The radii of the circular ends of a frustum

Solution:

Radii of circular ends of frustum

$r_{1}=14 \mathrm{~cm}, r_{2}=6 \mathrm{~cm}$

Slant height

$I=10 \mathrm{~cm}, h=?$

$l=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}$

$l^{2}=h^{2}+\left(r_{1}-r_{2}\right)^{2}$

{squaring on both sides}

$h^{2}=l^{2}-\left(r_{1}-r_{2}\right)^{2}$

$h^{2}=(10)^{2}-(14-6)^{2}$

 

$h^{2}=100-64$

$h^{2}=36$

$h=6 \mathrm{~cm}$

 

Hence, the correct answer is choice (a).