Question

The height of a trapezoid can be expressed as x-4, while the bases can be expressed as x+4 and x+9. If the area of the trapezoid is 99cm^2, find the length of the larger base?

Answer 1

Check the picture below.

`\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{bases}{parallel~sides}\\ \cline{1-1} h= x - 4\\ a = x+4\\ b=x+9\\ A=99 \end{cases}\implies 99=\cfrac{(x-4)[(x+4)+(x+9)]}{2} \\\\\\ 99=\cfrac{(x-4)[2x+13]}{2}\implies 198=\stackrel{\mathbb{FOIL}}{2x^2+5x-52}\implies 0=2x^2+5x-250`

`\bf 0=(2x+25)(x-10)\implies x= \begin{cases} ~~\begin{matrix} -25 \\[-0.6em]\cline{1-1}\\[-5pt]\end{matrix}~~\\ 10 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{larger base}}{x+9}\implies 10+9\implies 19`

keeping in mind that "x" cannot be equal to -25, since that'd give us negative values on either base and the bases are a positive value.

Answer 2

19 cm

Explanation:

Since we want to find the length of the longer base, let that be represented by z. Then we have ...

z = x+9

x = z-9

The other dimensions of the trapezoid are then ...

shorter base = x +4 = (z -9) +4 = z -5

height = x -4 = (z -9) -4 = z -13

__

The formula for the area of a trapezoid is ...

A = (1/2)(b1 + b2)h = (1/2)(z +(z -5))(z -13) = (1/2)(2z -5)(z -13)

The area is 99 cm², so we have ...

99 = 1/2(2z² -31z +65)

2z² -31z -133 = 0 . . . . multiply by 2 and subtract 198

(2z +7)(z -19) = 0 . . . . factor

This has one positive solution: z = 19

The length of the larger base is 19 cm.