← Back to the Triangle Theorems Centroid Circumcenter worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-CO.C.10, G-C.A.3 · Level: Grade 10 Geometry
Problem 1 Third angle of a triangle
The interior angles add to 180°: 180° − 47° − 68° = 65°.
65°
Problem 2 Exterior angle
By the Exterior Angle Theorem, m∠ACD = 52° + 79° = 131°.
Check: ∠ACB = 180° − 131° = 49°, and 52 + 79 + 49 = 180.
131°
Problem 3 Base angles
The two base angles are congruent and the three angles add to 180°, so each base angle is (180° − 38°) ÷ 2 = 71°.
71° each
Problem 4 Converse of the isosceles theorem
Congruent angles at B and C mean the opposite sides AC and AB are congruent: 3x − 4 = x + 10, so 2x = 14 and x = 7.
AB = 3(7) − 4 = 17, and AC = 7 + 10 = 17.
x = 7; AB = 17
Problem 5 Midsegment length
A midsegment is half the length of the third side: MN = ½(26.4) = 13.2.
MN = 13.2 cm
Problem 6 Midsegment with algebra
BC = 2MN: 8x − 10 = 2(2x + 7) = 4x + 14, so 4x = 24 and x = 6.
MN = 2(6) + 7 = 19 and BC = 8(6) − 10 = 38.
x = 6; MN = 19; BC = 38
Problem 7 Angle sum with algebra
(2x + 5) + (3x − 10) + (x + 35) = 180 gives 6x + 30 = 180, so x = 25.
The angles are 2(25) + 5 = 55°, 3(25) − 10 = 65° and 25 + 35 = 60°. Check: 55 + 65 + 60 = 180.
x = 25; 55°, 65°, 60°
Problem 8 Exterior angle with algebra
The exterior angle equals the sum of the remote angles: 8x − 9 = (3x + 14) + (2x + 25) = 5x + 39, so 3x = 48 and x = 16.
The exterior angle is 8(16) − 9 = 119°, and the remote angles are 62° and 57°, which add to 119°.
x = 16; 119°
Problem 9 Parts of a median
The centroid is two thirds of the way from the vertex: AG = ⅔(27) = 18, and GD = 27 − 18 = 9 (half of AG).
AG = 18, GD = 9
Problem 10 Centroid ratio with algebra
AG = 2GD: 3x − 2 = 2(x + 4) = 2x + 8, so x = 10.
AG = 28 and GD = 14, so the median is 28 + 14 = 42.
x = 10; AG = 28; median 42
Problem 11 Centroid coordinates
Average the coordinates: x = (1 + 7 + 4) ÷ 3 = 4 and y = (2 − 1 + 8) ÷ 3 = 3.
Centroid (4, 3)
Problem 12 Circumcenter of a right triangle
The right angle is at P, so the hypotenuse is QR. Its length is √(10² + 24²) = √676 = 26.
The circumcenter is the midpoint of QR: (5, 12). The radius is half the hypotenuse, 13. Check: distance to P = √(25 + 144) = 13.
Circumcenter (5, 12); radius 13
Problem 13 Inscribed circle radius
Area = ½ · 9 · 12 = 54. Semiperimeter = (9 + 12 + 15) ÷ 2 = 18. Then 54 = r · 18, so r = 3.
Check with the right-triangle formula: (9 + 12 − 15) ÷ 2 = 3.
r = 3
Problem 14 Name the point of concurrency
- Circumcenter, equidistant from the vertices.
- Incenter, equidistant from the sides.
- Centroid, the balance point.
- Orthocenter.
(a) circumcenter, (b) incenter, (c) centroid, (d) orthocenter
Problem 15 Fire station location
The station must be the circumcenter. The right angle is at A, so the circumcenter is the midpoint of hypotenuse BC: (6, 8).
BC = √(12² + 16²) = 20, so the distance is 10 miles. Check: distance to A = √(36 + 64) = 10.
Station at (6, 8); 10 miles from each town
Problem 16 Balance point of a plate
A uniform triangular plate balances at its centroid. x = (0 + 9 + 3) ÷ 3 = 4 and y = (0 + 0 + 12) ÷ 3 = 4.
The support goes at (4, 4)
Problem 17 Perimeter from midsegments
Each midsegment is half of the side it is parallel to: 9, 13 and 11. The perimeter is 9 + 13 + 11 = 33.
The perimeter of △ABC is 18 + 26 + 22 = 66, so the small triangle has half the perimeter.
Sides 9, 13, 11; perimeter 33, half of 66
Problem 18 Prove the exterior angle theorem
| Statement | Reason |
|---|---|
| 1. m∠A + m∠B + m∠ACB = 180° | Triangle Angle Sum Theorem |
| 2. ∠ACB and ∠ACD form a linear pair | Definition of linear pair |
| 3. m∠ACB + m∠ACD = 180° | Linear pairs are supplementary |
| 4. m∠A + m∠B + m∠ACB = m∠ACB + m∠ACD | Substitution |
| 5. m∠ACD = m∠A + m∠B | Subtraction Property of Equality |
m∠ACD = m∠A + m∠B
Problem 19 Prove the isosceles triangle theorem
| Statement | Reason |
|---|---|
| 1. AB ≅ AC | Given |
| 2. Draw the bisector AD of ∠BAC, so ∠BAD ≅ ∠CAD | Definition of angle bisector |
| 3. AD ≅ AD | Reflexive Property |
| 4. △ABD ≅ △ACD | SAS |
| 5. ∠B ≅ ∠C | CPCTC |
∠B ≅ ∠C
Problem 20 Midsegment on the coordinate plane
M = ((2 − 4) ÷ 2, (10 + 2) ÷ 2) = (−1, 6) and N = ((2 + 10) ÷ 2, (10 + 0) ÷ 2) = (6, 5).
Slope of MN = (5 − 6) ÷ (6 − (−1)) = −1/7. Slope of BC = (0 − 2) ÷ (10 − (−4)) = −2/14 = −1/7. Equal slopes, so MN ∥ BC.
Lengths: MN = √(7² + 1²) = √50 and BC = √(14² + 2²) = √200 = 2√50. So MN = ½BC.
Both slopes are −1/7; MN = √50 = ½ · BC