Cho hình chóp đều S.ABCD có tất...

C

- Hướng dẫn giải

Đáp án C

Vì $\left(P\right)//BD$nên $\left(P\right)\cap \left(SBD\right)=IE//BD$

Ta có $BD\perp SO,BD\perp AC\Rightarrow BD\perp \left(SAC\right)\Rightarrow BD\perp AM$

$\Rightarrow IE\perp AM$

Ta có $\frac{SM}{MC}.\frac{AC}{AO}.\frac{OK}{KS}=1\iff \mathrm{2.2.}\frac{OK}{KS}=1\iff \frac{OK}{KS}=\frac{1}{4}$

$\frac{SM}{MC}.\frac{IE}{BD}=\frac{SK}{SO}=\frac{4}{5}\Rightarrow IE=\frac{4}{5}BD=\frac{4a\sqrt{2}}{5}$

Vì $A{C}^2=S{A}^2+S{C}^2$ nên $\Delta SAC$ vuông tại S

$SM=\frac{2}{3}a;AM=\sqrt{a^2+{\left(\frac{2}{3}a\right)}^2}=\frac{a\sqrt{13}}{3}$

$S_{AEMI}=\frac{1}{2}EI.AM=\frac{1}{2}.\frac{4a\sqrt{2}}{5}.\frac{a\sqrt{13}}{3}=\frac{2\sqrt{26}{a}^2}{15}$