A frustum

A frustum of a pyramid consists of a square base of length 10 cm and a top square of length 7 cm. The height of the frustum is 6 cm. Calculate the surface area and volume.

Correct answer:

V =  438 cm³
S =  359.28 cm²

Step-by-step explanation:

$$\begin{gathered} a=10\text{cm} \\ b=7\text{cm} \\ h=6\text{cm} \\ {S}_1=a^2=10^2=100{\text{cm}}^2 \\ {S}_2=b^2=7^2=49{\text{cm}}^2 \\ V=\frac{h}{3}\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=\frac{6}{3}\cdot (100+\sqrt{100\cdot 49}+49)=438{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} x=(a-b)/2=(10-7)/2=\frac{3}{2}=1.5\text{cm} \\ {h}^2=s^2-x^2 \\ s=\sqrt{h^2+x^2}=\sqrt{6^2+1.5^2}\doteq 6.1847\text{cm} \\ {S}_3=\frac{1}{2}\cdot (a+b)\cdot s=\frac{1}{2}\cdot (10+7)\cdot 6.1847\doteq 52.5696{\text{cm}}^2 \\ S=S_1+S_2+4\cdot S_3=100+49+4\cdot 52.5696=359.28{\text{cm}}^2 \end{gathered}$$

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