RD Sharma - Mathematics

Book: RD Sharma - Mathematics

Chapter: 20. Mensuration-I (Area of a Trapezium and a Polygon)

Subject: Maths - Class 8th

Q. No. 5 of Exercise 20.3

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5

Find the area of the following regular hexagon.

NQ = 23 cm given


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 =


Area of rectangle = length × breadth


Area of regular hexagon =


Area of given hexagon = = 432 cm2


Therefore area of given hexagon is 432 cm2


Chapter Exercises

Exercise 20.1
Exercise 20.2
Exercise 20.3

More Exercise Questions

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.
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.
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?
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.
5

Find the area of the following regular hexagon.

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