1) Using the given length and width of the rectangle, find its area. 2) Then using the base and height of the triangle, find its area. 3) Since the areas are equal, set the expressions equal to each other, and solve for x. 4) Using the value of x you found, find the length and width of the rectangle and find its perimeter.
1) The area of the rectangle is A = LW Area of Rectangle = (x + 2)x = x^2 + 2x
2) The area of the triangle is A = (1/2)bh Area of Triangle = (1/2)(24)x = 12x
3) Set the areas equal and solve for x
x^2 + 2x = 12x
x^2 - 10x = 0
x(x - 10) = 0
x = 0 or x = 10
Since a width cannot be 0, we discard x = 0, and keep x = 10.
4) The length is x + 2 = 10 + 2 = 12 The width is x = 10
The perimeter is 2(L + W) = 2(10 + 2) = 2(22) = 44
