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.

∴ BD = DC =

b2

According to Pythagoras theorem,

AD² = AC² – CD²

or, AD = √ (AC² - CD²) = √ [a² - (

b2
)²]

= √ (

4a² - b²4
) =
√ (4a² - b²)2

Now, area of triangle (A) =

12
× base × height =
12
× b ×
√ (4a² - b²)2
=
b √ (4a² - b²)4

Thus, Area of isosceles triangle (A) =
b √ (4a² - b²)4