Geometry check: ready for Algebra II?

For students who finished Geometry (usually grades 9–10) or are deciding whether to start Algebra II.

14 items drawn from the Geometry final exam. About 30–40 minutes. Paper and pencil, no calculator unless an item says so. Work every item first, then check each answer and mark it.

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1. Write the contrapositive of this true statement, and tell whether the contrapositive is true: "If a quadrilateral is a rhombus, then its diagonals are perpendicular."

Answer: If the diagonals of a quadrilateral are not perpendicular, then the quadrilateral is not a rhombus. It is true, because a conditional and its contrapositive are logically equivalent and the original statement is true.
Lesson 1.3 Conditional Statements

2. The supplement of an angle measures 18^\circ more than three times its complement. Find the measure of the angle.

Answer: Let the angle measure x^\circ. Then 180 - x = 3(90 - x) + 18, so 180 - x = 288 - 3x, 2x = 108 and x = 54. The angle measures 54^\circ. (Check: supplement 126^\circ, complement 36^\circ, and 3(36) + 18 = 126.)
Lesson 2.4 Angle Pairs and the Vertical Angles Theorem

3. Grace has three straight sticks that are 4.2 cm, 5.9 cm and 10.3 cm long. Can she lay them end to end to form a triangle? Explain why or why not.

Answer: No. By the Triangle Inequality Theorem, the two shorter sides must add to more than the longest side, but 4.2 + 5.9 = 10.1 < 10.3.
Lesson 5.6 The Triangle Inequality and the Hinge Theorem

4. The frame of a stained-glass window is a regular polygon, and each of its interior angles measures 144^\circ. How many sides does the frame have?

Answer: Each exterior angle measures 180^\circ - 144^\circ = 36^\circ, and the exterior angles add to 360^\circ, so the frame has 360 \div 36 = 10 sides.
Lesson 6.1 Angles of Polygons

5. In rectangle ABCD, AC = 5x - 7 and BD = 3x + 11. Find AC.

Answer: The diagonals of a rectangle are congruent, so 5x - 7 = 3x + 11 and x = 9. Then AC = 5(9) - 7 = 38.
Lesson 6.4 Rectangles, Rhombuses and Squares

6. Find the image of P(3, -5) under a rotation of 90^\circ counterclockwise about the origin.

Answer: The rule is (x, y) \to (-y, x), so P'(5, 3).
Lesson 7.3 Rotations and Rotational Symmetry

7. A dilation with center at the origin maps A(4, -6) to A'(10, -15). (a) Find the scale factor k. (b) Find the image of B(-2, 8) under the same dilation.

Answer: (a) k = \dfrac{10}{4} = \dfrac{-15}{-6} = 2.5. (b) B' = (2.5 \cdot (-2),\ 2.5 \cdot 8) = (-5, 20).
Lesson 7.5 Dilations

8. At the same time of day, Samuel, who is 6 ft tall, casts a shadow 4.5 ft long, and the church steeple casts a shadow 63 ft long. How tall is the steeple?

Answer: The sun's rays make equal angles with the level ground and both Samuel and the steeple stand vertically, so the triangles are similar by AA. \dfrac{h}{63} = \dfrac{6}{4.5}, so h = 84 ft.
Lesson 8.3 Proving Triangles Similar: AA

9. Explain why, in every right triangle that has a 34^\circ acute angle, the ratio \dfrac{\text{opposite leg}}{\text{adjacent leg}} for that angle is the same number. (This is why \tan 34^\circ has one value.)

Answer: Any two such right triangles have a pair of congruent right angles and a pair of congruent 34^\circ angles, so they are similar by AA. Corresponding sides of similar triangles are proportional, so \dfrac{a}{a'} = \dfrac{b}{b'} for the opposite legs a, a' and adjacent legs b, b', which rearranges to \dfrac{a}{b} = \dfrac{a'}{b'}. The ratio depends only on the angle, so \tan 34^\circ \approx 0.6745 in every such triangle.
Lesson 9.5 The Tangent Ratio

10. A shed roof rises 9 ft over a horizontal run of 14 ft. To the nearest tenth of a degree, what angle does the roof make with the horizontal?

Answer: \tan \theta = \dfrac{9}{14}, so \theta = \tan^{-1}\left(\dfrac{9}{14}\right) \approx 32.7^\circ.
Lesson 9.6 The Sine and Cosine Ratios

11. A sailor climbs to a lookout 60 ft above the sea. Taking the earth's radius as 3960 miles, how far away is the horizon, to the nearest tenth of a mile? (The line of sight to the horizon is tangent to the earth; 1 mile = 5280 ft.)

Answer: The height is \dfrac{60}{5280} = \dfrac{1}{88} mile. The line from the lookout through the earth's center is a secant with outside part \dfrac{1}{88} and whole length \dfrac{1}{88} + 7920. By the Tangent-Secant Theorem, d^2 = \dfrac{1}{88}\left(\dfrac{1}{88} + 7920\right) \approx 90.0001, so d \approx 9.49, about 9.5 miles.
Lesson 10.6 Segment Lengths in Circles

12. Two stained-glass panes are similar. Their areas are 48 square inches and 108 square inches. The smaller pane is 10 inches tall. How tall is the larger pane?

Answer: The ratio of areas is \dfrac{108}{48} = \dfrac{9}{4}, so the scale factor is \sqrt{\dfrac{9}{4}} = \dfrac{3}{2}. The larger pane is 10 \times \dfrac{3}{2} = 15 inches tall.
Lesson 11.6 Areas of Similar Figures and Composite Regions

13. A grain silo is a cylinder with radius 6 ft and height 20 ft, topped by a cone with the same radius and a height of 5 ft. Find the volume of the silo in terms of \pi.

Answer: Cylinder: \pi(6^2)(20) = 720\pi. Cone: \dfrac{1}{3}\pi(6^2)(5) = 60\pi. Total: 780\pi cubic feet (about 2450.4 cubic feet).
Lesson 12.5 Volume of Pyramids and Cones

14. Find the coordinates of the point P on \overline{AB} with AP : PB = 1 : 2, where A(-2, 3) and B(10, -6).

Answer: P is one-third of the way from A to B: P = \left(-2 + \dfrac{1}{3}(12),\ 3 + \dfrac{1}{3}(-9)\right) = (2, 0).
Lesson 13.1 Coordinate Tools: Distance, Midpoint, Slope and Partitions

4-week summer review path

Four short days a week, 30–45 minutes: the 16 Geometry lessons that appear most often on its exams and reviews, in course order. Do the lessons you missed on the test first.

Geometry check: ready for Algebra II?

4-week summer review path

Week 1

Week 2

Week 3

Week 4