The altitude drawn to the base of an isosceles triangle is 8 cm and its perimeter is 64 cm. The area (in cm2) of the triangle is

Answer: Option D

Let AB = AC = a cm
BD = DC = b cm
∴ Altitude of isosceles triangle is also median
In right ΔADC
AD² = a² - b²
64 = a² - b² . . . . . . (i)
Perimeter = 64
a + a + 2b = 64
2a + 2b = 64
a + b = 32 . . . . . (ii)
On dividing $$=\frac{a^2-b^2}{a+b}=\frac{64}{32}=2$$
∴ a² - b² = (a + b)(a - b)
a - b = 2
∴ a + b = 32
On solving a = 17, b = 15
Area of ΔABC = $$\frac{1}{2}$$ × AD × BC
= $$\frac{1}{2}$$ × 8 × 30
= 120 cm