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Ex 11.2, 8 - ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm

Ex 11.2, 8 - Chapter 11 Class 7 Perimeter and Area - Part 2
Ex 11.2, 8 - Chapter 11 Class 7 Perimeter and Area - Part 3 Ex 11.2, 8 - Chapter 11 Class 7 Perimeter and Area - Part 4

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Ex 11.2, 8 ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of Δ ABC. What will be the height from C to AB i.e., CE? Finding area of ∆ABC Using BC as base and AD as height Here, Base = BC = 9 cm Height = AD = 6 cm Area of ∆ABC = 1/2 × Base × Height = 1/2 × BC × AD = 1/2 × 9 × 6 = 9 × 3 = 27 cm2 Now, we have to find EC which is height corresponding to base AB So, finding Area using EC as height and AB as base Base = AB = 7.5 cm Height = EC = ? Area of ∆ABC = 1/2 × Base × Height 27 = 1/2 × 7.5 × EC (27 × 2)/7.5 = EC (27 × 2 × 10)/75 = EC (9 × 2 × 10)/25 = EC (9 × 2 × 2)/5 = EC 36/5 = EC 7.2 = EC EC = 7.2 cm ∴ Area of ∆ABC is 27 cm2 and Height CE is 7.2 cm

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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.