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Ex 7.2, 10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order. [Hint : Area of a rhombus = 1/2 (product of its diagonals)] Let the vertices be A (3, 0) , B (4, 5) C (−1, 4) , D (−2, −1) We know that Area of Rhombus = 1/2 (Product of diagonals) = 1/2 × AC × BD We need to find AC & BD using distance formula Finding AC AC = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) Here, x1 = 3 , y1 = 0 x2 = −1, y2 = 4 Putting values AC = √((−1−3 )2+(4 −0)2) = √((−4 )2+(4)2) = √((4 )2+(4)2) = √(2(4)2) = √2 × 4 = 4√2 Finding BD BD = √((𝑥2 −𝑥1)2+(𝑦2 −𝑦1)2) Here, x1 = 4 y1 = 5 x2 = −2 y2 = −1 Putting values AC = √((−2 −4 )2+(−1−5)2) = √((−6 )2+(−6)2) = √((6)2+(6)2) = √(2(6)2) = √2 × 6 = 6√2 Now, Area of Rhombus = 1/2 × AC × BD = 1/2 × 4√2 × 6√2 = 2√2 × 6√2 = (2 × 6) × (√2 × √2) = (12) × (2) = 24 square units Hence, Area of rhombus = 24 square units

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.