Quadrilateral 81033

The foundations of a regular truncated quadrilateral pyramid are squares. The lengths of the sides differ by 6 dm. Body height is 7 dm. The body volume is 1813 dm³. Calculate the lengths of the edges of both bases.

Correct answer:

a =  19 dm
b =  13 dm

Step-by-step explanation:

$$\begin{gathered} a=6+b\text{dm} \\ V=1813{\text{dm}}^3 \\ h=7\text{dm} \\ V=\frac{h}{3}\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2) \\ 3\cdot V/h=S_1+\sqrt{S_1\cdot S_2}+S_2 \\ 3\cdot V/h=a^2+\sqrt{a^2\cdot b^2}+b^2 \\ 3\cdot V/h=a^2+a\cdot b+b^2 \\ 3\cdot V/h=a^2+a\cdot (a-6)+(a-6{)}^2 \\ 3\cdot 1813/7=a^2+a\cdot (a-6)+(a-6{)}^2 \\ -3a^2+18a+741=0 \\ 3a^2-18a-741=0 \\ 3...\text{ prime number} \\ 18=2\cdot 3^2 \\ 741=3\cdot 13\cdot 19 \\ \text{GCD}(3,18,741)=3 \\ {a}^2-6a-247=0 \\ p=1;q=-6;r=-247 \\ D=q^2-4pr=6^2-4\cdot 1\cdot (-247)=1024 \\ D>0 \\ {a}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{6\pm \sqrt{1024}}{2} \\ {a}_{1,2}=\frac{6\pm 32}{2} \\ {a}_{1,2}=3\pm 16 \\ {a}_1=19 \\ {a}_2=-13 \\ a=a_1=19\text{dm}=19\text{dm} \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} b=a-6=19-6=13=13\text{dm} \\ \text{ Verifying Solution: } \\ {S}_1=a^2=19^2=361{\text{dm}}^2 \\ {S}_2=b^2=13^2=169{\text{dm}}^2 \\ {V}_2=\frac{h}{3}\cdot (S_1+\sqrt{S_1\cdot S_2}+S_2)=\frac{7}{3}\cdot (361+\sqrt{361\cdot 169}+169)=1813{\text{dm}}^3 \end{gathered}$$

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