In Exercise 20.3, we shall discuss problems based on the area of a polygon. Our expert tutors at BYJUâ€™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.

## Download the Pdf of RD Sharma Solutions for Class 8 Maths Exercise 20.3 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

### Access answers to RD Sharma Maths Solutions For Class 8 Exercise 20.3 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

**1.** **Find the area of the pentagon shown in fig. 20.48, ifÂ ADÂ = 10 cm,Â AGÂ = 8 cm,Â AHÂ = 6 cm,Â AFÂ = 5 cm,Â BFÂ = 5 cm,Â CGÂ = 7 cm andÂ EHÂ = 3 cm.
**

**Solution:**

GH = AG â€“ AH = 8 â€“ 6 = 2 cm

HF = AH â€“ AF = 6 â€“ 5 = 1 cm

GD = AD â€“ AG = 10 â€“ 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 =Â 1/2 Ã— base Ã— altitude

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

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

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

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

âˆ´ Area of given pentagon = 52.5 cm^{2}

**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) Ã— altitude + Length Ã— Breadth

Area of figure =Â 1/2 (18 + 7) Ã— 8 + 18 Ã— 18

Area of figure =Â 1/2 (25) Ã— 8 + 18 Ã— 18

Area of figure =

âˆ´ Area of figure is 424 cm^{2}

**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) Ã— altitude + Length Ã— Breadth

Area of given figure =Â 1/2 (15 + 6) Ã— 8 + 15 Ã— 20

Area of given figure =Â 84 + 300 = 384

âˆ´ Area of figure is 384 cm^{2}

**Figure (iii)**

Using Pythagoras theorem in the right angled triangle,

5^{2}Â = 4^{2}Â + x^{2}

x^{2}Â = 25 â€“ 16

x^{2}Â = 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) Ã— altitude + Length Ã— Breadth

Area of given figure =Â 1/2 (14 + 6) Ã— 3 + 4 Ã— 6

Area of given figure =Â 30 + 24 = 54

âˆ´ Area of figure is 54 cm^{2}

**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 Jyotiâ€™s diagram =Â 2 Ã— 1/2 (Sum of lengths of parallel sides) Ã— altitude

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

Area of figure = 45 Ã— 7.5 = 337.5

Therefore, Area of figure = 337.5 cm^{2}

We also know that,

Area of Pentagon = Area of triangle + area of rectangle

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

Area of Pentagon =Â 1/2 Ã— 15 Ã— 15 + 15 Ã— 15

Area of Pentagon =Â 112.5 + 225 = 337.5

âˆ´ Area of pentagon is 337.5 m^{2}

**4.** **Find the area of the following polygon, ifÂ ALÂ = 10 cm,Â AMÂ = 20 cm,Â ANÂ = 50 cm.Â AOÂ = 60 cm andÂ ADÂ = 90 cm.**

**Solution:**

Given that,

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

LM = AM â€“ AL = 20 â€“ 10 = 10 cm

MN = AN â€“ AM = 50 â€“ 20 = 30 cm

OD = AD â€“ AO = 90 â€“ 60 = 30 cm

ON = AO â€“ AN = 60 â€“ 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 =Â 1/2 Ã— base Ã— altitude

Area of trapezium =Â 1/2 (Sum of lengths of parallel sides) Ã— altitude

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

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

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

âˆ´ Area of given hexagon is 5050 cm^{2}

**5.** **Find the area of the following regular hexagon.
**

**Solution:**

Given that,

NQ = 23 cm

NA = BQ = 10/2 = 5 cm

MR = OP = 13 cm

In the right triangle BPQ

PQ^{2} = BQ^{2} + BP^{2}

Substituting the values

(13)^{2} = (5)^{2} + BP^{2}

169 = 25 + BP^{2}

So we get

BP^{2} = 169 – 25 = 144

BP = 12 cm

Here

PR = MO = 2Â Ã— 12 = 24 cm

Area of rectangle RPOM = RPÂ Ã— PO = 24Â Ã— 13 = 321 cm^{2}

Area of triangle PRQ = 1/2Â Ã— PRÂ Ã— BQ

= 1/2Â Ã— 24Â Ã— 5

= 60 cm^{2}

Area of triangle MON = 60 cm^{2}

Area of hexagon = 312 + 60 + 60 = 432 cm^{2}

âˆ´ Area of given hexagon is 432 cm^{2}