Can these side lengths form a valid triangle? Show your inequality test below:
Triangle possible?
YES NO
16. Sides: \(9\text{ cm}, 15\text{ cm}, 17\text{ cm}\) 2 pts
Can these side lengths form a valid triangle? Show your inequality test below:
Triangle possible?
YES NO
\(d = r \cdot t\)
17. Find Total Distance: 3 pts
A delivery driver travels at an average speed of \(52\text{ mph}\) for \(3.5\text{ hours}\). How many miles does she travel?
Distance (\(d\)) =
18. Find Average Rate: 3 pts
A train travels a total distance of \(360\text{ miles}\) in \(4.5\text{ hours}\). What was the train's average speed in miles per hour?
Rate (\(r\)) =
\(C = \pi d\) | \(C = 2\pi r\)
19. Diameter Given: 3 pts
A circular fountain has a diameter of \(14\text{ meters}\). Calculate its circumference using \(\pi \approx \frac{22}{7}\).
Circumference (\(C\)) =
20. Radius Given: 3 pts
A bicycle tire has a radius of \(15\text{ inches}\). Calculate its circumference using \(\pi \approx 3.14\). Round to the nearest tenth if needed.
Circumference (\(C\)) =
Remember to check all unit labels (cm, ft, mph, m) and inverse signs! Page 3 of 3
34567891011
Answer: \(y \le 7\)
11. \(-3m \ge 18 \implies m \le -6\) SIGN FLIPPED (\(\ge \to \le\))
Divide by \(-3\): \(m \le \frac{18}{-3} \implies m \le -6\)
Dividing by negative flips \(\ge\) to \(\le\)! Solid circle at -6.
-10-9-8-7-6-5-4-3-2
Answer: \(m \le -6\)
12. \(\frac{k}{-2} < 5 \implies k > -10\) SIGN FLIPPED (\(< \to >\))
Step 1: \(\frac{k}{-2} < 9 - 4 \implies \frac{k}{-2} < 5\)
Step 2: Multiply by \(-2\) flips: \(k > 5 \cdot (-2) = -10\)
-14-13-12-11-10-9-8-7-6
Answer: \(k > -10\)
Include Correct Units
13. Diagram Problem (4 pts) 2 pts P, 2 pts A
Perimeter: \(P = 14 + 10 + 11 = 35\text{ cm}\)
Area: \(A = \frac{1}{2} \cdot 14 \cdot 8 = 7 \cdot 8 = 56\text{ cm}^2\)
Common Student Pitfall: Height (8 cm) is inside the triangle and NOT part of perimeter. Area requires squared units (\(\text{cm}^2\)).
P = 35 cm
A = 56 cm²
14. Working Backward for Height (3 pts) 3 pts
Formula: \(A = \frac{1}{2}bh\)
Substitute: \(54 = \frac{1}{2}(12)h \implies 54 = 6h\)
Divide both sides by 6: \(h = \frac{54}{6} = 9\text{ m}\)
Check: \(\frac{1}{2} \cdot 12 \cdot 9 = 6 \cdot 9 = 54\text{ m}^2\) \(\checkmark\)
Height (\(h\)) = 9 m
Continue to Page 3 for Inequality Theorem, Distance, & Circumference Answer Key → Page 2 of 3
Teacher Solutions & Grading Guide
Answer Key
Covers Triangle Theorem, Distance (\(d=rt\)) & Circle Circumference (\(C=\pi d\)) Page 3 Total: 16 Points (Overall: 30 Pts)
Short + Middle > Longest
15. Sides: \(8\text{ ft}, 13\text{ ft}, 22\text{ ft}\) 2 pts
Sum of two shorter sides:
\(8 + 13 = 21\)
Compare to longest side: \(21 \ngtr 22\) (False)
Verdict: NO (Impossible)
16. Sides: \(9\text{ cm}, 15\text{ cm}, 17\text{ cm}\) 2 pts
Sum of two shorter sides:
\(9 + 15 = 24\)
Compare to longest side: \(24 > 17\) (True)
Verdict: YES (Valid Triangle)
Units Required
17. Find Total Distance 3 pts
Formula: \(d = r \cdot t\)
Substitute: \(d = (52\text{ mph})(3.5\text{ hr})\)
Calculation: \(52 \times 3.5 = 156 + 26 = 182\)
\(d = 182\text{ miles}\)
18. Find Average Rate 3 pts
Formula: \(d = r \cdot t \implies 360 = r \cdot (4.5)\)
Solve: \(r = \frac{360}{4.5} = \frac{3600}{45}\)
Divide: \(r = 80\)
\(r = 80\text{ mph}\)
Section 6.4
19. Diameter Given (\(d = 14\text{ m}\)) 3 pts
Formula: \(C = \pi d\)
Substitute \(\pi = \frac{22}{7}\): \(C = \frac{22}{7} \cdot 14\)
Simplify: \(C = 22 \cdot 2 = 44\text{ m}\)
(Using 3.14: \(C \approx 43.96\text{ m}\) also full credit)
\(C = 44\text{ meters}\)
20. Radius Given (\(r = 15\text{ in}\)) 3 pts
Formula: \(C = 2\pi r\) (or \(d = 2(15) = 30\text{ in}\))
Substitute \(\pi \approx 3.14\): \(C = 2(3.14)(15)\)
Multiply: \(C = 30 \cdot 3.14 = 94.2\text{ in}\)
\(C = 94.2\text{ inches}\)
Grading Criteria: 1 pt formula/setup, 1 pt computation, 1 pt units/signs. Total: 30 Points. Page 3 of 3
