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Advanced Geometry : Solid Geometry

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Example Questions

Example Question #31 : Solid Geometry

In the figure below, a cone is placed on top of a cylinder so that they share the same base. Find the surface area of the figure.

Possible Answers:

2395.96

1909.34

2509.88

2261.95

Correct answer:

2261.95

Explanation:

First, find the lateral surface area of the cone.

$\text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})$

Plug in the given slant height and radius.

$\text{Lateral Area}=(18)(12)\pi=216\pi$

Next, find the surface area of the cylinder:

$\text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2$

The surface area of the cylinder is the sum of the lateral surface area of the cylinder and its two bases. However, since one base of the cylinder is covered up by the cone, we will need to subtract that area out of the total surface area of the cylinder.

$\text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2$

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

$\text{Surface Area of Cylindrical Portion}=2\pi(12)(15)+\pi(12)^2=504\pi$

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

$\text{Surface Area of Figure}=216\pi+504\pi=720\pi=2261.95$

Make sure to round to places after the decimal.

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Example Question #32 : Solid Geometry

In the figure below, a cone is placed on top of a cylinder so that they share the same base. Find the surface area of the figure.

Possible Answers:

1851.25

1669.02

1790.71

2010.77

Correct answer:

1790.71

Explanation:

First, find the lateral surface area of the cone.

$\text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})$

Plug in the given slant height and radius.

$\text{Lateral Area}=(17)(10)\pi=170\pi$

Next, find the surface area of the cylinder:

$\text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2$

$\text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2$

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

$\text{Surface Area of Cylindrical Portion}=2\pi(10)(15)+(10)^2\pi=400\pi$

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

$\text{Surface Area of Figure}=170\pi+400\pi=570\pi=1790.71$

Make sure to round to places after the decimal.

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Example Question #33 : Solid Geometry

In the figure below, a cone is placed on top of a cylinder so that they share the same base. Find the surface area of the figure.

Possible Answers:

7277.90

7409.57

8917.13

6918.71

Correct answer:

7277.90

Explanation:

First, find the lateral surface area of the cone.

$\text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})$

Plug in the given slant height and radius.

$\text{Lateral Area}=(25\sqrt3)(22)\pi=(550\sqrt3)\pi$

Next, find the surface area of the cylinder:

$\text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2$

$\text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2$

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

$\text{Surface Area of Cylindrical Portion}=2\pi(22)(20)+\pi(22)^2=1364\pi$

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

$\text{Surface Area of Figure}=(550\sqrt3)\pi+1364\pi=7277.90$

Make sure to round to places after the decimal.

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Example Question #31 : Solid Geometry

In the figure below, a cone is placed on top of a cylinder so that they share the same base. Find the surface area of the figure.

Possible Answers:

2035.75

2077.08

2166.39

2635.29

Correct answer:

2035.75

Explanation:

First, find the lateral surface area of the cone.

$\text{Lateral Surface Area}=(\text{slant height})(\pi)(\text{radius})$

Plug in the given slant height and radius.

$\text{Lateral Area}=(16)(12)\pi=192\pi$

Next, find the surface area of the cylinder:

$\text{Surface Area of Cylinder}=2\pi(\text{radius})(\text{height})+2\pi(\text{radius})^2$

$\text{Surface Area of Cylindrical Portion}=2\pi(\text{radius})(\text{height})+\pi(\text{radius})^2$

Plug in the given height and radius to find the surface area of the cylindrical portion of the figure:

$\text{Surface Area of Cylindrical Portion}=2\pi(12)(13)+\pi(12)^2=456\pi$

To find the surface area of the figure, add together the lateral area of the cone with the surface area of the cylindrical portion of the figure.

$\text{Surface Area of Figure}=192\pi+456\pi=648\pi=2035.75$

Make sure to round to places after the decimal.

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Example Question #31 : How To Find The Surface Area Of A Cone

Cone

Figure NOT drawn to scale.

The base of the above cone has circumference 50. Give the surface area of the cone.

Round to the nearest whole number.

Possible Answers:

306

301

712

737

699

Correct answer:

699

Explanation:

The surface area of a cone can be calculated using the formula

$S = \pi r (r+l)$

where is the radius of the base and is the slant height. The slant height is known to be 20.

The radius of the circular base can be found by dividing its circumference 50 by $2 \pi$ :

$r = \frac{50}{2 \pi} \approx \frac{50}{2 \cdot 3.1416} \approx 7.958$

Set r = 7.958 and l = 20 , and evaluate:

$S = \pi r (r+l)$

$S \approx 3.1416 \cdot 7.958 \cdot (7.958 + 20)$

$\approx 3.1416 \cdot 7.958 \cdot 27.958$

$\approx 699$

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Example Question #36 : Solid Geometry

Cone

Figure NOT drawn to scale.

The base of the above cone has area 50. Give the surface area of the cone.

Round to the nearest whole number.

Possible Answers:

737

699

306

301

712

Correct answer:

301

Explanation:

The area and the radius of the circular base are related by the formula

$\pi r^{2}= A$

The area of the circle is 50, so set A = 50 , then solve for :

$\pi r^{2}= 50$

Divide by $\pi$ :

$\frac{\pi r^{2}}{\pi}= \frac{50}{\pi}$

$r^{2} \approx \frac{50}{3.1416}$

$r^{2} \approx 15.9155$

Take the square root of both sides:

$r \approx \sqrt{15.9155}$

$r \approx 3.9894$

The surface area of the cone is the sum of the areas of the base and the lateral area. The lateral area of the cone can be found from the radius and the slant height using the formula

$L = \pi r l$

Set $r \approx 3.9894$ and l = 20 :

$L \approx 3.1416 \cdot 3.9894 \cdot 20$

$L \approx 251$

Add the area of the base, 50:

$S \approx 251 + 50 = 301$

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Example Question #37 : Solid Geometry

Cone

Figure NOT drawn to scale.

The base of the above cone has area 50. Give the surface area of the cone.

Round to the nearest whole number.

Possible Answers:

699

737

301

712

306

Correct answer:

306

Explanation:

The area and the radius of the circular base are related by the formula

$\pi r^{2}= A$

The area of the circle is 50, so set A = 50 , then solve for :

$\pi r^{2}= 50$

Divide by $\pi$ :

$\frac{\pi r^{2}}{\pi}= \frac{50}{\pi}$

$r^{2} \approx \frac{50}{3.1416}$

$r^{2} \approx 15.9155$

Take the square root of both sides:

$r \approx \sqrt{15.9155}$

$r \approx 3.9894$

Now examine the figure below.

Cone

The slant height can be calculated from the height and the radius by way of the Pythagorean Theorem:

$l = \sqrt{h^{2}+r^{2}}$

$l = \sqrt{20^{2}+ 3.9894^{2}}$

$\approx \sqrt{400 + 15.9153 }$

$\approx \sqrt{4 15.9153 }$

$\approx 20.3940$

The surface area of the cone is the sum of the areas of the base and the lateral area. The lateral area of the cone can be found from the radius and the slant height using the formula

$L = \pi r l$

Set $r \approx 3.9894$ and l = 20.3940 :

$L \approx 3.1416 \cdot 3.9894 \cdot 20.3940$

$L \approx 256$

Add the area of the base, 50, to this lateral area to obtain the surface area:

$S \approx 256 + 50 = 306$ .

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Example Question #38 : Solid Geometry

Cone

Figure NOT drawn to scale.

The base of the above cone has circumference 50. Give the surface area of the cone.

Round to the nearest whole number.

Possible Answers:

301

712

699

306

737

Correct answer:

737

Explanation:

The surface area of a cone can be calculated using the formula

$S = \pi r (r+l)$

where is the radius of the base and is the slant height.

The radius of the circular base can be found by dividing its circumference 50 by $2 \pi$ :

$r = \frac{50}{2 \pi} \approx \frac{50}{2 \cdot 3.1416} \approx 7.958$

Examine the figure below.

Cone

The slant height can be calculated from the height and the radius by way of the Pythagorean Theorem:

$l = \sqrt{h^{2}+r^{2}}$

$l = \sqrt{20^{2}+ 7.958^{2}}$

$\approx \sqrt{400 + 63.3298}$

$\approx \sqrt{4 63.3298}$

$\approx 21.5251$

Set $r \approx 7.958$ and $l \approx 21.5251$ in the surface area formula:

$S = \pi r (r+l)$

$S \approx 3.1416 \cdot 7.958 \cdot ( 7.958+21.5251 )$

$\approx 3.1416 \cdot 7.958 \cdot 29.4831$

$\approx 737$

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Example Question #1 : How To Find The Length Of An Edge

What is the length of one edge of a regular tetrahedron whose volume equals $\small \small \small \frac{4}{3\sqrt{2}} cm^3$ ?

Possible Answers:

None of the above.

$\small \small \small \frac{2}{\sqrt{2}} cm$

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

$\small \small \frac{\sqrt{2}}2{} cm$

$\small \small 2 cm$

Correct answer:

$\small \small 2 cm$

Explanation:

The formula for the volume of a tetrahedron is:

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

When $\small V= \frac{4}{3\sqrt{2}}$ we have $\small \frac{4}{3\sqrt{2}}=\frac{a^3}{6\sqrt{2}}$ .

Multiplying the left side by $\small \frac{2}{2}$ gives us,

$\small \frac{8}{6\sqrt{2}}=\frac{a^3}{6\sqrt{2}}$ , or $\small 8=a^3$ .

Finally taking the third root of both sides yields $\small 2=a$

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Example Question #1 : Tetrahedrons

A regular tetrahedron has a total surface area of $\small 16\sqrt{3} ft^2$ . What is the combined length of all of its edges?

Possible Answers:

None of the above.

$\small 4 ft$

$\small 20 ft$

$\small 12 ft$

$\small 24 ft$

Correct answer:

$\small 24 ft$

Explanation:

A regular tetrahedron has four faces of equal area made of equilateral triangles.

Therefore, we know that one face will be equal to:

$\small \left(\frac{1}{4}\right)\left(16\sqrt{3}\right) ft^2$ , or $\small 4\sqrt{3} ft^2$

Since the surface of one face is an equilateral triangle, and we know that,

$\small Area=\left(\frac{1}{2}\right)b\times h$ , the problem can be expressed as:

$\small 4\sqrt{3}=\left(\frac{1}{2}\right)b\times h$

In an equilateral triangle, the height $\small h$ , is equal to $\small \frac{1}{2}(b)\sqrt{3}$ so we can substitute for $\small h$ like so:

$\small 4\sqrt{3}=\left(\frac{1}{2}\right)b \times \left(\frac{1}{2}\right)b\sqrt{3}$

Solving for $\small b$ gives us the length of one edge.

$\small 4\sqrt{3}=\frac{b^2\sqrt{3}}{4}$

$\small (4)(4\sqrt{3})=(4)\left(\frac{b^2\sqrt{3}}{4}\right)$

$\small 16\sqrt{3}=b^2\sqrt{3}$

$\small 16=b^2$

$\small b=\pm 4$

However, we know that the edge of the tetrahedron is a positive number so $\small b=4$ .

Since the base $\small b$ is the same as one edge of the tetrahedron, and a tetrahedron has six edges we multiply $\small 6 \times 4$ to arrive at $\small 24 ft^2$

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Advanced Geometry : Solid Geometry

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

Example Question #31 : Solid Geometry

Example Question #32 : Solid Geometry

Example Question #33 : Solid Geometry

Example Question #31 : Solid Geometry

Example Question #31 : How To Find The Surface Area Of A Cone

Example Question #36 : Solid Geometry

Example Question #37 : Solid Geometry

Example Question #38 : Solid Geometry

Example Question #1 : How To Find The Length Of An Edge

Example Question #1 : Tetrahedrons

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