Isosceles Right Triangle Area?

In a right triangle, the longest side is the hypotenuse and an isosceles triangle has two equal sides. Since it is not possible to have two hypotenuse in a right triangle, we let be $x$ be the length of each equal side. Consider the diagram,

$4^2=x^2+x^2$

$16=2x^2$

$8=x^2$

It follows that,

$A=\dfrac{1}{2}x^2=\dfrac{1}{2}(8)=$ $\boxed{4}$

Since it's an isosceles triangle, so the two sides of the triangle will be equal. In right angled triangle, the hypotenuse is the longest side i.e. the hypotenuse will be $4$ .

So the other two equal sides will be $2\sqrt{2}$ ... So area of triangle will be

$A=\dfrac{1}{2}.b.h$ = $\dfrac{1}{2} . 2\sqrt{2} . 2\sqrt{2}$

= $\boxed{4}$

The right isosceles triangle is nothing but a triangle inscribed in a semicircle of diameter of $4$ and its altitude equal radius of semicircle i.e $2$

$Area = \dfrac {1}{2}\times (diameter) \times (radius) = \dfrac {1}{2}\times(4)\times (2) = 4$