Question

In isosceles right triangle ABC, point is on hypotenuse \overline{BC} such that \overline{AD} is an altitude of \triangle ABC and DC = 5. What is the area of triangle ABC?

Answer 1

Answer:

Area of triangle is 25.

Step-by-step explanation:

We have been given an isosceles right triangle

Isosceles triangle is the triangle having two sides equal.

Figure is shown in attachment

By Pythagoras theorem

$BC^2=AC^2+AB^2$

AD is altitude which divides the triangle into two parts

DC=5 implies BC =10 since D equally divides BC

Let AC=a implies AB=a being Isosceles

On substituting the values in the Pythagoras theorem:

$10^2=a^2+a^2$

$100=2a^2$

$\Rightarrow a^2=50$

$\Rightarrow a=\pm5\sqrt{2}$

WE can find area of right triangle by considering height AB and AD

Area of triangle ABC is:

$\frac{1}{2}\cdot BC\cdot AD$     (1)

$\Rightarrow \frac{1}{2}\cdot 10\cdot AD$

And other method of area of triangle is:

$\frac{1}{2}\cdot AB\cdot BC$       (2)

Equating (1) and (2) we get:

$\frac{1}{2}\cdot 10\cdot AD=\frac{1}{2}\cdot a\cdot a$

$\Rightarrow AD=\frac{a^2}{10}$

$\Rightarrow AD=\frac{50}{10}=5$

Using area of triangle is: $\frac{1}{2}\cdot BC\cdot AD$

Now, the area of triangle ABC=$\frac{1}{2}\cdot 5\cdot 10$

$\Rightarrow 25$