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Ex 11.2, 11 - Diagram of the adjacent picture frame has outer

Ex 11.2, 11 - Chapter 11 Class 8 Mensuration - Part 2
Ex 11.2, 11 - Chapter 11 Class 8 Mensuration - Part 3 Ex 11.2, 11 - Chapter 11 Class 8 Mensuration - Part 4 Ex 11.2, 11 - Chapter 11 Class 8 Mensuration - Part 5 Ex 11.2, 11 - Chapter 11 Class 8 Mensuration - Part 6

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Ex 11.2, 11 Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same. The picture frame has 4 sections marked as 1 , 2 , 3 and 4 Given that width of each section is same Let width of section = x cm Let’s first find x BC = x + 20 cm + x 28 = x + 20 + x 28 – 20 = 2x 8 = 2x 2x = 8 x = 8/2 x = 4 cm So, width of section is 4 cm Now, By symmetry, Area of Section 1 = Area of Section 3 Area of Section 2 = Area of Section 4 So, width of section is 4 cm Area of section 1 Here, Parallel sides are AB & EF Height is 4 cm Area of section 1 = Area of trapezium = 1/2 × Sum of parallel sides × Height = 1/2 × (24 + 16) × 4 = 1/2 × 40 × 4 = 20 × 4 = 80 cm2 Area of section 2 Here, Parallel sides are BC & FG Height is 4 cm Area of section 2 = Area of trapezium = 1/2 × Sum of parallel sides × Height = 1/2 × (28 + 20) × 4 = 1/2 × 48 × 4 = 24 × 4 = 96 cm2 Thus, Area of section 1 = Area of section 3 = 80 cm2 Area of section 2 = Area of section 4 = 96 cm2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.