 # RD Sharma Solutions for Class 8 Chapter 20 Mensuration - I (Area of a Trapezium and a Polygon) Exercise 20.3

In Exercise 20.3, we shall discuss problems based on the area of a polygon. Our expert tutors at BYJU’S have solved the problems from the textbook to ensure that the students are thorough with their basic concepts and help clear their doubts. RD Sharma Class 8 Solutions is the best study material where our experts have designed the solutions in a simple manner by using shortcut methods to solve problems. Students can refer to the RD Sharma textbook and download the pdf easily from the links provided below.

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1. Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm. Solution:

GH = AG – AH = 8 – 6 = 2 cm

HF = AH – AF = 6 – 5 = 1 cm

GD = AD – AG = 10 – 8 = 2 cm

From the figure we can write,

Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of triangle CGD + Area of triangle AHE + Area of triangle EGD

We know that,

Area of right angled triangle = 1/2 × base × altitude

Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude

Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 × AH × EH + 1/2 × HD × EH

Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 4 × 3

Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5

∴ Area of given pentagon = 52.5 cm2

2. Find the area enclosed by each of the following figures [fig. 20.49 (i)-(ii)] as the sum of the areas of a rectangle and a trapezium. Solution:

Figure (i)

From the figure we can write,

Area of figure = Area of trapezium + Area of rectangle

Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth

Area of figure = 1/2 (18 + 7) × 8 + 18 × 18

Area of figure = 1/2 (25) × 8 + 18 × 18

Area of figure = ∴ Area of figure is 424 cm2

Figure (ii)

From the figure we can write,

Area of figure = Area of trapezium + Area of rectangle

Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth

Area of given figure = 1/2 (15 + 6) × 8 + 15 × 20

Area of given figure = 84 + 300 = 384

∴ Area of figure is 384 cm2

Figure (iii)

Using Pythagoras theorem in the right angled triangle,

52 = 42 + x2

x2 = 25 – 16

x2 = 9

x = 3 cm

From the figure we can write,

Area of figure = Area of trapezium + Area of rectangle

Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth

Area of given figure = 1/2 (14 + 6) × 3 + 4 × 6

Area of given figure = 30 + 24 = 54

∴ Area of figure is 54 cm2

3. There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some another way of finding its area? Solution:

From the figure we can write,

Area of figure = Area of trapezium + Area of rectangle

Area of Jyoti’s diagram = 2 × 1/2 (Sum of lengths of parallel sides) × altitude

Area of figure = 2 × 1/2 × (15 + 30) × 7.5

Area of figure = 45 × 7.5 = 337.5

Therefore, Area of figure = 337.5 cm2

We also know that,

Area of Pentagon = Area of triangle + area of rectangle

Area of Pentagon = 1/2 × Base × Altitude + Length × Breadth

Area of Pentagon = 1/2 × 15 × 15 + 15 × 15

Area of Pentagon = 112.5 + 225 = 337.5

∴ Area of pentagon is 337.5 m2

4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm. AO = 60 cm and AD = 90 cm. Solution:

Given that,

AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm

LM = AM – AL = 20 – 10 = 10 cm

MN = AN – AM = 50 – 20 = 30 cm

OD = AD – AO = 90 – 60 = 30 cm

ON = AO – AN = 60 – 50 = 10 cm

DN = OD + ON = 30 + 10 = 40 cm

OM = MN + ON = 30 + 10 = 40 cm

LN = LM + MN = 10 + 30 = 40 cm

From the figure we can write,

Area of figure = Area of triangle AMF + Area of trapezium FMNE + Area of triangle END + Area of triangle ALB + Area of trapezium LBCN + Area of triangle DNC

We know that,

Area of right angled triangle = 1/2 × base × altitude

Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude

Area of given hexagon = 1/2 × AM × FM + 1/2 (MF + OE) × OM + 1/2 × OD × OE + 1/2 × AL × BL + 1/2 × (BL + CN) × LN + 1/2 × DN × CN

Area of given hexagon = 1/2 × 20 × 20 + 1/2 (20 + 60) × 40 + 1/2 × 30 × 60 + 1/2 × 10 × 30 + 1/2 × (30 + 40) × 40 + 1/2 × 40 × 40

Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050

∴ Area of given hexagon is 5050 cm2

5. Find the area of the following regular hexagon. Solution:

Given that,

NQ = 23 cm

ND + QC = 23 -13 = 10

ND = QC = 5 cm

OM = 24 cm

Area of given hexagon = 2 times Area of triangle MNO + Area of rectangle MOPR

Area of right angled triangle = 1/2 × base × altitude

Area of rectangle = length × breadth

Area of regular hexagon = 1/2 × OM × ND + MO × OP + 1/2 × RP × QC

Area of regular hexagon = 1/2 × 24 × 5 + 24 × 13 + 1/2 × 24 × 5

Area of given hexagon = 1/2 × 120 + 312 + 1/2 × 120 = 432

∴ Area of given hexagon is 432 cm2