Triangle Theorems and Points of Concurrency: Geometry Math Test Answer Key – PDF Download

Question 1: Angle sum with variables (2 points)

1. \((2x+4)+(3x-1)+(x+15)=180\) gives \(6x+18=180\), so \(6x=162\) and \(x=27\).
2. 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)

1. \(7x-4=(3x+8)+(2x+16)\) gives \(7x-4=5x+24\), so \(2x=28\) and \(x=14\).
2. \(7(14)-4=94^{\circ}\). 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)

1. Base angles are congruent: \(4x+6=6x-14\) gives \(x=10\), and \(m\angle B=4(10)+6=46^{\circ}\).
2. \(m\angle A=180-46-46=88^{\circ}\).

Scoring: 1 point for \(x=10\) and \(46^{\circ}\) with the isosceles reason. 1 point for \(88^{\circ}\).

Question 4: Midsegment theorem (2 points)

1. \(BC=2\cdot DE\): \(5x+4=2(3x-2)\) gives \(5x+4=6x-4\), so \(x=8\). \(DE=3(8)-2=22\text{ ft}\) and \(BC=5(8)+4=44\text{ ft}\).
2. \(AD=\frac{1}{2}(36)=18\) 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)

1. \(BG=2\cdot GE\): \(3x+5=2(x+4)\) gives \(3x+5=2x+8\), so \(x=3\).
2. \(BG=14\) and \(GE=7\), so \(BE=21\text{ in}\).

Scoring: 1 point for the 2 : 1 equation and \(x=3\). 1 point for \(BE=21\).

Question 6: Name the center (2 points)

1. Incenter (intersection of the angle bisectors).
2. 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)

1. 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 \(7.5\).
2. \((\frac{0+12+0}{3},\frac{0+0+9}{3})=(4,3)\).

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)

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)

1. I lies on the angle bisectors: \(m\angle IAB=\frac{64}{2}=32^{\circ}\) and \(m\angle IBA=\frac{48}{2}=24^{\circ}\).
2. \(m\angle AIB=180-32-24=124^{\circ}\).

Scoring: 1 point for both half angles. 1 point for \(124^{\circ}\).

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