Cho hình chóp S.ABC có SA vuông góc với (ABC), tam...

A

- Hướng dẫn giải

Ta có \({V_{S.ABC}} = \frac{1}{3}{S_{ABC}}.SA = \frac{{5{a^3}}}{6}.\)

\(S{B^2} = S{C^2} = S{A^2} + A{B^2} = 26{a^2}\)

Khi đó: \(\frac{{{V_{S.ADE}}}}{{{V_{S.ABC}}}} = \frac{{SD}}{{SB}}.\frac{{SE}}{{SC}} = \frac{{SD.SB}}{{S{B^2}}}.\frac{{SE.SC}}{{S{C^2}}} = \frac{{S{A^2}}}{{S{B^2}}}.\frac{{S{A^2}}}{{S{C^2}}} = \frac{{S{A^4}}}{{S{B^4}}} = \frac{{625}}{{676}}.\)

\( \Rightarrow {V_{S.ADE}} = \frac{{625}}{{676}}{V_{S.ABC}} = \frac{{625}}{{676}}.\frac{{5{a^3}}}{6} = \frac{{3125{a^3}}}{{4056}}\)

\( \Rightarrow {V_{A.BCED}} = {V_{S.ABC}} - {V_{S.ADE}} = \frac{{5{a^3}}}{6} - \frac{{3125{a^3}}}{{4056}} = \frac{{85{a^3}}}{{1352}}.\)