Question

finding a missing dimension
the congruent sides of an isosceles triangle are each 1 unit longer than the length of the shortest side of the triangle. the perimeter of the triangle is the same as the perimeter of a square whose side length is 2 units shorter than the length of the shortest side of the triangle. what is the length of the shortest side of the triangle?
10 units

Explanation:

Step1: Let the shortest side length

Let the length of the shortest side of the isosceles triangle be $x$ units. Then the length of each of the congruent sides is $x + 1$ units. The side - length of the square is $x-2$ units.

Step2: Calculate the perimeter of the triangle

The perimeter of the isosceles triangle $P_{triangle}=x+(x + 1)+(x + 1)=3x+2$ units.

Step3: Calculate the perimeter of the square

The perimeter of the square $P_{square}=4(x - 2)=4x-8$ units.

Step4: Set up the equation

Since the perimeters are equal, we set up the equation $3x + 2=4x-8$.

Step5: Solve the equation

Subtract $3x$ from both sides: $3x+2-3x=4x - 8-3x$, which gives $2=x - 8$. Then add 8 to both sides: $2 + 8=x-8+8$, so $x = 10$.

Answer:

10