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

Made by

Davneet Singh

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

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