← Back to the Constructing Triangles worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: 7.G.A.2 · Level: Grade 7
Problem 1 Three Poles, First Test
The two shorter sides add to 5 + 6 = 11, which is less than 12. They cannot reach across.
No triangle
Problem 2 Three Poles, Second Test
Check all pairs: 7 + 9 = 16 > 12, 7 + 12 = 19 > 9 and 9 + 12 = 21 > 7. The longest-side check is enough.
Yes, a triangle can be formed
Problem 3 Find the Third Angle
48 + 67 = 115, and 180 − 115 = 65.
65°
Problem 4 Which Set Works
(a) 8 + 8 = 16 > 15: yes. (b) 6 + 9 = 15 < 16: no. (c) 10 + 10 = 20, not greater than 20: no.
Only (a) forms a triangle
Problem 5 Range of the Third Side
The third side must be less than 7 + 11 = 18, and it must be greater than 11 − 7 = 4 so that 7 and the new side reach across 11.
Greater than 4 in and less than 18 in
Problem 6 A Flat Triangle
6 + 9 = 15, which is equal to the third side, not greater. The two shorter sides lie exactly along the longest side, so the “triangle” collapses to a straight segment with no area.
No triangle, only a flat segment
Problem 7 Count the Triangles
(a) Angles add to 180°, but no side is given, so infinitely many triangles fit. (b) 70 + 80 + 50 = 200, not 180, so none. (c) SAS: exactly one. (d) 4 + 4 = 8 < 9, so none.
(a) many, (b) none, (c) exactly one, (d) none
Problem 8 ASA Conditions
Third angle: 180 − 35 − 80 = 65. Two angles with the side between them (ASA) fix one triangle.
65°; unique
Problem 9 Classify by Angles
112 + 41 = 153, and 180 − 153 = 27. One angle is 112°, which is greater than 90°.
27°; obtuse
Problem 10 Base of an Isosceles Triangle
The base must be less than 9 + 9 = 18, and it must be positive. Also 9 + base > 9 holds for any positive base.
1, 2, 3, …, 17 inches (any length from 0 to 18 exclusive)
Problem 11 Longest Side Known
Call the third side x. Since 14 is longest, x ≤ 14. The two shorter sides must exceed 14: 6 + x > 14, so x > 8.
9, 10, 11, 12, 13 or 14 inches
Problem 12 Two Triangles from SSA
The side BC = 5 is longer than the height 4 and shorter than AC = 8, so it can swing to two points on line AB.
Distance from the foot of the height to B: 5² − 4² = 9, so 3 units. The foot is 8 cos 30° = 4√3 ≈ 6.93 from A.
AB = 6.93 − 3 ≈ 3.93 or AB = 6.93 + 3 ≈ 9.93.
Two triangles, AB ≈ 3.93 and AB ≈ 9.93
Problem 13 Same Angles, Different Sizes
Each set is ASA, so each triangle is unique, but the given sides differ. The third angle is 70° in both. Zane’s triangle is an enlargement of Maya’s with scale factor 8 ÷ 4 = 2.
No, they have the same shape but not the same size
Problem 14 Draw and Compare
Angle C = 180 − 40 − 65 = 75°. The longest side lies opposite the largest angle, which is C = 75°, so the longest side is AB, the side of 3 in. The shortest is BC, opposite A = 40°.
Angle C = 75°; AB is the longest side
Problem 15 Poles for a Tent Frame
9 + 12 = 21, which is not greater than 21. The poles would lie flat.
No triangular frame is possible
Problem 16 Roof Truss Beams
The beam must be greater than 14 − 10 = 4 and less than 10 + 14 = 24. Whole numbers from 5 to 23: 23 − 5 + 1 = 19.
19 different lengths
Problem 17 Hiking Triangle
The return leg is less than 3 + 4.5 = 7.5 and greater than 4.5 − 3 = 1.5.
Greater than 1.5 mi and less than 7.5 mi
Problem 18 Sail Angles
74 + 58 = 132, and 180 − 132 = 48. All three angles (74°, 58°, 48°) are less than 90°.
48°; acute
Problem 19 Stained Glass Piece
Third angle: 180 − 40 − 85 = 55°.
The conditions are ASA, so the triangle is unique. Steps: draw a 5 in segment, mark 40° at one end and 85° at the other with a protractor, draw both rays and mark where they meet.
55°; unique
Problem 20 Surveyor and the Tree
BC = 3: shorter than the height 5, so it cannot reach the line, giving none.
BC = 5: equal to the height, giving exactly one (a right triangle).
BC = 7: between 5 and 10, giving two triangles.
BC = 12: longer than AC = 10, giving exactly one.
3 ft: none; 5 ft: one; 7 ft: two; 12 ft: one