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

Area of triangle is 25.

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√(2)`

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

Area of triangle ABC is:

`(1)/(2)\cdot BC\cdot AD` (1)

`\Rightarrow (1)/(2)\cdot 10\cdot AD`

And other method of area of triangle is:

`(1)/(2)\cdot AB\cdot BC` (2)

Equating (1) and (2) we get:

`(1)/(2)\cdot 10\cdot AD=(1)/(2)\cdot a\cdot a`

`\Rightarrow AD=(a^2)/(10)`

`\Rightarrow AD=(50)/(10)=5`

Using area of triangle is:
`(1)/(2)\cdot BC\cdot AD`

Now, the area of triangle ABC=
`(1)/(2)\cdot 5\cdot 10`

`\Rightarrow 25`