Geometry • Unit 1

Review Sheet for the Unit 1 Assessment

Twelve questions in the same format and order as the assessment: right triangles, distance, word problems with a required sketch, translations by vector, and reflections across all four lines.

Directions

Before you start

  • Show all work. An answer with no supporting equation earns no credit on the assessment.
  • Round every answer to two decimal places unless the answer is exact.
  • Questions 5 and 6 require a labeled sketch before any algebra.
  • For every transformation question, state the coordinates of the corner points of the new image.
  • Attempt all twelve questions before opening a single solution. Give yourself about 40 minutes.
  • Each question group has a Khan Academy 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

\[a^2+b^2=c^2\]
\[d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

Transformation rules are not provided on the assessment. 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 & 2 — Solve for the identified variable

Stuck? — solve for the hypotenuse Pythagorean theorem — worked examples Khan Academy Open the video →
Stuck? — solve for a leg Pythagorean theorem 2 — solving for a leg Khan Academy Open the video →

Round each answer to two decimal places.

1

Solve for \(x\).

7x15
Show solution

The 15 is the hypotenuse, so this is a subtraction problem.

\[7^2+x^2=15^2\;\Rightarrow\;49+x^2=225\;\Rightarrow\;x^2=176\]
\[x=\sqrt{176}\approx 13.27\]
2

Solve for \(y\).

118y
Show solution

Both legs are known, so the two squares are added.

\[8^2+11^2=y^2\;\Rightarrow\;64+121=y^2\;\Rightarrow\;185=y^2\]
\[y=\sqrt{185}\approx 13.60\]

Questions 3 & 4 — Find the distance between the two points

Stuck? — count, then solve Example finding distance with the Pythagorean theorem Khan Academy Open the video →
Stuck? — use the formula Distance formula Khan Academy Open the video →

Either method is acceptable: build a right triangle on the grid, or read the coordinates and use the distance formula.

3

Find the distance between the two plotted points.

-10-10-8-8-6-6-4-4-2-2224466881010xy(−4, 3)(2, −5)
Show solution

The points are \((-4,3)\) and \((2,-5)\). The horizontal leg is 6 units and the vertical leg is 8 units.

\[d=\sqrt{(-4-2)^2+(3-(-5))^2}=\sqrt{36+64}=\sqrt{100}\]
\[d=10.00\text{ units}\]

This is a 6–8–10 triangle, a multiple of the 3–4–5 triple.

4

Find the distance between the two plotted points.

-10-10-8-8-6-6-4-4-2-2224466881010xy(−6, −2)(3, 4)
Show solution

The points are \((-6,-2)\) and \((3,4)\).

\[d=\sqrt{(-6-3)^2+(-2-4)^2}=\sqrt{81+36}=\sqrt{117}\]
\[d\approx 10.82\text{ units}\]

Questions 5 & 6 — Draw a picture, then solve

Stuck? — word problems Multi-step word problem with the Pythagorean theorem Khan Academy Open the video →

A labeled sketch is worth points on its own. Draw the right angle, label the two known measurements, and mark the unknown before substituting.

5

The ladder

A 15-foot ladder is leaned against a building so that the top of the ladder reaches exactly 12 feet up the building. How far from the base of the building is the base of the ladder?

Show solution

The ladder is the hypotenuse. The building is the vertical leg and the ground is the horizontal leg.

x12 ft15 ft
\[x^2+12^2=15^2\;\Rightarrow\;x^2+144=225\;\Rightarrow\;x^2=81\]
\[x=\sqrt{81}=9.00\text{ feet}\]

The base of the ladder sits 9 feet from the building — a 9–12–15 triangle.

6

The rectangle

What is the length of a diagonal of a rectangle whose width is 22 inches and length is 36 inches?

Show solution

A diagonal cuts the rectangle into two right triangles. The width and length are the legs; the diagonal is the hypotenuse.

36 in22 ind
\[22^2+36^2=d^2\;\Rightarrow\;484+1296=d^2\;\Rightarrow\;1780=d^2\]
\[d=\sqrt{1780}\approx 42.19\text{ inches}\]

Questions 7 & 8 — Translations

Stuck? — draw the image Drawing the image of a translation Khan Academy Open the video →
Stuck? — find the vector Determining a translation between points Khan Academy Open the video →

Sketch the translation of the figure by the given vector, then state the coordinates of the corner points of the new image.

7

Translate by the vector \([4,\,-2]\).

-10-10-8-8-6-6-4-4-2-2224466881010xyABC
Show solution

The rule is \((x,y)\rightarrow(x+4,\;y-2)\): right 4 units, down 2 units.

PreimageImage
A(−6, 2)A′(−2, 0)
B(−2, 5)B′(2, 3)
C(−1, 1)C′(3, −1)
-10-10-8-8-6-6-4-4-2-2224466881010xyABCA'B'C'
8

Translate by the vector \([-3,\,-5]\).

-10-10-8-8-6-6-4-4-2-2224466881010xySTUV
Show solution

The rule is \((x,y)\rightarrow(x-3,\;y-5)\): left 3 units, down 5 units.

PreimageImage
S(2, 3)S′(−1, −2)
T(6, 3)T′(3, −2)
U(6, 7)U′(3, 2)
V(2, 7)V′(−1, 2)
-10-10-8-8-6-6-4-4-2-2224466881010xySTUVS'T'U'V'

Questions 9 – 12 — Reflections

Stuck? — over the axes Example reflecting a quadrilateral over the x-axis Khan Academy Open the video →
Stuck? — over \(y=x\) Reflecting over \(y=x\) Khan Academy Open the video →
Stuck? — over \(y=-x\) Reflecting over \(y=-x\) Khan Academy Open the video →

Sketch the reflection of the image over the given line, then state the coordinates of the corner points of the new image.

9

Reflect over the x-axis.

-10-10-8-8-6-6-4-4-2-2224466881010xyABC
Show solution

Rule: \((x,y)\rightarrow(x,-y)\). Each y-coordinate changes sign.

\(A(-7,6)\rightarrow A'(-7,-6)\), \(B(-3,8)\rightarrow B'(-3,-8)\), \(C(-2,3)\rightarrow C'(-2,-3)\).

-10-10-8-8-6-6-4-4-2-2224466881010xyABCA'B'C'
10

Reflect over the line \(y=-x\).

y = −x-10-10-8-8-6-6-4-4-2-2224466881010xyAB
Show solution

Rule: \((x,y)\rightarrow(-y,-x)\). The coordinates switch places and both change sign.

\(A(2,6)\rightarrow A'(-6,-2)\) and \(B(7,3)\rightarrow B'(-3,-7)\).

y = −x-10-10-8-8-6-6-4-4-2-2224466881010xyABA'B'
11

Reflect over the y-axis.

-10-10-8-8-6-6-4-4-2-2224466881010xyABCD
Show solution

Rule: \((x,y)\rightarrow(-x,y)\). Each x-coordinate changes sign.

\(A(3,2)\rightarrow A'(-3,2)\), \(B(8,2)\rightarrow B'(-8,2)\), \(C(6,7)\rightarrow C'(-6,7)\), \(D(2,6)\rightarrow D'(-2,6)\).

-10-10-8-8-6-6-4-4-2-2224466881010xyABCDA'B'C'D'
12

Reflect over the line \(y=x\).

y = x-10-10-8-8-6-6-4-4-2-2224466881010xyABC
Show solution

Rule: \((x,y)\rightarrow(y,x)\). The coordinates switch places.

\(A(-5,-1)\rightarrow A'(-1,-5)\), \(B(-1,-3)\rightarrow B'(-3,-1)\), \(C(-2,-7)\rightarrow C'(-7,-2)\).

y = x-10-10-8-8-6-6-4-4-2-2224466881010xyABCA'B'C'

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.

#AnswerSkillRevisit
1\(x\approx 13.27\)Solving for a legLesson 1
2\(y\approx 13.60\)Solving for the hypotenuseLesson 1
310.00 unitsDistance from a graphLesson 1 Day 2
4\(\approx 10.82\) unitsDistance from a graphLesson 1 Day 2
59.00 feetApplication with sketchLesson 1 Day 3
6\(\approx 42.19\) inchesApplication with sketchLesson 1 Day 3
7A′(−2, 0), B′(2, 3), C′(3, −1)Translation by vectorLesson 3
8S′(−1, −2), T′(3, −2), U′(3, 2), V′(−1, 2)Translation by vectorLesson 3
9A′(−7, −6), B′(−3, −8), C′(−2, −3)Reflection across the x-axisLesson 4 Day 1
10A′(−6, −2), B′(−3, −7)Reflection across \(y=-x\)Lesson 4 Day 2
11A′(−3, 2), B′(−8, 2), C′(−6, 7), D′(−2, 6)Reflection across the y-axisLesson 4 Day 1
12A′(−1, −5), B′(−3, −1), C′(−7, −2)Reflection across \(y=x\)Lesson 4 Day 2