CBSE NCERT Solutions Chapter 6 Triangles Exercise 6.5 Question 7
7. Prove that the sum of squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given: ABCD is a rhombus. Diagonals AC and BD of rhombus intersect at point O.
To Prove:
Proof:
In
{Diagonals of rhombus intersect at 90
}
Therefore,
{By, pythagoras theorem} (1)
Similarly,
{By, pythagoras theorem} (2)
Similarly,
{By, pythagoras theorem} (3)
Similarly,
{By, pythagoras theorem} (4)
Adding (1), (2), (3) and (4), we get
(5)
But, we have OA = OC and OB = OD {Diagonals of rhombus bisect each other} (6)
Putting (6) in (5), we get
{2OA = AC and 2OB= BD}
Hence Proved
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