Texas Grab-and-Go Substitute Packet · Teacher Answer Key
Answer Key: The Pythagorean Theorem & Distance
Course: Geometry (Grades 9–12)
Subject: Mathematics
For teacher use only
Math
How to use this key
Answers are grouped by section and question number, with full worked steps. Accept any correct method that reaches the right value — a student may use a² + b² = c² directly or the distance formula, since the distance formula is the theorem. Where an answer is not a perfect square, both the exact root and a rounded decimal (nearest tenth) are given; accept either as long as the exact form is shown. Watch-for notes flag common misconceptions. A basic or scientific calculator is allowed, so grade the reasoning and steps, not just the arithmetic. Estimated grading time: ~6–8 min per packet.
Start Warm-Up: Retrieval ~1 min
1 Square, add, and square-root.
5² = 25 and 12² = 144; 25 + 144 = 169; √169 = 13. Full credit
for showing 5 × 5 and 12 × 12. Watch-for: a student who writes 5² = 10 (multiplying by
2 instead of squaring) or 12² = 24. Squaring means the number times itself.
2 Vertical and horizontal distances.
Vertical from (2, 3) to (2, 8): 8 − 3 = 5 units (x stays the same).
Horizontal from (2, 3) to (6, 3): 6 − 2 = 4 units (y stays the same). Watch-for:
subtracting the wrong pair of coordinates; on a vertical segment the y-values change, on a
horizontal segment the x-values change. This sets up the legs used in the distance
formula.
Build Study the Math ~1 min
3 New hypotenuse with legs a = 6, b = 12.
6² + 12² = c² → 36 + 144 = c² → 180 = c². c = √180 = √(36 × 5) = 6√5
(exact) ≈ 13.4 units (rounded to the nearest tenth). Accept √180 left
unsimplified as long as the decimal ≈ 13.4 is given. Watch-for: stopping at c² = 180
and forgetting the square root, or writing c = 18 by mis-simplifying.
Apply Use What You Know ~3–4 min
4 Find a missing side.
4a (missing hypotenuse): a² + b² = c² → 9² + 12² = c² → 81 + 144 = 225 →
c = √225 = 15 (a whole number; exact and rounded agree).
4b (missing leg): the hypotenuse is known, so subtract: a² = c² − b² = 13² − 5² = 169 − 25 = 144 → a = √144 = 12. Watch-for: a student who adds (13² + 5² = 194) instead of subtracting — remind them the hypotenuse is already the largest side, so you take away the known leg's square. Check: 5, 12, 13 is a Pythagorean triple.
4b (missing leg): the hypotenuse is known, so subtract: a² = c² − b² = 13² − 5² = 169 − 25 = 144 → a = √144 = 12. Watch-for: a student who adds (13² + 5² = 194) instead of subtracting — remind them the hypotenuse is already the largest side, so you take away the known leg's square. Check: 5, 12, 13 is a Pythagorean triple.
5 Distance PQ for P(1, 2) and Q(7, 10).
Horizontal change x₂ − x₁ = 7 − 1 = 6; vertical change y₂ − y₁ =
10 − 2 = 8. d = √[6² + 8²] = √[36 + 64] = √100 = 10 units (exact and rounded agree).
Alternate: subtracting in the other order, √[(1 − 7)² + (2 − 10)²] =
√[(−6)² + (−8)²] = √100 = 10 — squaring removes the sign. Watch-for: a student who
forgets the square root and answers 100.
6 Ladder: 17-ft ladder, base 8 ft from the wall.
The ladder is the hypotenuse (c = 17); the ground distance is a leg
(b = 8); the wall height is the other leg (a). a² = c² − b² = 17² − 8² = 289 − 64 = 225 →
a = √225 = 15 feet up the wall (exact and rounded agree; 8-15-17 is a triple).
Watch-for: treating 17 as a leg and adding (17² + 8² = 353 → ≈18.8), which mislabels
the hypotenuse. The ladder, leaning across from the right angle, must be the longest side.
7 Error analysis: added the legs instead of using the theorem.
The mistake: the student added the legs (6 + 8 = 14) instead of squaring
them, adding the squares, and taking the square root. The Pythagorean Theorem uses
a² + b² = c², not a + b = c.
Correct work: c² = 6² + 8² = 36 + 64 = 100 → c = √100 = 10 (not 14).
Full credit requires (1) naming that they added the legs instead of squaring/adding/ rooting, and (2) the correct hypotenuse of 10. Sense-check to offer: the hypotenuse must be longer than each leg but shorter than the two legs added together — 10 sits between 8 and 14, while the student's 14 equals the sum, which only happens for a flat (degenerate) triangle.
Correct work: c² = 6² + 8² = 36 + 64 = 100 → c = √100 = 10 (not 14).
Full credit requires (1) naming that they added the legs instead of squaring/adding/ rooting, and (2) the correct hypotenuse of 10. Sense-check to offer: the hypotenuse must be longer than each leg but shorter than the two legs added together — 10 sits between 8 and 14, while the student's 14 equals the sum, which only happens for a flat (degenerate) triangle.
8 Converse test on sides 10, 24, 26.
Longest side c = 26. Two shorter sides: 10 and 24.
a² + b² = 10² + 24² = 100 + 576 = 676. c² = 26² = 676. Since
676 = 676, a² + b² = c², so the triangle IS a right triangle (10-24-26 is a
Pythagorean triple, twice 5-12-13). Watch-for: choosing the wrong side as c;
the longest side must be the hypotenuse in the test.
Block Extra Applied Task block only · ~2 min
9 Composite distance: R(0, 0), S(3, 4), T(10, 4) (km).
9a. RS: d = √[(3 − 0)² + (4 − 0)²] = √[9 + 16] = √25 = 5 km.
9b. ST: d = √[(10 − 3)² + (4 − 4)²] = √[49 + 0] = √49 = 7 km (a horizontal segment, so the vertical change is 0).
9c. Total path RS + ST = 5 + 7 = 12 km. Straight line RT: d = √[(10 − 0)² + (4 − 0)²] = √[100 + 16] = √116 = 2√29 (exact) ≈ 10.8 km (rounded). The path (12 km) is longer than the straight line (≈10.8 km). Reasoning students should give: a straight line is the shortest distance between two points, so bending at S makes the trip longer than going directly from R to T. Watch-for: adding RS + ST and thinking it should equal RT (that only happens when the three points are collinear, which they are not here).
9b. ST: d = √[(10 − 3)² + (4 − 4)²] = √[49 + 0] = √49 = 7 km (a horizontal segment, so the vertical change is 0).
9c. Total path RS + ST = 5 + 7 = 12 km. Straight line RT: d = √[(10 − 0)² + (4 − 0)²] = √[100 + 16] = √116 = 2√29 (exact) ≈ 10.8 km (rounded). The path (12 km) is longer than the straight line (≈10.8 km). Reasoning students should give: a straight line is the shortest distance between two points, so bending at S makes the trip longer than going directly from R to T. Watch-for: adding RS + ST and thinking it should equal RT (that only happens when the three points are collinear, which they are not here).
Explain Justify Your Strategy (Q10) ~1–2 min
10 Is a 7-9-12 triangle a right triangle?
Model answer. Claim: No — the triangle is not a right triangle.
Evidence: the longest side is 12, so c = 12. The two shorter sides give
a² + b² = 7² + 9² = 49 + 81 = 130, while c² = 12² = 144. Since 130 ≠ 144,
the sides do not satisfy a² + b² = c².
Reasoning: the converse of the Pythagorean Theorem says a triangle is right
only when the squares of the two shorter sides add up to the square of the longest
side. Here the sum (130) is less than 144, so the triangle is not right (in fact it
is obtuse, since a² + b² < c²). Having three different side lengths only makes it
scalene; scalene says nothing about the angles, so the classmate's reason does not
prove a right angle.
Note: Full credit requires the correct "no," the computation 130 vs. 144, and a statement that the converse (not "three different sides") is what decides it. Students are not required to name it obtuse, but it is a strong extension.
Note: Full credit requires the correct "no," the computation 130 vs. 144, and a statement that the converse (not "three different sides") is what decides it. Students are not required to name it obtuse, but it is a strong extension.
Justification scoring rubric (3 points)
| Score | Claim | Evidence | Reasoning |
|---|---|---|---|
| 3 | States clearly it is not a right triangle. | Correctly computes a² + b² = 130 and c² = 144 (c = longest = 12) and notes 130 ≠ 144. | Cites the converse and explains that "three different sides" (scalene) does not prove a right angle. |
| 2 | Correct "not right" answer. | One square or the comparison shown, with a minor slip. | Reasoning present but does not address the classmate's flawed reason or the converse by name. |
| 1 | An answer with weak or no support. | Evidence largely missing or incorrect (e.g., wrong side chosen as c). | Little or flawed reasoning. |
| 0 | No/incorrect claim. | No evidence. | No reasoning. |
Close ACE ~1 min
ACE Articulate / Connect / Extend.
Articulate: Accept any correct explanation: "The hypotenuse is the side
across from the right angle and is the longest side. You must take the
square root at the end because a² + b² = c² gives you c² (the square of the side),
not c itself; √(c²) = c undoes the squaring to get the actual length."
Connect: Q5 (distance PQ) — legs are the horizontal change 6 and vertical change 8; also Q9a/9b/9c. Any of these earns credit if the legs are named. Q6's ladder and Q3 use the theorem directly rather than via coordinates.
Extend: Any genuine "can't-measure-directly" distance — e.g., the diagonal of a rectangular room or TV screen, the straight-line distance across a park cut by two streets meeting at a right angle, a ramp's length from its run and rise, a guy-wire on a pole, or the diagonal of a baseball diamond. Must involve a right triangle where two sides are known and the third is found.
Connect: Q5 (distance PQ) — legs are the horizontal change 6 and vertical change 8; also Q9a/9b/9c. Any of these earns credit if the legs are named. Q6's ladder and Q3 use the theorem directly rather than via coordinates.
Extend: Any genuine "can't-measure-directly" distance — e.g., the diagonal of a rectangular room or TV screen, the straight-line distance across a park cut by two streets meeting at a right angle, a ramp's length from its run and rise, a guy-wire on a pole, or the diagonal of a baseball diamond. Must involve a right triangle where two sides are known and the third is found.
Challenge Early Finisher (optional)
EF Build Pythagorean triples from 3-4-5; test 5-6-8.
(1) New triples by scaling 3-4-5: × 2 → 6-8-10; × 3 → 9-12-15.
(2) Proofs: 6² + 8² = 36 + 64 = 100 = 10² ✔; 9² + 12² = 81 + 144 = 225 = 15² ✔.
(3) Why scaling works: multiplying every side by k multiplies both sides of a² + b² = c² by k² ((ka)² + (kb)² = k²(a² + b²) = k²c² = (kc)²), so the equation still holds — the new triangle is similar to the original and keeps its right angle.
Converse on 5-6-8: longest side 8, so 5² + 6² = 25 + 36 = 61, but 8² = 64. Since 61 ≠ 64, 5-6-8 is NOT a right triangle (it is close, but not exact). Full credit requires two correct scaled triples with proofs, a reason scaling preserves the right angle, and the correct "not right" conclusion for 5-6-8.
(2) Proofs: 6² + 8² = 36 + 64 = 100 = 10² ✔; 9² + 12² = 81 + 144 = 225 = 15² ✔.
(3) Why scaling works: multiplying every side by k multiplies both sides of a² + b² = c² by k² ((ka)² + (kb)² = k²(a² + b²) = k²c² = (kc)²), so the equation still holds — the new triangle is similar to the original and keeps its right angle.
Converse on 5-6-8: longest side 8, so 5² + 6² = 25 + 36 = 61, but 8² = 64. Since 61 ≠ 64, 5-6-8 is NOT a right triangle (it is close, but not exact). Full credit requires two correct scaled triples with proofs, a reason scaling preserves the right angle, and the correct "not right" conclusion for 5-6-8.
Watch Common misconceptions
- Adding the legs instead of using the theorem. The Q7 error: writing c = a + b (6 + 8 = 14) instead of c = √(a² + b²) = 10. Anchor: you must square the legs, add the squares, and then take the square root. The sum of the legs is always more than the hypotenuse.
- Mislabeling the hypotenuse. The Q6 trap: treating the longest given length (the ladder, 17) as a leg and adding. The hypotenuse is across from the right angle and is the longest side; when it is known you subtract (a² = c² − b²). Reinforce on Q4b, Q6, and Q8.
- Forgetting to square-root. Students stop at c² = 180 (Q3) or d² = 100 (Q5) and report 180 or 100. The theorem gives the square of the side; the final step is √ to get the length itself.
- Squaring vs. doubling. On Q1 and throughout, some write 5² = 10 or 12² = 24. Squaring means the number times itself (5 × 5 = 25), not times 2.
- Dropping the exact root. When a² + b² is not a perfect square (Q3 → √180, Q9c → √116), require the exact radical and the rounded decimal; a bare decimal loses the exact value, and a bare "180" is an unfinished answer.
- Choosing the wrong side as c in the converse. On Q8 and Q10, the longest side must be squared alone on one side of the test. Testing 7² + 12² vs. 9² would give a false result — always add the two shorter sides and compare to the longest.
- Assuming scalene means right (Q10). Three different side lengths make a triangle scalene, which says nothing about its angles. Only the converse (a² + b² = c²) decides whether it is right.
Teacher follow-up based on likely errors
If many students miss Q7, do a quick 5-minute practice: for legs 3 and 4, contrast 3 + 4 = 7 with √(3² + 4²) = 5, and have students explain why the sum of the legs cannot be the hypotenuse. If Q6 shows mislabeled hypotenuses, sketch a ladder and label the longest side as the hypotenuse before setting up. The Q8 and Q10 converse items reveal who can pick the longest side as c and who confuses "scalene" with "right." The Q5 and Q9 items show who connects the distance formula back to a² + b² = c². The exact-vs-rounded answers on Q3 and Q9c show who finishes with both a radical and a decimal — all good warm-up discussions next class.
HS_GEOM_Pythagorean_01 — The Pythagorean Theorem & Distance on the Coordinate Plane
Teacher Answer Key