i hate word problems...please help last one!!!?

Question

The base of a triangle is numerically 8 less than its area, and the height is numerically 12 less than its area. Find the area of the triangle.


Please enter your answer as a number without the units.

Answer 1

Let A be area:

Base, b = (A - 8)
Height, h = (A - 12)

Area of triangle given by:

A = (1/2) b h

Substitute into the equation and you get:

A = (1/2)(A-8)(A-12) -->
2A = A^2 - 12A - 8A + 96 -->
2A = A^2 - 20A + 96 -->
A^2 - 22A + 96 = 0 -->


Apply quadratic formula to find A

A = [-(-22) +- SQRT{(-22)^2 - 4*1*96}]/[2*1]
A = [22 +- SQRT(484 - 384)]/2
A = [22 +- SQRT(100)]/2
A = (22 +- 10)/2

A = 32/2 = 16
A = 12/2 = 6

But, you can't have negative base or height, therefore you need to eliminate A=6. You now have:

h = 16 - 8 = 8
b = 16 - 2 = 4

Check if this works by plugging into triangle area equation:

A = (1/2) b h = (1/2)*4*8 = (1/2)*32 = 16

Answer 2

Area of triangle(A) = ½ * Base * Height
=> A = ½ * (A - 8) * (A - 12)
=> 2A = A² - 12A - 8A + 96
=> A² - 22A + 96 = 0
=> (A - 6)(A - 16) = 0
=> A = 6 Or A = 16
Area cannot be 6 as it will make Base and Height negative.
Hence Area is 16

Answer 3

a = Area = (1/2)(base)(height)

a = (1/2)(a - 8)(a - 12) = (1/2)(a² - 20a + 96)
2a = a² - 20a + 96
a² - 22a + 96 = 0
(a - 6)(a - 16) = 0
a = 6, 16

Since length and height need to be positive the smaller solution is eliminated.

a = 16