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GED Math : Triangles

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Example Question #6 : 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 #1 : Pythagorean Theorem

Triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Possible Answers:

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

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

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

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

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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Example Question #8 : Pythagorean Theorem

A right triangle has one leg of length 42 inches; its hypotenuse has length 70 inches. What is the area of this triangle?

Possible Answers:

$1,470 \textrm{ in}^{2}$

$2,352\textrm{ in}^{2}$

$1,176 \textrm{ in}^{2}$

$2,940\textrm{ in}^{2}$

Correct answer:

$1,176 \textrm{ in}^{2}$

Explanation:

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

$b = \sqrt{70^{2}-42^{2}} = \sqrt{4,900-1,764} = \sqrt{3,136}= 56$ inches.

Use the area formula for a triangle, with the legs as the base and height:

$A = \frac{1}{2}bh = \frac{1}{2} \cdot 42 \cdot 70 = 1,176$ square inches.

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Example Question #9 : Pythagorean Theorem

The area of a right triangle is 136 square inches; one of its legs measures 8 inches. What is the length of its hypotenuse? (If not exact, give the answer to the nearest tenth of an inch.)

Possible Answers:

$33.1\textrm{ in}$

$34\textrm{ in}$

$34.9\textrm{ in}$

$36\textrm{ in}$

Correct answer:

$34.9\textrm{ in}$

Explanation:

The area of a triangle is calcuated using the formula

$A = \frac{1}{2} bh$

In a right triangle, the bases can be used for base and height, so solve for by substitution:

$136= \frac{1}{2} \cdot 8h$

136=4h

$136 \div 4 =4h \div 4$

34 = h

The legs measure 8 and 34 inches, respectively; use the Pythagorean Theorem to find the length of the hypotenuse:

$c=\sqrt{8^{2}+34^{2}} = \sqrt{64+1,156}= 34.9$ inches

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Example Question #10 : Pythagorean Theorem

Triangle

Refer to the above diagram. Which of the following expressions gives the length of $\overline{AC}$ ?

Possible Answers:

$AC = \sqrt{ x^{2}+10x }$

$AC = 2x^{2}+20x +100$

$AC = \sqrt{ 2x^{2}+20x +100 }$

$AC = x^{2}+10x$

Correct answer:

$AC = \sqrt{ 2x^{2}+20x +100 }$

Explanation:

By the Pythagorean Theorem,

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(x+10)^{2} + x^{2}}$

$AC = \sqrt{ x^{2}+ 2 \cdot x \cdot 10 +10^{2} + x^{2}}$

$AC = \sqrt{ x^{2}+20x +100+ x^{2}}$

$AC = \sqrt{ 2x^{2}+20x +100 }$

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Example Question #11 : Pythagorean Theorem

Thingy

Note: figure NOT drawn to scale

The square in the above diagram has area 2,500. What is ?

Possible Answers:

$x = 33\frac{1}{3}$

$x = 22 \frac{1}{2}$

x = 25

x = 30

Correct answer:

x = 30

Explanation:

The area of the square is 2,500; its sidelength is the square root of this:

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

This means that the right triangle in the diagram has legs and 40 and hypotenuse 50, so by the Pythagorean Theorem:

$x = \sqrt{50^{2} - 40^{2}} = \sqrt{2,500-1,600} = \sqrt{900} = 30$

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Example Question #12 : Pythagorean Theorem

Square

The above figure shows Square SUAR with sidelength 2; is the midpoint of $\overline{SU}$ ; is the midpoint of $\overline{RS}$ . To the nearest tenth, what is the perimeter of Quadrilateral QARE ?

Possible Answers:

7.2

6.2

6.4

6.7

Correct answer:

6.7

Explanation:

The perimeter of Quadrilateral QARE is the sum of the lengths of $\overline{QA}, \overline{AR}, \overline{RE}, \overline{EQ}$ .

$\overline{AR}$ is a side of the square and has length 2.

is the midpoint of $\overline{RS}$ , so $\overline{RE}$ is half a side of the square; its length is 1.

The other two sides are hypotenuses of right triangles and can be found using the Pythagorean Theorem.

and are the midpoints of $\overline{SU}$ and $\overline{RS}$ , respectively,both of which are segments with length 2; therefore,

ES = SQ = QU = 1 .

$\overline{QE}$ is therefore the hypotenuse of a triangle with legs 1 and 1, and its length is

$QE = \sqrt{1^{2}+ 1^{2}} = \sqrt{1 + 1} = \sqrt{2}$ .

Similarly, $\overline{QA}$ is the hypotenuse of a triangle with legs 1 and 2, and its length is

$QE = \sqrt{1^{2}+ 2^{2}} = \sqrt{1 + 4} = \sqrt{5}$ .

Add these:

$P = 1 + 2 + \sqrt{2} + \sqrt{5} \approx 1 + 2 + 1.414 + 2.236 \approx 6.650$

This rounds to 6.7.

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Example Question #13 : Pythagorean Theorem

What is the area of a right triangle if the hypotenuse is 10, and one of the side lengths is 6?

Possible Answers:

Correct answer:

Explanation:

To determine the other side length, we will need to use the Pythagorean Theorem.

a^2+b^2 = c^2

Substitute the hypotenuse and the known side length as either or .

a^2+ 6^2 = 10^2

a^2+36 = 100

Subtract 36 from both sides and reduce.

a^2 = 64

Square root both sides and reduce.

a=8

The length and width of the triangle are now known.

Write the formula for the area of a triangle.

$A=\frac{1}{2}bh$

Substitute the dimensions.

$A=\frac{1}{2}(8)(6) = 24$

The answer is .

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Example Question #14 : Pythagorean Theorem

If the hypotenuse of a right triangle is 5, and a side length is 2, what is the area?

Possible Answers:

$\frac{\sqrt{29}}{2}$

$\frac{\sqrt{21}}{2}$

$3\sqrt7$

$\sqrt{21}$

Correct answer:

$\sqrt{21}$

Explanation:

To find the other side length, we will need to first use the Pythagorean Theorem.

a^2+b^2 = c^2

Substitute the side and hypotenuse.

a^2+2^2 = 5^2

Solve for the missing side.

a^2+4=25

a^2 =21

$a=\sqrt{21}$

Write the formula for the area of a triangle.

$A=\frac{BH}{2}$

Substitute the sides.

$A=\frac{2\sqrt{21}}{2} = \sqrt{21}$

The answer is: $\sqrt{21}$

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Example Question #15 : Pythagorean Theorem

If the hypotenuse of a right triangle if 7, and a side length is 5, what must be the length of the missing side?

Possible Answers:

$6\sqrt2$

$\frac{7}{5}$

$\frac{5}{7}$

$2\sqrt{6}$

$\sqrt{74}$

Correct answer:

$2\sqrt{6}$

Explanation:

Write the formula for the Pythagorean Theorem.

a^2+b^2 =c^2

Substitute the values into the equation.

a^2+5^2 =7^2

a^2 +25 = 49

Subtract 25 from both sides.

a^2 +25 -25= 49-25

a^2 = 24

Square root both sides.

$a = \sqrt{24} = \sqrt{4}\cdot \sqrt{6} = 2\sqrt{6}$

The answer is: $2\sqrt{6}$

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GED Math : Triangles

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #6 : Pythagorean Theorem

Example Question #1 : Pythagorean Theorem

Example Question #8 : Pythagorean Theorem

Example Question #9 : Pythagorean Theorem

Example Question #10 : Pythagorean Theorem

Example Question #11 : Pythagorean Theorem

Example Question #12 : Pythagorean Theorem

Example Question #13 : Pythagorean Theorem

Example Question #14 : Pythagorean Theorem

Example Question #15 : Pythagorean Theorem

All GED Math Resources

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