Question

The perimeter of an equilateral triangle is 10 inches more than the perimeter of a​ square, and the side of the triangle is 6 inches longer than the side of the square. Find the side of the triangle.​ (Hint: An equilateral triangle has three sides the same​ length.)

Answer 1

Let \( s \) be the side length of the square. The perimeter of the square is given by:

\[
P_{square} = 4s
\]

Let \( t \) be the side length of the equilateral triangle. The perimeter of the triangle is given by:

\[
P_{triangle} = 3t
\]

According to the problem, the perimeter of the triangle is 10 inches more than the perimeter of the square:

\[
P_{triangle} = P_{square} + 10
\]

Substituting the perimeter formulas into this equation gives:

\[
3t = 4s + 10
\]

The problem also states that the side of the triangle is 6 inches longer than the side of the square:

\[
t = s + 6
\]

Now, we can substitute the expression for \( t \) into the first equation. Replacing \( t \) in the perimeter equation:

\[
3(s + 6) = 4s + 10
\]

Expanding the left side leads to:

\[
3s + 18 = 4s + 10
\]

Now, we will rearrange the equation to isolate \( s \):

\[
3s + 18 - 10 = 4s
\]
\[
3s + 8 = 4s
\]
\[
8 = 4s - 3s
\]
\[
8 = s
\]

Now that we have found \( s \) (the side of the square), we can find \( t \) (the side of the triangle):

\[
t = s + 6 = 8 + 6 = 14
\]

Thus, the side length of the triangle is:

\[
\boxed{14 \text{ inches}}
\]

To verify, we check the perimeters. The perimeter of the square:

\[
P_{square} = 4s = 4 \times 8 = 32 \text{ inches}
\]

And the perimeter of the triangle:

\[
P_{triangle} = 3t = 3 \times 14 = 42 \text{ inches}
\]

Now checking the condition:

\[
P_{triangle} = P_{square} + 10 \\
42 = 32 + 10
\]

Since both sides match, the answer is confirmed correct. The side of the triangle is \( \boxed{14} \) inches.