Middle School Math Quiz: Apply Pythagorean Theorem To Problems

20 questions · exam conditions

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Apply Pythagorean Theorem To ProblemsQuestion 1 of 20

An isosceles right triangle has legs of length $x$ . If the area of the triangle is 18 square units, what is the length of the hypotenuse?

6 units

6\sqrt{2}

units

9 units

9\sqrt{2}

units

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Middle School Math Quiz

Middle School Math Quiz: Apply Pythagorean Theorem To Problems

Practice Apply Pythagorean Theorem To Problems in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Apply Pythagorean Theorem To Problems, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

An isosceles right triangle has legs of length $x$ . If the area of the triangle is 18 square units, what is the length of the hypotenuse?

6 units
$6\sqrt{2}$ units (correct answer)
9 units
$9\sqrt{2}$ units

Explanation: The area of the triangle is

\frac{1}{2}x^2 = 18

, so

x^2 = 36

and

x = 6

. In an isosceles right triangle, if the legs have length 6, the hypotenuse has length

6\sqrt{2}

. Choice A gives the leg length instead of hypotenuse. Choice C incorrectly adds 6 + 3. Choice D uses

x = 9

instead of

x = 6

, giving

9\sqrt{2}

Question 2

A square has a diagonal that measures $8\sqrt{2}$ centimeters. What is the perimeter of the square?

16 centimeters
32 centimeters (correct answer)
64 centimeters
$16\sqrt{2}$ centimeters

Explanation: In a square with side length

s

, the diagonal equals

s\sqrt{2}

. Given diagonal =

8\sqrt{2}

, we have

s\sqrt{2} = 8\sqrt{2}

, so

s = 8

. The perimeter is

4s = 4(8) = 32

centimeters. Choice A gives the side length instead of perimeter. Choice C squares the side length:

8^2 = 64

. Choice D incorrectly uses

2 \cdot 8\sqrt{2} = 16\sqrt{2}

Question 3

A right triangle has legs of length 6 cm and 8 cm. What is the length of the hypotenuse?

10 cm (correct answer)
14 cm
100 cm
12 cm

Explanation: This question tests applying the Pythagorean theorem

a^2 + b^2 = c^2

to find the unknown hypotenuse in a right triangle with given legs. For a right triangle with legs a=6 cm and b=8 cm, and hypotenuse c (the longest side opposite the 90° angle), plug the legs into the formula:

6^2 + 8^2 = 36 + 64 = 100 = c^2

, so c =

\sqrt{100}

= 10 cm. In this specific problem, the right triangle has legs of 6 cm and 8 cm, so the hypotenuse is found using c =

\sqrt{(6^2 + 8^2)} = \sqrt{(36 + 64)} = \sqrt{100}

= 10 cm. The correct setup involves identifying the legs as a and b, squaring them, adding, and taking the square root for c, which matches choice A. A common error is adding without squaring, like 6 + 8 = 14, or forgetting the square root and choosing 100 cm. To solve: (1) identify the right triangle, (2) label legs as 6 cm and 8 cm, hypotenuse as c, (3) note both legs known, hypotenuse unknown, (4) set up

6^2 + 8^2 = c^2

, (5) calculate 36 + 64 = 100, c =

\sqrt{100}

= 10 cm, (6) verify it's reasonable as 10 cm is longer than both legs. Common mistakes include confusing legs with hypotenuse or arithmetic errors like

6^2 = 36

but

8^2 = 68

incorrectly.

Question 4

In a right triangle, the hypotenuse is 13 units and one leg is 5 units. What is the length of the other leg?

8 units
18 units
144 units
12 units (correct answer)

Explanation: This problem tests applying the Pythagorean theorem

a^2 + b^2 = c^2

to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or even 3D space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using

a^2 + b^2 = c^2

; for example, to find the hypotenuse, plug in the legs like

6^2 + 8^2 = 36 + 64 = 100 = c^2

, so

c = \sqrt{100} = 10

, or to find a leg, rearrange to

a^2 = c^2 - b^2

, such as if c=13 and b=5, then

a^2 = 169 - 25 = 144

, so a=12. In real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base=9 ft and ladder=15 ft, and solve with

9^2 + h^2 = 15^2

to get h=12 ft. In this specific problem, the right triangle has hypotenuse 13 units and one leg 5 units, so the other leg is found using

a^2 = 13^2 - 5^2 = 169 - 25 = 144

, giving a =

\sqrt{144}

= 12 units. The correct setup involves identifying the hypotenuse as c=13 and leg b=5, rearranging the formula to

a^2 = c^2 - b^2

, and calculating the square root to get 12 units. Common errors include subtracting without squaring like 13-5=8, mistaking 5² as 10, forgetting the square root and saying 144 units, or confusing the hypotenuse and using it as a leg. To solve: (1) identify the right triangle with the 90° angle implied, (2) label legs a (unknown) and b=5, hypotenuse c=13 as the longest, (3) note hypotenuse and one leg known, other leg unknown, (4) set up

a^2 = c^2 - b^2

, (5) calculate 169 - 25 = 144,

\sqrt{144}=12

, (6) verify

5^2 + 12^2 = 25 + 144 = 169 = 13^2

. Common mistakes: using a + b = c forgetting squares, wrong side as hypotenuse like treating 5 as c, arithmetic errors in subtraction or square root, or negative length.

Question 5

A right triangle has a hypotenuse of 26 cm and one leg of 24 cm. What is the length of the other leg?

676 cm
10 cm (correct answer)
2 cm
50 cm

Explanation: Using the Pythagorean theorem,

a^2+b^2=c^2

, with hypotenuse

c=26

and one leg

a=24

b^2=26^2-24^2=676-576=100

, so

b=\sqrt{100}=10

cm. Choice A is wrong because it stops after computing

26^2=676

, without subtracting

24^2

or taking the square root. Choice C is wrong because it subtracts the two side lengths directly,

26-24=2

, instead of using the Pythagorean theorem. Choice D is wrong because it adds the two given values,

26+24=50

, instead of applying the theorem correctly.

Question 6

A right triangle has a hypotenuse of 17 units and one leg of 8 units. What is the length of the other leg?

9 units
289 units
25 units
15 units (correct answer)

Explanation: This question tests applying the Pythagorean theorem a² + b² = c² to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or 3D problems like space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using a² + b² = c²; for example, finding the hypotenuse by plugging in the legs like 6² + 8² = 36 + 64 = 100 = c², so c = √100 = 10, or finding a leg by rearranging to a² = c² - b², such as if c = 13 and b = 5, then a² = 169 - 25 = 144, so a = 12; in real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base = 9 ft and ladder = 15 ft, and solve 9² + h² = 15² to get h = 12 ft. In this specific problem, the right triangle has a hypotenuse of 17 units and one leg of 8 units, so the other leg is found using a² = 17² - 8² = 289 - 64 = 225, a = √225 = 15 units. The correct setup identifies c = 17 and b = 8, rearranges to a² = c² - b², calculates squares, subtracts, and takes square root to get 15 units, matching choice B. A common error might be adding squares instead of subtracting, like 289 + 64 = 353, or not squaring and doing 17 - 8 = 9, or choosing 289 without square root. To solve these problems, follow these steps: (1) identify the right triangle with 90° angle, (2) label legs and hypotenuse correctly, (3) identify knowns and unknown, (4) set up a² = c² - b², (5) calculate accordingly, (6) verify 15 is between 8 and 17. Common mistakes: forgetting squares, wrong hypotenuse, arithmetic slips, negative length.

Question 7

A ramp forms a right triangle with the ground. The ramp is 10 ft long (the hypotenuse) and reaches a platform that is 6 ft high. How far is the bottom of the ramp from the platform (horizontal distance)?

8 ft (correct answer)
4 ft
100 ft
16 ft

Explanation: This question tests applying the Pythagorean theorem a² + b² = c² to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or 3D problems like space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using a² + b² = c²; for example, finding the hypotenuse by plugging in the legs like 6² + 8² = 36 + 64 = 100 = c², so c = √100 = 10, or finding a leg by rearranging to a² = c² - b², such as if c = 13 and b = 5, then a² = 169 - 25 = 144, so a = 12; in real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base = 9 ft and ladder = 15 ft, and solve 9² + h² = 15² to get h = 12 ft. In this specific ramp problem, the ramp is 10 ft (hypotenuse) and height is 6 ft (one leg), so the horizontal distance is b = √(10² - 6²) = √(100 - 36) = √64 = 8 ft. The correct setup identifies c = 10 ft and a = 6 ft, rearranges to b² = c² - a², calculates, and takes square root to get 8 ft, matching choice B. A common error might be adding squares instead, like 100 + 36 = 136, or not squaring and doing 10 - 6 = 4, or choosing 100 without root. To solve these problems, follow these steps: (1) identify right triangle from ramp and ground, (2) label legs as height and base, hypotenuse as ramp, (3) knowns: hypotenuse and height, unknown base, (4) set up b² = c² - a², (5) calculate, (6) verify 8 ft makes sense with 10 ft ramp. Common mistakes: no squares, wrong subtraction, arithmetic errors, negative value.

Question 8

A student walks 3 blocks east and then 4 blocks north, making a right angle turn. About how many blocks is the straight-line distance from the starting point to the ending point?

12 blocks
1 block
5 blocks (correct answer)
7 blocks

Explanation: This question tests applying the Pythagorean theorem a² + b² = c² to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or 3D problems like space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using a² + b² = c²; for example, finding the hypotenuse by plugging in the legs like 6² + 8² = 36 + 64 = 100 = c², so c = √100 = 10, or finding a leg by rearranging to a² = c² - b², such as if c = 13 and b = 5, then a² = 169 - 25 = 144, so a = 12; in real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base = 9 ft and ladder = 15 ft, and solve 9² + h² = 15² to get h = 12 ft. In this specific problem, walking 3 blocks east and 4 blocks north forms a right triangle, so the straight-line distance is the hypotenuse d = √(3² + 4²) = √(9 + 16) = √25 = 5 blocks. The correct setup treats the directions as legs a = 3 and b = 4, with d as c, squares them, adds, and takes square root to get 5 blocks, matching choice A. A common error might be just adding 3 + 4 = 7 without squaring, or subtracting to get 1, or not recognizing the right angle turn implies a right triangle. To solve these problems, follow these steps: (1) identify the right triangle from perpendicular paths, (2) label legs as east and north distances, hypotenuse as straight line, (3) identify known legs and unknown hypotenuse, (4) set up a² + b² = c², (5) calculate squares, add, square root, (6) verify 5 is longer than 4 but shorter than path walked. Common mistakes: a + b = c, arithmetic like 3² = 6, forgetting square root, wrong setup.

Question 9

A right triangle has legs of lengths 6 cm and 8 cm. What is the length of the hypotenuse?

10 cm (correct answer)
100 cm
14 cm
9 cm

Explanation: This question tests applying the Pythagorean theorem

a^2 + b^2 = c^2

to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or 3D problems like space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using

a^2 + b^2 = c^2

; for example, finding the hypotenuse by plugging in the legs like

6^2 + 8^2 = 36 + 64 = 100 = c^2

, so

c = \sqrt{100} = 10

, or finding a leg by rearranging to

a^2 = c^2 - b^2

, such as if c = 13 and b = 5, then

a^2 = 169 - 25 = 144

, so a = 12; in real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base = 9 ft and ladder = 15 ft, and solve

9^2 + h^2 = 15^2

to get h = 12 ft. In this specific problem, the right triangle has legs of 6 cm and 8 cm, and we need to find the hypotenuse using

6^2 + 8^2 = c^2

, giving 36 + 64 = 100, so

c = \sqrt{100} = 10

cm. The correct setup involves identifying the legs as a = 6 cm and b = 8 cm, with c as the hypotenuse, then calculating the squares, adding them, and taking the square root to get 10 cm, which matches choice A. A common error might be adding without squaring, like 6 + 8 = 14, or forgetting the square root and choosing 100 cm, or confusing legs and hypotenuse by using the wrong sides in the formula. To solve these problems, follow these steps: (1) identify the right triangle with a 90° angle, (2) label the sides with legs a and b forming the right angle and hypotenuse c opposite it as the longest side, (3) identify the knowns (two legs) and unknown (hypotenuse), (4) set up

a^2 + b^2 = c^2

, (5) calculate by squaring the knowns, adding, and taking the square root, (6) verify it makes sense, like 10 cm being longer than both legs. Common mistakes include using a + b = c without squaring, misidentifying the hypotenuse, arithmetic errors in squaring or adding, or taking a negative square root.

Question 10

A ramp forms a right triangle with the ground. The ramp is 13 feet long (the hypotenuse), and it rises 5 feet vertically. How far is the bottom of the ramp from the point directly below the top (the horizontal distance)?

8 ft
144 ft
12 ft (correct answer)
18 ft

Explanation: This problem tests applying the Pythagorean theorem a² + b² = c² to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or even 3D space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using a² + b² = c²; for example, to find the hypotenuse, plug in the legs like 6² + 8² = 36 + 64 = 100 = c², so c = √100 = 10, or to find a leg, rearrange to a² = c² - b², such as if c=13 and b=5, then a² = 169 - 25 = 144, so a=12. In real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base=9 ft and ladder=15 ft, and solve with 9² + h² = 15² to get h=12 ft. In this specific ramp problem, the ramp is 13 ft (hypotenuse) and rises 5 ft vertically, so the horizontal distance is found using h² = 13² - 5² = 169 - 25 = 144, giving h = √144 = 12 ft. The correct setup involves identifying the right triangle with vertical rise as one leg, horizontal as unknown leg, ramp as hypotenuse, rearranging the formula, and calculating to get 12 ft. Common errors include subtracting wrong like 13-5=8, mistaking 5² as 10, forgetting square root and saying 144 ft, or confusing vertical and horizontal. To solve: (1) identify right triangle with 90° at ground, (2) label legs horizontal h and vertical 5 ft, hypotenuse 13 ft, (3) note hypotenuse and one leg known, other unknown, (4) set up h² = c² - b², (5) calculate 169 - 25 = 144, √144=12, (6) verify 12 ft reasonable as 5² + 12² = 25 + 144 = 169 = 13². Common mistakes: a + b = c like 13+5=18, wrong hypotenuse, arithmetic errors, negative length.

Question 11

A 15-foot ladder leans against a vertical wall. The bottom of the ladder is 9 feet from the wall on level ground. About how high up the wall does the ladder reach?

24 ft
10 ft
12 ft (correct answer)
6 ft

Explanation: This problem tests applying the Pythagorean theorem a² + b² = c² to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or even 3D space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using a² + b² = c²; for example, to find the hypotenuse, plug in the legs like 6² + 8² = 36 + 64 = 100 = c², so c = √100 = 10, or to find a leg, rearrange to a² = c² - b², such as if c=13 and b=5, then a² = 169 - 25 = 144, so a=12. In real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base=9 ft and ladder=15 ft, and solve with 9² + h² = 15² to get h=12 ft. In this specific ladder problem, the ladder is 15 ft (hypotenuse) and base is 9 ft from the wall, so the height is found using h² = 15² - 9² = 225 - 81 = 144, giving h = √144 = 12 ft. The correct setup involves recognizing the right triangle formed by the wall, ground, and ladder, with c=15 ft and b=9 ft, rearranging to h² = c² - b², and calculating to get 12 ft. Common errors include adding instead of subtracting like 225 + 81, mistaking 9² as 18, forgetting the square root and saying 144 ft, or not identifying the right triangle setup. To solve: (1) identify the right triangle with 90° at the wall base, (2) label legs base=9 ft and height h, hypotenuse ladder=15 ft, (3) note hypotenuse and one leg known, other leg unknown, (4) set up h² = c² - b², (5) calculate 225 - 81 = 144, √144=12, (6) verify 12 ft is reasonable as 9² + 12² = 81 + 144 = 225 = 15². Common mistakes: using a + b = c like 9+15=24, confusing hypotenuse, arithmetic errors in squares or subtraction, or negative values.

Question 12

Refer to the figure below. A right triangle has legs of length $x$ and $x+7$ , and a hypotenuse of length $13$ . What is the value of $x$ ?

$3$
$5$ (correct answer)
$6$
$12$

Explanation: Using

x^2 + (x+7)^2 = 13^2

x^2 + x^2 + 14x + 49 = 169

, giving

2x^2 + 14x - 120 = 0

, or

x^2 + 7x - 60 = 0

. Factoring:

(x-5)(x+12)=0

, so

x = 5

. Legs are

5

and

12

, hypotenuse

13

. Choice A gives legs

3

and

10

\sqrt{109} \ne 13

. Choice C gives legs

6

and

13

: not correct. Choice D is the other leg's length, not

x

Question 13

Refer to the figure below. A wire is stretched from the top of a $24$ -foot pole to a point on the ground $18$ feet from the base of the pole. A second wire runs from the same point on the ground to the top of a $10$ -foot pole located $8$ feet from that point (on the opposite side). What is the total length of both wires (to the nearest tenth)?

$30.0$ ft
$42.8$ ft (correct answer)
$12.8$ ft
$60.0$ ft

Explanation: First wire:

\sqrt{24^2 + 18^2} = \sqrt{576+324} = \sqrt{900} = 30

ft. Second wire:

\sqrt{10^2 + 8^2} = \sqrt{100+64} = \sqrt{164} \approx 12.8

ft. Total:

30 + 12.8 \approx 42.8

ft. Choice A is only the first wire. Choice C is only the second wire. Choice D adds the given ground/pole lengths incorrectly.

Question 14

Refer to the triangle below. Triangle ABC is a right triangle with the right angle at C. If AC = 12 and BC = 9, what is the length of the altitude from C to the hypotenuse AB?

$\frac{108}{15}$ units
$\frac{36}{5}$ units (correct answer)
$\frac{108}{5}$ units
$\frac{45}{4}$ units

Explanation: First find the hypotenuse:

AB = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15

. The area can be calculated two ways:

\frac{1}{2} \cdot 12 \cdot 9 = 54

and

\frac{1}{2} \cdot 15 \cdot h = \frac{15h}{2}

. Setting equal:

54 = \frac{15h}{2}

, so

h = \frac{108}{15} = \frac{36}{5}

. Choice A stops at

\frac{108}{15}

without simplifying. Choice C incorrectly uses

\frac{108}{5}

by using 5 instead of 15. Choice D uses incorrect formula setup.

Question 15

A right triangle has sides in the ratio 5:12:13. If the shortest side is 15 units long, what is the area of the triangle?

90 square units
180 square units
270 square units (correct answer)
195 square units

Explanation: Since the sides are in ratio 5:12:13 and the shortest side is 15, the scale factor is

\frac{15}{5} = 3

. The three sides are

5 \times 3 = 15

12 \times 3 = 36

, and

13 \times 3 = 39

. The area is

\frac{1}{2} \times 15 \times 36 = \frac{540}{2} = 270

square units. Choice A uses the original 5:12 ratio:

\frac{1}{2} \times 5 \times 36 = 90

. Choice B uses

\frac{1}{2} \times 15 \times 24 = 180

. Choice D uses

\frac{1}{2} \times 15 \times 26 = 195

Question 16

A baseball diamond is a square with 90-foot sides. A player runs from home plate to second base. How much shorter is this distance compared to running from home plate to first base and then to second base?

$90(\sqrt{2} - 1)$ feet
$90\sqrt{2}$ feet
45 feet
$90(2 - \sqrt{2})$ feet (correct answer)

Explanation: When you see a question about distances on a square field, you're dealing with the Pythagorean theorem and comparing direct versus indirect paths. First, let's find the distance from home plate to second base. Since a baseball diamond is a square with 90-foot sides, home plate and second base are diagonally opposite corners. Using the Pythagorean theorem:

d = \sqrt{90^2 + 90^2} = \sqrt{2 \cdot 90^2} = 90\sqrt{2}

feet. The indirect route (home to first base, then first base to second base) covers two sides of the square:

90 + 90 = 180

feet. The difference between these distances is:

180 - 90\sqrt{2} = 90(2 - \sqrt{2})

feet. Looking at the wrong answers: Choice A gives

90(\sqrt{2} - 1)

, which would result from incorrectly calculating

90\sqrt{2} - 90

instead of

180 - 90\sqrt{2}

. Choice B is

90\sqrt{2}

, which is just the diagonal distance itself, not the difference between routes. Choice C suggests 45 feet, which might come from thinking the difference is simply half a side length. The correct answer is D:

90(2 - \sqrt{2})

feet. Strategy tip: In geometry problems involving squares and diagonals, always identify whether you need the Pythagorean theorem for the diagonal, and be careful about what the question is actually asking for—sometimes it's the distance itself, sometimes it's a comparison between distances.

Question 17

In the coordinate plane, point A is at $(3, 4)$ and point B is at $(7, 1)$ . What is the distance between points A and B?

5 units (correct answer)
7 units
25 units
$\sqrt{7}$ units

Explanation: Use the distance formula, which is based on the Pythagorean theorem:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

. Here:

d = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

. Choice B adds the horizontal and vertical distances:

4 + 3 = 7

. Choice C forgets to take the square root of 25. Choice D incorrectly uses

\sqrt{4 + 3} = \sqrt{7}

Question 18

A rectangular garden has a diagonal walkway that measures 25 feet. If the garden is 15 feet wide, what is the length of the garden?

20 feet (correct answer)
29 feet
40 feet
10 feet

Explanation: Using the Pythagorean theorem:

a^2 + b^2 = c^2

. Here, width = 15 feet, diagonal = 25 feet, and we need to find length. So

15^2 + length^2 = 25^2

, which gives us

225 + length^2 = 625

. Therefore

length^2 = 400

, so

length = 20

feet. Choice B incorrectly adds 15 + 25 - 11. Choice C uses

15 + 25 = 40

. Choice D uses

25 - 15 = 10

Question 19

In a right triangle, the hypotenuse is 13 m and one leg is 5 m. What is the length of the other leg?

18 m
8 m
12 m (correct answer)
144 m

Explanation: This question tests applying the Pythagorean theorem a² + b² = c² to find an unknown leg in a right triangle with given hypotenuse and one leg. For a right triangle with hypotenuse c=13 m and leg b=5 m, rearrange to a² = c² - b²: a² = 169 - 25 = 144, so a = √144 = 12 m. In this specific problem, with hypotenuse 13 m and one leg 5 m, the other leg is found using a = √(13² - 5²) = √(169 - 25) = √144 = 12 m. The correct setup involves identifying the hypotenuse as the longest side, subtracting the squared leg from the squared hypotenuse, and taking the square root, matching choice B. A common error is subtracting incorrectly or using addition like 13 + 5 = 18, or not taking the square root and choosing 144 m. To solve: (1) identify the right triangle, (2) label hypotenuse as 13 m, one leg as 5 m, other as a, (3) note hypotenuse and one leg known, other leg unknown, (4) set up a² = 13² - 5², (5) calculate 169 - 25 = 144, a = √144 = 12 m, (6) verify it's reasonable as 12 m is between 5 m and 13 m. Common mistakes include treating the given leg as hypotenuse or arithmetic errors like 13² = 179.

Question 20

On a coordinate plane, point $A$ is at $(0,0)$ and point $B$ is at $(9,12)$ . What is the distance from $A$ to $B$ ?

225
21
3
15 (correct answer)

Explanation: This problem tests applying the Pythagorean theorem a² + b² = c² to find an unknown side in right triangles, such as 2D problems like ladders or diagonals, or even 3D space diagonals. For a right triangle with legs a and b, and hypotenuse c (the longest side opposite the 90° angle), if two sides are known, the third can be found using a² + b² = c²; for example, to find the hypotenuse, plug in the legs like 6² + 8² = 36 + 64 = 100 = c², so c = √100 = 10, or to find a leg, rearrange to a² = c² - b², such as if c=13 and b=5, then a² = 169 - 25 = 144, so a=12. In real-world scenarios, identify the right triangle, like a ladder against a wall forming a right angle, assign values such as base=9 ft and ladder=15 ft, and solve with 9² + h² = 15² to get h=12 ft. In this specific coordinate problem, points A(0,0) and B(9,12) form a right triangle with legs 9 and 12 (differences in x and y), so distance d = √(9² + 12²) = √(81 + 144) = √225 = 15. The correct setup involves recognizing the distance formula as Pythagorean theorem with Δx=9 and Δy=12 as legs, calculating to get 15. Common errors include adding coordinates like 0+9=9, mistaking differences, forgetting square root and saying 225, or not squaring. To solve: (1) identify right triangle from axes, (2) label legs Δx=9 and Δy=12, hypotenuse d, (3) note legs known, d unknown, (4) set up d² = Δx² + Δy², (5) calculate 81 + 144 = 225, √225=15, (6) verify 15 is reasonable. Common mistakes: a + b = d like 9+12=21, arithmetic errors, wrong differences, negative values.