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:
(ii) (a) Given:
(i) Given:
Area = 21 in²
One side = 7 in
-
Area = 28 in²
One side = 7 in
Answer: Missing side = 4 in
Answer: Missing side = 5 in
(iv) Given
Answer: Missing side = 2 in
-
Area = 29 m²
-
One side = 4 m
Answer: Missing side = 7.25 m
(b) Given:
(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 )
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:
(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:
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 plotPlot 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?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
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.
✔️ 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:
(ii) From the figure:
(iii)
Base = 4 cm
Height = 3 cm
Given from the figure
area of
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:
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
-
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
So visually and mathematically, it is one–fourth of the 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)
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.
Draw the river as two parallel horizontal lines and label it River.
-
Mark a point below the river and label it House.
-
Mark another point below the river (but on the other side) and label it Water Tank.
-
Reflect the Water Tank across the river line to get T′ (reflected tank) above the river.
-
Join House to T′ with a straight line.
-
Mark the point where this line meets the river as Touch point.
-
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:
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:
Since the area of a triangle is
both combined regions have equal area.
Answer:
The blue region occupies half of the rectangle’s area.
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:
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.
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
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
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 .
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 .
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:
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:
-
Take a rectangle.
-
Draw both diagonals.
-
Cut along one diagonal to get two congruent triangles.
-
Rotate one triangle and join it to the other along equal sides.
-
The new figure formed is a rhombus.
Reason:
3. Find the areas of the following figures:
(i) Given:
1/2 x 16 x (10+7)8 x 17
Answer: 136 ft²
(ii) Given:
Formula:
Answer: 420 m²
(iii) Given:
1/2 x 10 (14+6) 5 x 20 sq inches
Answer: 100 in²
(iv) Given:
Answer: 120 ft²
4. [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a
rectangle using dissection.
Method:
-
Drop perpendiculars from the shorter parallel side to the longer base.
-
Two equal right triangles are formed on the sides.
-
Cut these triangles.
-
Shift them to the middle.
-
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:
-
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:
-
Take a rectangle of area 144 cm²
(example: 12 cm × 12 cm).
-
Cut one triangle from a corner.
-
Shift it to the opposite side.
-
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:
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:
- 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(△
- 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.
What do you think is the area of your classroom? Areas of classroom, house, etc., are generally measured in ft² or m²
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:
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:
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
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