Question

In the isosceles right triangle above $\overline{AB}$=10 feet. Find the length, in feet, of $\overline{AC}$.

• A. 5
• B. 10
• C. 20
• D. $\sqrt{20}$
• E. $10\sqrt{2}$

Answer 1

The correct option is E $10\sqrt{2}$

In an isosceles triangle two sides of the triangle are equal

$\therefore AB=BC=10$

It is a right angle triangle, so

$A{C}^2=B{C}^2+A{B}^2$

$\Rightarrow A{C}^2={10}^2+{10}^2$

$\Rightarrow A{C}^2=100+100$

$\Rightarrow A{C}^2=200$

$\Rightarrow AC=\sqrt{2\times 100}$

$\Rightarrow AC=10\sqrt{2}$