If in a triangle, angles are in the ratio 1 : 1 : 2 and the length of its longest side is $$6\surd2$$ cm, then what is the Area (in cm$$^2$$) of the triangle?

If in a triangle, angles are in the ratio 1 : 1 : 2.

As we know the sum of all the three angles of a triangle is $$180^{\circ\ }$$.

Let's assume the angles are y, y and 2y respectively.

$$y+y+2y = 180^{\circ\ }$$

$$4y = 180^{\circ\ }$$

$$y = 45^{\circ\ }$$

So the angles of the given triangles are $$45^{\circ\ },\ 45^{\circ\ }and\ 90^{\circ\ }$$.

It is an isolated triangle where AB=AC=a (Because $$\angle\ B=\angle\ C$$.) and BC will be the longest side.

the length of its longest side is $$6\surd2$$ cm.

BC = $$6\surd2$$

Now put perpendicular from A on BC. It will divide the triangle into two equal parts.

In triangle ABD, $$\angle B\ =\ \angle\ BAD\ $$, then side AD and BD will also be equal.

So AD = $$3\sqrt{\ 2}$$

Area of the triangle = $$\frac{1}{2}\times\ BC\times\ AD$$

= $$\frac{1}{2}\times\ (3\sqrt{\ 2}+3\sqrt{\ 2}) \times\ 3\sqrt{\ 2}$$

= $$\frac{1}{2}\times\ 6\sqrt{\ 2}\times\ 3\sqrt{\ 2}$$

= $$3\sqrt{\ 2}\times\ 3\sqrt{\ 2}$$

= 18 cm$$^2$$