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GED Math : Pythagorean Theorem

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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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Example Question #271 : 2 Dimensional Geometry

Determine the hypotenuse of a right triangle if the side legs are 2,4 respectively.

Possible Answers:

$2\sqrt5$

$4\sqrt5$

$8\sqrt2$

Correct answer:

$2\sqrt5$

Explanation:

Write the Pythagorean Theorem to find the hypotenuse.

a^2+b^2 =c^2

Substitute the dimensions.

2^2+4^2 =c^2

4+16 =c^2

20=c^2

Square root both sides.

$\sqrt{20}=\sqrt{c^2}$

$c= \sqrt{20} = \sqrt{4} \times \sqrt{5}$

The answer is: $2\sqrt5$

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Example Question #17 : Pythagorean Theorem

A car left City A and drove straight east for miles then it drove straight north for miles, where it stopped. In miles, what is the shortest distance between the car and City A?

Possible Answers:

160

120

100

140

Correct answer:

100

Explanation:

Start by drawing out what the car did.

You'll notice that a right triangle will be created as shown by the figure above. Thus, the shortest distance between the car and City A is also the hypotenuse of the triangle. Use the Pythagorean Theorem to find the distance between the car and City A.

$\text{Distance}=\sqrt{60^2+80^2}=100$

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Example Question #18 : Pythagorean Theorem

You want to build a garden in the shape of a right triangle. If the two arms will be 6ft and 8ft, what does the length of the hypotenuse need to be?

Possible Answers:

14ft

12.5ft

10ft

9ft

Correct answer:

10ft

Explanation:

You want to build a garden in the shape of a right triangle. If the two arms will be 6ft and 8ft, what does the length of the hypotenuse need to be?

To find the length of a hypotenuse of a right triangle, simply use the Pythagorean Theorem.

a^2+b^2=c^2

Where a and b are the arm lengths, and c is the hypotenuse.

Plug in our knowns and solve.

(6ft)^2+(8ft)^2=c^2

36ft^2+64ft^2=c^2

100ft^2=c^2

$c=\sqrt{100ft^2}=10ft$

Note that we could also have found c by identifying a Pythagorean Triple:

3x-4x-5x

3(2)-4(2)-5(2)

6-8-10

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Example Question #19 : Pythagorean Theorem

Right triangle 2

Note: Figure NOT drawn to scale.

Refer to the above diagram.

Which is true of ?

Possible Answers:

21< x < 22

19< x < 20

18 < x < 19

20< x < 21

Correct answer:

18 < x < 19

Explanation:

By the Pythagorean Theorem, if and are the lengths of the shorter two sides, or legs, of a right triangle, and is the length of the longest side, or hypotenuse, of the triangle, then

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

Set a = 10, b = 16, c = x :

$x ^{2}= 10^{2}+16^{2} = 10\cdot 10 + 16 \cdot 16 = 100 + 256 = 356$

Since we are trying to determine between which two of the consecutive integers in the set $\left \{ 18, 19, 20, 21, 22 \right \}$ falls, it suffices to find out between which two of their squares $x^{2}$ falls. Each square is the product of the integer and itself, so:

$18^{2} = 18 \cdot 18 = \textbf{324}$

$19^{2} = 19 \cdot 19 =\textbf{361}$

$20^{2} = 20 \cdot 20 = 400$

$21^{2}= 21 \cdot 21 = 441$

$22^{2} = 22 \cdot 22 = 484$

324 < 356< 361 ,

or, substituting,

$18 ^{2} < x^{2} < 19 ^{2}$

It follows that

18 < x < 19 .

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Example Question #20 : Pythagorean Theorem

You are visiting a friend who has right-triangular shaped pool. You are seeing who can swim around the perimeter of the pool fastest. If the long side is 20 meters, and second shortest side is 15 meters long, how long is the shortest side?

Possible Answers:

$5\sqrt{7}m$

$12\sqrt{5}m$

$7\sqrt{5}m$

$25\sqrt{7}m$

Correct answer:

$5\sqrt{7}m$

Explanation:

Let's begin by recalling Pythagorean Theorem

a^2+b^2=c^2

So, we know that c is our hypotenuse or longest side.

a and b are our shorter sides. It doesn't really matter which one is which.

Let's plug in and solve!

$(15m)^2+b^2=(20m)^2\rightarrow 225m^2+b^2=400m^2$

$b^2=400m^2-225m^2\rightarrow b^2=175m^2$

$b^2=175m^2\rightarrow b=\sqrt{175m^2}=\sqrt{25}\sqrt{7}m=5\sqrt{7}m$

So, our answer is $5\sqrt{7}m$

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GED Math : Pythagorean Theorem

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

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

Example Question #271 : 2 Dimensional Geometry

Example Question #17 : Pythagorean Theorem

Example Question #18 : Pythagorean Theorem

Example Question #19 : Pythagorean Theorem

Example Question #20 : Pythagorean Theorem

All GED Math Resources

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