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

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Solution

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The side of square = 'a'
Let the Diagonals be $\frac{^{'} 2 d_{1}^{'}}{^{'} 2 a_{2}^{'}}$ &
$∴$ Area of square (ABCD)
$\Rightarrow (a^{2})$
By Pythagoras Theorem, we get
$d_{1}^{2} + d_{2}^{2} = a^{2} . . . (1)$
Now, Area of Rhombus $= A_{2} = \frac{1}{2} \times 2 d_{1} \times 2 d_{2} = 2 d_{1} d_{2} . . . (2)$
$\Rightarrow (d_{1} - d_{2})^{2} = d_{1}^{2} + d_{2}^{2} - 2 d_{1} d_{2}$
$d_{1}^{2} + d_{2}^{2} - 2 d_{1} d_{2} \Rightarrow d_{1}^{2} + d_{2}^{2} = (d_{1} - d_{2})^{2} + 2 d_{1} d_{2}$
$a^{2} = (d_{1} - d_{2})^{2} + A_{2}$
(from (1) & (2))
$∴$ sum of square= M+
sum of Rhombus
$(∵ M = (d_{1} - d_{2})^{2})$
(also M is +ve)
$∴$ we can say that square has
larger are as compared to Rhombus.

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