Truncated pyramid

Find the volume and surface area of a regular quadrilateral truncated pyramid if base lengths a1 = 17 cm, a2 = 5 cm, height v = 8 cm.

Correct answer:

V =  1064 cm³
S =  754 cm²

Step-by-step explanation:

$$\begin{gathered} a_1=17\text{ cm } \\ {a}_2=5\text{ cm } \\ v=8\text{ cm } \\ {S}_1=a_1^2=17^2=289{\text{cm}}^2 \\ {S}_2=a_2^2=5^2=25{\text{cm}}^2 \\ V={\displaystyle \frac{1}{3}}\cdot v\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)={\displaystyle \frac{1}{3}}\cdot 8\cdot (289+\sqrt{289\cdot 25}+25)=1064{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} h_2=\sqrt{v^2+((a_1-a_2)\mathrm{/}2{)}^2}=\sqrt{8^2+((17-5)\mathrm{/}2{)}^2}=10\text{ cm } \\ {S}_3={\displaystyle \frac{a_1+a_2}{2}}\cdot h_2={\displaystyle \frac{17+5}{2}}\cdot 10=110{\text{cm}}^2 \\ S=S_1+S_2+4\cdot S_3=289+25+4\cdot 110=754{\text{cm}}^2 \end{gathered}$$

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