Consider a square and a triangle. The sides of the square are x cm long. The base length and height of the triangle are equal, and are twice as long as the sides of the square. The area of the triangle is 9cm^{2 }larger than the area of the square.

Find the perimeter of the square

Guest May 7, 2017

#1**+1 **

Consider a square and a triangle. The sides of the square are x cm long. The base length and height of the triangle are equal, and are twice as long as the sides of the square. The area of the triangle is 9cm2 larger than the area of the square.

Find the perimeter of the square

The base and height of the triangle = 2x

So.....

Area of the triangle = Area of square + 9cm^2

(1/2)B*H = x^2 + 9

(1/2) (2x) (2x) = x^2 + 9

2x^2 = x^2 + 9 subtract x^2, 9 from both sides

x^2 - 9 = 0 factor

(x + 3) ( x - 3) = 0

The first factor produces a negative solution [reject ]

Setting the second factor to 0 we have

x - 3 = 0 add 3 to both sides

x = 3 cm this is the side of the square........and its perimeter is 4 times this = 12 cm

Check.....area of triangle = 2x^2 = 2(3)^2 = 18 cm^2

Area of square = 9 cm ^2

And the triangle has an area that is 9 cm greater than that of the square

CPhill
May 7, 2017