ii  Area of Isosceles Triangle

Activity

The given figure is an isosceles triangle. Let, AB = AC = a and BC = b. Let us draw AD perpendicular to BC from A which bisects base BC. The perpendicular drawn from the vertex to the base of an isosceles triangle bisects the base.

Since

BD = DC =  b 2

Using Pythagoras theorem in right triangle:

AD2 = AC2 − CD2
AD =  AC2 − CD2

Substitute values:

AD =  a2 −  ( b 2 ) 2
AD =  a2 −  b2 4
AD =  4a2 − b2 4
AD =  4a2 − b2 2

Now, Area of triangle:

Area =  1 2  × base × height
=  1 2  × b ×  4a2 − b2 2
=  b  4a2 − b2 4
Thus, Area of isosceles triangle (A) = b  4a2 − b2 4