Quadrilateral 5814

Calculate the surface area and volume of a regular quadrilateral truncated pyramid if the edges of the base are 87 cm and 64 cm and the wall height is 49 cm.

Correct answer:

V =  273610.6227 cm³
S =  26463 cm²

Step-by-step explanation:

$$\begin{gathered} a=87\text{cm} \\ b=64\text{cm} \\ s=49\text{cm} \\ x=(a-b)/2=(87-64)/2=\frac{23}{2}=11.5\text{cm} \\ h=\sqrt{s^2-x^2}=\sqrt{49^2-11.5^2}\doteq 47.6314\text{cm} \\ {S}_1=a^2=87^2=7569{\text{cm}}^2 \\ {S}_2=b^2=64^2=4096{\text{cm}}^2 \\ V=h/3\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=47.6314/3\cdot (7569+\sqrt{7569\cdot 4096}+4096)=273610.6227{\text{cm}}^3=2.736\cdot 10^5{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} S_3=(a+b)\cdot s/2=(87+64)\cdot 49/2=\frac{7399}{2}=3699.5{\text{cm}}^2 \\ S=S_1+S_2+4\cdot S_3=S_1+S_2+4\cdot \frac{7399}{2}=7569+4096+4\cdot 3699.5=26463{\text{cm}}^2 \end{gathered}$$

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