


Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 6.5
Ex 6.5, 2 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 3 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 4 Deleted for CBSE Board 2023 Exams
Ex 6.5, 5 Deleted for CBSE Board 2023 Exams
Ex 6.5, 6 Deleted for CBSE Board 2023 Exams
Ex 6.5, 7 Important Deleted for CBSE Board 2023 Exams You are here
Ex 6.5, 8 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 9 Deleted for CBSE Board 2023 Exams
Ex 6.5, 10 Deleted for CBSE Board 2023 Exams
Ex 6.5, 11 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 12 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 13 Deleted for CBSE Board 2023 Exams
Ex 6.5, 14 Deleted for CBSE Board 2023 Exams
Ex 6.5, 15 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 16 Deleted for CBSE Board 2023 Exams
Ex 6.5, 17 (MCQ) Deleted for CBSE Board 2023 Exams
Last updated at Aug. 4, 2021 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
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, = = = =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