Dimensions of the trapezoid

One of the bases of the trapezoid is one-fifth larger than its height, the second base is 1 cm larger than its height. Find the dimensions of the trapezoid if its area is 115 cm²

Correct result:

a =  12 cm
c =  11 cm
h =  10 cm

Solution:

$$\begin{gathered} a=h+h\mathrm{/}5 \\ c=1+h=1+10=1 \\ S=115{\text{cm}}^2 \\ S={\displaystyle \frac{a+c}{2}}\cdot h \\ 2\ast S=((h+h\mathrm{/}5)+(1+h))\ast h \\ 2\cdot 115=((h+h\mathrm{/}5)+(1+h))\cdot h \\ -2.2h^2-h+230=0 \\ 2.2h^2+h-230=0 \\ a=2.2;b=1;c=-230 \\ D=b^2-4ac=1^2-4\cdot 2.2\cdot (-230)=2025 \\ D>0 \\ {h}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-1\pm \sqrt{2025}}{4.4}} \\ {h}_{1,2}=-0.22727273\pm 10.2272727273 \\ {h}_1=10 \\ {h}_2=-10.4545454545 \\ \text{ Factored form of the equation: } \\ 2.2(h-10)(h+10.4545454545)=0 \\ h=h_1=10 \\ a=h+h\mathrm{/}5=10+10\mathrm{/}5=12\text{ cm} \end{gathered}$$

Our quadratic equation calculator calculates it.

$c=1+h=1+10=11\text{ cm}$

$$\begin{gathered} h=10=10\text{ cm } \\ \text{ Verifying Solution: } \\ {S}_1={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{12+11}{2}}\cdot 10=115{\text{cm}}^2 \end{gathered}$$

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