallel sides= 14 in and 6 in

1/2 x 10 (14+6)

5 x 20 sq inches

Answer: 100 in²

(iv) Given:

• Parallel sides = 12 ft, 18 ft

• Height = 8 ft

Answer: 120 ft²

4. [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

Method:

1. Drop perpendiculars from the shorter parallel side to the longer base.

2. Two equal right triangles are formed on the sides.

3. Cut these triangles.

4. Shift them to the middle.

5. A rectangle is formed.

Why area is equal:
No part is added or removed — only rearranged.

 5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area —

Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH? [Hint: If ∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ, then the trapezium and rectangle have equal areas.]

Construction idea:

• Drop perpendiculars from A and B to DC.

• Mark points I and J on AD and BC respectively.

• Cut triangles:

  • ∆AHI ≅ ∆DGI

  • ∆BEJ ≅ ∆CFJ

• Rearrange these congruent triangles to form rectangle EFGH.

Key idea:
Equal triangles removed and added ⇒ area preserved.

6. Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm2.

Steps:

1. Take a rectangle of area 144 cm²
   (example: 12 cm × 12 cm).

2. Cut one triangle from a corner.

3. Shift it to the opposite side.

4. The new figure is a trapezium.

 Area remains 144 cm².

7. A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.

A regular hexagon can be divided into 6 congruent equilateral triangles.

From the figure:

• Trapezium = 3 triangles

• Rhombus = 2 triangles

• Equilateral triangle = 1 triangle

Answer: 3 : 2 : 1

8. ZYXW is a trapezium with ZY‖WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ∆ZWB.

Given:

• ZY ∥ WX

• A is midpoint of XY

Proof idea:

• Join ZB.

• Triangles formed have the same height.

• Bases lie on the same line.

• Hence areas are equal.

Explanation:

• The area of trapezium ZYXW is the sum of the areas of △ZYA and trapezium ZAWX.
• △ZYA and △ZXA share the same height (the perpendicular distance between ZY and WX) and have equal bases (AY = AX, since A is the midpoint of XY).
   
  Therefore, Area(△ZYA) = Area(△
  ZXA).
• Area of trapezium ZYXW = Area(△ZYA) + Area(trapezium ZAWX) = Area(△
  ZXA) + Area(trapezium ZAWX) = Area(△ZWA).
• Points W, X, and B are collinear, so the area of △ZWA is equal to the area of △ZWB, as they share the same height from Z to the line WXB and have the same base WA = WB (implied by the diagram where X is between W and B, and A is the midpoint).
• Therefore, Area(trapezium ZYXW) = Area(△ZWB).

Areas in Real Life

What do you think is the area of an A4 sheet? Its sidelengths are 21 cm and 29.7 cm. Now find its area

Given:

• Length = 29.7 cm

• Breadth = 21 cm

Formula:
Area = length × breadth

Calculation:
Area = 29.7 × 21
Area = 623.7 cm²

Area of an A4 sheet = 623.7 cm²

What do you think is the area of the tabletop that you use at school or at home? You could perhaps try to visualise how many A4 sheets can fit on your table. The dimensions of furniture like tables and chairs are sometimes measured in inches (in) and feet (ft). 1 in = 2.54 cm 1 ft = 12 in

Suppose a tabletop is about:

• Length = 120 cm

• Breadth = 60 cm

Area = 120 × 60 = 7200 cm²

Now compare with A4 sheets:

Number of A4 sheets ≈
7200 ÷ 623.7 ≈ 11–12 A4 sheets

About 11–12 A4 sheets can fit on the table

Express the following lengths in centimeters: (i) 5 in (ii) 7.4 in

Given:
1 inch = 2.54 cm

(i) 5 in

5 × 2.54 = 12.7 cm

(ii) 7.4 in

7.4 × 2.54 = 18.796 cm

Express the following lengths in inches: (i) 5.08 cm (ii) 11.43 cm

(i) 5.08 cm

(ii) 11.43 cm

How many cm² is 1 in² ? So, 1 in² = 2.54² cm² = 6.4516 cm²

1 in = 2.54 cm

So,
1 in² = 2.54 × 2.54 = 6.4516 cm²

1 in² = 6.4516 cm²

How many cm² is 10 in² ? 10 in² = 10 × 6.4516 cm² = 64.516 cm².

10 in² = 10 × 6.4516 = 64.516 cm²

Convert 161.29 cm² to in². Every 6.4516 cm² gives an in². Hence, 161.29 cm² = 161.29 /6.4516 in² Evaluate the quotient.

 Answer: 25 in²

What do you think is the area of your classroom? Areas of classroom, house, etc., are generally measured in ft² or m²

• Suppose classroom size:

  • Length = 8 m

  • Breadth = 6 m

  Area = 8 × 6 = 48 m²

   Classrooms are measured in m² or ft²

How many in² is 1 ft² ?

1 ft = 12 in

So,
1 ft² = 12 × 12
= 144 in²

What do you think is the area of your school? Make an estimate and compare it with the actual data. Larger areas of land are also measured in acres. 1 acre = 43,560 ft2. Besides these units, different parts of India use different local units for measuring area, such as bigha, gaj, katha, dhur, cent, ankanam, etc. 

Suppose school land ≈ 2 acres

Given:
1 acre = 43,560 ft²

So,
2 acres = 87,120 ft²

Find out the local unit of area measurement in your region.

Some commonly used local units:

• Bigha

• Gaj

• Katha

• Dhur

• Cent

• Ankanam

(Varies from state to state.)

What do you think is the area of your village/town/city? Make an estimate and compare it with the actual data. Larger areas are measured in km².

Example:

• Small town ≈ 20 km²

Cities and towns are measured in km².

How many m² is a km²  ?

1 km = 1000 m

So,
1 km² = 1000 × 1000
= 1,000,000 m²

How many times is your village/town/city bigger than your school?

Example:

• City area = 20 km² = 20,000,000 m²

• School area = 5,000 m²

Ratio = 20,000,000 ÷ 5,000 = 4000 times

Find the city with the largest area in (i) India, and (ii) the world

(i) India ➡ Delhi (largest by area among Indian cities)

(ii) World ➡ Chongqing, China

Find the city with the smallest area in (i) India, and (ii) the world

(i) India ➡ Panaji (Goa)

(ii) World➡ Vatican City