a, Diện tích xung quanh :
\(S_{xp}=2.0,1.\left(0,9+0,6\right)\\ =0,3\left(m^2\right)\)
Thể tích :
\(V=0,9.0,6.0,1=0,054\left(m^3\right)\)
b, Đổi \(\dfrac{4}{5}m=8dm\)
Diện tích xung quanh :
\(S_{xp}=2.\dfrac{3}{4}.\left(8+\dfrac{2}{3}\right)=13\left(dm^2\right)\)
Thể tích :
\(8.\dfrac{2}{3}.\dfrac{3}{4}=4\left(dm^3\right)\)
a: \(S_{XQ}=\left(0.9+0.6\right)\cdot2\cdot0.1=1.5\cdot0.2=0.3\left(m^2\right)\)
\(V=0.9\cdot0.6\cdot0.1=0.054\left(m^3\right)\)
b: \(S_{Xq}=\left(\dfrac{4}{5}\cdot10+\dfrac{2}{3}\right)\cdot2\cdot\dfrac{3}{4}=13\left(dm^2\right)\)
\(V=\dfrac{4}{5}\cdot10\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}=8\cdot\dfrac{1}{2}=4\left(dm^3\right)\)