# Class 8 RD Sharma Solutions – Chapter 20 Area Of Trapezium And Polygon- Exercise 20.3

### Question 1. Find the area of the pentagon shown in figure below, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm.

**Solution:**

Given:AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm BF = 5 cm, CG = 7 cm, EH = 3 cm

From given data,

FG = AG â€“ AF = 8 â€“ 5 = 3 cm

And,

GD = AD â€“ AG = 10 â€“ 8 = 2 cm

From given figure:

Area of Pentagon = (Area of triangle AFB) + (Area of trapezium FBCG) +

(Area of triangle CGD) + (Area of triangle ADE)

= (0.5 x AF x BF) + [0.5 x (BF + CG) x (FG)] + (0.5 x GD x CG) + (1/2 x AD x EH).

= (0.5 x 5 x 5) + [0.5 x (5 + 7)x (3) + (0.5 x 2 x 7) + (0.5 x 10 x 3)

= 12.5 + 18 + 7 + 15 = 52.5 cm

^{2 }

### Questions 2. Find the area enclosed by each of the following fig(Fig. (i)-(iii)J as the sum of the areas of a rectangle and a trapezium.

**Solution:**

(i)Figure can be divided into 2 parts a square and a trapezium as shown:Area = (Area of square) + (Area of trapezium)

= (side)

^{2}+ (0.5 x (sum of parallel sides) x height= (18 x 18) + 0.5 x (18 + 7) Ã— (8)

= 324+100

= 424 cm

^{2}

(ii)Figure can be divided into 2 parts a rectangle and a trapezium as shown:From the figure:

Height of trapezium = 28 – 20 = 8 cm

Area = (Area of rectangle) + (Area of trapezium)

= (length x breadth) + (0.5 x (sum of parallel sides) x height)

= (20 x 15) + [0.5 x (15 + 6) Ã— (8)]

= 300 + 84

= 384 cm

^{2}

(iii)Figure can be divided into 2 parts one trapezium and one rectangle:Pythagoras theorem in one of the right angle triangle:

5

^{2}= 4^{2}+ (base)^{2}base

^{2}= 25 – 16base = âˆš9 = 3 cm

Therefore,

The height of trapezium = 3 cm

One side of trapezium = 6 + 4 + 6 = 14 cm

Area = (Area of rectangle) + (Area of trapezium)

= (length x breadth) + (0.5 x (sum of parallel sides) x height)

= (6 x 4) + (0.5 x (14 + 6) x (3))

= 24 + 30

= 54 cm

^{2}

### Question 3. There is a pentagonal shaped park as shown in Fig. 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:**

Jyoti and Kavita divided the park in two different ways.

(i)Jyoti divided the park into two equal trapeziums:From the figure the park divided into equal trapezium having height 7.5 m and sides 30 m and 15 m

Area of the park = 2 x (Area of a trapezium)

= 2 x (0.5 x (sum of parallel sides) x height)

= 2 x (0.5 x (30 + 15) x (7.5))

= 337.5 m

^{2}

(ii)Kavita divided the park into a rectangle and a triangle:From diagram,

The height of the triangle = 30 – 15 = 15 m

Area of the park = (Area of square) + (Area of triangle)

= (15 x 15) + (0.5 x 15 x 15)

= 225 + 112.5

= 337.5 m

^{2}

### Question 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:AL = 10 cm, AM = 20 cm, AN = 50 cm

AO = 60 cm, AD = 90 cm

From given data,

MO = AO â€“ AM = 60 â€“ 20 = 40 cm

OD = AD â€“ A0 = 90 â€“ 60 = 30 cm

ND = AD â€“ AN = 90 â€“ 50 = 40 cm

LN = AN â€“ AL = 50 â€“ 10 = 40 cm

Area of Polygon = (Area of triangle AMF) + (Area of trapezium MOEF) +

(Area of triangle DNC) + (Area of trapezium NLBC) +

(Area of triangle ALB)

= (0.5 x AM x MF) + [0.5 x (MF + OE) x OM] + (0.5 x OD x OE) +

(0.5 x DN x NC) + [0.5 x (LB + NC) x NL] + (0.5 x AL x LB)

= (0.5 x 20 x 20) + [0.5 x (20 + 60) x (40)] +(0.5 x 30 x 60) +

(0.5 x 40 x 40) +[0.5 x (30 + 40) x (40)] + (0.5 x 10 x 30)

= 200 + 1600 + 900 + 800 + 1400 +150 = 5050 cm

^{2}

### Question 5. Find the area of the following regular hexagon.

**Solution:**

As it is a regular Hexagon so all sides are 13 cm and AN = BQ

From the figure,

As the diagonal QN is 23 cm,

QB + BA + AN = QN

AN + 13 + AN = 23

2AN = 23 â€“ 13 = 10

AN = 5 cm

Hence, AN = BQ = 5 cm

Applying Pythagoras theorem in triangle MAN:

MN

^{2 }= AN^{2 }+ AM^{2}169 = 25 + AM

^{2}AMÂ² = 169 – 25

AM = âˆš144

AM = 12cm

From figure,

OM = RP = 2 Ã— AM = 2 x 12 = 24 cm

This Hexagon can be divided into 3 parts 2 triangles and one rectangle therefore,

Area of the regular hexagon = (area of triangle MON) + (area of rectangle MOPR) +

(area of triangle RPQ)

= (0.5 x OM x AN) + (RP X PO) + (0.5 x RP x BQ)

= (0.5 x 24 x 5) + (24 x 13) + (0.5 x 24 x 5)

= 60 + 312 + 60

= 432 cm

^{2}

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