Isosceles trapezoid area

Calculate the area of an isosceles trapezoid ABCD, whose longer base measures 48 cm, the shorter base measures 3/4 of the longest base, and the leg of the trapezoid measures 2/3 of the longer base. The result is rounded to the nearest hundredth.

Final Answer:

S =  1320.16 cm²

Step-by-step explanation:

$$\begin{gathered} a=48\text{cm} \\ c=\frac{3}{4}\cdot a=\frac{3}{4}\cdot 48=\frac{3\cdot 48}{4}=\frac{144}{4}=36\text{cm} \\ b=\frac{2}{3}\cdot a=\frac{2}{3}\cdot 48=\frac{2\cdot 48}{3}=\frac{96}{3}=32\text{cm} \\ d=b=32\text{cm} \\ x=(a-c)/2=(48-36)/2=6\text{cm} \\ v=\sqrt{b^2-x^2}=\sqrt{32^2-6^2}=2\sqrt{247}\text{cm}\doteq 31.4325\text{cm} \\ S=\frac{a+c}{2}\cdot v=\frac{48+36}{2}\cdot 31.4325=84\sqrt{247}=1320.16{\text{cm}}^2 \end{gathered}$$

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