The-base-of-a-pyramid-is-an-equilateral-triangle-of-side-length-6-cm-The-other-edges-of-the-pyramid-are-each-of-length-15-cm-Find-the-volume-of-the-pyramid-

Question Number 17169 by Tinkutara last updated on 01/Jul/17

$$\mathrm{The}\:\mathrm{base}\:\mathrm{of}\:\mathrm{a}\:\mathrm{pyramid}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{of}\:\mathrm{side}\:\mathrm{length}\:\mathrm{6}\:\mathrm{cm}.\:\mathrm{The}\:\mathrm{other} \\ $$$$\mathrm{edges}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pyramid}\:\mathrm{are}\:\mathrm{each}\:\mathrm{of}\:\mathrm{length} \\ $$$$\sqrt{\mathrm{15}}\:\mathrm{cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pyramid}. \\ $$

Answered by ajfour last updated on 02/Jul/17

Commented by ajfour last updated on 02/Jul/17

V=∫dV=∫ ((√3)/4)x^2 dy cot θ= (y/((x/(√3))))=((√3)/(2(√3)))=(1/2) ⇒ y=(x/(2(√3))) and dy=(dx/(2(√3))) so V=∫_0 ^( 6) (((√3)/4)x^2 )((dx/(2(√3)))) =(1/8)∫_0 ^( 6) x^2 dx =(1/8)((x^3 /3))∣_0 ^6 =9 cm^3 .

$$\mathrm{V}=\int\mathrm{dV}=\int\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \mathrm{dy} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{cot}\:\theta=\:\frac{\mathrm{y}}{\left(\mathrm{x}/\sqrt{\mathrm{3}}\right)}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{y}=\frac{\mathrm{x}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\mathrm{and}\:\:\mathrm{dy}=\frac{\mathrm{dx}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{so}\:\:\:\:\mathrm{V}=\int_{\mathrm{0}} ^{\:\:\mathrm{6}} \:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{x}^{\mathrm{2}} \right)\left(\frac{\mathrm{dx}}{\mathrm{2}\sqrt{\mathrm{3}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\:\mathrm{6}} \mathrm{x}^{\mathrm{2}} \mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}}\right)\mid_{\mathrm{0}} ^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{9}\:\mathrm{cm}^{\mathrm{3}} . \\ $$

Commented by Tinkutara last updated on 02/Jul/17

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Answered by mrW1 last updated on 02/Jul/17

Area of base = (1/2)×6×((6(√3))/2)=9(√3) Height of pyramid =(√(((√(15)))^2 −((2/3)×((6(√3))/2))^2 ))=(√(15−12))=(√3) V=(1/3)×9(√3)×(√3)=9 cm^3

$$\mathrm{Area}\:\mathrm{of}\:\mathrm{base}\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{6}×\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{9}\sqrt{\mathrm{3}} \\ $$$$\mathrm{Height}\:\mathrm{of}\:\mathrm{pyramid}\:=\sqrt{\left(\sqrt{\mathrm{15}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{15}−\mathrm{12}}=\sqrt{\mathrm{3}} \\ $$$$\mathrm{V}=\frac{\mathrm{1}}{\mathrm{3}}×\mathrm{9}\sqrt{\mathrm{3}}×\sqrt{\mathrm{3}}=\mathrm{9}\:\mathrm{cm}^{\mathrm{3}} \\ $$

Commented by Tinkutara last updated on 09/Jul/17

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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