Ex 9.1, 9 - Top surface of a raised platform is in shape of a regular - Ex 9.1

part 2 - Ex 9.1, 9 - Ex 9.1 - Serial order wise - Chapter 9 Class 8 Mensuration
part 3 - Ex 9.1, 9 - Ex 9.1 - Serial order wise - Chapter 9 Class 8 Mensuration
part 4 - Ex 9.1, 9 - Ex 9.1 - Serial order wise - Chapter 9 Class 8 Mensuration
part 5 - Ex 9.1, 9 - Ex 9.1 - Serial order wise - Chapter 9 Class 8 Mensuration

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Ex 9.1, 9 Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.Area of octagonal surface = Area of trapezium ABCH + Area of rectangle HCDG + Area of Trapezium GDEF ∴ Area = (2 × Area of trapezium ABCH) + Area of rectangle HCDG Area of Trapezium ABCH In trapezium ABCH, AB and HC are parallel sides. Since sides of regular octagon are equal AB = BC = CD = DE = EF = GF = HG = HA = 5 m So, AB = 5 m HC = 11 m and height = 4 m Now, Area = 1/2 × (Sum of Parallel sides) × Height = 1/2 × (AB + HC) × Height Now, Area = 𝟏/𝟐 × (Sum of Parallel sides) × Height = 1/2 × (AB + HC) × Height = 𝟏/𝟐 × (5 + 11) × 4 = 1/2 × 16 × 4 = 8 × 4 = 32 m2 Area of Rectangle HCDG Area = Length × Breadth Putting Length = HC = 11 m Breadth = CD = 5m Area = 11 × 5 = 55 m2 Now, ∴ Area of octagonal surface = (2 × Area of Trapezium) + Area of rectangle = (2 × 32) + 55 = 64 + 55 = 119 m2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo