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the congruent sides of an isosceles triangle are each 1 unit longer tha…

Question

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? units

Explanation:

Step1: Define variables

Let the length of the shortest side of the triangle be $x$ units. Then the length of each congruent side is $x + 1$ units.

Step2: Calculate the perimeter of the triangle

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

Step3: Calculate the perimeter of the square

The side length of the square is $x - 2$ units (since the side length of the square is 2 units shorter than the shortest side of the triangle). So the perimeter of the square $P_{square}=4(x - 2)=4x-8$.

Step4: Set up the equation

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

Step5: Solve for $x$

Subtract $3x$ from both sides: $2=x - 8$. Then add 8 to both sides: $x = 10$.

Answer:

10