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HiSET: Math : HiSet: High School Equivalency Test: Math

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Example Questions

Example Question #1 : Tangent

Right

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Possible Answers:

$\frac{32}{\tan 35^{\circ}}$

None of the other choices gives the correct response.

$32 \tan 35^{\circ}$

$64 \tan 35^{\circ}$

$\frac{64}{\tan 35^{\circ}}$

Correct answer:

$\frac{32}{\tan 35^{\circ}}$

Explanation:

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= \frac{1}{2}(AB)(AC)$

We are given that AB = 8 . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = \frac{AB}{AC}$

Setting AB = 8 and $m \angle C = 35 ^{\circ }$ , we get

$\tan 35^{\circ } = \frac{8}{AC}$

Solve for by first, finding the reciprocal of both sides:

$\frac{1}{\tan 35^{\circ }} = \frac{AC}{8}$

Now, multiply both sides by 8:

$\frac{1}{\tan 35^{\circ }} \cdot 8 = \frac{AC}{8} \cdot 8$

$\frac{8}{\tan 35^{\circ }} = AC$

Now, set AB = 8 and $AC = \frac{8}{\tan 35^{\circ }}$ in the area formula:

$a= \frac{1}{2}(AB)(AC)$

$= \frac{1}{2}(8)\left ( \frac{8}{\tan 35^{\circ}} \right )$

$= \frac{1}{2} \left (\frac{8}{1} \right )\left ( \frac{8}{\tan 35^{\circ}} \right )$

$= \frac{32}{\tan 35^{\circ}}$ ,

the correct choice.

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Example Question #1 : Problems Involving Right Triangle Trigonometry

$\cos N = \frac{x}{3}$

Evaluate $\tan N$ in terms of .

Possible Answers:

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{3}{x}$

$\frac{3}{\sqrt{x^{2}-9}}$

$\frac{\sqrt{x^{2}-9}}{x}$

$\frac{x}{\sqrt{x^{2}-9}}$

Correct answer:

$\frac{\sqrt{x^{2}-9}}{x}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring . The cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} = \frac{\sqrt{x^{2}-9}}{x}$ .

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Example Question #4 : Problems Involving Right Triangle Trigonometry

$\sin a^{\circ } = \frac{4}{7}$

$\tan a^{\circ } = ?$

Possible Answers:

$\frac{\sqrt{33}}{7}$

$\frac{4}{\sqrt{33}}$

$\frac{7}{4}$

$\frac{7}{\sqrt{33}}$

$\frac{\sqrt{33}}{4}$

Correct answer:

$\frac{4}{\sqrt{33}}$

Explanation:

The sine of an angle is defined to be the ratio of the length of the opposite leg of a right triangle to the length of its hypotenuse. Therefore, we can set O = 4, H = 7 . By the Pythagorean Theorem:
the adjacent leg of the triangle has measure

$A = \sqrt{H^{2}-O^{2}} = \sqrt{7^{2}-4^{2}} = \sqrt{49-16} = \sqrt{33}$

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, which is

$\tan a ^{\circ }= \frac{O}{A} = \frac{4}{\sqrt{33}}$ ,

the correct response.

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Example Question #101 : Measurement And Geometry

What is the volume of a cylinder with a diameter of cm and a height of 15.5 cm?

Possible Answers:

$181.25\pi$

$387.5 \pi$

$155\pi$

$300\pi$

Correct answer:

$387.5 \pi$

Explanation:

Step 1: Find the diameter.

If we are given the diameter, the length of the radius is one-half the diameter.

So, the radius is $\dfrac {1}{2}\times 10=5$

Step 2: Recall the volume formula...

Volume formula of cylinder is $\pi r^2 h$ .

$V=\pi (5)^2(15.5)\rightarrow \pi(25)(15.5)\implies 387.5 \pi$

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Example Question #1 : Pyramids

What is the area of the base of the pyramid with volume $12 \;\textup{m}^3$ and height $1\;\textup{m}$ ?

Possible Answers:

$32\;\textup{m}^2$

$42\;\textup{m}^2$

$19\;\textup{m}^2$

$36\;\textup{m}^2$

$23\;\textup{m}^2$

Correct answer:

$36\;\textup{m}^2$

Explanation:

The formula for the volume of a pyramid is

$V = \frac{lwh}{3}$

The height of the pyramid is $1\;\textup{m}$ , so

h=1 .

The volume of the pyramid is $12 \;\textup{m}^3$ .

Thus,

$12=\frac{lw}{3}$

$lw=12\times3=36$ .

Note, the area of the base of the pyramid is

A = lw .

Thus,

A = lw = 36 .

Hence,

$A = 36 \;\textup{m}^2$

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Example Question #1 : Use Volume Formulas To Solve Problems

A pyramid with a square base is inscribed inside a right cone with radius 24 and height 10.

Give the volume of the pyramid.

Possible Answers:

$7,680\sqrt{2}$

7,680

$3,840\sqrt{2}$

3,840

None of the other choices gives the correct response.

Correct answer:

3,840

Explanation:

The volume of a pyramid with height and a base of area can be determined using the formula

$V = \frac{1}{3}Bh$ .

The height of the inscribed pyramid is equal to that of the cone, so we can set h= 10 . The base of the pyramid is a square inscribed inside a circle, so the length of each diagonal of the square is equal to the diameter of the circle. See the figure below, which shows the bases of the pyramid and the cone.

Circle in square

The circle has radius 24, so its diameter - and the lengths of the diagonals - is twice this, or 48.

The area of a square - which is also a rhombus - is equal to half the product of the lengths of its diagonals, so

$B = \frac{1}{2}d_{1}d_{2} = \frac{1}{2} (48) (48) = 1,152$

Substituting for and , we get

$V = \frac{1}{3} (1,152)(10) = 3,840$ .

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Example Question #1 : Use Volume Formulas To Solve Problems

The base of a right pyramid with height 6 is a regular hexagon with sides of length 6.

Give its volume.

Possible Answers:

$108 \sqrt{2}$

$54 \sqrt{3}$

$108 \sqrt{3}$

$54 \sqrt{2}$

108

Correct answer:

$108 \sqrt{3}$

Explanation:

The regular hexagonal base can be divided by its diameters into six equilateral triangles, as seen below:

Hexagon 1

Each of the triangles has as its sidelength that of the hexagon. If we let this common sidelength be , each of the triangles has area

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

the total area of the base is six times this.

Substituting 6 for , the area of each triangle is

$A = \frac{6^{2}\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9 \sqrt{3}$

The total area of the base is six times this, or

$B = 6 A =6 \cdot 9 \sqrt{3} = 54 \sqrt{3}$

The volume of a pyramid with height and a base of area can be determined using the formula

$V = \frac{1}{3}Bh$ .

Set $B = 54 \sqrt{3}$ and h = 6 ;

$V = \frac{1}{3} (54\sqrt{3}) (6) = 108 \sqrt{3}$

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Example Question #2 : Use Volume Formulas To Solve Problems

A right pyramid and a right rectangular prism both have square bases. The base of the pyramid has sides that are 20% longer than those of the bases of the prism; the height of the pyramid is 20% greater than that of the prism.

Which of the following is closest to being correct?

Possible Answers:

The volume of the pyramid is 61.6% less than that of the prism.

The volume of the pyramid is 42.4% less than that of the prism.

The volume of the pyramid is 82.9% less than that of the prism.

The volume of the pyramid is 74.4% less than that of the prism.

The volume of the pyramid is 33.3% less than that of the prism.

Correct answer:

The volume of the pyramid is 42.4% less than that of the prism.

Explanation:

The volume of a right prism with height and bases of area can be determined using the formula

V = Bh .

Since its base is a square, if we let be the length of one side, then $B = s^{2}$ , and

$V =s^{2}h$

The volume of a right pyramid with height and a base of area can be determined using the formula

$V '=\frac{1}{3} B'h'$ .

Since its base is also a square, if we let be the length of one side, then $B' = s'^{2}$ , and

$V '=\frac{1}{3} s'^{2}h'$ .

The height of the pyramid is 20% greater than the height of the prism - this is 120% of , so h'= 1.2h . Similarly, the length of a side of the base of the pyramid is 20% greater than that of a base of the prism, so s' = 1.2s . Substitute in the pyramid volume formula:

$V '=\frac{1}{3} (1.2s )^{2} \cdot 1.2h$

$=\frac{1}{3} \cdot 1.2^{2} \cdot s ^{2} \cdot 1.2h$

$=\frac{1}{3} \cdot 1.44 \cdot 1.2 \cdot s ^{2} h$

$=0.576 s ^{2} h$

We can substitute , the volume of the prism, for $s^{2}h$ . This yields

V'=0.576 V

The volume of the pyramid is equal to 57.6 % of that of the prism, or, equivalently, $(100 - 57.6) \% = 42.4 \%$ less.

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Example Question #161 : Hi Set: High School Equivalency Test: Math

A triangular pyramid in coordinate space has its vertices at the origin, (c,0,0) , (0,c,0) , and (0,0,c) . In terms of , give its volume.

Possible Answers:

$\frac{1}{9}c^{3}$

$\frac{1}{6}c^{3}$

$\frac{1}{4}c^{3}$

$\frac{1}{8}c^{3}$

$\frac{1}{12}c^{3}$

Correct answer:

$\frac{1}{6}c^{3}$

Explanation:

The pyramid in question can be seen in the diagram below:

Pyramid 1

This pyramid can be seen as having as its base the triangle on the -plane with vertices at the origin, (c,0,0) , and (0,c,0) ; this is a right triangle with two legs of length , so its area is half their product, or $B = \frac{1}{2}c^{2}$ .

The altitude (perpendicular to the base) is the segment from the origin to (0,0,c) , which has length (the height of the pyramid) h = c .

Setting $B = \frac{1}{2}c^{2}$ and h = c in the formula for the volume of a pyramid:

$V = \frac{1}{3} Bh$

$V = \frac{1}{3}\cdot \frac{1}{2}c^{2} \cdot c= \frac{1}{6}c^{3}$ , the correct response.

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Example Question #111 : Measurement And Geometry

About the -axis, rotate the triangle with its vertices at (5, 0) , (0, 6 ) , and the origin. What is the volume of the solid of revolution formed?

Possible Answers:

$60 \pi$

$50 \pi$

None of the other choices gives the correct response.

$66 \pi$

$55 \pi$

Correct answer:

$50 \pi$

Explanation:

When this triangle is rotated about the -axis, the resulting solid of revolution is a cone whose base has radius r= 5 , and which has height h= 6 . Substitute these values into the formula for the volume of a cone:

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

$= \frac{1}{3} \pi (5^{2}) (6)$

$= 25 \cdot 6 \cdot \frac{1}{3} \pi$

$= 50 \pi$

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HiSET: Math : HiSet: High School Equivalency Test: Math

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Tangent

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #4 : Problems Involving Right Triangle Trigonometry

Example Question #101 : Measurement And Geometry

Example Question #1 : Pyramids

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #1 : Use Volume Formulas To Solve Problems

Example Question #2 : Use Volume Formulas To Solve Problems

Example Question #161 : Hi Set: High School Equivalency Test: Math

Example Question #111 : Measurement And Geometry

All HiSET: Math Resources

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