RD Sharma Solutions Class 8 Area Of Trapezium And Polygon Exercise 20.3

RD Sharma Class 8 Solutions Chapter 20 Ex 20.3 PDF Free Download

RD Sharma Solutions Class 8 Chapter 20 Exercise 20.3

Exercise 20.3

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.

Answer:

Given:

AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm

BF = 5 cm, CG = 7 cm, EH = 3 cm

Therefore, FG = AG – AF = 8 – 5 = 3 cm

And, GD = AD – AG = 10 – 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)= (\(\frac{1}{2}\)x AF x BF) + [\(\frac{1}{2}\) x (BF + CG) x (FG)] + (\(\frac{1}{2}\) x GD x CG) + (\(\frac{1}{2}\) x AD x EH).

= (\(\frac{1}{2}\) x 5 x 5) + [\(\frac{1}{2}\) x (5 + 7) x (3)] + (\(\frac{1}{2}\) x 2 x 7) + (\(\frac{1}{2}\) x 10 x 3)

= (\(\frac{25}{2}\)) + (\(\frac{36}{2}\)) + (\(\frac{14}{2}\)) + (\(\frac{30}{2}\))

= 12.5 + 18 + 7 + 15 = 52.5 cm2

2. Find the area enclosed by each of the following figures (Fig. 20.49 (i)-(iii)J as the sum of the areas of a rectangle and a trapezium:

Answer:

(i) From the figure:

Area of the complete figure = (Area of square ABCF) + (Area of trapezium CDEF)

=(AB x BC)+[ \(\frac{1}{2}\) x (FC +ED) x (Distance between FC and ED)]

=(18 x 18)+[ \(\frac{1}{2}\) x (18 + 7) x (8)]

= 324 + 100 = 424 cm2

(ii) From the figure:

AB = AC – BC = 28 – 20 = 8 cm

So that area of the complete figure = (area of rectangle BCDE) + (area of trapezium ABEF)

=(BC x CD) + [\(\frac{1}{2}\) x (BE + AF) x (AB)]

=(20 x 15) + [\(\frac{1}{2}\) x(15 + 6) x (8)]

= 300 + 84 = 384 cm2

(iii) From the figure:

EF = AB = 6 cm

Now, using the Pythagoras theorem in the right angle triangle CDE:

52 = 42 + CE2

CE2 = 25-16 = 9

CE = \(\sqrt{9}\) = 3 cm

And, GD = GH + HC + CD = 4 + 6 + 4 = 14 cm

Area of the complete figure = (Area of rectangle ABCH)+(Area of trapezium GDEF)

=(AB x BC)+[ \(\frac{1}{2}\) x (GD + EF) x (CE)]

=(6 x 4) + [\(\frac{1}{2}\) x (14 + 6) x (3)]

= 24 + 30 = 54 cm2

3. There is a pentagonally 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?

Answer:

A pentagonal park is given below:

Jyoti and Kavita divided it in two different ways.

(i) Jyoti divided is into two trapeziums. It is clear that the park is divided in two equal trapeziums whose parallel sides are 30 m and 15 m.

And, the distance between the two parallel lines: \(\frac{15}{2}\) = 7.5 m

Therefore, Area of the park = 2 x (Area of a trapezium) = 2 x [\(\frac{1}{2}\) x (30 + 15) x (7.5)] = 337.5 m2

(ii) Kavita divided the park into a rectangle and a triangle. Here, the height of the triangle = 30 – 15 = 15 m

Therefore, Area of the park = (Area of square with sides 15 cm)+(Area of triangle with base 15 m and altitude 15 m]

=(15 x 15) + (\(\frac{1}{2}\) x 15 x 15)

= 225 + 112. 5 = 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.

Answer:

Given: AL = 10 cm, AM=20 cm, AN=50 cm

AO = 60 cm, AD = 90 cm

Hence, we have the following: MO = AO – AM = 60 – 20 = 40 cm

OD = AD – A0 = 90 – 60 = 30 cm

ND = AD – AN = 90 – 50 = 40 cm

LN = AN – AL = 50 – 10 = 40 cm

From given figure:

Area of Polygon = (Area of triangle AMF) + (Area of trapezium MOEF) + (Area of triangle DNC ) + (Area of trapezium NLBC) + (Area of triangle ALB)

= (\(\frac{1}{2}\) x AM x MF) + [\(\frac{1}{2}\) x (MF + OE) x OM] + (\(\frac{1}{2}\) x OD x OE) + (\(\frac{1}{2}\) x DN x NC) + [\(\frac{1}{2}\) x (LB + NC) x NL] + (\(\frac{1}{2}\) x AL x LB )

=(\(\frac{1}{2}\) x 20 x 20) + [\(\frac{1}{2}\) x (20 + 60) x (40)] + (\(\frac{1}{2}\) x 30 x 60) + (\(\frac{1}{2}\) x 40 x 40) + [\(\frac{1}{2}\) x (30 + 40) x (40)] + (\(\frac{1}{2}\) x 10 x 30 )

= 200 + 1600 + 900 + 800 + 1400 +150 = 5050 cm2

5. Find the area of the following regular hexagon.

Answer:

The given figure is :

Join QN

It is given that the hexagon is regular. So, all its sides must be equal to 13 cm.

Also, AN = BQ

QB + BA + AN = QN

AN + 13 + AN = 23

2AN = 23 – 13 = 10

AN = \(\frac{10}{2}\) = 5 cm

Hence, AN = BQ = 5 cm

Now, in the right angle triangle MAN:

MN2 = AN2 + AM2

132 = 52 + AM2

AM2 = 169 – 25 = 144

AM = \(\sqrt{144}\) = 12cm.

Therefore, OM = RP = 2 x AM = 2 x 12 = 24 cm

Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR) + (area of triangle RPQ)

= (\(\frac{1}{2}\) x OM x AN) + (RP x PO) + (\(\frac{1}{2}\) x RP x BQ)

= (\(\frac{1}{2}\) x 24 x 5) + (24 x 13) + (\(\frac{1}{2}\) x 24 x 5)

= 60 + 312 + 60 = 432 cm2.

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