# The legs of right triangle ABC have lengths 3 and 4. What is the perimeter of a right triangle with each side twice the length of its corresponding side in triangle ABC?

$2 \left(3\right) + 2 \left(4\right) + 2 \left(5\right) = 24$

#### Explanation:

Triangle ABC is a 3-4-5 triangle - we can see this from using the Pythagorean Theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${3}^{2} + {4}^{2} = {5}^{2}$

$9 + 16 = 25$

25=25 color(white)(00)color(green)root

So now we want to find the perimeter of a triangle that has sides twice that of ABC:

$2 \left(3\right) + 2 \left(4\right) + 2 \left(5\right) = 6 + 8 + 10 = 24$