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Question

A square and a rhombus stand on the same base. Prove that the square has a larger area.

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Solution

REF.Image
The side of square = 'a'
Let the Diagonals be 2d12a2 &
Area of square (ABCD)
(a2)
By Pythagoras Theorem, we get
d21+d22=a2...(1)
Now, Area of Rhombus=A2=12×2d1×2d2=2d1d2...(2)
(d1d2)2=d21+d222d1d2
d21+d222d1d2d21+d22=(d1d2)2+2d1d2
a2=(d1d2)2+A2
(from (1) & (2))
sum of square= M+
sum of Rhombus
(M=(d1d2)2)
(also M is +ve)
we can say that square has
larger are as compared to Rhombus.

1146104_1143937_ans_cdb9b446af96475d8bccfaaa0b1a70cb.jpg

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