In the given right angled isosceles triangle ABC, if AC = 10cm then the area of triangle (in $c m^{2}$ ) is
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Solution

In right angled isosceles triangle ABC, B = $90^{\circ}$

AB=BC

So, $θ$ = $α$

$cos$ $θ$ = $\frac{1}{\sqrt{2}}$ at = $45^{\circ}$

So, $\frac{A B}{A C}$ = $\frac{1}{\sqrt{2}}$

AB= $\frac{A C}{\sqrt{2}}$

AB=5 $\sqrt{2}$ cm

And $sin θ$ = $\frac{1}{\sqrt{2}}$ at = $45^{\circ}$

So, $\frac{B C}{A C}$ = $\frac{1}{\sqrt{2}}$

BC=5 $\sqrt{2}$ cm

Area of triangle= $\frac{1}{2}$ (Base $\times$ Height)

= $\frac{1}{2}$ (5 $\times$ 5 $\sqrt{4}$ )

=25 $c m^{2}$

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