Unit 2 Review Sheet | Geometry | Welch Teaching & Coaching Website

Directions

Before you start

Show all work. An answer with no supporting equation or proportion earns no credit on the assessment.
Round every answer to two decimal places unless the answer is exact.
When you solve for a variable, substitute it back and state the length or angle the question actually asked for.
Questions 11 and 12 require a written sentence of justification, not just a number.
Attempt all twelve questions before opening a single solution. Give yourself about 45 minutes.
Each question group has a video attached. Use it only after you have tried the questions — and note that the videos do not appear on the printed copy.

Formulas provided

\[m\angle A+m\angle B+m\angle C=180^\circ\]

\[a^2+b^2=c^2\]

The Isosceles Triangle Theorem, the side-splitter theorem, and the geometric mean relationships are not provided. Know them.

Self-check plan: mark each question you had to guess on. Any guessed question points to the lesson listed beside it in the answer key at the bottom of the page — reread that section of the unit content summary before the assessment.

Questions 1 – 3 — Corresponding parts of congruent figures

Stuck? — naming corresponding parts Figuring out all the angles for congruent triangles Khan Academy Open the video →

Stuck? — writing the statement Finding congruent triangles Khan Academy Open the video →

Use the letter order in each congruence statement. The figures are not drawn to scale.

1

Name two corresponding parts.

Given \(\triangle ABC\cong\triangle XYZ\), find \(XZ\) and \(m\angle Z\).

Show solution

Match by position: \(A\leftrightarrow X\), \(B\leftrightarrow Y\), \(C\leftrightarrow Z\). Then \(\overline{XZ}\) corresponds to \(\overline{AC}\), which is marked \(15\).

\[XZ=AC=15\]

\(\angle Z\) corresponds to \(\angle C\), which is not labeled — so find it from the angle sum first.

\[m\angle C=180-47-68=65^\circ\;\Rightarrow\;m\angle Z=65^\circ\]

2

Solve for three variables.

Given \(\triangle DEF\cong\triangle GHJ\), find \(x\), \(y\), and \(z\).

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\(\overline{DE}\) matches \(\overline{GH}\), \(\overline{EF}\) matches \(\overline{HJ}\), \(\overline{FD}\) matches \(\overline{JG}\). Set each pair equal.

\[4x+1=21\;\Rightarrow\;4x=20\;\Rightarrow\;x=5\]

\[3y-2=19\;\Rightarrow\;3y=21\;\Rightarrow\;y=7\]

\[2z+3=15\;\Rightarrow\;2z=12\;\Rightarrow\;z=6\]

3

Solve for \(x\), then find an angle.

Given \(\triangle ABC\cong\triangle DEF\), find \(x\) and then \(m\angle F\).

Show solution

\(\angle E\) corresponds to \(\angle B\), so the expression at \(E\) equals \(85^\circ\).

\[2x+9=85\;\Rightarrow\;2x=76\;\Rightarrow\;x=38\]

\(\angle F\) corresponds to \(\angle C\), which comes from the angle sum in \(\triangle ABC\).

\[m\angle F=m\angle C=180-51-85=44^\circ\]

Questions 4 – 7 — Isosceles triangles, medians, and altitudes

Stuck? — isosceles examples Equilateral and isosceles example problems Khan Academy Open the video →

Stuck? — solving for x Another isosceles example problem Khan Academy Open the video →

Stuck? — medians Triangle medians and centroids Khan Academy Open the video →

Decide what the marks are telling you before writing any equation.

4

Find the base angles and \(y\).

\(\triangle RST\) has \(\overline{RS}\cong\overline{RT}\) and a vertex angle of \(38^\circ\). Find the measure of each base angle, then find \(y\).

Show solution

Subtract the vertex angle from \(180\), then split what is left in half.

\[\text{base angle}=\frac{180-38}{2}=\frac{142}{2}=71^\circ\]

\[4y+3=71\;\Rightarrow\;4y=68\;\Rightarrow\;y=17\]

5

Use the converse.

In \(\triangle ABC\), \(\angle B\cong\angle C\). Find \(x\) and the length of \(\overline{AB}\).

Show solution

The arcs mark congruent angles, so this is the converse of the Isosceles Triangle Theorem: the sides opposite those angles are congruent, which makes the two legs equal.

\[5x-8=2x+7\;\Rightarrow\;3x=15\;\Rightarrow\;x=5\]

\[AB=5(5)-8=17\qquad AC=2(5)+7=17\]

Both expressions give \(17\), which confirms the algebra.

6

Use a median.

\(\overline{PM}\) is a median of \(\triangle PQR\). Find \(x\), \(QM\), and \(QR\).

Show solution

A median lands on the midpoint, so \(QM=MR\).

\[5x-3=2x+9\;\Rightarrow\;3x=12\;\Rightarrow\;x=4\]

\[QM=5(4)-3=17\qquad MR=2(4)+9=17\]

\[QR=17+17=34\]

The question asks for all three. Answering only \(x\) leaves points on the table.

7

Median, altitude, both, or neither?

In each triangle, \(\overline{AD}\) is drawn from \(A\) to \(\overline{BC}\). Classify each one.

Show solution

Tick marks mean midpoint, so median. A square corner means perpendicular, so altitude.

a. tick marks only → median
b. right angle only → altitude
c. tick marks and a right angle → both
d. no marks → neither

Questions 8 & 9 — Similar figures and scale factor

Stuck? — finding the scale factor Identifying scale factors Khan Academy Open the video →

Stuck? — similarity examples Similarity example problems Khan Academy Open the video →

Find the scale factor from the one complete pair, then use it everywhere else.

8

Scale factor and perimeter.

Quadrilateral \(ABCD\sim\) quadrilateral \(EFGH\). Find the scale factor from \(ABCD\) to \(EFGH\), find \(FG\), and find the perimeter of \(EFGH\) given that the perimeter of \(ABCD\) is \(42\).

Show solution

Letter order pairs \(\overline{AB}\) with \(\overline{EF}\) and \(\overline{BC}\) with \(\overline{FG}\).

\[k=\frac{EF}{AB}=\frac{20}{12}=\frac{5}{3}\approx 1.67\]

\[FG=9\times\frac{5}{3}=15\]

Perimeter scales by \(k\), not by \(k^2\) — that exponent belongs to area.

\[P_{EFGH}=42\times\frac{5}{3}=70\]

9

Find two missing sides.

\(\triangle ABC\sim\triangle RST\). Find \(ST\) and \(TR\).

Show solution

The complete pair is \(\overline{AB}\) with \(\overline{RS}\).

\[k=\frac{RS}{AB}=\frac{15}{10}=1.5\]

\[ST=14\times 1.5=21\qquad TR=16\times 1.5=24\]

Questions 10 & 11 — Parallel lines inside triangles

Stuck? — using the theorem Using the Side Splitter Theorem MrPilarski Open the video →

Stuck? — more practice Using the Triangle Proportionality Theorem to solve for unknowns Mathispower4u Open the video →

Question 11 asks whether two segments are parallel. Compute both ratios and answer in a sentence.

10

Solve for \(x\).

In \(\triangle ABC\), \(\overline{DE}\parallel\overline{BC}\). Find \(x\).

Show solution

Both pieces are labeled on both sides, so use piece over piece.

\[\frac{10}{15}=\frac{12}{x}\;\Rightarrow\;10x=180\;\Rightarrow\;x=18\]

Check: \(\tfrac{10}{15}=\tfrac{2}{3}\) and \(\tfrac{12}{18}=\tfrac{2}{3}\).

11

Is \(\overline{ST}\parallel\overline{QR}\)?

Point \(S\) is on \(\overline{PQ}\) and point \(T\) is on \(\overline{PR}\). Decide whether \(\overline{ST}\parallel\overline{QR}\), and justify the answer in a sentence.

Show solution

Compare the two ratios. Nothing is being solved for here.

\[\frac{PS}{SQ}=\frac{6}{9}=0.667\qquad\frac{PT}{TR}=\frac{8}{14}\approx 0.571\]

The ratios are not equal, so \(\overline{ST}\) is not parallel to \(\overline{QR}\). By the converse of the Triangle Proportionality Theorem, the segment would be parallel only if the two sides were divided in the same ratio, and they are not.

“Not parallel” is a complete answer. The two computed ratios are the justification.

Question 12 — Similarity in right triangles

Stuck? — where it comes from Pythagorean Theorem proof using similarity Khan Academy Open the video →

Stuck? — a harder one Challenging similarity problem Khan Academy Open the video →

Write the whole hypotenuse on the figure before computing either leg.

12

Find every missing length, then check it two ways.

The altitude to the hypotenuse divides it into pieces of \(18\) and \(32\). Find \(x\), \(y\), and \(z\). Then find the area of the whole triangle two different ways and confirm they agree.

Show solution

First the whole hypotenuse: \(AB=18+32=50\).

\[x=\sqrt{18\cdot 32}=\sqrt{576}=24\]

\[y=\sqrt{18\cdot 50}=\sqrt{900}=30\]

\[z=\sqrt{32\cdot 50}=\sqrt{1600}=40\]

Now the two area computations. Using the two legs, which meet at the right angle:

\[A=\tfrac12(30)(40)=600\]

Using the hypotenuse as the base and the altitude as the height:

\[A=\tfrac12(50)(24)=600\]

They agree. Notice \(30\)–\(40\)–\(50\) is the \(3\)–\(4\)–\(5\) triple times ten, so the Pythagorean check passes as well.

Answer Key & Reteach Guide

Check answers only after attempting all twelve. The last column names the lesson to revisit for any question that was missed.

#	Answer	Skill	Revisit
1	\(XZ=15\), \(m\angle Z=65^\circ\)	Corresponding parts	Lesson 1
2	\(x=5\), \(y=7\), \(z=6\)	Corresponding sides with algebra	Lesson 1
3	\(x=38\), \(m\angle F=44^\circ\)	Corresponding angles with algebra	Lesson 1
4	base angles \(71^\circ\), \(y=17\)	Isosceles Triangle Theorem	Lesson 2 Day 1
5	\(x=5\), \(AB=17\)	Converse of the Isosceles Triangle Theorem	Lesson 2 Day 1
6	\(x=4\), \(QM=17\), \(QR=34\)	Definition of a median	Lesson 2 Day 2
7	a. median b. altitude c. both d. neither	Medians versus altitudes	Lesson 2 Day 2
8	\(k=\tfrac53\), \(FG=15\), perimeter \(=70\)	Scale factor and perimeter	Lesson 3
9	\(ST=21\), \(TR=24\)	Applying the scale factor	Lesson 3
10	\(x=18\)	Triangle Proportionality Theorem	Lesson 4
11	Not parallel: \(\tfrac69\approx0.667\neq\tfrac{8}{14}\approx0.571\)	Converse of the side-splitter	Lesson 4
12	\(x=24\), \(y=30\), \(z=40\); area \(=600\) both ways	Geometric mean in right triangles	Lesson 5