CBSE Class 8 Mathematics Part II Chapter 7 Questions and Answers

Class 8 Maths

Area Class 8 Ganita Prakash Part 2 Chapter 7 NCERT Solutions

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Figure it Out (Page 150)

1. Identify the missing sidelengths.

Solution:

(i)

Area of rectangle ABCD = 21 in²
7 × BC = 21 → BC = 21 / 7 = 3 in
AD = BC = 3 in
AE = AD + DE = 3 + 4 = 7 in
Area of rectangle GAEF = 28 in²
FE × AE = 28 → FE × 7 = 28 → FE = 28 / 7 = 4 in
GA = FE = 4 in
HA = HG + GA = 3 + 4 = 7 in
Area of IJAH = 35 in²
HI × HA = 35 → HI × 7 = 35 → HI = 35 / 7 = 5 in
AJ = HI = 5 in
AK = AJ + JK = 5 + 2 = 7 in
Area of AKLM = 14 in²
AK × KL = 14 → 7 × KL = 14 → KL = 14 / 7 = 2 in
Therefore, the missing sidelength is 2 in.

(ii)

AB = HE = 4 m
Area of rectangle HEFG = 11 m²
HE × HG = 11 → 4 × HG = 11 → HG = 11 / 4 = 2.75 m
Area of rectangle ABEH = 29 m²
AB × AH = 29 → 4 × AH = 29 → AH = 29 / 4 = 7.25 m
Area of rectangle ACDH = 50 m²
AH × HD = 50 → 7.25 × HD = 50 → HD = 50 / 7.25 = 5000 / 725 ≈ 6.9 m
HD = HE + ED → 6.9 = 4 + ED → ED = 6.9 – 4 = 2.9 m
Thus: AH = 7.25 m, HG = 2.75 m, and HD = 6.9 m.

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.
[Hint: There is a relation between the areas of EFGH, the path, and ABCD.]
Solution:
To find the area of the path, we need:
• Length (L) and breadth (B) of the outer rectangle ABCD.
• Length (l) and breadth (b) of the inner rectangle EFGH.
Area of path = Area of outer rectangle − Area of inner rectangle
Area of path = (L × B) − (l × b).
Example: Let L = 20 m, B = 12 m, l = 14 m, b = 8 m.
Area of path = (20 × 12) – (14 × 8) = 240 − 112 = 128 m².
Formula: (L × B) − (l × b).

(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.]
Solution:
If only the width of the path is given, we cannot find its area. We also need the length (l) and breadth (b) of the inner rectangle (park EFGH).
Let width of path = x.
Outer length = l + 2x, Outer breadth = b + 2x.
Area of path = (l + 2x)(b + 2x) − (l × b)
= lb + 2lx + 2bx + 4x² − lb = 2x(l + b) + 4x².
Example: Let l = 10 m, b = 6 m, x = 2 m.
Outer dimensions: Length = 14 m, Breadth = 10 m.
Area of path = (14 × 10) – (10 × 6) = 140 − 60 = 80 m².

(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:
No, the area of the path does not change when the outer rectangle is moved because the area depends only on the dimensions of the two rectangles, not on their relative positions.

3. The figure shows a plot with sides 14 m and 12 m, 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:

Length (l) of plot = 14 m, Breadth (b) of plot = 12 m.
Other measurement needed: Width of the crosspath (w).
Let width w = 2 m.
Area of horizontal path = 14 × 2 = 28 m²
Area of vertical path = 12 × 2 = 24 m²
Area of common central square = 2 × 2 = 4 m²
Area of the path = 28 + 24 − 4 = 48 m².
Formula: Area of path = lw + bw − w².

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:
Width of tube = 1 unit.
Total length of the spiral centerline/segments = 20 + 20 + 20 + 15 + 15 + 10 + 10 + 5 + 5 = 120 units.
Area of the spiral tube = length × width = 120 × 1 = 120 sq. units.

For the bent tube of width 1 unit:
Area = (5 + 5) × 1 = 10 sq. units.
Length of straight tube having same area = 10 / 1 = 10 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.

Solution:
Let the initial side of the square be s.
Total area = s².
Area of region 3 = s² / 2.
Area of region 1 = s² / 4.
Area of region 2 = s² / 4.

When the side is doubled to 2s, the new total area becomes (2s)² = 4s².
New area of region 3 = 4s² / 2 = 2s² (4 times the original).
New area of region 1 = 4s² / 4 = s² (4 times the original).
New area of region 2 = 4s² / 4 = s² (4 times the original).
Therefore, the area of each region increases by 4 times (quadruples).

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:
Do it yourself.

Figure it Out (Page 157)

1. Find the areas of the following triangles:

Solution:
(i) Area of △ABC = (1/2) × base × height = (1/2) × BC × AE = (1/2) × 4 cm × 3 cm = 6 cm².
(ii) Area of △DEF = (1/2) × EF × ND = (1/2) × 5 cm × 3.2 cm = 8 cm².
(iii) Area of ∆NAT = (1/2) × AT × NA = (1/2) × 3 cm × 4 cm = 6 cm².

2. Find the length of the altitude BY.

Solution:
Area of △ABC = (1/2) × BC × AX = (1/2) × 6 × 4 = 12 sq. units.
Also, Area of △ABC = (1/2) × AC × BY
12 = (1/2) × 8 × BY
12 = 4 × BY → BY = 12 / 4 = 3 cm.

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:
In isosceles triangle SUB with SU = SB and SE ⊥ UB:
△SEU ≅ △SEB (by RHS congruence rule).
Area of △SEU = Area of △SEB = 24 sq. units.
Area of △SUB = 24 + 24 = 48 sq. units.

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

Solution:
(i) Let ABCD be a rectangle with length a and breadth b.
(ii) Mark E as the midpoint of AD.
(iii) Draw a line perpendicular to AD through E.
(iv) Mark a point F on this line such that FE = b.
(v) Join F to B and C to form △FBC.
Verification:
Area of rectangle ABCD = a × b = ab.
Area of △FBC = (1/2) × base × height = (1/2) × a × 2b = ab.
Thus, Area of △FBC = Area of rectangle ABCD.

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

Solution:
(i) Consider triangle △ABC with base BC = b and height AD = h.
(ii) Mark M as the midpoint of AD.
(iii) Draw lines through B and C perpendicular to BC.
(iv) Through point M, draw a line parallel to BC, meeting the perpendiculars at F and G.
(v) The figure BCGF is the desired rectangle.
Verification:
Area of rectangle BCGF = b × (h/2) = bh / 2.
Area of △ABC = (1/2) × b × h = bh / 2.
Thus, Area of rectangle BCGF = Area of △ABC.

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?

Solution:

Let the side of each square be a.
(i) Area of red region (△HCD) = (1/2) × DC × HC = (1/2) × a × 2a = a².
Given a² = 49 sq. units.
Area of blue region (△ADL) = (1/2) × AL × AD = (1/2) × (a/2) × a = a² / 4 = 49 / 4 = 12.25 sq. units.

(ii) Total area of blue + red = a² + (a² / 4) = 5a² / 4.
5a² / 4 = 180 → a² = (180 × 4) / 5 = 144 sq. units.
The area of each square is 144 sq. units.

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]

Solution:

A median divides a triangle into two triangles of equal area.
In △XYN, NM is a median → Area(△XMN) = Area(△YMN) = (1/2) Area(△XYN).
In △XYZ, YN is a median → Area(△XYN) = (1/2) Area(△XYZ).
∴ Area(△XMN) = (1/2) × (1/2) Area(△XYZ) = 1/4 Area(△XYZ).

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:

Reflect the position of the water tank across the river line. Join the house to this reflected point with a straight line. The intersection of this line with the bank of the river gives the optimal point on the river to minimize total walking distance.

Figure it Out (Page 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.

Solution:
Area of △ABC = (1/2) × AC × BM = (1/2) × 22 × 3 = 33 cm².
Area of △ADC = (1/2) × AC × DN = (1/2) × 22 × 3 = 33 cm².
Area of quadrilateral ABCD = 33 + 33 = 66 cm².

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

Solution:
Area of rectangle ABCD = AB × AD = 18 × 10 = 180 cm².
Area of unshaded △AFE = (1/2) × AE × AF = (1/2) × 10 × 6 = 30 cm².
Area of unshaded △EBC = (1/2) × BE × BC = (1/2) × 8 × 10 = 40 cm².
Area of shaded region = 180 − (30 + 40) = 180 − 70 = 110 cm².

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

Solution:
A regular hexagon can be partitioned into 6 identical equilateral triangles. We only need the length of one side of the regular hexagon (or the distance from center to a vertex) to find its total area.

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

Solution:

Let length = a, breadth = b. Area of rectangle = ab.
Triangles AOD and BOC have base a and altitudes x and y such that x + y = b.
Combined area of blue region = (1/2)ax + (1/2)ay = (1/2)a(x + y) = (1/2)ab.
Fraction = ((1/2)ab) / ab = 1/2.

5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral.

Solution:
Let ABCD be the given quadrilateral. Connect the midpoints P, Q, R, and S of the four sides AB, BC, CD, and DA in order. The resulting Varignon parallelogram PQRS has an area exactly half that of ABCD.

Figure it Out (Page 162)

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) All these parallelograms have the same area because they share the same base and lie between the same parallel lines (same perpendicular height).
(ii) Their perimeters are not the same. As the parallelogram tilts further, the slant side length increases. Figure (a) has the minimum perimeter, and Figure (g) has the maximum perimeter.

2. Find the areas of the following parallelograms:

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

3. Find QN.

Solution:
Area of parallelogram PQRS = SR × QM = 12 × 6 = 72 cm².
Also, Area = PS × QN = 7.6 × QN.
7.6 × QN = 72 → QN = 72 / 7.6 ≈ 9.47 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.]

Solution:
Area of rectangle = 5 × 4 = 20 cm².
For the parallelogram, the perpendicular height corresponding to base 5 cm is strictly less than the slanted side (4 cm).
Therefore, Area of parallelogram < 20 cm².
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?

Solution:
Method 1: Construct a rectangle sharing the same base and same altitude as the triangle. Since Area(triangle) = (1/2)bh and Area(rectangle) = bh, the rectangle has twice the area.
Method 2: Duplicate the triangle and join the two copies along a matching side to form a parallelogram, which can then be restructured into an equal-area rectangle.

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

Solution:
Keep the base b of the triangle the same, and construct a rectangle with height equal to half the triangle's altitude (h/2). Area = b × (h/2) = (1/2)bh.

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.]

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

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.

Solution:
Let side length = a.
(i) Area of triangle = (√3 / 4)a² ≈ 0.433a². Area of square = a².
Since 0.433 < 1, the square has the greater area.

(ii) Two triangles = 2 × (√3 / 4)a² = (√3 / 2)a² ≈ 0.866a².
Since 0.866 < 1, the square still has the greater area.

Figure it Out (Page 169 – 170)

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

Solution:
Area of rhombus = (1/2) × d₁ × d₂ = (1/2) × 20 × 15 = 150 cm².

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

Solution:
Divide the rectangle into two equal halves parallel to one side. Draw diagonals in both halves to create 4 congruent right triangles, and reassemble their hypotenuses outward to form a rhombus of equal area.

3. Find the areas of the following figures:

Solution:
Area of trapezium = (1/2) × (a + b) × h
(i) (1/2) × (10 + 7) × 16 = 17 × 8 = 136 ft².
(ii) (1/2) × (36 + 24) × 14 = 60 × 7 = 420 m².
(iii) (1/2) × (14 + 6) × 10 = 20 × 5 = 100 in².
(iv) (1/2) × (18 + 12) × 8 = 30 × 4 = 120 ft².

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

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.]

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

Solution:
A regular hexagon consists of 6 congruent equilateral triangles.
• Equilateral triangle = 1 triangle unit.
• Rhombus = 2 triangle units.
• Trapezium = 3 triangle units.
Ratio (Trapezium : Equilateral triangle : Rhombus) = 3 : 1 : 2.

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:
In △ZAY and △BAX:
∠ZAY = ∠BAX (Vertically opposite angles)
AY = AX (A is midpoint of XY)
∠ZYB = ∠XBA (Alternate interior angles, since ZY ‖ WB)
∴ △ZAY ≅ △BAX (by ASA congruence rule).
Area(△ZAY) = Area(△BAX).
Area(trapezium ZYXW) = Area(quadrilateral ZAXW) + Area(△ZAY)
= Area(quadrilateral ZAXW) + Area(△BAX) = Area(△ZWB).
Hence, Area of trapezium ZYXW = Area of triangle ZWB.