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GED Math : 2-Dimensional Geometry

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Example Questions

Example Question #2 : Similar Triangles And Proportions

Triangles

Note: Figures NOT drawn to scale.

Refer to the above figures. Given that $\Delta ABC \sim \Delta DE F$ , evaluate .

Possible Answers:

$93\frac{1}{3}$

$52 \frac{1}{2}$

Correct answer:

Explanation:

By the Pythagorean Theorem, since is the hypotenuse of a right triangle with legs 6 and 8, its measure is

$AC = \sqrt{6^{2} + 8 ^{2}} = \sqrt{36 + 64 } = \sqrt{100} = 10$ .

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{DF}{AC} = \frac{70}{10} = 7$ .

Likewise,

$\frac{EF}{BC} = 7$

$\frac{EF}{6} = 7$

EF = 42

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Example Question #2 : Similar Triangles And Proportions

Triangles

Note: Figures NOT drawn to scale.

Refer to the above figures. Given that $\Delta ABC \sim \Delta DE F$ , give the area of $\Delta DE F$ .

Possible Answers:

810

405

486

607.5

Correct answer:

486

Explanation:

Corresponding angles of similar triangles are congruent, so, since $\angle B$ is right, so is $\angle E$ . This makes $\overline{DE}$ and $\overline{EF}$ the legs of a right triangle, so its area is half their product.

By the Pythagorean Theorem, since is the hypotenuse of a right triangle with legs 6 and 8, its measure is

$AC = \sqrt{6^{2} + 8 ^{2}} = \sqrt{36 + 64 } = \sqrt{100} = 10$ .

The similarity ratio of $\Delta DE F$ to $\Delta ABC$ is

$\frac{DF}{AC} = \frac{45}{10} = 4.5$ .

This can be used to find and :

$\frac{DE}{AB} = 4.5$

$\frac{DE}{8} = 4.5$

$DE = 8 \cdot 4.5 = 36$

$\frac{EF}{BC} = 4.5$

$\frac{EF}{6} = 4.5$

$EF = 6 \cdot 4.5 = 27$

The area of $\Delta DE F$ is therefore

$\frac{1}{2} \cdot DE \cdot EF = \frac{1}{2} \cdot 36 \cdot 27 =486$ .

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Example Question #1 : Similar Triangles And Proportions

In the figure below, the two triangles are similar. Find the value of .

Possible Answers:

Correct answer:

Explanation:

Since the two triangles are similar, we know that their corresponding sides must be in the same ratio to each other. Thus, we can write the following equation:

$\frac{12}{8}=\frac{9}{x}$

Now, solve for .

12x=72

x=6

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Example Question #261 : 2 Dimensional Geometry

The two legs of a right triangle measure 30 and 40. What is its perimeter?

Possible Answers:

120

135

140

130

125

Correct answer:

120

Explanation:

By the Pythagorean Theorem, if a ,b are the legs of a right triangle and is its hypotenuse,

$c^{2} = a^{2} + b^{2}$

Substitute a = 30, b= 40 and solve for :

$c^{2} = 30 ^{2} + 40^{2} = 900 + 1,600 = 2,500$

$c = \sqrt{2,500} = 50$

The perimeter of the triangle is

P = 30 + 40 + 50 = 120

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Example Question #262 : 2 Dimensional Geometry

A right triangle has legs 30 and 40. Give its perimeter.

Possible Answers:

600

1,200

120

Correct answer:

120

Explanation:

The hypotenuse of the right triangle can be calculated using the Pythagorean theorem:

$c = \sqrt{30^{2}+40^{2}} = \sqrt{900+1600} = \sqrt{2,500} = 50$

Add the three sides:

P = 30 + 40 + 50 = 120

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Example Question #3 : Pythagorean Theorem

A right triangle has one leg measuring 14 inches; its hypotenuse is 50 inches. Give its perimeter.

Possible Answers:

$114\textrm{ in}$

$78 \textrm{ in}$

$112 \textrm{ in}$

$64\textrm{ in}$

Correct answer:

$112 \textrm{ in}$

Explanation:

The Pythagorean Theorem can be used to derive the length of the second leg:

$b = \sqrt{50^{2}-14^{2}} = \sqrt{2,500 - 196} = \sqrt{2,304} = 48$ inches

Add the three sides to get the perimeter.

P = 14 + 48 + 50 = 112 inches.

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Example Question #3 : Pythagorean Theorem

Whicih of the following could be the lengths of the sides of a right triangle?

Possible Answers:

10 inches, 1 foot, 14 inches

2 feet, 32 inches, 40 inches

7 inches, 2 feet, 30 inches

15 inches, 3 feet, 40 inches

Correct answer:

2 feet, 32 inches, 40 inches

Explanation:

A triangle is right if and only if it satisfies the Pythagorean relationship

$a ^{2}+ b^{2}= c^{2}$

where is the measure of the longest side and a,b are the other two sidelengths. We test each of the four sets of lengths, remembering to convert feet to inches by multiplying by 12.

7 inches, 2 feet, 30 inches:

2 feet is equal to 24 inches. The relationship to be tested is

$7^{2}+ 24 ^{2} = 30^{2}$

49 + 576 = 900

625 = 900 - False

10 inches, 1 foot, 14 inches:

1 foot is equal to 12 inches. The relationship to be tested is

$10^{2}+ 12^{2}= 14^{2}$

100 + 144 = 196

244 = 196 - False

15 inches, 3 feet, 40 inches:

3 feet is equal to 36 inches. The relationship to be tested is

$15^{2}+ 36^{2} = 40^{2}$

225 + 1,296 = 1,600

1,521 = 1,600 - False

2 feet, 32 inches, 40 inches:

2 feet is equal to 24 inches. The relationship to be tested is

$24 ^{2}+ 32^{2} = 40^{2}$

576+ 1,024= 1,600

1,600 =1,600 - True

The correct choice is 2 feet, 32 inches, 40 inches.

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Example Question #4 : Pythagorean Theorem

An isosceles right triangle has hypotenuse 80 inches. Give its perimeter. (If not exact, round to the nearest tenth of an inch.)

Possible Answers:

$193.1 \textrm{ in}$

$136.6\textrm{ in}$

$153.1 \textrm{ in}$

$96.6\textrm{ in}$

Correct answer:

$193.1 \textrm{ in}$

Explanation:

Each leg of an isosceles right triangle has length that is the length of the hypotenuse divided by $\sqrt{2}$ . The hypotenuse has length 80, so each leg has length

$\frac{80}{\sqrt{2}} = \frac{80\cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}= \frac{80 \sqrt{2}}{2} = 40 \sqrt{2}$ .

The perimeter is the sum of the three sides:

$80 + 40 \sqrt{2}+ 40\sqrt{2}= 80 + 80\sqrt{2}$ inches.To the nearest tenth:

$80 + 80\sqrt{2} \approx 80 + 80 \times 1.414 \approx 80 + 113.1 = 193.1$ inches.

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Example Question #5 : Pythagorean Theorem

Triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. AC = 5, BC = 3, CD = 4, CE = 12 . Give the perimeter of Quadrilateral ABDE .

Possible Answers:

Correct answer:

Explanation:

The perimeter of Quadrilateral ABDE is the sum of the lengths of $\overline{AB}$ , $\overline{DE}$ , $\overline{BD}$ , and $\overline{EA}$ .

The first two lengths can be found by subtracting known lengths:

AB = AC - BC = 5 - 3 = 2

DE = CE - DE = 12 - 4 = 8

The last two segments are hypotenuses of right triangles, and their lengths can be calculated using the Pythagorean Theorem:

$\overline{BD }$ is the hypotenuse of a triangle with legs BC = 3, CD = 4 ; it measures

$BD = \sqrt{3^{2}+4^{2}} = \sqrt{9 + 16}= \sqrt{25} = 5$

$\overline{EA}$ is the hypotenuse of a triangle with legs AC = 5, CE = 12 ; it measures

$EA= \sqrt{5^{2}+12^{2}} = \sqrt{25+144}= \sqrt{169} = 13$

Add the four sidelengths:

P = AB + BD + DE + EA = 2+ 5 + 8 + 13 = 28

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Example Question #6 : Pythagorean Theorem

Triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Possible Answers:

$X = 4\frac{2}{13}$

$X = 4\frac{8}{13}$

$X = 3\frac{11}{13}$

$X = 4\frac{7}{13}$

Correct answer:

$X = 4\frac{8}{13}$

Explanation:

The altitude perpendicular to the hypotenuse of a right triangle divides that triangle into two smaller triangles similar to each other and the large triangle. Therefore, the sides are in proportion. The hypotenuse of the triangle is equal to

$\sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169} =13$

We can set up a proportion statement by comparing the large triangle to the smaller of the two in which it is divided. The sides compared are the hypotenuse and the longer side:

$\frac{X}{5} = \frac{12}{13}$

Solve for :

$\frac{X}{5} \cdot 5 = \frac{12}{13} \cdot 5$

$X = \frac{60}{13} = 4 \frac{8}{13}$

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GED Math : 2-Dimensional Geometry

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #2 : Similar Triangles And Proportions

Example Question #2 : Similar Triangles And Proportions

Example Question #1 : Similar Triangles And Proportions

Example Question #261 : 2 Dimensional Geometry

Example Question #262 : 2 Dimensional Geometry

Example Question #3 : Pythagorean Theorem

Example Question #3 : Pythagorean Theorem

Example Question #4 : Pythagorean Theorem

Example Question #5 : Pythagorean Theorem

Example Question #6 : Pythagorean Theorem

All GED Math Resources

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