Pyramid cut

We cut the regular square pyramid with a parallel plane to the two parts (see figure). The volume of the smaller pyramid is 20% of the volume of the original one. The bottom of the base of the smaller pyramid has a content of 10 cm2. Find the area of the original pyramid.

Result

S1 =  29.24 cm2

Solution:

S2=10 cm2 S2=a22 a2=S2=1010 cm3.1623 cm  V2=20100V1  13S2 h2=2010013S1 h1 S2 h2=0.20 S1 h1  a1:h1=a2:h2 a1:a2=h1:h2  S2=a1/a2 0.20 S1 a23=a13 0.20  a1=a23/0.203 a1=a2 1/0.203=3.1623 1/0.2035.4074 cm  S1=a12=5.4074229.240229.24 cm2S_{2}=10 \ \text{cm}^2 \ \\ S_{2}=a_{2}^2 \ \\ a_{2}=\sqrt{ S_{2} }=\sqrt{ 10 } \doteq \sqrt{ 10 } \ \text{cm} \doteq 3.1623 \ \text{cm} \ \\ \ \\ V_{2}=\dfrac{ 20 }{ 100 } V_{1} \ \\ \ \\ \dfrac{ 1 }{ 3 } S_{2} \ h_{2}=\dfrac{ 20 }{ 100 } \dfrac{ 1 }{ 3 } S_{1} \ h_{1} \ \\ S_{2} \ h_{2}=0.20 \ S_{1} \ h_{1} \ \\ \ \\ a_{1}:h_{1}=a_{2}:h_{2} \ \\ a_{1}:a_{2}=h_{1}:h_{2} \ \\ \ \\ S_{2}=a_{1}/a_{2} \cdot \ 0.20 \ S_{1} \ \\ a_{2}^3=a_{1}^3 \cdot \ 0.20 \ \\ \ \\ a_{1}=\sqrt[3]{ a_{2}^3 / 0.20 } \ \\ a_{1}=a_{2} \cdot \ \sqrt[3]{ 1 / 0.20 }=3.1623 \cdot \ \sqrt[3]{ 1 / 0.20 } \doteq 5.4074 \ \text{cm} \ \\ \ \\ S_{1}=a_{1}^2=5.4074^2 \doteq 29.2402 \doteq 29.24 \ \text{cm}^2



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