Question

the base of a solid is the region enclosed by a circle having a radius of 7 cm. find the volume of the solid if all plane sections perpendicular to a fixed diameter of the base are equilateral triangles

Answer 1

Answer:

$V=621.8\ cm^3$

Step-by-step explanation:

we know that

If the base of the solid is a circle and all plane sections perpendicular to a fixed diameter of the base are equilateral triangles

then

The solid is a cone

The volume of a cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=7\ cm$

Find out the height of the cone

Remember that

If all plane sections perpendicular to a fixed diameter of the base are equilateral triangles

then

The length side of each equilateral triangle is equal to the diameter of the base

Applying the Pythagoras Theorem find out the height of the cone

$D^{2}=r^{2} +h^{2}$

we have

$D=(2)7=14\ cm$ ----> the diameter is two times the radius

$r=7\ cm$

substitute and solve for h

$14^{2}=7^{2} +h^{2}$

$h^{2}=14^{2}-7^{2}$

$h^{2}=147$

$h=\sqrt{147}\ cm$

Find the volume

$V=\frac{1}{3}\pi (7)^{2}\sqrt{147}$

$V=198.03\pi\ cm^3$

assume

$\pi =3.14$

$V=198.03(3.14)=621.8\ cm^3$