Area Class 8 Solutions Maths Ganita Prakash Part 2 Chapter 7

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Class 8 Maths Ganita Prakash Part 2 Chapter 7 Solutions

Ganita Prakash Class 8 Chapter 7 Solutions Area

Class 8 Maths Ganita Prakash Part 2 Chapter 7 Area Solutions Question Answer

7.1 Rectangle and Squares

Figure It Out (Pages 150-152)

Question 1.
Identify the missing sidelengths.

Solution:
(i) After naming the figure
In rectangle ABCD,
Area of rectangle = Length × Breadth
7 × BC = 21
⇒ BC = 3 in

∴ AF = AD + DF
= 3 in + 4 in
= 7 in
In the rectangle EFAG,
EF × AF = 28 in²
⇒ EF × 7 in = 28 in²
⇒ EF = 4 in
∴ HA = HG + GA
= 3 in + 4 in
= 7 in
In rectangle HIJA,
Area = HA × AJ
⇒ 35 in² = 7 in × AJ
⇒ AJ = 5 in
∴ AK = AJ + JK
= 5 in + 2 in
= 7 in
In rectangle KLMA,
Area = KL × LM
⇒ x in × 7 in = 14 in²
⇒ x = 2 in
Thus, the missing sidelength = 2 in

(ii) After naming the figure,

In rectangle ABGH,
Area = AB × AH
AB × 4 m = 29 m²
AB = \(\frac {29}{4}\) m or 7.25 m
Area of rectangle HGDC = Area of rectangle ABDC – Area of rectangle ABGH
= 50 m² – 29 m²
= 21 m²
In rectangle HGDC,
CD × GD = 21 m²
⇒ \(\frac {29}{4}\)m × GD = 21 m²
⇒ GD = \(\frac {84}{29}\) m or 2.9 m
In rectangle BEFG,
BG × BE = 11 m²
⇒ 4 m × M = 11 m²
⇒ BE = \(\frac {11}{4}\)m or 2.75 m
Thus, AB = \(\frac {29}{4}\) m; BE = \(\frac {11}{4}\) m and GD = \(\frac {84}{29}\) m

Question 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.]
(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.]
(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?

Solution:
(i) Measurements needed:
Length of outer rectangle ABCD = A
Width of outer rectangle ABCD = B
Length of inner rectangle EFGH = a
Width of inner rectangle EFGH = b
Let A = 10, B = 8, a = 6, b = 4.
Calculation: Area of path = Area of ABCD – Area of EFGH
= 10 × 8 – 6 × 4
= 80 – 24
= 56 m²
Formula:
Area of path = (A × B) – (a × b)

(ii) Yes, if the width of the path is uniform along each side, we can find its area, but we also need the dimensions of either the outer or inner rectangle.
If the width of the path d = 2 m (Uniform on all sides)
Length of inner path EFGH = l = 16 m
Width of inner path EFGH = w = 11 m
Length of outer rectangle = l + 2d
= 16 + 2(2)
= 20 m
Width of outer rectangle = w + 2d
= 11 + 2(2)
= 15 m
Breaking the path into rectangles
Now there are 4 rectangles
Left rectangle = w × d
= 11 × 2
= 22 m²
Right rectangle = w × d
= 11 × 2
= 22 m²
Top rectangle = (l + 2d) d
= 20 × 2
= 40 m²
Bottom rectangle = (l + 2d) d
= 20 × 2
= 40 m²
Total area of path = 22 + 22 + 40 + 40 = 124 m²
Formula:
Area of path = 2d(l + w) + 4d
where d – width of path
l – length of inner path
w – width of inner path

(iii) No, the area of the path does not change.
Reason: The area of the path depends only on:
Area of outer rectangle ABCD.
Area of inner rectangle EFGH.

Question 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.

Solution:
Measurements needed:
Length of plot = 14 m
Width of plot = 12 m
Width of horizontal path = x₁
Width of vertical path = x₂
Assign values:
Let x₁ = 2 m
x₂ = 2 m
Now area of horizontal path = 14 × 2 = 28 m
Area of vertical path = 12 × 2 = 24 m
Area of overlapping square = 2 × 2 = 4 m²
∴ Area of cross path = 28 + 24 – 4 = 48 m²
Formula = Area of cross path = (L × w₁) + (W × w₂) – (w₁ × w₂)
Here, L = length of plot
W = Width of plot
w₁ = Width of horizontal path
w₂ = Width of vertical path

Question 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?
Solution:
After naming the figure,

The area of the spiral tube = Area of the rectangle, ABEC + Area of the rectangle, DEGF + Area of the rectangle, GHIJ + Area of the rectangle, JKML + Area of the rectangle, NOPL + Area of the rectangle, PQRS + Area of the rectangle, STUV + Area of the rectangle, VWYX + Area of the rectangle, XZA₁B₁
= AC × AB + EG × DE + IH × JI + LJ × LM + NO × NL + PQ × PS + UT × ST + VX × VW + ZA₁ × A₁B₁
= 20 × 1 + 18 × 1 + 20 × 1 + 13 × 1 + 15 × 1 + 8 × 1 + 10 × 1 + 3 × 1 + 5 × 1
= 20 + 18 + 20 + 13 + 15 + 8 + 10 + 3 + 5
= 112 sq. units
Thus, the area of the spiral tube is 112 sq. units.
Let the length of the straight tube be x.
The area of the bent tube on the left = Area of rectangle, BACD + Area of rectangle, BGFE
= AC × CD + BG × BE
= 5 × 1 + 4 × 1
= 9 sq. units

Area of straight tube = x × 1
Area of bent tube = 9 sq. units
According to the question, both are the same.
So x × 1 = 9
⇒ x = 9 units.

Question 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.

Solution:
Let ‘a’ be the side of the square.
Area of triangle, DCB (region 3)




Thus, the increase in the areas of regions 1, 2, and 3 is 4 times.
Reason: If the sidelength of the square is doubled, then the area becomes 4 times.

Question 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.
Solution:
1. Let us take a square of cardboard (8 cm × 8 cm).
2. Draw two perpendicular lines inside the square (not through the center), dividing it into 4 rectangular pieces.
3. Cut along these lines to get 4 pieces.
4. Rearrange these 4 pieces.
Place them at the four corners of a larger imaginary square.
The pieces should be arranged so that they form a square with a hole in the middle.

Triangles

Figure It Out (Pages 157-159)

Question 1.
Find the areas of the following triangles:

Solution:
(i) Area of triangle ABC = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × BC × AE
= \(\frac {1}{2}\) × 4 cm × 3 cm
= 6 cm²
Thus, the area of the triangle, ABC = 6 cm²

(ii) Area of triangle DEF = \(\frac {1}{2}\) × EF × ND
= \(\frac {1}{2}\) × 5 cm × 3.2 cm
= 5 × 1.6 cm²
= 8 cm²
Thus, the area of the triangle DEF = 8 cm²

(iii) Area of triangle = \(\frac {1}{2}\) × base × height
Area of ∆NAT = \(\frac {1}{2}\) × AT × NA
= \(\frac {1}{2}\) × 3 cm × 4 cm
= 6 cm²
Thus, the area of the triangle, NAT = 6 cm²

Question 2.
Find the length of the altitude BY.

Solution:
Area of ∆AXC = \(\frac {1}{2}\) × XC × AX
= \(\frac {1}{2}\) × (XB + 6) × 4
= 2 × (XB + 6)
= 2 × XB + 12 sq. units
Area of ∆AXB = \(\frac {1}{2}\) × XB × AX
= \(\frac {1}{2}\) × XB × 4
= 2 × XB sq. units
Area of ∆ABC = \(\frac {1}{2}\) × AC × BY
= \(\frac {1}{2}\) × 8 × BY
= 4BY
∴ Area of ∆AXC = Area of ∆AXB + Area of ∆ABC
2XB + 12 = 2XB + 4BY
⇒ 4BY = 12
⇒ BY = 3 units
Thus, the length of the altitude BY is 3 units.

Question 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.

Solution:
Given,
The area of ∆SEB = 24 sq. units
Given that ∆SUB is an isosceles triangle.
SU and SB are equal sides, and UB is the base.
∴ SE is perpendicular to UB
⇒ UE = EB
SE is the common base of ∆SUE and ∆SEB.
∴ Area of ∆SEB = 24 sq. units = Area of ∆SEU
∴ The area of triangle SUB = Area of ∆SEU + Area of ∆SEB
= 24 + 24
= 48 sq. units
Thus, the area of ∆SUB is 48 sq. units.

In the Sulba-Sutras, 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 Sulba-Sutras give solutions to many such problems. Such problems are also posed and solved in Euclid’s Elements. Here are two problems of this kind.

Question 4.
[Sulba-Sutras] Give a method to transform a rectangle into a triangle of equal area.
Solution:
1. Let us take a rectangle ABCD, with length a and breadth b.

2. Now mark the midpoint E of side CD.
Draw a line perpendicular to CD passing through E.
Mark a point M on it such that ME = b.
3. Draw a triangle using base = AB (same as the rectangle’s length) and join M to A and B.

Question 5.
[Sulba-Sutras] Give a method to transform a triangle into a rectangle of equal area.
Solution:
1. Take a triangle ABC with base b and height h.

2. Find the midpoint M of the height.
3. Draw a line parallel to the base through M.
This line intersects the sides of the triangle.
4. Create a rectangle using.
Length = Same as the triangle’s base = b
Width = Half of the triangle’s height = \(\frac {h}{2}\)

Question 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 black region?
(ii) In another version of this figure, if the total area enclosed by the black and red regions is 180 sq. units, then what is the area of each square?
Solution:
Given that ABCD, BCEF, and BFGH are identical.
The area of the red region (∆HBI + IBCD) = 49 sq. units.
Let the side of each square be a.
∴ IB = \(\frac{A B}{2}=\frac{a}{2}\) units
Let ‘a’ units be the side of the square.
(i) Area of the red region ∆HDC = \(\frac {1}{2}\) × DC × DC
(∴ HC = HB + BC = a + a = 2a)
= \(\frac {1}{2}\) × a × 2a
⇒ \(\frac {1}{2}\) × 2a² = 49
⇒ a = 7 units

∴ The area of the black region, ∆IAD = \(\frac {1}{2}\) × AI × AD
= \(\frac{1}{2} \times \frac{7}{2} \times 7\)
= \(\frac {49}{4}\) sq. units
= 12.25 sq. units
Thus, the area of the black region is 12.25 square units.

(ii) Given, the total area enclosed by the black and red regions = Area of ∆HDC + Area of ∆AID = 180 sq. units
Let ‘a’ be the side of the square.

∴ The area of each square = a²
= (12)²
= 144 sq. units
Thus, the area of each square is 144 sq. units

Question 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 XY]

Solution:
Let O be the midpoint of YZ, then join M to O and N to O.

According to mid point theorem,
MN = \(\frac {1}{2}\) YZ, and MN is parallel to YZ.
The triangle XYZ is divided into four equal triangles.
So, Area of ∆XMN = \(\frac {1}{4}\) × Area of ∆XYZ.

Question 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.

Solution:
Let P be the point on the bank of the river, then the shortest path from H to P to T is created by the point P such that ∆HPT has minimum area.

Shortest path
House (H) → P (Point on river) → Water tank

Area of a Polygon

Figure It Out (Page 160)

Question 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.

Solution:
Area of the quadrilateral ABCD = Area of triangle CAD + Area of triangle ACB
Area of ∆ACB = \(\frac {1}{2}\) × AC × BM
= \(\frac {1}{2}\) × 22 cm × 3 cm
= 33 cm²
Area of ∆CAD = \(\frac {1}{2}\) × AC × DN
= \(\frac {1}{2}\) × 22 cm × 3 cm
= 33 cm²
∴ The area of the quadrilateral ABCD = 33 cm² + 33 cm² = 66 cm²

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

Solution:
The area of the shaded region = Area of the rectangle ABCD – (Area of triangle AEF + Area of triangle EBC)
Area of the rectangle ABCD = Length × Breadth
= AB × AD
= 18 cm × 10 cm [∵ AB = AE + EB = 10 cm + 8 cm = 18 cm; AD = AF + FD = 6 cm + 4 cm = 10 cm]
= 180 cm²
Area of the triangle AEF = \(\frac {1}{2}\) × AE × AF
= \(\frac {1}{2}\) × 10 cm × 6 cm
= 30 cm²
Area of the triangle EBC = \(\frac {1}{2}\) × EB × BC
= \(\frac {1}{2}\) × 8 cm × 10 cm
= 40 cm²
∴ The area of the shaded region = 180 cm² – (30 cm² + 40 cm²)
= 180 cm² – 70 cm²
= 110 cm²

Question 3.
What measurements would you need to find the area of a regular hexagon?
Solution:
Minimum measurement needed
Side length (l) of the hexagon
Area of regular hexagon = \(\frac{3 \sqrt{3} a^2}{2}\)
where l is the side length.

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

Solution:
Let l be the length and b be the breadth of the rectangle ABCD.

Total area of rectangle, ABCD = DC × BC = l × b sq. units
Area of ∆AOB = \(\frac {1}{2}\) × AB × OE = \(\frac {1}{2}\) × l × x sq. units
Area of ∆DOC = \(\frac {1}{2}\) × DC × OF = \(\frac {1}{2}\) × l × y sq. units
∴ The area of the red region = Area of ∆AOB + Area of ∆DOC
= \(\frac {1}{2}\) × l × x + \(\frac {1}{2}\) × l × y
= \(\frac {1}{2}\) × x × (x + y) sq. units
= \(\frac {1}{2}\) × l × b sq. units [∴ x + y = b]
∴ Area of red region = \(\frac {1}{2}\) × Area of rectangle
Thus, the required fraction is \(\frac {1}{2}\).

Question 5.
Give a method to obtain a quadrilateral whose area is halfthat of a given quadrilateral.
Solution:
Let ABCD be a given quadrilateral.
Mark mid points of AB, BC, CD, and DA as P, Q, R, and S.
Join midpoints, then PQRS is the required quadrilateral with half the area of the given quadrilateral ABCD.

Parallelogram

Figure It Out (Pages 162-164)

Question 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?
Solution:
(i) (a) Area of parallelogram = base × height
= 5 × 3
= 15 sq. units
(b) Area of parallelogram = 5 × 3 = 15 sq. units
(c) Area of parallelogram = 5 × 3 = 15 sq. units
(d) Area of parallelogram = 5 × 3 = 15 sq. units
(e) Area of parallelogram = 5 × 3 = 15 sq. units
(f) Area of parallelogram = 5 × 3 = 15 sq. units
(g) Area of parallelogram = 5 × 3 = 15 sq. units
All parallelograms have equal areas.
(ii) The perimeters of these parallelograms are different even though their areas are the same.
Figure (d) has the minimum perimeter, and Figure (g) has the maximum perimeter.

Question 2.
Find the areas of the following parallelograms:

Solution:
Area of the parallelogram = base × height
(i) Here, base = 7 cm and height = 4 cm
Area of the parallelogram = 7 cm × 4 cm = 28 cm²
(ii) Here, base = 5 cm and height = 3 cm
Area of the parallelogram = 5 cm × 3 cm = 15 cm²
(iii) Here, base = 5 cm and height = 4.8 cm
Area of the parallelogram = 5 cm × 4.8 cm = 24 cm²
(iv) Here, base = 2 cm and height = 4.4 cm
Area of the parallelogram = 2 cm × 4.4 cm = 8.8 cm²

Question 3.
Find QN.

Solution:
In ∆PNQ, ∠PNQ = 90°
PN = 7.6 cm
PQ = 12 cm
By Pythagoras theorem
PQ² = PN² + NQ²
⇒ (12)² = (7.6)² + NQ²
⇒ QN² = (12)² – (7.6)²
⇒ QN² = 144 – 57.76
⇒ QN² = 86.24
⇒ QN = 9.28 cm

Question 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.]

Solution:
For rectangle:
l = 5, w = 4, all angles = 90°
∴ Area = l × w
= 5 × 4
= 20 cm²
For parallelogram:
Base (b) = 5 cm, one slanted side = 4 cm
Height will be less than 4 cm because the side is slanted.
Area = 5 × 4 < 20 cm²
Hence, the rectangle has a greater area than the parallelogram.

Question 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?
Solution:
Given: Triangle with area A
Required: Rectangle with area = A
Method 1: If the triangle has base b and height h
Area of triangle = \(\frac {1}{2}\) × b × h = A
To get a rectangle with an area of 2A.
Take length = b, width = h
Area of a rectangle = b × h
= 2 × (\(\frac {1}{2}\) × b × h)
= 2A
Steps:
1. Measure the base and height of the given triangle.
2. Construct a rectangle with these measurements as length and width.
Method 2: Scaling method:
1. Take the rectangle.
2. Create a rectangle with base = (base of triangle) and height = height of triangle.
3. This rectangle automatically has twice the area of the rectangle.

Question 6.
[Sulba-Sutras] Give a method to obtain a rectangle of the same area as a given triangle.
Solution:
Given: A triangle with base b and height h.
Required: Rectangle with the same area
Area of triangle = \(\frac {1}{2}\)bh
To get a rectangle with the same area
Rectangle length = \(\frac {b}{2}\)(half the triangle’s base)
Rectangle width = h (same as the triangle’s height)
Area = \(\frac {b}{2}\) × h = \(\frac {1}{2}\) × b × h

Question 7.
[Sulba-Sutras] 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.]
Solution:
Given: Isosceles triangle ABC, where AB = AC, and AD is the altitude from A to BC.
Method: Since the triangle is isosceles:
AD is perpendicular to BC. D is the midpoint of BC (property of an isosceles triangle).
AD bisects the triangle into two congruent right triangles: ∆ADB and ∆ADC.
Dissection Process:
Step 1: The altitude AD divides the isosceles triangle into two congruent right triangles ∆ADB and ∆ADC.
Step 2: Each of these right triangles can be made into half of a rectangle.
Step 3: Assembly:
Take triangle ∆ADB
Take triangle ∆ADC
Rotate one triangle 180°
Arrange them so that:
The two equal sides (AB and AC) form opposite sides of a rectangle.
The altitude AD appears twice, forming the other pair of opposite sides.
Step 4: The resulting figure is a rectangle with:
Length = BC (base of the isosceles triangle)
Width = \(\frac {AD}{2}\) (half the altitude) or alternatively.
Length = AB (= AC, the equal sides)
Width related to the base.

Question 8.
[Sulba-Sutras] Give a method to convert a rectangle into an isosceles triangle by dissection.
Solution:
This is the reverse of Question 7.
Method:
Given: Rectangle PQRS with length l and width w.

Required: Isosceles triangle with the same area
Dissection Steps:
Step 1: Take a rectangle PQRS with PQ = l and PS = w.
Step 2: Mark the midpoint M of side PQ.
Step 3: From M, draw lines to the bottom corners R and S.
Step 4: Cut the rectangle into three pieces:
Triangle PMS (left); Triangle QMR (right); Central region (if any)
Step 5: Rearrange:
Flip the triangle PMS and attach it along MS to form one half of the triangle.
Flip triangle QMR and attach it along MR to form the other half.
These create an isosceles triangle.

Question 9.
Which has a greater area — an equilateral triangle or a square of the same side length as the triangle? Which has a greater area: two identical equilateral triangles together or a square of the same side length as the triangle? Give reasons.
Solution:
Area of equilateral triangle = \(\frac{\sqrt{3}}{4} a^2\)
Area of square = a²

⇒ \(\frac{\sqrt{3}}{4} a^2\) < a²
So, the area of a square is greater than the area of an equilateral triangle of the same side length.
Area of two identical equilateral triangles = \(\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{4} a^2\)
= \(\frac{2 \sqrt{3}}{4} a^2\)
= \(\frac{\sqrt{3}}{2} a^2\)
Area of square of side length a = a²
Clearly, \(\frac{\sqrt{3}}{2} a^2\) < a²
So, the area of a square is greater than the area of two identical equilateral triangles.

Rhombus & Trapezium

Figure It Out (Pages 169-170)

Question 1.
Find the area of a rhombus whose diagonals are 20 cm and 15 cm.
Solution:
Given, first diagonal = 20 cm
second diagonal = 15 cm
The area of a rhombus = \(\frac {1}{2}\) × (Product of diagonals)
= \(\frac {1}{2}\) × First diagonal × second diagonal
= \(\frac {1}{2}\) × 20 cm × 15 cm
= 150 cm²
Thus, the area of a rhombus is 150 cm²

Question 2.
Give a method to convert a rectangle into a rhombus of equal area using dissection.
Solution:
This is the reverse of the rhombus to rectangle dissection.
Method:
Given: Rectangle PQRS with length l and width W
Required: Rhombus with the same area = l × W
Dissection Process:
Step 1: The rhombus will have diagonals d₁ and d₂ such that: \(\frac {1}{2}\) × d₁ × d₂ = l × w
So, d₁ × d₂ = 2lw.
Step 2: Choose convenient diagonal lengths:
Let d₁ = 2l (twice the rectangle length)
Then d₂ = w (same as rectangle width)
Check: \(\frac {1}{2}\) × 2l × w = lw
Or
Let d₁ = 2w (twice the rectangle width)
Then d₂ = l (same as rectangle length)
Step 3: Dissection process (reverse of textbook method):
Divide the rectangle into two halves
Mark the center point O.
Cut and rotate pieces to form two isosceles triangles.
Arrange these triangles to share a common diagonal.
This creates a rhombus.

Question 3.
Find the areas of the following figures:

Solution:
The area of the trapezium = \(\frac {1}{2}\) × (Sum of parallel sides) × (Distance between them) = \(\frac {1}{2}\) × (a + b) × h
(i) Here, a = 10 ft, b = 7 ft and h = 16 ft
Area of trapezium = \(\frac {1}{2}\) × (10 + 7) × 16
= 17 × 8
= 136 ft²

(ii) Here, a = 36 m, b = 24 m and h = 14 m
Area of trapezium = \(\frac {1}{2}\) × (36 + 24) × 14
= 60 × 7
= 420 m²

(iii) Here, a = 14 in, b = 6 in and h = 10 in
Area of trapezium = \(\frac {1}{2}\) × (14 + 6) × 10
= 20 × 5
= 100 in²

(iv) Here, a = 18 ft, b = 12 ft and h = 8 ft
Area of trapezium = \(\frac {1}{2}\) × (18 + 12) × 8
= 30 × 4
= 120 ft²

Question 4.
[Sulba-Sutras] Give a method to convert an isosceles trapezium to a rectangle using dissection.
Solution:
An isosceles trapezium has special properties that make dissection simpler.

Properties of Isosceles Trapezium ABCD:
AB || CD (parallel sides)
AD = BC (non-parallel sides are equal)
∠A = ∠B and ∠D = ∠C (base angles are equal)
Dissection Method:
Step 1: Draw perpendiculars from C and D to AB, meeting at points P and Q, respectively.
This creates rectangle PQDC in the middle.
Two congruent right triangles: ΔAPD and ΔBQC.
Step 2: Since the trapezium is isosceles:
ΔAPD ≅ ΔBQC (congruent triangles)
AP = BQ
Step 3: Rearrangement:
Cut triangle ΔAPD
Rotate it and attach it to the right side (next to ΔBQC)
The two triangles together form a rectangle with a width = height of the trapezium
Step 4: Combine: The central rectangle PQDC
The rectangle formed from the two triangles.
These can be joined to form one large rectangle.
Resulting Rectangle:
Length = CD + AP
= CD + (AB – CD)/2
= (AB + CD)/2
Width = h (height of trapezium)
Area = (AB + CD)/2 × h = \(\frac {1}{2}\)h(AB + CD)
This matches the trapezium area formula!

Question 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 ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ, then the trapezium and rectangle have equal areas.]
Solution:
Given: Trapezium ABCD with AB || CD
Required: Find the positions of the vertices E, F, G, and H to form a rectangle EFGH with equal area
Using the Hint: If ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ, then areas are equal.
Method:
Step 1: The rectangle EFGH should have:
EF as one side (top side)
GH is the opposite parallel side (bottom side)
EH and FG are the other pair of sides.
Step 2: Position the rectangle such that:
Points I and J are strategically chosen on the trapezium.
ΔAHI (part outside rectangle on left) is cut and moved to fill ΔDGI.
ΔBEJ (part outside rectangle on right) is cut and moved to fill ΔCFJ.
Step 3: For congruency:
Mark I on side AD
Mark J on side BC
Choose positions such that:
HI = GI (making ΔAHI ≅ ΔDGI possible)
EJ = FJ (making ΔBEJ ≅ ΔCFJ possible)
Step 4: The height of the rectangle = h (height of the trapezium)
Step 5: The length of rectangle = (a + b)/2
where a and b are parallel sides
This ensures: Area of rectangle = h × (a + b)/2 = Area of trapezium
Practical construction:
Draw the trapezium ABCD
Calculate required rectangle length = (AB + CD)/2
Mark points H and E on AB such that the central portion has this length.
Draw perpendiculars to get rectangle EFGH.
Verify that the triangular pieces outside match those inside.
This construction beautifully demonstrates area conservation through dissection!

Question 6.
Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm².
Solution:
Area of the trapezium = \(\frac {1}{2}\) × (10 + 8) × 16 = 144 cm²
Area of square = 16 × 9 = 144 cm²

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

Solution:
Here total area of hexagon = \(6 \times \frac{\sqrt{3}}{4} a^2=\frac{3 \sqrt{3}}{2} a^2\)
Equilateral triangle:
Area = \(\frac{\sqrt{3}}{4} a^2\)
Rhombus:
Area = \(2 \times \frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{2} a^2\)
Trapezium:
Remaining Area = Total area – Triangle area – Rhombus area
= \(\frac{3 \sqrt{3}}{4}-\frac{\sqrt{3}}{4} a^2-\frac{\sqrt{3}}{2} a^2\)
= \(\frac{3 \sqrt{3}}{4} a^2\)
Ratio of areas = Triangle : Rhombus : Trapezium
= \(\frac{\sqrt{3}}{4} a^2: \frac{\sqrt{3}}{2} a^2: \frac{3 \sqrt{3}}{4} a^2\)
= 1 : 2 : 3

Question 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.

Solution:
∠ZAY = ∠BAX (Vertically opposite angles)

AY = AX (∵ A is mid point of XY)
∠YZB = ∠XBZ (∵ ZY || XB alternate interior angles are equal)
∠ZYA = ∠BXA (∵ they are alternate interior angles)
So ∆ZAY ≅ ∆BAX
By the AAA congruence.
So Area of ∆ZAY = Area of ∆BAY
Thus, Area of trapezium ZYXW = Area of triangle ZWB

Areas in Real Life (Pages 170-171)

Question 1.
What do you think is the area of an A₄ sheet?
Its sidelengths are 21 cm and 29.7 cm. Now find its area.
Solution:
We know that the size of an A₄ sheet is rectangular.
∴ The area of an A₄ sheet = Length × Breadth
= 21 cm × 29.7 cm
= 623.7 cm²

Question 2.
Express the following lengths in centimeters:
(i) 5 in
(ii) 7.4 in
Solution:
We know that,
1 in = 2.54 cm
(i) 5 in = 5 × 2.54 cm = 12.7 cm
(ii) 7.4 in = 7.4 × 2.54 cm = 18.796 cm

Question 3.
Express the following lengths in inches:
(i) 5.08 cm
(ii) 11.43 cm
Solution:
We know that,
2.54 cm = 1 in
∴ 1 cm = \(\frac {1}{2.54}\) in
(i) 5.08 cm = 5.08 × \(\frac {1}{2.54}\) in = 2 in
(ii) 11.43 cm = 11.43 × \(\frac {1}{2.54}\) in = 4.5 in

Question 4.
How many in² is 1 ft²?
Solution:
We know that,
1 ft = 12 in
∴ 1 ft² = (12 in)²
= 12² in²
= 144 in²

Question 5.
How many m² is a km²?
Solution:
We know that,
1 km = 1000 m
1 km² = 1000 m × 1000 m
= 1000000 m²
= 10⁶ m²
Thus, 1 km² = 1000000 m²

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