A square and an equilateral triangle have the same perimeter. The area of the square is 13 more than the length of a side of the triangle.

Question

What is the length of a side in the triangle (in inches)? What is the area of the square (in square inches)?

Answer 1

If a is the side length of the square and b the side length of the equilateral triangle then by assumptions we have 4a=3b and a^2=13+b. Solving this system we obtain that a=13/3 and b=52/9.

Answer 2

They tell us the the perimeters of the two shapes are equal, so we know that P_t (the perimeter of the triangle) is the same as P_s, the perimeter of the square. So,

P_t = P_s.

We get the perimeter of a shape by adding together all of its sides, so the square gives us

P_s = 4s,

and the triangle gives us

P_t = 3t.

Since P_t = P_s, we know

4s = 3t, OR t = (4/3)s.

The other piece of information we're given is that the area of the square is 13 more than the length of a side of the triangle, or

A_s= t + 13.

We get the area of a square by multiplying the two sides, so

A_s = s^2.

And we already found that t = (4/3)s, so we can rewrite this new equation as:

s^2 = (4/3)s + 13.

This is a quadratic equation, so we'll use the quadratic formula to solve it! I'll multiply everything by 3 to get rid of the fraction (I just think it makes everything neater) and put all of our terms on one side. That gives us:

3s^2 - 4s - 39 =0 , where

a = 3

b= -4

c = -39.

Plugging these into the quadratic formula, we get s = 13/3, or 4 1/3.

From before, we found

t = (4/3)s, so plug in our 13/3 and we get that t = (52/9) in.

Finally, since we were given that

A_s= t + 13,

we can plug in t to find

A_s= (52/9) + 13 = (169/9) sq. in.

Just to be sure, we also found that s = (13/3), so

A_s = s^2 = (13/3)^2 = (169/9) sq in. It checks out!

Hope that helps!