RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

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Class 8 : RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

The document RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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PAGE NO 20.28:

Question 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.
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

ANSWER:

The given figure is:  
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

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

∴ FG  =  AG - AF  =  8 - 5  =  3 cm 

And, GD  =  AD - AG  =  10 - 8  =  2 cm

From given figure:Area of Pantagon  =  (Area of triangle AFB)  +  (Area of trapezium FBCG)  +  (Area of triangle CGD)  +  (Area of triangle ADE)

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

=  (1/2  ×  5  ×  5)  +  [1/2  ×  (5  +  7)  ×  (3)]  +  (1/2  ×  2  ×  7)  +  (1/2  ×  10  ×  3)

=  (25/2)  +  [36/2]  +  (14/2)  +  (30/2)

= 12.5  +  18  +  7  +  15

= 52.5 cm² 


Question 2:

Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

ANSWER:

RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev
(i)The given figure can be divided into a rectangle and a trapezium as shown below:  
From the above firgure:

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

= (AB × BC) + [1/2 × (FC + ED) × (Distance between FC and ED)]

= (18 × 18) + [1/2 × (18 + 7) × (8)]

= 324 + 100

= 424 cm²

(ii)The given figure can be divided in the following manner:  

From the above figure:

AB  =  AC-BC = 28-20 = 8 cm

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

= (BC × CD) + [1/2 × (BE + AF) × (AB)]

= (20 × 15) + [1/2 × (15 + 6) × (8)]

= 300 + 84

= 384 cm²

(iii)The given figure can be divided in the following manner:
From the above figure:

EF  =  AB  =  6 cm

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

52  =  42 + CE2

CE2  =  25-16 = 9

CE  =  √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 × BC) + [1/2 × (GD + EF) × (CE)]

= (6 × 4) + [1/2 × (14 + 6) × (3)]

= 24 + 30 = 54 cm² 


Question 3:

There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways.

RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev
 Find the area of this park using both ways. Can you suggest some another way of finding its areas?

ANSWER:

A pentagonal park is given below:  

RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRevRD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

Jyoti and Kavita divided it in two different ways. 

 (i) Jyoti divided is into two trapeziums as shown below: 

Jyoti and Kavita divided it in two different ways. 

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: 15/2 = 7.5 m

∴ Area of the park = 2 × (Area of a trapazium)

= 2 × [1/2 × (30 + 15) × (7.5)] = 337.5 m²

(ii)Kavita divided the park into a rectangle and a triangle, as shown in the figure.  

Here, the height of the triangle  =  30-15 = 15 m

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

= (15 × 15) + (1/2 × 15 × 15)

= 225 + 112.5

= 337.5 m²

 

PAGE NO 20.29:

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.
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

ANSWER:

The given polygon is:  The given polygon is:  
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev
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-AO = 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 EOD) + (Area of triangle DNC) + (Area of trapezium NLBC) + (Area of triangle ALB)

=(1/2 × AM × MF) + [1/2 × (MF + OE) × (OM)] + (1/2 × OD × OE) + (1/2 × DN × NC) + [1/2 × (LB + NC) × (NL)] + (1/2 × AL × LB)

= (1/2 × 20 × 20) + [1/2 × (20 + 60) × (40)] + (1/2 × 30 × 60) + (1/2 × 40 × 40) + [1/2 × (30 + 40) × (40)] + (1/2 × 10 × 30)

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

= 5050 cm2


Question 5:

Find the area of the following regular hexagon.
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev

ANSWER:

The given figure is: The given figure is:
Join QN.
RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev
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 

=  10AN  =  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 =RD Sharma Solutions - Chapter 20 - Mensuration - I (Part - 3), Class 8, Maths Class 8 Notes | EduRev = 12cm

∴ OM  =  RP  =  2 × AM  =  2 × 12  =  24 cm

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

= (1/2 × OM × AN) + (RP × PO) + (1/2 × RP × BQ)

= (1/2 × 24 × 5) + (24 × 13) + (1/2 × 24 × 5)

= 60 + 312 + 60

= 432 cm² 

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