RRL Basics

What is the length of the smaller base of an isosceles trapezoid and the height, if a = 9 dm, the side is 6 dm and the angle ACB is 90 degrees?

Final Answer:

c =  1 dm
h =  4.4721 dm

Step-by-step explanation:

$$\begin{gathered} a=9\text{dm} \\ r=6\text{dm} \\ \angle ACB=90{}^{\circ } \\ AC=\sqrt{a^2-r^2}=\sqrt{9^2-6^2}=3\sqrt{5}\text{dm}\doteq 6.7082\text{dm} \\ {S}_1=\frac{r\cdot AC}{2}=\frac{6\cdot 6.7082}{2}=9\sqrt{5}{\text{dm}}^2\doteq 20.1246{\text{dm}}^2 \\ {S}_1=\frac{a\cdot h}{2} \\ h=\frac{2\cdot S_1}{a}=\frac{2\cdot 20.1246}{9}=2\sqrt{5}\text{dm}\doteq 4.4721\text{dm} \\ x=\sqrt{r^2-h^2}=\sqrt{6^2-4.4721^2}=4\text{dm} \\ a=x+c+x \\ c=a-2\cdot x=9-2\cdot 4=1\text{dm} \end{gathered}$$

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