This geometry triangle theorems test answer key gives the full solution and the scoring guide for each of the 9 test questions on triangle theorems and points of concurrency.
Use this key to score the triangle theorems test out of 20 points. Each solution lists the theorem used and a scoring guide. Award partial credit exactly as described and accept any equivalent correct method.
Geometry Triangle Theorems Test Answer Key: questions 1–5
Question 1: Angle sum with variables (2 points)
- \((2x+4)+(3x-1)+(x+15)=180\) gives \(6x+18=180\), so \(6x=162\) and \(x=27\).
- The angles are \(58^{\circ}\), \(80^{\circ}\), and \(42^{\circ}\). The largest is \(80^{\circ}\).
Scoring: 1 point for a correct equation and \(x=27\). 1 point for \(80^{\circ}\).
Question 2: Exterior angle equation (2 points)
- \(7x-4=(3x+8)+(2x+16)\) gives \(7x-4=5x+24\), so \(2x=28\) and \(x=14\).
- \(7(14)-4=94^{\circ}\). Check: \(50+44=94\).
Scoring: 1 point for using the Exterior Angle Theorem to find \(x=14\). 1 point for \(94^{\circ}\).
Question 3: Isosceles triangle angles (2 points)
- Base angles are congruent: \(4x+6=6x-14\) gives \(x=10\), and \(m\angle B=4(10)+6=46^{\circ}\).
- \(m\angle A=180-46-46=88^{\circ}\).
Scoring: 1 point for \(x=10\) and \(46^{\circ}\) with the isosceles reason. 1 point for \(88^{\circ}\).
Question 4: Midsegment theorem (2 points)
- \(BC=2\cdot DE\): \(5x+4=2(3x-2)\) gives \(5x+4=6x-4\), so \(x=8\). \(DE=3(8)-2=22\text{ ft}\) and \(BC=5(8)+4=44\text{ ft}\).
- \(AD=\frac{1}{2}(36)=18\) and \(AE=\frac{1}{2}(40)=20\). Perimeter \(=18+20+22=60\text{ ft}\).
Scoring: 1 point for \(x=8\) with \(DE\) and \(BC\). 1 point for the three sides and perimeter \(60\) ft.
Question 5: Centroid segments (2 points)
- \(BG=2\cdot GE\): \(3x+5=2(x+4)\) gives \(3x+5=2x+8\), so \(x=3\).
- \(BG=14\) and \(GE=7\), so \(BE=21\text{ in}\).
Scoring: 1 point for the 2 : 1 equation and \(x=3\). 1 point for \(BE=21\).
Geometry Triangle Theorems Test Answer Key: questions 6–9
Question 6: Name the center (2 points)
- Incenter (intersection of the angle bisectors).
- Circumcenter (intersection of the perpendicular bisectors).
Scoring: 1 point for each correct center. Accept the matching description of the lines as support.
Question 7: Centers on the plane (3 points)
- The triangle is right-angled at A, so the circumcenter is the midpoint of the hypotenuse \(\overline{BC}\): \((\frac{12+0}{2},\frac{0+9}{2})=(6,4.5)\). The hypotenuse is \(\sqrt{12^{2}+9^{2}}=15\), so the radius is \(7.5\).
- \((\frac{0+12+0}{3},\frac{0+0+9}{3})=(4,3)\).
Scoring: 1 point for the circumcenter \((6,4.5)\), 1 point for the radius \(7.5\), 1 point for the centroid \((4,3)\).
Question 8: Prove the angle sum (3 points)
| Statements | Reasons |
|---|---|
| Line \(DE\) passes through C and \(DE\parallel AB\) | Given |
| \(\angle DCA\cong\angle A\) and \(\angle ECB\cong\angle B\) | Alternate interior angles are congruent |
| \(m\angle DCA+m\angle ACB+m\angle ECB=180^{\circ}\) | Angles along a straight line add to 180 degrees |
| \(m\angle A+m\angle ACB+m\angle B=180^{\circ}\) | Substitution |
Scoring: 1 point for the parallel lines with the alternate interior angle reason. 1 point for the straight angle statement. 1 point for the substitution that reaches the conclusion.
Question 9: Incenter angles (2 points)
- I lies on the angle bisectors: \(m\angle IAB=\frac{64}{2}=32^{\circ}\) and \(m\angle IBA=\frac{48}{2}=24^{\circ}\).
- \(m\angle AIB=180-32-24=124^{\circ}\).
Scoring: 1 point for both half angles. 1 point for \(124^{\circ}\).
Scoring summary
20 points: add the points of every question of this geometry triangle theorems test answer key.
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Geometry Triangle Theorems and Points of Concurrency test answer key
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