Geometry Theorems About Triangles Worksheet Solutions (PDF)
This triangle theorems answers for Geometry covers theorems about triangles (Common Core G-CO.C.10), with a free PDF to print.
Geometry · Common Core State Standards: G-CO.C.10
Triangle theorems answers: exercises 1 to 10
Exercise 1 – Third angle
\(180^{\circ} – 52^{\circ} – 67^{\circ} = 61^{\circ}\)
61°
Exercise 2 – Right triangle
\(90^{\circ} – 38^{\circ} = 52^{\circ}\)
52°
Exercise 3 – Equilateral triangle
Each side is \(54 \div 3 = 18\).
\(3x – 6 = 18\), so \(x = 8\).
Each angle is \(180^{\circ} \div 3 = 60^{\circ}\).
x = 8; each angle 60°
Exercise 4 – Exterior angle
The exterior angle equals the sum of the remote interior angles.
\(48^{\circ} + 65^{\circ} = 113^{\circ}\)
113°
Exercise 5 – Remote angle
\(140^{\circ} – 85^{\circ} = 55^{\circ}\)
55°
Exercise 6 – Isosceles apex
The base angles are congruent and the three angles sum to 180°.
\((180^{\circ} – 40^{\circ}) \div 2 = 70^{\circ}\)
70° each
Exercise 7 – Isosceles base
Both base angles are 54°.
\(180^{\circ} – 2 \cdot 54^{\circ} = 72^{\circ}\)
72°
Exercise 8 – Angles in a ratio of 1 : 2 : 3
\(6x = 180\), so \(x = 30\).
30°, 60°, 90°
Exercise 9 – Angles with x
\((2x + 10) + (3x – 5) + (x + 25) = 180\)
\(6x + 30 = 180\)
\(x = 25\)
Angles: 60°, 70°, 50°. Check: 60 + 70 + 50 = 180.
x = 25; 60°, 70°, 50°
Exercise 10 – Exterior angle with x
\((2x + 11) + (3x + 13) = 7x – 16\)
\(5x + 24 = 7x – 16\)
\(40 = 2x\)
\(x = 20\)
\(7(20) – 16 = 124^{\circ}\)
x = 20; 124°
Exercises 11 and up
Exercise 11 – Midsegment length
A midsegment is half the length of the third side.
\(DE = 18 \div 2 = 9\)
DE = 9
Exercise 12 – Midsegment with x
\(BC = 2 \cdot DE\), so \(5x – 6 = 2(2x + 3)\).
\(5x – 6 = 4x + 6\), so \(x = 12\).
\(DE = 27\) and \(BC = 54\).
x = 12; DE = 27; BC = 54
Exercise 13 – Midsegment on the plane
\(M(4, 1)\) and \(N(1, 4)\).
\(MN = \sqrt{3^{2} + 3^{2}} = 3\sqrt{2}\).
\(BC = \sqrt{6^{2} + 6^{2}} = 6\sqrt{2}\).
MN is half of BC.
M(4, 1), N(1, 4); MN = 3√2 = ½ BC
Exercise 14 – Centroid and a median
The centroid is two thirds of the way from the vertex.
\(AG = \frac{2}{3} \cdot 15 = 10\), and the rest is \(15 – 10 = 5\).
AG = 10; 5 units
Exercise 15 – Centroid coordinates
\(\left( \frac{0 + 6 + 3}{3}, \frac{0 + 0 + 9}{3} \right) = (3, 3)\)
(3, 3)
Exercise 16 – Can it be a triangle?
The two shorter sides add to \(5 + 7 = 12\), which is less than 13.
No
Exercise 17 – Third side range
\(14 – 9 = 5\) and \(14 + 9 = 23\).
5 < x < 23
Exercise 18 – Garden angles
The parts total \(2 + 3 + 5 = 10\), so one part is \(180^{\circ} \div 10 = 18^{\circ}\).
Angles: 36°, 54°, 90°.
36°, 54°, 90°; a right triangle
Exercise 19 – Roof truss
\((180^{\circ} – 110^{\circ}) \div 2 = 35^{\circ}\)
35°
Exercise 20 – Walking paths
\(200 – 120 = 80\) and \(200 + 120 = 320\).
The third path must be between 80 and 320 feet, so 70 feet is too short.
No; between 80 ft and 320 ft
Key facts behind this triangle theorems answers
- Triangle angle sum: 180 degrees.
- Exterior angle = sum of the two remote interior angles.
- Isosceles triangle: base angles are congruent; equilateral angles are 60 degrees each.
- Midsegment: parallel to the third side and half its length.
- Medians meet at the centroid, two thirds of the way from each vertex.
How to use this triangle theorems answers
Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.
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