Question

A triangle has a vertical height 2cm longer than it's base lengths. it's area is 20cm^2. Find the base and the height of the triangle.

Answer 1

Let the base be b. The the height of the triangle is b+2.
(base)(height)
The area of any triangle is A = ----------------------
2
(b)(b+2)
Here, A = 20 cm^2 = --------------
2

so 40 cm^2 = b^2 + 2b, or b^2 + 2b - 40 = 0

solving this quadratic with the quadratic formula:

-2 plus or minus sqrt( 4-4(1)(-40) )
b = --------------------------------------------------
2

= -1 plus or minus sqrt( 4 + 160) = -1 plus or minus sqrt 164

Note that sqrt(164) = sqrt(4)sqrt(41) = 2sqrt(41)

so b = -1 plus or minus 2sqrt(41). We must eliminate the negative root here. So, we get b = -1 + sqrt(41). Then the height is h = -1 + 2 + sqrt(41), or

b = -1 + sqrt(41) and h = b+2 = 1 + sqrt(41) (answer)