Quadrilateral 81385

A regular quadrilateral pyramid with base edge length a = 15cm and height v = 21cm is given. We draw two planes parallel to the base, dividing the height of the pyramid into three equal parts. Calculate the ratio of the volumes of the 3 bodies created.

Final Answer:

r = 1:7:19

Step-by-step explanation:

$$\begin{gathered} a_3=15\text{cm} \\ {v}_3=21\text{cm} \\ {V}_3=\frac{1}{3}\cdot a_3^2\cdot v_3=\frac{1}{3}\cdot 15^2\cdot 21=1575{\text{cm}}^3 \\ {v}_1=\frac{1}{3}\cdot v_3=\frac{1}{3}\cdot 21=\frac{1\cdot 21}{3}=\frac{21}{3}=7\text{cm} \\ {a}_1=\frac{v_1}{v_3}\cdot a_3=\frac{7}{21}\cdot 15=\frac{7\cdot 15}{21}=\frac{105}{21}=5\text{cm} \\ {V}_1=\frac{1}{3}\cdot a_1^2\cdot v_1=\frac{1}{3}\cdot 5^2\cdot 7=\frac{175}{3}=58\frac{1}{3}\doteq 58.3333{\text{cm}}^3 \\ {v}_2=\frac{2}{3}\cdot v_3=\frac{2}{3}\cdot 21=\frac{2\cdot 21}{3}=\frac{42}{3}=14\text{cm} \\ {a}_2=\frac{v_2}{v_3}\cdot a_3=\frac{14}{21}\cdot 15=\frac{14\cdot 15}{21}=\frac{210}{21}=10\text{cm} \\ {V}_2=\frac{1}{3}\cdot a_2^2\cdot v_2=\frac{1}{3}\cdot 10^2\cdot 14=\frac{1400}{3}=466\frac{2}{3}\doteq 466.6667{\text{cm}}^3 \\ A=V_1=58.3333=\frac{175}{3}=58\frac{1}{3}\doteq 58.3333{\text{cm}}^3 \\ B=V_2-V_1=466.6667-58.3333=\frac{1225}{3}=408\frac{1}{3}\doteq 408.3333{\text{cm}}^3 \\ C=V_3-V_2=1575-466.6667=\frac{3325}{3}=1108\frac{1}{3}\doteq 1108.3333{\text{cm}}^3 \\ {r}_1=A:B=58.3333:408.3333=\frac{1}{7}\doteq 0.1429=1:7 \\ {r}_2=B:C=408.3333:1108.3333=\frac{7}{19}\doteq 0.3684=7:19 \\ r=1:7:19 \end{gathered}$$

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