← Back to the Points Lines Planes Segments Angles worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-CO.A.1 · Level: Grade 10 Geometry
Problem 2 Finding a midpoint length
A midpoint creates congruent halves, so MS = RM = 18.6. Then RS = 18.6 + 18.6 = 37.2.
MS = 18.6 cm and RS = 37.2 cm
Problem 3 Complement and supplement
Complement: 90° − 37° = 53°. Supplement: 180° − 37° = 143°.
Complement 53°, supplement 143°
Problem 4 Angles in a linear pair
A linear pair is supplementary: m∠BOC = 180° − 118° = 62°.
62°
Problem 5 Vertical angles
The vertical angle ∠QOS is equal to ∠POR, so it measures 74°. The other two angles each form a linear pair with a 74° angle: 180° − 74° = 106°.
74°, 106° and 106°
Problem 6 Segment addition with algebra
Segment Addition: (3x + 2) + (5x − 7) = 51, so 8x − 5 = 51, 8x = 56 and x = 7.
AB = 3(7) + 2 = 23 and BC = 5(7) − 7 = 28. Check: 23 + 28 = 51.
x = 7, AB = 23, BC = 28
Problem 7 Angle addition with algebra
(2x + 10) + (3x − 8) = 112 gives 5x + 2 = 112, so x = 22.
m∠AOB = 2(22) + 10 = 54° and m∠BOC = 3(22) − 8 = 58°. Check: 54 + 58 = 112.
x = 22, 54° and 58°
Problem 8 Angle bisector equation
A bisector gives equal halves: 4x + 3 = 6x − 17, so 20 = 2x and x = 10.
Each half is 4(10) + 3 = 43°, so m∠AOC = 2 · 43° = 86°.
x = 10 and m∠AOC = 86°
Problem 9 Midpoint on a number line
Midpoint: (−7 + 12) ÷ 2 = 5 ÷ 2 = 2.5. Length: 12 − (−7) = 19.
Midpoint at 2.5; DE = 19
Problem 10 A supplement and a complement
Let the angle be x. Then 180 − x = 3(90 − x) + 25 = 295 − 3x, so 2x = 115 and x = 57.5.
Check: complement 32.5, and 3(32.5) + 25 = 122.5, which is the supplement 180 − 57.5.
57.5°
Problem 11 Vertical angles with algebra
Vertical angles are equal: 5x − 12 = 3x + 22, so 2x = 34 and x = 17.
Each vertical angle is 5(17) − 12 = 73°. The neighbors measure 180° − 73° = 107°.
x = 17; 73° and 107°
Problem 12 Linear pair with algebra
(7x + 9) + (5x + 15) = 180 gives 12x + 24 = 180, so 12x = 156 and x = 13.
The angles are 7(13) + 9 = 100° and 5(13) + 15 = 80°, and 100 + 80 = 180.
100° and 80°
Problem 13 A right angle split in two
Perpendicular lines form a 90° angle, so (2x + 7) + (3x − 2) = 90. Then 5x + 5 = 90 and x = 17.
∠ABC = 2(17) + 7 = 41° and ∠CBD = 3(17) − 2 = 49°.
41° and 49°
Problem 14 Always, sometimes or never
- Always: parallel lines are defined as coplanar lines that never meet.
- Sometimes: three points can lie on one line, but most triples do not.
- Never: a ray has no second endpoint, so no point is halfway to anything.
- Sometimes: vertical angles are equal, so they add to 180° only when each measures 90°.
(a) always, (b) sometimes, (c) never, (d) sometimes
Problem 15 Hiking trail midpoint
TC = 2.4 + 1.75 = 4.15 miles. The midpoint is 4.15 ÷ 2 = 2.075 miles from T.
The lookout is 2.4 miles from T, and 2.4 − 2.075 = 0.325, so the fountain is before the lookout, on the trailhead side.
TC = 4.15 mi; the fountain is 0.325 mi from the lookout, toward the trailhead
Problem 16 Robot sensor beam
The bisector makes m∠AOB = 128° ÷ 2 = 64°, and also m∠BOC = 64°.
Ray OD lies inside ∠BOC, so by Angle Addition m∠DOC = 64° − 19° = 45°.
m∠AOB = 64° and m∠DOC = 45°
Problem 17 Four collinear points
x + (2x − 4) + (x + 10) = 62 gives 4x + 6 = 62, so x = 14.
AB = 14, BC = 24 and CD = 24. The midpoint M is 31 from A. Since B is 14 from A and C is 38 from A, and 14 < 31 < 38, M is between B and C, 17 units from B.
x = 14; AB = 14, BC = 24, CD = 24; yes, M is between B and C
Problem 18 Linear pair, vertical angles, bisector
(3x + 5) + (2x + 15) = 180 gives 5x + 20 = 180 and x = 32.
∠1 = 3(32) + 5 = 101° and ∠2 = 2(32) + 15 = 79°. Vertical angles are equal, so ∠3 = 101° and ∠4 = 79°. A bisector of ∠2 makes two angles of 79° ÷ 2 = 39.5°.
101°, 79°, 101°, 79°; bisected pieces 39.5°
Problem 19 Highway mile markers
Rest stop: (14.6 + 31.2) ÷ 2 = 45.8 ÷ 2 = 22.9. The gas station is at 22.9 + 3.5 = 26.4. Its distance to Exit 2 is 31.2 − 26.4 = 4.8.
Rest stop at marker 22.9; gas station is 4.8 miles from Exit 2
Problem 20 Bisector on a straight angle
∠COD = ∠BOC = (x + 18)°. The three angles fill a straight angle: (4x − 6) + 2(x + 18) = 180, so 6x + 30 = 180 and x = 25.
∠AOB = 94°, ∠BOC = 43° and ∠COD = 43°. Check: 94 + 43 + 43 = 180.
x = 25; 94°, 43°, 43°