Question

Someone please help me with this math problem?

Answer 1

Answer:

The length of the shortest side of the triangle is 10 units.

Step-by-step explanation:

Let a be the shortest side of the isosceles triangle and b be the two congruent sides.

The congruent sides b are each one unit longer than the shortest side. Hence:

$b=a+1$

The perimeter of the isosceles triangle is given by:

$\displaystyle P_{\Delta}=b+b+a=2b+a$

This is equivalent to the perimeter of a square whose side lengths are two units shorter than the shortest side of the triangle. Let the side length of the square be s. Hence:

$s=a-2$

The perimeter of the square is:

$\displaystyle P_{\text{square}}=4s=4(a-2)$

Since the two perimeters are equivalent:

$2b+a=4(a-2)$

Substitute for b:

$2(a+1)+a=4(a-2)$

Solve for a. Distribute:

$2a+2+a=4a-8$

Simplify:

$3a+2=4a-8$

Hence:

$a=10$

The length of the shortest side of the triangle is 10 units.