Geometry Quiz: Congruence And Similarity To Solve Problems

11 questions · exam conditions

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Congruence And Similarity To Solve ProblemsQuestion 1 of 11

Triangle $PQR$ has a right angle at $Q$ . If triangle $PQR$ is similar to triangle $STU$ with similarity ratio $3:2$ , and the area of triangle $STU$ is 24 square units, what is the length of the hypotenuse of triangle $PQR$ if the hypotenuse of triangle $STU$ is 8 units?

10

units

12

units

16

units

18

units

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Geometry Quiz

Geometry Quiz: Congruence And Similarity To Solve Problems

Practice Congruence And Similarity To Solve Problems 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 Congruence And Similarity To Solve Problems, 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

$10$ units
$12$ units (correct answer)
$16$ units
$18$ units

Explanation: Since the triangles are similar with ratio

3:2

, corresponding linear dimensions are in the ratio

3:2

. The hypotenuse of

\triangle STU

is 8 units, so the hypotenuse of

\triangle PQR

8 \cdot \frac{3}{2} = 12

units. Note that areas are in the ratio of the square of the similarity ratio, which would be

9:4

, making the area of

\triangle PQR

equal to

24 \cdot \frac{9}{4} = 54

square units, but this doesn't affect the hypotenuse calculation.

Question 2

A right triangle has legs of length $a$ and $b$ and hypotenuse of length $c$ . A second triangle is formed by connecting the midpoints of the three sides of the original triangle. If the area of the original triangle is 60 square units, what is the area of the triangle formed by connecting the midpoints?

$15$ square units (correct answer)
$20$ square units
$30$ square units
$45$ square units

Explanation: The triangle formed by connecting the midpoints of any triangle is similar to the original triangle with a similarity ratio of

1:2

. This follows from the theorem that a line connecting midpoints of two sides of a triangle is parallel to the third side and half its length. Since areas of similar figures are in the ratio of the square of the similarity ratio, the area ratio is

(1/2)^2 = 1/4

. Therefore, the area of the midpoint triangle is

60 \cdot \frac{1}{4} = 15

square units.

Question 3

Triangle $ABC$ is a right triangle with the right angle at $C$ . If $AC = 9$ and $BC = 12$ , and triangle $ABC$ is similar to triangle $PQR$ where the shortest side of triangle $PQR$ is 6, what is the length of the longest side of triangle $PQR$ ?

$8$
$15$
$12$
$10$ (correct answer)

Explanation: When you encounter similar triangles, the key insight is that all corresponding sides are proportional by the same scale factor. This means if you know one pair of corresponding sides, you can find the scale factor and determine all other sides. First, let's find all sides of triangle

ABC

. Since it's a right triangle with legs

AC = 9

and

BC = 12

, you can use the Pythagorean theorem to find the hypotenuse:

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

. So triangle

ABC

has sides 9, 12, and 15, where 9 is the shortest side and 15 is the longest. Since triangle

PQR

is similar to triangle

ABC

, and the shortest side of triangle

PQR

is 6, you can find the scale factor by comparing shortest sides:

\frac{6}{9} = \frac{2}{3}

. This means each side of triangle

PQR

\frac{2}{3}

the length of the corresponding side in triangle

ABC

. Therefore, the longest side of triangle

PQR

15 \times \frac{2}{3} = 10

. Looking at the wrong answers: A) 8 represents the middle side of triangle

PQR

(

12 \times \frac{2}{3}

), not the longest. B) 15 is the longest side of the original triangle

ABC

, not the similar triangle. C) 12 is the middle side of triangle

ABC

, showing confusion about which triangle we're analyzing. Remember: when working with similar triangles, always identify the scale factor first by comparing corresponding sides, then apply it consistently to find unknown measurements.

Question 4

In right triangle $RST$ with right angle at $S$ , the altitude from $S$ to hypotenuse $RT$ creates two smaller triangles. If $RS = 12$ and $ST = 16$ , what is the length of the altitude from $S$ to $RT$ ?

$\frac{192}{5}$
$\frac{96}{5}$
$\frac{48}{5}$ (correct answer)
$\frac{240}{5}$

Explanation: When you encounter a right triangle with an altitude drawn to the hypotenuse, you're dealing with geometric mean relationships. This altitude creates two smaller triangles that are all similar to each other and to the original triangle. To find the altitude from

S

to hypotenuse

RT

, you can use the area method, which is often the most straightforward approach. First, calculate the area of triangle

RST

using the two legs as base and height: Area

= \frac{1}{2} \times 12 \times 16 = 96

. Next, find the hypotenuse using the Pythagorean theorem:

RT = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20

. Now use the fact that the area can also be calculated as

\frac{1}{2} \times \text{base} \times \text{height}

, where the base is the hypotenuse and the height is the altitude:

96 = \frac{1}{2} \times 20 \times h

. Solving for

h

h = \frac{96 \times 2}{20} = \frac{192}{20} = \frac{48}{5}

. Looking at the wrong answers: (A)

\frac{192}{5}

occurs if you forget to divide by 2 when using the area formula. (B)

\frac{96}{5}

results from dividing the area by the hypotenuse without accounting for the factor of

\frac{1}{2}

. (D)

\frac{240}{5}

comes from incorrectly using

12 \times 16 = 192

as the area instead of

\frac{1}{2} \times 12 \times 16

. Remember: when finding altitude to the hypotenuse, the area method (using both leg calculations and hypotenuse calculation) is typically the most reliable approach.

Question 5

Triangle $ABC$ has vertices $A(0, 0)$ , $B(8, 0)$ , and $C(0, 6)$ . A line parallel to $BC$ intersects $AB$ at point $P(5, 0)$ and $AC$ at point $Q$ . What are the coordinates of point $Q$ ?

$\left(0, \frac{8}{15}\right)$
$\left(0, \frac{15}{8}\right)$
$\left(0, \frac{15}{4}\right)$ (correct answer)
$\left(0, \frac{4}{15}\right)$

Explanation: When you encounter parallel lines intersecting the sides of a triangle, you're dealing with the Side-Splitter Theorem (also called the Triangle Proportionality Theorem). This theorem states that if a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally. First, let's find the slope of line

BC

. With

B(8, 0)

and

C(0, 6)

, the slope is

\frac{6-0}{0-8} = -\frac{3}{4}

. Since line

PQ

is parallel to

BC

, it has the same slope. Now apply the proportionality relationship. Point

P(5, 0)

divides

AB

in the ratio

AP:PB = 5:3

(since

AP = 5

and

PB = 3

). By the Side-Splitter Theorem, point

Q

must divide

AC

in the same ratio

5:3

. Since

AC

has length 6 (from

(0,0)

(0,6)

), and

Q

divides it in ratio

5:3

, we have

AQ = \frac{5}{5+3} \times 6 = \frac{5}{8} \times 6 = \frac{15}{4}

. Therefore,

Q = \left(0, \frac{15}{4}\right)

, which is answer C. Answer A gives

\frac{8}{15}

, which incorrectly flips the ratio. Answer B gives

\frac{15}{8}

, which uses the correct numbers but in the wrong fraction form. Answer D gives

\frac{4}{15}

, which represents the complement ratio error. Strategy tip: For parallel line problems in triangles, always identify the ratio on one side first, then apply that same ratio to find the corresponding division on the other side.

Question 6

In triangle $ABC$ , point $D$ is on side $AB$ and point $E$ is on side $AC$ such that $DE \parallel BC$ . If $AD = 6$ , $DB = 9$ , and $AE = 8$ , what is the perimeter of triangle $ADE$ given that the perimeter of triangle $ABC$ is 45?

$18$ (correct answer)
$24$
$27$
$30$

Explanation: Since

DE \parallel BC

, by the theorem about parallel lines in triangles,

\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}

. We have

AD = 6

and

DB = 9

, so

AB = 15

. The ratio is

\frac{AD}{AB} = \frac{6}{15} = \frac{2}{5}

. Since

AE = 8

, we get

AC = \frac{8}{2/5} = 20

, so

EC = 12

. Also,

DE = \frac{2}{5} \cdot BC

. The perimeter of

\triangle ABC

AB + AC + BC = 15 + 20 + BC = 45

, so

BC = 10

and

DE = 4

. The perimeter of

\triangle ADE

6 + 8 + 4 = 18

Question 7

A right triangle has sides in the ratio $5:12:13$ . If this triangle is similar to another right triangle whose shortest side is 15 units, what is the difference between the longest and shortest sides of the second triangle?

$36$ units
$27$ units
$30$ units
$24$ units (correct answer)

Explanation: When you encounter similar triangles with given ratios, you're working with proportional relationships where corresponding sides maintain the same scale factor throughout. The first triangle has sides in the ratio

5:12:13

. Since

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

, this confirms it's a right triangle where 5 is the shortest side, 12 is the middle side, and 13 is the hypotenuse (longest side). For the similar triangle with shortest side 15 units, you need to find the scale factor. Since the shortest sides correspond, set up the proportion:

\frac{15}{5} = 3

. This means the second triangle is 3 times larger than the first. Multiply each side of the original triangle by 3:

Shortest side: $5 \times 3 = 15$
Middle side: $12 \times 3 = 36$
Longest side: $13 \times 3 = 39$

The difference between longest and shortest sides is

39 - 15 = 24

units. Looking at the wrong answers: (A) 36 represents the middle side length, not the difference between longest and shortest. (B) 27 might result from incorrectly calculating

39 - 12 = 27

, using the wrong sides. (C) 30 could come from miscalculating the scale factor or the proportional relationships. Strategy tip: With similar triangles, always identify the scale factor first by comparing corresponding sides, then apply it consistently to all dimensions. Double-check by verifying the ratios remain constant across both triangles.

Question 8

In the diagram, $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ . Additionally, $\overline{FG} \parallel \overline{BC}$ with $F$ on $\overline{AD}$ and $G$ on $\overline{AE}$ . Given $AF = 3$ , $FD = 2$ , $DB = 5$ , and $AG = 4.5$ , what is the length of $\overline{GC}$ ?

$3$
$7.5$
$10.5$ (correct answer)
$15$

Explanation: Since

\overline{FG}\parallel\overline{BC}

\frac{AF}{AB}=\frac{AG}{AC}

. Here

AB = AF+FD+DB = 3+2+5 = 10

, so

\frac{3}{10}=\frac{4.5}{AC}

, giving

AC=15

. Then

GC = AC - AG = 15 - 4.5 = 10.5

. Choice A comes from computing only

AE - AG

with wrong

AE

. Choice B comes from stopping at

AE

. Choice D is

AC

itself, forgetting to subtract

AG

Question 9

Refer to the figure below. In $\triangle PQR$ , segment $\overline{ST}$ is drawn with $S$ on $\overline{PQ}$ and $T$ on $\overline{PR}$ . Given $PS = 4$ , $SQ = 6$ , $PT = 5$ , and $TR = 7$ , which statement is true?

$\overline{ST} \parallel \overline{QR}$ because $\frac{PS}{SQ} = \frac{PT}{TR}$ .
$\overline{ST} \parallel \overline{QR}$ because $\frac{PS}{PQ} = \frac{PT}{PR}$ .
$\overline{ST}$ is not parallel to $\overline{QR}$ because $\frac{PS}{SQ} \neq \frac{PT}{TR}$ . (correct answer)
$\overline{ST}$ is not parallel to $\overline{QR}$ because $PS + PT \neq SQ + TR$ .

Explanation: By the converse of the Side-Splitter Theorem,

\overline{ST} \parallel \overline{QR}

iff

\frac{PS}{SQ}=\frac{PT}{TR}

. Here

\frac{4}{6}=\frac{2}{3}

but

\frac{5}{7} \neq \frac{2}{3}

, so the segments are not parallel. Choice A is false because the ratios are not actually equal. Choice B misstates a valid form and is false numerically. Choice D uses an irrelevant criterion (sums, not ratios).

Question 10

In the figure shown, $\overline{DE} \parallel \overline{BC}$ in $\triangle ABC$ , with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$ . If $AD = 8$ , $AB = 20$ , and $BC = 15$ , what is the length of $\overline{DE}$ ?

$6$ (correct answer)
$7.5$
$9$
$10$

Explanation: Since

\overline{DE} \parallel \overline{BC}

\triangle ADE \sim \triangle ABC

with ratio

\frac{AD}{AB}=\frac{8}{20}=\frac{2}{5}

. Therefore

DE = \frac{2}{5}\cdot 15 = 6

. Choice B is half of

BC

, a common guess. Choice C uses ratio

\frac{AD}{DB}=\frac{8}{12}

multiplied by

BC

inaccurately. Choice D uses ratio

\frac{AD}{DB}=\frac{2}{3}

giving

\frac{2}{3}\cdot 15 = 10

, mistakenly using part-to-part instead of part-to-whole.

Question 11

In triangle $DEF$ , point $G$ lies on $DE$ such that $DG = 4$ and $GE = 12$ . Point $H$ lies on $DF$ such that $GH \parallel EF$ . If $DF = 20$ , what is the length of $FH$ ?

$5$
$15$ (correct answer)
$16$
$12$

Explanation: Since

GH \parallel EF

, by the theorem about parallel lines in triangles,

\frac{DG}{DE} = \frac{DH}{DF}

. We have

DG = 4

and

GE = 12

, so

DE = 16

. Therefore,

\frac{DG}{DE} = \frac{4}{16} = \frac{1}{4}

. Since

DF = 20

, we get

\frac{DH}{20} = \frac{1}{4}

, so

DH = 5

. Therefore,

FH = DF - DH = 20 - 5 = 15