Please helpme solve this...?

Question

The base of a triangle is 6 cm greater than the height. The area is 56 cm squared (56^2). Find the height and the length of the base.

H=?

L=?

Thank you for helping!

Answer 1

x = length

y = height

x - 6 = y (solve by using substitution)

xy/2 = 56

xy/2 = 56

x(x - 6)/2 = 56

x*x - x*6 = 2(56)

x^2 - 6x = 112

x^2 - 6x - 112 = 0

x^2 + 8x - 14x - 112 = 0

(x^2 + 8x) - (14x + 112) = 0

x(x + 8) - 14(x + 8) = 0

(x + 8)(x - 14) = 0

x + 8 = 0

x = -8

x - 14 = 0

x = 14

(the length should be 14 because the value must be positive.)

x - 6 = y

14 - 6 = y

y = 8

∴ x (length) = 14 , y (height) = 8

Answer 2

L= 18.6

H= 12.6

The area is 56cm^2 i think not 56^2

Answer 3

I don't think you mean 56^2.......you mean (I believe) 56 cm squared. Cm squared is a measurement of area in centimeters. The area of a triangle is 1/2BH, where L=Base and H=height. So:

A = 1/2 LH

56 = 1/2 (6 + H) H ....Base = L = 6+H

56 = (6H + H^2)/2 ....multiply both sides by 2

112 = 6H + H^2

H^2 + 6H - 112 = 0

(H + 14)(H - 8) = 0

H = -14, +8

We throw out -14 (triangle can't have negative sides)

H = 8,  L =  H + 6 = 14

proof

1/2(8)(14) = 56