Geometry Quiz: Solving Right Triangles Pythagorean Theorem Trigonometry

20 questions · exam conditions

0:00

Solving Right Triangles Pythagorean Theorem TrigonometryQuestion 1 of 20

In the plane, right triangle $\triangle GHI$ is shown. The right angle is explicitly marked at $H$ ( $\angle GHI = 90^\circ$ ). The hypotenuse is $\overline{GI}$ and is labeled $15$ . The leg $\overline{GH}$ is labeled $9$ . The leg $\overline{HI}$ is unlabeled.

Which method should be used to find the length of $\overline{HI}$ ?

(Diagram is not drawn to scale. No acute angle measures are given.)

Use the Pythagorean Theorem with

15

and

9

Use

\sin(90^\circ)=\dfrac{9}{15}

Use

\tan(9^\circ)=\dfrac{HI}{15}

Use a

30

60

90

triangle relationship.

← Back to quizzes

Geometry Quiz

Geometry Quiz: Solving Right Triangles Pythagorean Theorem Trigonometry

Practice Solving Right Triangles Pythagorean Theorem Trigonometry in Geometry 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 Solving Right Triangles Pythagorean Theorem Trigonometry, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

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

Which method should be used to find the length of $\overline{HI}$ ?

(Diagram is not drawn to scale. No acute angle measures are given.)

Use the Pythagorean Theorem with $15$ and $9$ . (correct answer)
Use $\sin(90^\circ)=\dfrac{9}{15}$ .
Use $\tan(9^\circ)=\dfrac{HI}{15}$ .
Use a $30$ - $60$ - $90$ triangle relationship.

Explanation: Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse GI = 15 and one leg GH = 9. Since we have the hypotenuse and one leg, and need the other leg HI, the Pythagorean theorem applies. The equation is HI² = 15² - 9². This setup correctly finds HI, matching the method in choice A. A common misconception is assuming a special triangle like 30-60-90 without angle information, as in choice D. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.

Question 2

In the plane, right triangle $\triangle JKL$ is shown. The right angle is explicitly marked at $K$ ( $\angle JKL = 90^\circ$ ). The hypotenuse is $\overline{JL}$ (explicitly identified) and is labeled $10$ . The acute angle at $J$ is marked and labeled $35^\circ$ . The leg $\overline{JK}$ is unlabeled.

What is the length of $\overline{JK}$ ?

(Diagram is not drawn to scale. No other angles are marked.)

$10\sin(35^\circ)$
$10\cos(35^\circ)$ (correct answer)
$\dfrac{10}{\sin(35^\circ)}$
$10\tan(35^\circ)$

Explanation: Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse

\overline{JL} = 10

and the acute angle at

J = 35^\circ

. Since we have the hypotenuse and an angle, and need the adjacent leg

\overline{JK}

, trigonometric ratios apply. The setup is

\cos(35^\circ) = \frac{\text{JK}}{10}

. This gives

\text{JK} = 10 \cos(35^\circ)

, matching choice B. A common misconception is using sine for the adjacent side, as in choice A, confusing opposite and adjacent. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.

Question 3

In the diagram, $\triangle RST$ is a right triangle with the right angle explicitly marked at $S$ . The hypotenuse is $\overline{RT}$ . The legs are labeled $RS=5$ and $ST=12$ . Which method should be used to find the length of $\overline{RT}$ ?

(Diagram is not drawn to scale; no acute angles are labeled.)

Use the Pythagorean Theorem because two side lengths are known. (correct answer)
Use sine because an acute angle and the opposite side are known.
Use cosine because an acute angle and the adjacent side are known.
Use tangent because an acute angle and two legs are known.

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, both legs are given as 5 and 12. The Pythagorean theorem applies because both legs are known, and we need the hypotenuse. The equation is RT = √(5² + 12²). This justifies using the theorem over trigonometry, matching choice A. A common misconception is opting for sine when no angles are given, as in choice B. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.

Question 4

In the right triangle $\triangle VWX$ shown, $\angle W$ is a right angle and the hypotenuse is $\overline{VX}$ . The acute angle at $V$ is $41^\circ$ , and $WX=10$ . What is the length of $\overline{VX}$ ?

$\dfrac{10}{\cos 41^\circ}$
$\dfrac{10}{\sin 41^\circ}$ (correct answer)
$10\sin 41^\circ$
$10\cos 41^\circ$

Explanation: This problem involves finding the hypotenuse using trigonometry. We are given the opposite side WX = 10 and the angle at V = 41°, and need to find the hypotenuse VX. Since we have the opposite side and need the hypotenuse, we use the sine ratio: sin(angle) = opposite/hypotenuse. Setting up the equation: sin(41°) = 10/VX, which rearranges to VX = 10/sin(41°). This is justified because sine relates the opposite side to the hypotenuse in a right triangle. A common mistake is using cosine (10/cos(41°)) which would be used if WX were adjacent to angle V. Before solving, identify the position of the known side relative to the given angle.

Question 5

In right triangle $\triangle DEF$ shown, $\angle E$ is a right angle. The hypotenuse is $DF=13$ and one leg is $DE=5$ . What is the length of leg $EF$ ?

$8$
$\sqrt{194}$
$\sqrt{144}$ (correct answer)
$18$

Explanation: This problem involves finding a missing leg when given the hypotenuse and one leg of a right triangle. We have hypotenuse DF = 13 and leg DE = 5, and need to find leg EF. The Pythagorean theorem applies in the form a² + b² = c², where c is the hypotenuse. Rearranging to find the missing leg: EF² = DF² - DE² = 13² - 5² = 169 - 25 = 144. Therefore, EF = √144 = 12. A common mistake is adding instead of subtracting when finding a leg (13² + 5² = 194). Remember: when finding a leg, subtract the known leg squared from the hypotenuse squared.

Question 6

Right triangle $\triangle STU$ is shown in the plane. The right angle is explicitly marked at $T$ ( $\angle STU = 90^\circ$ ). The hypotenuse is $\overline{SU}$ and is labeled $17$ . The leg $\overline{TU}$ is labeled $8$ . The acute angle at $S$ is unlabeled.

What is the measure of $\angle S$ ?

(Diagram is not drawn to scale. No other angles are marked.)

$\sin^{-1}\!\left(\dfrac{8}{17}\right)$ (correct answer)
$\cos^{-1}\!\left(\dfrac{8}{17}\right)$
$\tan^{-1}\!\left(\dfrac{17}{8}\right)$
$\sin\!\left(\dfrac{8}{17}\right)$

Explanation: Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the hypotenuse SU = 17 and the opposite leg to angle S, TU = 8. Since we have the opposite side and hypotenuse, and need the angle at S, inverse trigonometric ratios apply. The setup is angle S = sin⁻¹(8/17). This correctly finds the angle, matching choice A. A common misconception is using inverse cosine instead, as in choice B, which would apply to the adjacent side. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.

Question 7

A right triangle $\triangle MNO$ is shown with the right angle at $N$ . What is the length of hypotenuse $MO$ ?

$21$
$\sqrt{63}$
$\sqrt{225}$
$15$ (correct answer)

Explanation: This problem asks us to find the hypotenuse of a right triangle given both legs. We have a right angle at N, with legs MN = 9 and NO = 12. The Pythagorean theorem applies directly since we know both legs: MO² = MN² + NO². Setting up the calculation: MO² = 9² + 12² = 81 + 144 = 225, so MO = √225 = 15. The answer is 15 because this is a 3-4-5 right triangle scaled by 3 (9-12-15). A common error would be adding the legs directly (9 + 12 = 21) instead of using the Pythagorean theorem. Recognizing special right triangle ratios like 3-4-5 can help verify your answer quickly.

Question 8

In the coordinate plane, triangle $\triangle ABC$ is shown. $\angle B$ is a right angle (marked with a square). The segment $\overline{AB}$ is horizontal from $A(-1,2)$ to $B(5,2)$ , and $\overline{BC}$ is vertical from $B(5,2)$ to $C(5,10)$ . The hypotenuse is $\overline{AC}$ . The diagram is not drawn to scale. No other angles or lengths are marked.

What is the length of $\overline{AC}$ ?

$6+8=14$
$\sqrt{6^2+8^2}$ (correct answer)
$\sqrt{8^2-6^2}$
$\sqrt{6^2-8^2}$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, the coordinates provide the lengths of the legs AB = 6 and BC = 8 in right triangle ABC with right angle at B. The Pythagorean theorem applies because both legs are known, and we need the hypotenuse AC. The correct equation is AC = √(6² + 8²). This setup is justified as it directly relates the squares of the legs to the square of the hypotenuse in a right triangle. A common misconception is subtracting the squares, as in choice C, which might confuse finding a leg with finding the hypotenuse. To transfer this strategy, always choose the method—Pythagorean for sides or trig for angles—before computing.

Question 9

In the diagram, $\triangle PQR$ is a right triangle with the right angle explicitly marked at $Q$ . The hypotenuse is $\overline{PR}$ . The side $PQ$ is labeled $12$ , and the acute angle at $P$ is labeled $35^\circ$ . What is the length of $\overline{PR}$ ?

(Diagram is not drawn to scale; no other angles are marked.)

$12\cos 35^\circ$
$\dfrac{12}{\cos 35^\circ}$ (correct answer)
$\dfrac{12}{\sin 35^\circ}$
$12\tan 35^\circ$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, one leg is 12, and an adjacent acute angle is 35°. Cosine applies because it relates the adjacent side to the hypotenuse for the given angle. The equation is cos(35°) = 12 / PR, so PR = 12 / cos(35°). This matches choice B, as it correctly isolates the hypotenuse. A common misconception is using sine instead, leading to 12 / sin(35°) in choice C, which would be for the opposite side. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.

Question 10

In right triangle $\triangle ABC$ shown, $\angle C$ is a right angle. The legs are $AC=6$ and $BC=8$ . What is the length of hypotenuse $AB$ ?

$10$ (correct answer)
$14$
$\sqrt{28}$
$\sqrt{100}$

Explanation: This problem requires solving for the hypotenuse of a right triangle using the Pythagorean theorem. We are given the two legs: AC = 6 and BC = 8, and need to find the hypotenuse AB. Since we have both legs of a right triangle, the Pythagorean theorem applies: a² + b² = c². Setting up the equation: 6² + 8² = AB², which gives us 36 + 64 = 100, so AB = √100 = 10. The answer is justified because 10² = 100 = 36 + 64. A common error would be adding the legs directly (6 + 8 = 14) instead of using the Pythagorean theorem. When solving right triangles, always identify what's given and what method applies before computing.

Question 11

A right triangle $\triangle JKL$ is shown in the plane. $\angle K$ is marked as a right angle. The hypotenuse is $\overline{JL}$ and is labeled $25$ . The leg $\overline{KL}$ is labeled $7$ . The diagram is not drawn to scale. No other angles or lengths are marked.

What is the length of $\overline{JK}$ ?

$\sqrt{25^2-7^2}$ (correct answer)
$\sqrt{25^2+7^2}$
$25-7$
$\sqrt{7^2-25^2}$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, the hypotenuse JL = 25 and one leg KL = 7 are given in right triangle JKL with right angle at K. The Pythagorean theorem applies because we need the other leg JK, with the hypotenuse and one leg known. The correct equation is JK = √(25² - 7²). This is justified as rearranging the theorem isolates the unknown leg by subtracting squares. A distractor like choice B adds squares, which finds the hypotenuse instead of a leg. To transfer this strategy, always choose the method—Pythagorean for sides or trig for angles—before computing.

Question 12

A right triangle $\triangle KLM$ is shown in the plane. The right angle is explicitly marked at $L$ ( $\angle KLM = 90^\circ$ ). The legs are $\overline{KL}=6$ and $\overline{LM}=8$ . The slanted side $\overline{KM}$ is explicitly identified as the hypotenuse.

Which claim about the triangle is NOT justified?

(Diagram is not drawn to scale. No angle measures other than the right angle are marked.)

$KM=10$ .
$KL^2+LM^2=KM^2$ .
$\sin(\angle K)=\dfrac{8}{10}$ .
$\angle K=60^\circ$ . (correct answer)

Explanation: Solving right triangles involves using the Pythagorean theorem or trigonometry to find unknown sides or angles. In this problem, we are given the two legs KL = 6 and LM = 8. To evaluate the claims, the Pythagorean theorem and trig ratios apply to find the hypotenuse and angles. The equation for the hypotenuse is KM² = 6² + 8², giving KM = 10. Then, sin(∠K) = 8/10, so ∠K ≈ 53.13°, not 60°, making claim D unjustified. A common misconception is assuming a 60° angle based on side ratios without calculating, as in choice D. To transfer this strategy, always choose whether to use the Pythagorean theorem or trig ratios based on the given information before computing.

Question 13

In the diagram, $\triangle XYZ$ is a right triangle with the right angle explicitly marked at $Y$ . The hypotenuse is $\overline{XZ}$ . The leg $YZ$ is labeled $4$ , and the leg $XY$ is labeled $3$ . Which claim about the triangle is NOT justified?

(Diagram is not drawn to scale; no acute angle measures are given.)

$XZ=5$
$\sin(\angle X)=\dfrac{4}{5}$
$\cos(\angle X)=\dfrac{4}{5}$ (correct answer)
$\tan(\angle X)=\dfrac{4}{3}$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, both legs are given as 3 and 4. The Pythagorean theorem applies to verify the hypotenuse as 5, enabling trigonometric ratio checks. The equations are sin(X) = 4/5, cos(X) = 3/5, and tan(X) = 4/3. Choice C is not justified because it incorrectly states cos(X) = 4/5 instead of 3/5. A common misconception is swapping opposite and adjacent sides in cosine, leading to the error in choice C. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.

Question 14

In the diagram, $\triangle MNO$ is a right triangle with the right angle explicitly marked at $N$ . The hypotenuse is $\overline{MO}$ . The side $NO$ is labeled $7$ , and the acute angle at $O$ is labeled $28^\circ$ . Which expression represents the correct setup to find the length of the hypotenuse $\overline{MO}$ ?

(Diagram is not drawn to scale; no other angles are marked.)

$\sin 28^\circ=\dfrac{7}{MO}$
$\cos 28^\circ=\dfrac{7}{MO}$ (correct answer)
$\tan 28^\circ=\dfrac{MO}{7}$
$\sin 28^\circ=\dfrac{MO}{7}$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, the adjacent side is 7, and the acute angle is 28°. Cosine applies because it relates the adjacent side to the hypotenuse for the given angle. The equation is cos(28°) = 7 / MO. This matches choice B, as it sets up correctly for solving MO. A common misconception is using sine for the adjacent side, leading to sin(28°) = 7 / MO in choice A. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.

Question 15

In the right triangle shown, $\angle A$ is a right angle. The hypotenuse is $\overline{BC}$ . If $AB=8$ and $BC=17$ , what is the length of $AC$ ?

$\sqrt{353}$
$9$
$25$
$15$ (correct answer)

Explanation: This problem asks us to find a leg of a right triangle using the Pythagorean theorem. We're given leg AB = 8 and hypotenuse BC = 17, with angle A being the right angle. Since we need to find a leg when given the other leg and hypotenuse, we use: a² = c² - b². Setting up: AC² = BC² - AB² = 17² - 8² = 289 - 64 = 225, so AC = √225 = 15. The answer is 15, not √225, because we must simplify the square root. A student might add the squares instead (getting √353) by incorrectly treating this as finding a hypotenuse. Always check: is the unknown side the hypotenuse (longest side) or a leg before deciding whether to add or subtract.

Question 16

In the right triangle $\triangle JKL$ shown, $\angle K$ is a right angle and the hypotenuse is $\overline{JL}$ . The acute angle at $J$ is $58^\circ$ , and $JL=20$ . What is the length of $\overline{JK}$ ?

$20\sin 58^\circ$
$\dfrac{20}{\sin 58^\circ}$
$20\cos 58^\circ$ (correct answer)
$20\tan 58^\circ$

Explanation: This problem involves finding the adjacent side of a right triangle using trigonometry. We are given the hypotenuse JL = 20 and the angle at J = 58°, and need to find the adjacent side JK. Since we have the hypotenuse and need the adjacent side, we use the cosine ratio: cos(angle) = adjacent/hypotenuse. Setting up the equation: cos(58°) = JK/20, which gives us JK = 20·cos(58°). This is justified because cosine relates the adjacent side to the hypotenuse. A common mistake is using sine (20·sin(58°)) which would give the opposite side instead. Before solving, identify which side you need relative to the given angle to choose the correct trigonometric function.

Question 17

In the diagram, $\triangle JKL$ is a right triangle with the right angle explicitly marked at $K$ . The hypotenuse is $\overline{JL}$ . The leg lengths are labeled $JK=9$ and $JL=15$ . What is the length of $\overline{KL}$ ?

(Diagram is not drawn to scale; no other angles are marked.)

$\sqrt{306}$
$\sqrt{144}$ (correct answer)
$24$
$\sqrt{81+15}$

Explanation: Solving right triangles involves finding unknown sides or angles using the Pythagorean theorem or trigonometric ratios. In this problem, one leg is 9, and the hypotenuse is 15. The Pythagorean theorem applies because a leg and the hypotenuse are given, and we need the other leg. The equation is

KL = \sqrt{15^2 - 9^2}

. This yields

KL = \sqrt{144}

, matching choice B, as it simplifies correctly. A common misconception is adding instead of subtracting, leading to

\sqrt{15^2 + 9^2} = \sqrt{306}

in choice A. To transfer this strategy, always choose between the Pythagorean theorem or trigonometry based on whether sides or angles are provided before computing.

Question 18

Refer to the figure below. In right triangle $PQR$ with the right angle at $Q$ , $PQ = 5$ and $QR = 12$ . What is $\cos(R)$ ?

$\frac{5}{13}$
$\frac{12}{13}$ (correct answer)
$\frac{5}{12}$
$\frac{12}{5}$

Explanation: First, find

PR

using the Pythagorean theorem:

PR = \sqrt{25 + 144} = 13

. Since

P

and

R

are complementary,

\cos(R) = \sin(P) = \frac{QR}{PR} = \frac{12}{13}

. Alternatively,

\cos(R) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}

. Choice A is

\sin(R)

. Choice C is

\tan(P)

. Choice D is

\tan(R)

Question 19

In the right triangle shown, $\angle A$ is a right angle. The hypotenuse is $\overline{BC}$ . If $BC=30$ and $\angle C=22^\circ$ , what is the length of $AB$ ?

$30\cos(22^\circ)$
$\dfrac{30}{\sin(22^\circ)}$
$30\sin(22^\circ)$ (correct answer)
$\dfrac{30}{\cos(22^\circ)}$

Explanation: This problem requires finding the side opposite to a given angle in a right triangle. We have hypotenuse BC = 30, angle C = 22°, and angle A is the right angle. Since AB is opposite to angle C and BC is the hypotenuse, we use sine: sin(C) = opposite/hypotenuse = AB/BC. Rearranging: AB = BC × sin(C) = 30sin(22°). The answer is 30sin(22°) because sine relates the opposite side to the hypotenuse. A common error would be using cosine (30cos(22°)), which would give the adjacent side AC instead. When solving with trigonometry, always identify which side you need relative to the given angle: opposite uses sine, adjacent uses cosine.

Question 20

In the right triangle $\triangle PQR$ shown, $\angle Q$ is a right angle and the hypotenuse is $\overline{PR}$ . The acute angle at $P$ is $22^\circ$ , and $PQ=12$ . Which expression represents the correct setup to find the length of $\overline{QR}$ ?

$\tan 22^\circ=\dfrac{QR}{12}$ (correct answer)
$\sin 22^\circ=\dfrac{QR}{12}$
$\cos 22^\circ=\dfrac{QR}{12}$
$\tan 22^\circ=\dfrac{12}{QR}$

Explanation: This problem asks for the correct setup to find the opposite side using trigonometry. We are given the adjacent side PQ = 12 and the angle at P = 22°, and need to find the opposite side QR. Since we have the adjacent side and need the opposite side, we use the tangent ratio: tan(angle) = opposite/adjacent. The correct setup is: tan(22°) = QR/12, which can be rearranged to find QR. This is justified because tangent relates the opposite side to the adjacent side. A common mistake is using sine or cosine, which would require the hypotenuse. When setting up trigonometric equations, identify which sides you have and need relative to the given angle.