GED Math : Faces and Surface Area

Example Questions

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Example Question #1 : Faces And Surface Area

A circular swimming pool at an apartment complex has diameter 18 meters and depth 2.5 meters throughout.

The apartment manager needs to get the interior of the swimming pool painted. The paint she wants to use covers 40 square meters per can. How many cans of paint will she need to purchase?

You may use 3.14 for .

Explanation:

The pool can be seen as a cylinder with depth (or height) 2.5 meters and a base with diameter 18 meters - and radius half this, or 9 meters.

The bottom of the pool - the base of the cylinder - is a circle with radius 9 meters, so its area is

square meters.

Its side - the lateral face of the cylinder - has area

square meters.

Their sum - the total area to be painted - is  square feet. Since one can of paint covers 40 square meters, divide:

Nine cans of paint and part of a tenth will be required, so the correct response is ten.

Example Question #2 : Faces And Surface Area

A water tank takes the shape of a sphere whose exterior has radius 24 feet; the tank is six inches thick throughout. To the nearest hundred, give the surface area of the interior of the tank in square feet.

Use 3.14 for .

Explanation:

Six inches is equal to 0.5 feet, so the radius of the interior of the tank is

feet.

The surface area of the interior of the tank can be calculated using the formula

,

which rounds to 6,900 square feet.

Example Question #3 : Faces And Surface Area

Give the total surface area of the above cone to the nearest square meter.

Explanation:

The base is a circle with radius , and its area can be calculated using the area formula for a circle:

square meters.

To find the lateral area, we need the slant height of the cone. This can be found by way of the Pythagorean Theorem. Treating the height  and the radius  as the legs and slant height  as the hypotenuse, calculate:

meters.

The formula for the lateral area can be applied now:

Add the base and the lateral area to obtain the total surface area:

.

This rounds to 186 square meters.

Example Question #4 : Faces And Surface Area

A water tank takes the shape of a closed cylinder whose exterior has a height of 40 feet and a base with radius 15 feet; the tank is three inches thick throughout. To the nearest hundred, give the surface area of the interior of the tank in square feet.

Use 3.14 for .

Explanation:

Three inches is equal to 0.25 feet, so the height of the interior of the tank is

feet.

The radius of the interior of the tank is

feet.

The surface area of the interior of the tank can be determined by using this formula:

,

which rounds to 5,000 square feet.

Example Question #5 : Faces And Surface Area

Above is a diagram of a conic tank that holds a city's water supply.

The city wishes to completely repaint the exterior of the tank - sides and base. The paint it wants to use covers 40 square meters per gallon. Also, to save money, the city buys the paint in multiples of 25 gallons.

How many gallons will the city purchase in order to paint the tower?

Explanation:

The surface area of a cone with radius  and slant height  is calculated using the formula .

Substitute 35 for  and 100 for  to find the surface area in square meters:

square meters.

The paint covers 40 square meters per gallon, so the city needs

gallons of paint.

Since the city buys the paint in multiples of 25 gallons, it will need to buy the next-highest multiple of 25, or 375 gallons.

Example Question #6 : Faces And Surface Area

A regular octahedron has eight congruent faces, each of which is an equilateral triangle.

A given octahedron has edges of length three inches. Give the total surface area of the octahedron.

Explanation:

The area of an equilateral triangle is given by the formula

.

Since there are eight equilateral triangles that comprise the surface of the octahedron, the total surface area is

.

Substitute :

square inches.

Example Question #7 : Faces And Surface Area

A regular icosahedron has twenty congruent faces, each of which is an equilateral triangle.

A given regular icosahedron has edges of length two inches. Give the total surface area of the icosahedron.

Explanation:

The area of an equilateral triangle is given by the formula

.

Since there are twenty equilateral triangles that comprise the surface of the icosahedron, the total surface area is

.

Substitute :

square inches.

Example Question #8 : Faces And Surface Area

A regular tetrahedron has four congruent faces, each of which is an equilateral triangle.

A given tetrahedron has edges of length five inches. Give the total surface area of the tetrahedron.

Explanation:

The area of an equilateral triangle is given by the formula

.

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is

.

Substitute :

square inches.

Example Question #9 : Faces And Surface Area

A cube has a height of 9cm.  Find the surface area.

Explanation:

To find the surface area of a cube, we will use the following formula:

where l is the length, and w is the width of the cube.

Now, we know the height of the cube is 9cm. Because it is a cube, all lengths, widths, and heights are the same. Therefore, the length and the width are also 9cm.

Knowing this, we can substitute into the formula. We get

Example Question #10 : Faces And Surface Area

A sphere has a radius of 7in.  Find the surface area.

Explanation:

To find the surface area of a sphere, we will use the following formula:

where r is the radius of the sphere.

Now, we know the radius of the sphere is 7in.

So, we can substitute into the formula. We get

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