area of a right angled isosceles triangle.

Given the base of an isosceles right angle triangle is $30$ cm.It is required to find the area of the same.

Its quite clear that if the triangle is isosceles right angled triangle, and its base is 30, then the length of perpendicular would also be equal to 30.

By applying the formula for the area of a triangle i.e. $\frac 12$$.b.h, the area clearly comes out to be \frac 12.30.30=450.But the answer according to the book is wrong and is 225.Can anyone please explain me how am I wrong? • @Apurv How does \frac{1}{2} \times 30 \times 30 = 225 ? – SchrodingersCat Oct 23 '15 at 16:38 • Note that there is no such thing as "the base of an isosceles triangle" if the orientation of the triangle is not specified. In other words, this is an ambiguous question, at least as stated. – MPW Oct 26 '15 at 3:34 1 Answer The problem evidently intends the base to be the third side that is not necessarily equal to the other two. The height of the perpendicular, then, is 15, so the area is indeed$$\frac12\cdot 30 \cdot 15 = 225$\$

You have assumed that the base and one of the other sides constitute the pair of equal sides.

I think your interpretation isn't incorrect, but is not what the problem intended.