If a square and a rhombus stand on the same base, prove that the square has the larger area.

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Solution

Let the side of the square be 'a'.
Let the diagonals be $^{'} 2 d_{1} a n d^{'} 2 d_{2}^{'} .$
Area of the square, $A_{1} = a^{2}$
Using pythagoras theorem,
$d_{1}^{2} + d_{2}^{2} = a^{2}$
Area of the rhombus, $A^{2} = \frac{1}{2} . 2 d_{1} .2 d_{2} = 2 d_{1} d_{2}$
$(d_{1} - d_{2})^{2} = d_{1}^{2} + d_{2}^{2} - 2 d_{1} d_{2}$
$\Rightarrow d_{1}^{2} + d_{2}^{2} = (d_{1} - d_{2})^{2} + 2 d_{1} d_{2}$
$\Rightarrow$ Area of square = Positive quantity + Area of Rhombus
Area of Square > Area of Rhombus

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