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

  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.5,7 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, ∠ 𝐴𝑂𝐵=∠ 𝐵𝑂𝐶=∠ 𝐶𝑂𝐷=∠𝐷𝑂𝐴=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 𝐴𝐶)2+(1/2 𝐵𝐷)2 (AB)2 = 𝐴𝐶2/4+𝐵𝐷2/4 (AB)2 = (𝐴𝐶2 + 𝐵𝐷2)/4 4AB2 = AC2 + BD2 AB2 + AB2 + AB2 + AB2 = AC2 + BD2 AB2 + BC2 + CD2 + AD2 = AC2 + BD2 Hence proved

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