Math 1 Quiz: Cpctc

6 questions · exam conditions

0:00

CpctcQuestion 1 of 6

In triangles $ABC$ and $DEF$ , $\overline{AB} \cong \overline{DE}$ , $\overline{BC} \cong \overline{EF}$ , and $\angle B \cong \angle E$ . If $m\angle A = 3x + 15$ and $m\angle D = 5x - 5$ , what is the value of $x$ ?

x = 10

x = 15

x = 20

x = 25

← Back to quizzes

Math 1 Quiz

Math 1 Quiz: Cpctc

Practice Cpctc in Math 1 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 Cpctc, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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

$x = 10$ (correct answer)
$x = 15$
$x = 20$
$x = 25$

Explanation: By SAS, triangles ABC and DEF are congruent. By CPCTC, corresponding angles are congruent, so

\angle A \cong \angle D

. Therefore

3x + 15 = 5x - 5

, which gives

20 = 2x

, so

x = 10

. Choice B results from solving

3x + 15 = 5x + 5

(sign error). Choice C results from setting up

3x - 15 = 5x - 5

(sign error in first expression). Choice D results from incorrectly solving

3x + 15 = 5x - 5

15 + 5 = 5x - 3x

, getting

20 = 2x

but then making an arithmetic error.

Question 2

In right triangle $ABC$ with right angle at $C$ , the altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$ . If $\triangle ACD \cong \triangle CBD$ , what can be concluded about triangle $ABC$ ?

Triangle $ABC$ has $\angle A = 30°$ and $\angle B = 60°$
Triangle $ABC$ is isosceles with $AB = AC$
Triangle $ABC$ is isosceles with $AC = BC$ (correct answer)
Triangle $ABC$ has all angles equal to $60°$

Explanation: When you encounter a problem involving congruent triangles formed by an altitude to the hypotenuse of a right triangle, focus on what the congruence tells you about the original triangle's properties. Given that

\triangle ACD \cong \triangle CBD

, these triangles share the altitude

\overline{CD}

and both contain right angles (

\angle ADC = \angle BDC = 90°

). For the triangles to be congruent, their corresponding sides must be equal. Since

CD

is common to both triangles, we need

AC = BC

for congruence to hold. This makes triangle

ABC

isosceles with the two legs equal. You can verify this by noting that if

AC = BC

, then

\angle CAD = \angle CBD

(the base angles of the isosceles triangle), and the triangles become congruent by AAS (Angle-Angle-Side). Answer A is incorrect because while the triangle has equal base angles, they're each

45°

, not

30°

and

60°

. In an isosceles right triangle, the two acute angles are always

45°

each. Answer B is wrong because it states

AB = AC

. Since

AB

is the hypotenuse and

AC

is a leg, this would violate the fundamental property that the hypotenuse is the longest side in a right triangle. Answer D is impossible since we already have a

90°

angle at

C

, so the triangle cannot have all

60°

angles. Study tip: When an altitude creates congruent triangles in a right triangle, the original triangle is always isosceles with equal legs, making it a

45°

45°

90°

triangle.

Question 3

Two triangles $\triangle JKL$ and $\triangle MNO$ are congruent. If $J$ corresponds to $M$ , $K$ corresponds to $N$ , and $L$ corresponds to $O$ , and if $m\angle J = 4y + 10$ , $m\angle M = 6y - 20$ , $m\angle K = 3z + 5$ , what is the value of $z$ when $m\angle N = 65°$ ?

$z = 15$
$z = 20$ (correct answer)
$z = 25$
$z = 30$

Explanation: Since the triangles are congruent with the given correspondence, by CPCTC we have

\angle J \cong \angle M

and

\angle K \cong \angle N

. From

\angle J \cong \angle M

4y + 10 = 6y - 20

, so

30 = 2y

and

y = 15

. From

\angle K \cong \angle N

3z + 5 = 65

, so

3z = 60

and

z = 20

. Choice A results from solving

4y + 10 = 65

incorrectly. Choice C results from

z = (65 + 10)/3

. Choice D results from

z = (65 + 25)/3

Question 4

Given that $\triangle PQR \cong \triangle STU$ by AAS, where $PQ = 2x - 3$ , $ST = x + 7$ , and $QR = 3y + 1$ . If $TU = 19$ , what is the value of $xy$ ?

30
40
50
60 (correct answer)

Explanation: From the congruence statement △PQR ≅ △STU, corresponding sides are equal by CPCTC. PQ corresponds to ST, so 2x - 3 = x + 7, which gives x = 10. QR corresponds to TU, so 3y + 1 = 19, which gives 3y = 18, so y = 6. Therefore xy = 10 × 6 = 60.

Question 5

In parallelogram $WXYZ$ , diagonal $\overline{WY}$ is drawn. Given that $\triangle WXY \cong \triangle YZW$ , and $m\angle XWY = 35°$ , what is $m\angle WYZ$ ?

$145°$
$55°$
$70°$
$35°$ (correct answer)

Explanation: When you encounter congruent triangles in a parallelogram with a diagonal, you're dealing with corresponding angles and the properties of parallel sides. Given that

\triangle WXY \cong \triangle YZW

, the key insight is identifying which angles correspond to each other. In these congruent triangles, angle

XWY

in the first triangle corresponds to angle

ZYW

in the second triangle. Since corresponding angles in congruent triangles are equal, and we know

m\angle XWY = 35°

, we can conclude that

m\angle ZYW = 35°

. But wait - angle

ZYW

is the same as angle

WYZ

! They're just written in different notation. So

m\angle WYZ = 35°

, making D correct. Let's examine why the other answers are wrong. Choice A (

145°

) likely comes from incorrectly thinking this angle is supplementary to the given

35°

angle, but these angles aren't supplementary. Choice B (

55°

) might result from subtracting

35°

from

90°

, perhaps assuming a right angle exists where none is given. Choice C (

70°

) could come from doubling the given angle, which has no geometric basis in this problem. Study tip: When working with congruent triangles, always carefully match up corresponding vertices in the congruence statement. The order matters -

\triangle WXY \cong \triangle YZW

tells you that W↔Y, X↔Z, and Y↔W, which directly gives you the angle correspondences you need.

Question 6

Quadrilateral $PQRS$ has diagonals that bisect each other at point $T$ . If $\triangle PQT \cong \triangle RST$ , which statement must be true by CPCTC?

$\overline{PQ} \parallel \overline{RS}$ and $\overline{PS} \parallel \overline{QR}$
$\overline{PQ} \cong \overline{RS}$ and $\overline{PS} \cong \overline{QR}$ (correct answer)
$\overline{PT} \cong \overline{RT}$ and $\overline{QT} \cong \overline{ST}$
$\angle PTQ \cong \angle RTS$ and $\angle PTS \cong \angle RTQ$

Explanation: Since

\triangle PQT \cong \triangle RST

, by CPCTC we know that

\overline{PQ} \cong \overline{RS}

and

\overline{PS} \cong \overline{QR}

(noting that

\overline{PT} \cong \overline{RT}

from the given congruence). Choice A describes parallelism, which requires additional reasoning beyond CPCTC. Choice C states what we already know from the diagonals bisecting each other. Choice D describes vertical angles, which are always congruent regardless of triangle congruence.