GMAT Math : Geometry

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Example Question #3 : Calculating The Surface Area Of A Tetrahedron

Tetra_1

The above diagram shows a regular right triangular pyramid. Its base $\bigtriangleup ABC$ is an equilateral triangle; the other three faces are congruent isosceles triangles, with $\overline{VM}$ an altitude of $\bigtriangleup AVC$ . Give the surface area of the pyramid.

Possible Answers:

$36 +216 \sqrt{2}$

$324+ 36 \sqrt{3}$

$252 \sqrt{3}$

$252 \sqrt{2}$

$36 \sqrt{3}+216 \sqrt{2}$

Correct answer:

$36 \sqrt{3}+216 \sqrt{2}$

Explanation:

The base is an equilateral triangle with sidelength 12, so its area can be calculated as follows:

$\frac{12^{2}\cdot \sqrt{3}}{4} = \frac{144 \sqrt{3}}{4} = 36 \sqrt{3}$ .

Each of the three other faces is congruent, with base 12. The area of each is the product of its base and its height. To find the common height, we examine $\bigtriangleup VMC$ , which, since $\overline{VM}$ is an altitude of isosceles $\bigtriangleup AVC$ , is a right triangle with hypotenuse of length 18 and one leg of length $MC = \frac{1}{2} AC = \frac{1}{2} \cdot 12 = 6$ . We can find using the Pythagorean Theorem:

$VM = \sqrt{\left (VC \right )^{2} - \left ( CM\right )^{2}}$

$= \sqrt{18^{2} - 6^{2}}$

$= \sqrt{288}$

$= \sqrt{144} \cdot \sqrt{2}$

$=12\sqrt{2}$

The area of $\bigtriangleup AVC$ is half the product of this height and the base:

$\frac{1}{2} \cdot 12 \cdot 12 \sqrt{2} = 72 \sqrt{2}$

All three lateral faces have this area.

Now add the areas of the four faces:

$36 \sqrt{3}+72 \sqrt{2}+72 \sqrt{2}+72 \sqrt{2} = 36 \sqrt{3}+216 \sqrt{2}$

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Example Question #2 : Calculating The Surface Area Of A Tetrahedron

Tetra_1

The cube in the above figure has surface area 384. Give the surface area of the tetrahedron with vertices D,E,G,H , shown in red.

Possible Answers:

128

$96 + 32 \sqrt{3}$

$128 \sqrt{3}$

$128+ 32 \sqrt{3}$

$96 + 64\sqrt{3}$

Correct answer:

$96 + 32 \sqrt{3}$

Explanation:

The surface area formula can be used to find the length of each edge of the cube:

$6s^{2}=384$

$s^{2}= 64$

s= 8

Three faces of the tetrahedron - $\bigtriangleup DHE$ , $\bigtriangleup DHG$ , $\bigtriangleup EHG$ - are right triangles with legs of length 8, so the area of each is half the product of the lengths of their legs:

$\frac{1}{2} \cdot 8 \cdot 8 =32$ .

Each triangle is isosceles, so, by the 45-45-90 Theorem, each of their hypotentuses measures $\sqrt{2}$ times a leg, or $8\sqrt{2}$ . is therefore an equilateral triangle with sidelenghth $8\sqrt{2}$ . Its area can be found as follows:

$A = \frac{s^{2}\sqrt{3}}{4}$

$= \frac{(8\sqrt{2})^{2}\sqrt{3}}{4}$

$= \frac{8^{2}\cdot (\sqrt{2})^{2} \cdot \sqrt{3}}{4}$

$= \frac{64\cdot 2 \cdot \sqrt{3}}{4}$

$= 32 \sqrt{3}$

The total surface area is

$32+32+32+ 32 \sqrt{3} = 96 + 32 \sqrt{3}$

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Example Question #501 : Geometry

What is the surface area of a cylinder with a radius of 7 and a height of 3?

Possible Answers:

$\dpi{100} \small 140\pi$

$\dpi{100} \small 120\pi$

$\dpi{100} \small 80\pi$

$\dpi{100} \small 98\pi$

$\dpi{100} \small 42\pi$

Correct answer:

$\dpi{100} \small 140\pi$

Explanation:

All we really need here is to remember the formula for the surface area of a cylinder.

$\dpi{100} \small SA=2\pi r^{2}+2\pi rh=2\pi \left ( 49 \right )+2\pi \left ( 7 \right )\left ( 3 \right )=98\pi +42\pi=140\pi$

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Example Question #502 : Geometry

The height of a cylinder is twice the circumference of its base. The radius of the base is 9 inches. What is the surface area of the cylinder?

Possible Answers:

$810 \pi^{2} \textrm{ in}^{2}$

$810 \pi \textrm{ in}^{2}$

$1,620 \pi^{2} \textrm{ in}^{2}$

$\left ( 162 \pi + 648 \pi^{2} \right )\textrm{ in}^{2}$

$\left ( 648 \pi +162 \pi^{2} \right )\textrm{ in}^{2}$

Correct answer:

$\left ( 162 \pi + 648 \pi^{2} \right )\textrm{ in}^{2}$

Explanation:

The radius of the base is 9 inches, so its circumference is $2 \pi$ times this, or $2 \pi \cdot 9 = 18 \pi$ inches. The height is twice this, or $36 \pi$ inches.

Substitute $r = 9 , h = 36 \pi$ in the formula for the surface area of the cylinder:

$A = 2 \pi r (r+h)$

$A = 2 \pi \cdot 9 \cdot (9+36\pi ) = 162 \pi + 648 \pi^{2}$ square inches

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Example Question #3 : Calculating The Surface Area Of A Cylinder

Calculate the surface area of the following cylinder.

(Not drawn to scale.)

Possible Answers:

$56\pi$

$32\pi$

$154\pi$

$88\pi$

Correct answer:

$88\pi$

Explanation:

The equation for the surface area of a cylinder is:

$SA=2\pi rh+2\pi r^2$

we plug in our values: r=4, h=7 to find the surface area

$SA=2\pi (4)(7)+2\pi (4)^2$

$SA=2 \pi (28)+32\pi$

$SA=88\pi$

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Example Question #1 : Cylinders

Calculate the surface area of the following cylinder.

(Not drawn to scale.)

Possible Answers:

$288\pi$

$360\pi$

$130\pi$

$208\pi$

Correct answer:

$130\pi$

Explanation:

The equation for the surface area of a cylinder is

$SA=2\pi rh+2\pi r^2$

We plug in our values r=5, h=8 into the equation to find our answer.

Note: we were given the diameter of the cylinder (10), in order to find the radius we had to divide the diameter by two.

$SA=2\pi (5)(8)+2\pi (5)^2$

$SA=2\pi (40)+2\pi (25)$

$SA=130\pi$

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Example Question #5 : Calculating The Surface Area Of A Cylinder

A cylinder has a height of 9 and a radius of 4. What is the total surface area of the cylinder?

Possible Answers:

$52\pi$

$104\pi$

$132\pi$

$68\pi$

$88\pi$

Correct answer:

$104\pi$

Explanation:

We are given the height and the radius of the cylinder, which is all we need to calculate its surface area. The total surface area will be the area of the two circles on the bottom and top of the cylinder, added to the surface area of the shaft. If we imagine unfolding the shaft of the cylinder, we can see we will have a rectangle whose height is the same as that of the cylinder and whose width is the circumference of the cylinder. This means our formula for the total surface area of the cylinder will be the following:

$SA=2(\pi r^2)+2\pi rh$

$SA=2\pi (4)^2+2\pi (4)(9)$

$SA=32\pi+72\pi$

$SA=104\pi$

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Example Question #506 : Geometry

Grant is making a canister out of sheet metal. The canister will be a right cylinder with a height of 150 mm. The base of the cylinder will have a radius of $\frac{5}{\pi}$ mm. If the canister will have an open top, how many square millimeters of metal does Grant need?

Possible Answers:

$1525\text{ } mm^2$

$375+\frac{25}{\pi}\text{ } mm^2$

$1500+\frac{25}{\pi} \text{ } mm^2$

$30+\frac{25}{\pi}\text{ } mm^2$

$1500+\frac{25}{\pi}\text{ } mm^2$

Correct answer:

$1500+\frac{25}{\pi} \text{ } mm^2$

Explanation:

This question is looking for the surface area of a cylinder with only 1 base. Our surface area of a cylinder is given by:

$SA=2\pi rh+2\pi r^2$

However, because we only need 1 base, we can change it to:

$SA=2\pi rh+\pi r^2$

We know our radius and height, so simply plug them in and simplify.

$SA=2\pi \left( \frac{5}{\pi}\right)150+\pi\left(\frac{5}{\pi}\right)^2=1500+\frac{25}{\pi}$

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Example Question #4 : Cylinders

Find the surface area of a cylinder whose height is and radius is .

Possible Answers:

$18\pi$

$27\pi$

$45\pi$

$54\pi$

Correct answer:

$54\pi$

Explanation:

To find the surface area of a cylinder, you must use the following equation.

$SA=2\pi{r}h+2\pi{r^2}$

Thus,

$SA=2\pi(3)(6)+2\pi(3^2)=54\pi$

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Example Question #8 : Calculating The Surface Area Of A Cylinder

A right circular cylinder has bases of radius $\pi ^{x}$ ; its height is $\pi ^{y}$ . Give its surface area.

Possible Answers:

$2 \pi^{ x+y+1}$

$\pi^{ x+2y+1}$

$\pi ^{ 2x+1 } + \pi ^{x+y+1}$

$\pi ^{ 2y+1 } + \pi ^{x+y+1}$

$\pi^{2 x+y+1}$

Correct answer:

$\pi ^{ 2x+1 } + \pi ^{x+y+1}$

Explanation:

The surface area of a cylinder can be calculated from its radius and height as follows:

$A = \pi r ^{2} + \pi rh$

Setting $r = \pi^{x}$ and $h = \pi ^{x}$ :

$A = \pi \cdot ( \pi^{x}) ^{2} + \pi \cdot \pi^{x} \cdot \pi^{y}$

$A = \pi ^{1 } \cdot \pi^{2x} + \pi ^{1 } \cdot \pi^{x} \cdot \pi^{y}$

$A = \pi ^{1 +2x} + \pi ^{1+x+y}$ or $A = \pi ^{ 2x+1 } + \pi ^{x+y+1}$

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GMAT Math : Geometry

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #3 : Calculating The Surface Area Of A Tetrahedron

Example Question #2 : Calculating The Surface Area Of A Tetrahedron

Example Question #501 : Geometry

Example Question #502 : Geometry

Example Question #3 : Calculating The Surface Area Of A Cylinder

Example Question #1 : Cylinders

Example Question #5 : Calculating The Surface Area Of A Cylinder

Example Question #506 : Geometry

Example Question #4 : Cylinders

Example Question #8 : Calculating The Surface Area Of A Cylinder

All GMAT Math Resources

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