Ex 6.5, 7 - Prove that sum of squares of sides of rhombus - Ex 6.5

Ex 6.5, 7 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.5, 7 - Chapter 6 Class 10 Triangles - Part 3

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Question7 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Given:- Rhombus ABCD with diagonals AC & BD intersecting at O To prove: Sum of square of all sides = Sum of the squares of it s diagonals AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Proof: Since sides of a rhombus are equal AB = BC = CD = AD We know that, diagonals of a rhombus bisect each other a right angles . Therefore, = = = =90 Also AO = CO = 1/2 AC & BO = DO = 1/2 BD Now , AOB is a right angle triangle Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (AB)2 = (OA)2 + (OB)2 (AB)2 = (1/2 )2+(1/2 )2 (AB)2 = 2/4+ 2/4 (AB)2 = ( 2 + 2)/4 4AB2 = AC2 + BD2 AB2 + AB2 + AB2 + AB2 = AC2 + BD2 AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Hence proved

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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