Truncated pyramid

The truncated regular quadrilateral pyramid has a volume of 74 cm³, a height v = 6 cm, and an area of the lower base 15 cm² greater than the upper base's area. Calculate the area of the upper base.

Correct answer:

S₁ =  5.6185 cm²

Step-by-step explanation:

$$\begin{gathered} V=74{\text{cm}}^3 \\ v=6\text{cm} \\ {S}_2=S_1+15 \\ V=\frac{v}{3}\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2) \\ 3\cdot V/v=S_1+\sqrt{S_1\cdot (S_1+15)}+S_1+15 \\ 3\cdot V/v-15=2\cdot S_1+\sqrt{S_1\cdot (S_1+15)} \\ {S}_1=\frac{-15\cdot v^2-\sqrt{4\cdot v^2\cdot V^2-75\cdot v^4}+4\cdot v\cdot V}{2\cdot v^2}=\frac{-15\cdot 6^2-\sqrt{4\cdot 6^2\cdot 74^2-75\cdot 6^4}+4\cdot 6\cdot 74}{2\cdot 6^2}\doteq 5.6185{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ {S}_2=S_1+15=5.6185+15\doteq 20.6185{\text{cm}}^2 \\ {V}_2=\frac{v}{3}\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=\frac{6}{3}\cdot (5.6185+\sqrt{5.6185\cdot 20.6185}+20.6185)=74{\text{cm}}^3 \end{gathered}$$

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