Solid Geometry

Help Questions

Geometry › Solid Geometry

Questions 1 - 10

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

The slant height of a cone is ; the diameter of its base is one-fifth its slant height. Give the surface area of the cone in terms of .

$\frac{11\pi L^{2}}{100}$

$\frac{6\pi L^{2}}{25}$

$\frac{ \pi L^{2}}{10}$

$\frac{ \pi L^{2}}{5}$

$\frac{ \pi L^{2}}{20}$

Explanation

The formula for the surface area of a cone with base of radius and slant height is

$A = \pi r^{2} + \pi r L$ .

The diameter of the base is $\frac{1}{5}L$ ; the radius is half this, so

$r = \frac{1}{2} \cdot \frac{1}{5}L= \frac{1}{10}L$

Substitute in the surface area formula:

$A = \pi \left ( \frac{1}{10}L \right ) ^{2} + \pi \left ( \frac{1}{10}L \right ) L$

$A = \frac{1}{100} \pi L^{2} + \frac{1}{10} \pi L^{2}$

$A = \frac{1}{100} \pi L^{2} + \frac{10}{100} \pi L^{2}$

$A = \frac{11}{100} \pi L^{2} = \frac{11\pi L^{2}}{100}$

What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

The volume of a tetrahedron is found with the formula:

$\small V= \frac{a^3}{6\sqrt{2}}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

Find the surface area of a regular tetrahedron with a side length of $\frac{1}{2}$ .

$\frac{\sqrt3}{4}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{8}$

$\frac{\sqrt3}{6}$

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3\left(\frac{1}{2}\right)^2=\frac{\sqrt3}{4}$

What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

The volume of a tetrahedron is found with the formula:

$\small V= \frac{a^3}{6\sqrt{2}}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

Find the surface area of a regular tetrahedron with a side length of $\frac{1}{2}$ .

$\frac{\sqrt3}{4}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{8}$

$\frac{\sqrt3}{6}$

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3\left(\frac{1}{2}\right)^2=\frac{\sqrt3}{4}$

Find the surface area of a regular tetrahedron with a side length of $\frac{1}{2}$ .

$\frac{\sqrt3}{4}$

$\frac{\sqrt3}{2}$

$\frac{\sqrt3}{8}$

$\frac{\sqrt3}{6}$

Explanation

Use the following formula to find the surface area of a regular tetrahedron.

$\text{Surface Area}=\sqrt3 (side^2)$

Now, substitute in the value of the side length into the equation.

$\text{Surface Area}=\sqrt3\left(\frac{1}{2}\right)^2=\frac{\sqrt3}{4}$

What is the volume of a regular tetrahedron with edges of ?

$\small \frac{36}{\sqrt{2}} cm^3$

$\small \frac{216}{\sqrt{2}} cm^3$

$\small \frac{36}{\sqrt{2}} cm^3$

$\small 6\sqrt{2} cm^3$

$\small \sqrt{2} cm^3$

Explanation

The volume of a tetrahedron is found with the formula:

$\small V= \frac{a^3}{6\sqrt{2}}$ ,

where $\small a$ is the length of the edges.

When $\small a=6$ ,

$\small V=\frac{36}{\sqrt{2}}$ .

A regular tetrahedron has a surface area of $684cm^{2}$ . Each face of the tetrahedron has a height of . What is the length of the base of one of the faces?

Explanation

A regular tetrahedron has 4 triangular faces. The area of one of these faces is given by:

$\small A=\frac{1}{2}bh$

Because the surface area is the area of all 4 faces combined, in order to find the area for one of the faces only, we must divide the surface area by 4. We know that the surface area is $\small 684cm^2$ , therefore: