Gọi V,V1,V2 lần lượt là thể tích của khối tròn

* Hướng dẫn giải

Đáp án C

Đặt $BC=a,AC=b,AB=c,AH=h$.

Ta có: $\left\{\begin{array}{l}V=\frac{1}{3}BC.\pi h^2=\frac{\pi a{h}^2}{3}\\ V_1=\frac{1}{3}CA.\pi A{B}^2=\frac{\pi b{c}^2}{3}\\ V_2=\frac{1}{3}BA.\pi A{C}^2=\frac{\pi c{b}^2}{3}\end{array}\right.\Rightarrow \left\{\begin{array}{l}\frac{1}{V^2}=\frac{9}{{\pi }^2{a}^2{h}^4}\\ \frac{1}{V_1^2}=\frac{9}{{\pi }^2{b}^2{c}^4}\\ \frac{1}{V_2^2}=\frac{9}{{\pi }^2{c}^2{b}^4}\end{array}\right.$.

Suy ra: $\frac{1}{V_1^2}+\frac{1}{V_2^2}=\frac{9}{{\pi }^2{b}^2{c}^4}+\frac{9}{{\pi }^2{c}^2{b}^4}=\frac{9}{{\pi }^2{b}^2{c}^2}\left(\frac{1}{c^2}+\frac{1}{b^2}\right)$

$=\frac{9}{{\pi }^2{a}^2{h}^2}.\frac{1}{h^2}=\frac{9}{{\pi }^2{a}^2{h}^4}=\frac{1}{V^2}$.

Vậy $\frac{1}{V^2}=\frac{1}{V_1^2}+\frac{1}{V_2^2}\Rightarrow V=\frac{V_1.V_2}{\sqrt{V_1^2+V_2^2}}=\frac{3.4}{\sqrt{3^2+4^2}}=\frac{12}{5}$.