An isosceles triangle has its base length 2a and its height is h. On each side of the triangle a square is drawn external to the triangle. What is the area of the figure thus formed?

Given:

An isosceles triangle has its base length 2a & its height is h.

Formula used:

Area of Triangle = 1/2 × base × height

Area of Square = side²

Calculation:

According to the figure,

Area of isosceles triangle = 1/2 × b × h 

⇒ 1/2 × 2a × h  ( base = 2a)     .....(1)

Again, according to the figure,

By Pythagoras theorem

⇒ AC = √DC² + AD² 

⇒ AC = √a² + h²

Then, Area of square drawn on AB & AC 

Side² = (√a² + h²)² = (a² + h²)

Now, both side will be equal then, 

Area = 2(a² + h²)     .....(2)

Again, according to the figure,

Area of square drawn on BC which is (2a)²

Area of square = (2a)² = (4a²)    .....(3)

By adding above equation (1), (2) & (3) we get,

Area of the figure,

(1/2 × 2a × h) + 2(a² + h²) + (4a²)

⇒ ah + 2a² + 2h² + 4a²

⇒(2a² +4a²) + 2h² + ah

⇒ 6a² + 2h² + ah

∴ The area of the figure is  6a2 + 2h2 + ah