Each answer shows the theorem used, the algebra, and a check. When a problem asks for a center, compare your result with the figure: the centroid, incenter, and acute-triangle orthocenter always fall inside the triangle.
Problem 1: Find the third angle
The angles add up to \(180^{\circ}\): \(m\angle C=180-52-67=61^{\circ}\). Check: \(52+67+61=180\). Answer: \(61^{\circ}\).
Problem 2: Acute angles of a right triangle
\(m\angle B=180-90-35=55^{\circ}\). The acute angles of a right triangle add up to \(90^{\circ}\): \(35+55=90\). Answer: \(55^{\circ}\).
Problem 3: Base angles of an isosceles triangle
The base angles are congruent. They share \(180-40=140^{\circ}\), so each is \(140: 2=70^{\circ}\). Check: \(40+70+70=180\). Answer: \(70^{\circ}\) each.
Problem 4: Angle sum with variables
Set the sum equal to 180: \((x+10)+(2x-5)+(3x+7)=180\). Combine: \(6x+12=180\). Subtract 12: \(6x=168\). Divide: \(x=28\).
The angles are \(28+10=38^{\circ}\), \(2(28)-5=51^{\circ}\), and \(3(28)+7=91^{\circ}\). Check: \(38+51+91=180\). Answer: \(x=28\); \(38^{\circ}\), \(51^{\circ}\), \(91^{\circ}\).
Problem 5: Use the exterior angle
The exterior angle equals the sum of the remote interior angles: \(41+67=108^{\circ}\).
Check: the interior angle at C is \(180-41-67=72^{\circ}\), and \(180-72=108\). Answer: \(108^{\circ}\).
Problem 6: Missing remote angle
The exterior angle equals the sum of the remote angles: \(75+x=128\), so \(x=128-75=53^{\circ}\). Check: the adjacent interior angle is \(180-128=52^{\circ}\), and \(75+53+52=180\). Answer: \(53^{\circ}\).
Problem 7: Exterior angle with variables
Exterior angle \(=\) sum of remote angles: \(6x-4=(2x+10)+(3x+6)\). Simplify: \(6x-4=5x+16\). Subtract \(5x\): \(x-4=16\). Add 4: \(x=20\).
Exterior angle: \(6(20)-4=116^{\circ}\). Remote angles: \(50^{\circ}\) and \(66^{\circ}\), and \(50+66=116\). Answer: \(x=20\); \(116^{\circ}\).
Problem 8: Isosceles triangle with variables
Base angles are congruent: \(3x+4=5x-16\). Subtract \(3x\): \(4=2x-16\). Add 16: \(20=2x\), so \(x=10\).
Each base angle is \(3(10)+4=34^{\circ}\). Then \(m\angle A=180-34-34=112^{\circ}\). Answer: \(x=10\); base angles \(34^{\circ}\); \(m\angle A=112^{\circ}\).
Problem 9: Equal angles, equal sides
By the converse of the Isosceles Triangle Theorem, the sides opposite equal angles are congruent. \(\overline{QR}\) is opposite \(\angle P\) and \(\overline{PR}\) is opposite \(\angle Q\), so \(PR=QR\).
\(2x+1=4x-13\) gives \(14=2x\), so \(x=7\). \(PR=2(7)+1=15\text{ cm}\) and \(QR=4(7)-13=15\text{ cm}\). \(m\angle R=180-58-58=64^{\circ}\). Answer: \(x=7\), \(PR=15\text{ cm}\), \(m\angle R=64^{\circ}\).
Problem 10: Midsegment length
The midsegment is half the length of the parallel side: \(DE=\frac{1}{2}(26)=13\). Answer: \(DE=13\).
Problem 11: Midsegment with variables
\(BC=2\cdot DE\), so \(7x+2=2(4x-3)\). Distribute: \(7x+2=8x-6\). Subtract \(7x\): \(2=x-6\). Add 6: \(x=8\).
\(DE=4(8)-3=29\text{ in}\) and \(BC=7(8)+2=58\text{ in}\). Check: \(2\times 29=58\). Answer: \(x=8\), \(DE=29\text{ in}\), \(BC=58\text{ in}\).
Problem 12: Midsegment on a grid
\(D=(\frac{2+(-2)}{2},\frac{8+0}{2})=(0,4)\) and \(E=(\frac{2+10}{2},\frac{8+0}{2})=(6,4)\).
\(\overline{DE}\) is horizontal (both \(y\)-coordinates are 4) and \(\overline{BC}\) lies on the \(x\)-axis, so they are parallel. \(DE=6-0=6\) and \(BC=10-(-2)=12\), so \(DE=\frac{1}{2}BC\). Answer: \(D(0,4)\), \(E(6,4)\); parallel; \(6=\frac{1}{2}(12)\).
Problem 13: Segments of a median
The centroid is \(\frac{2}{3}\) of the way from the vertex: \(AG=\frac{2}{3}(21)=14\). The rest is \(GD=21-14=7\). Check: \(AG=2\cdot GD\). Answer: \(AG=14\), \(GD=7\).
Problem 14: Centroid with variables
\(BG=2\cdot GD\), so \(5x-1=2(2x+1)\). Distribute: \(5x-1=4x+2\). Subtract \(4x\): \(x-1=2\). Add 1: \(x=3\).
\(BG=5(3)-1=14\) and \(GD=2(3)+1=7\), so \(BD=14+7=21\text{ in}\). Answer: \(x=3\), \(BG=14\text{ in}\), \(BD=21\text{ in}\).
Problem 15: Centroid on a grid
The centroid has the average of the vertex coordinates.
\(x=\frac{1+7+4}{3}=\frac{12}{3}=4\) and \(y=\frac{2+2+8}{3}=\frac{12}{3}=4\). Answer: \((4,4)\).
Problem 16: Name the center
- Centroid.
- Circumcenter, the center of the circumscribed circle.
- Incenter, the center of the inscribed circle.
- Orthocenter.
Problem 17: Circumcenter of a right triangle
In a right triangle the circumcenter is the midpoint of the hypotenuse \(\overline{BC}\): \((\frac{12+0}{2},\frac{0+5}{2})=(6,2.5)\).
The hypotenuse is \(\sqrt{12^{2}+5^{2}}=13\), and the radius is half of it, \(6.5\). Check: \(\sqrt{6^{2}+2.5^{2}}=\sqrt{42.25}=6.5\). Answer: circumcenter \((6,2.5)\), radius \(6.5\).
Problem 18: Circumcenter by perpendicular bisectors
\(\overline{AB}\) is horizontal with midpoint \((3,0)\), so its perpendicular bisector is \(x=3\).
\(\overline{AC}\) has midpoint \((1,3)\) and slope \(\frac{6}{2}=3\), so the perpendicular slope is \(-\frac{1}{3}\): \(y-3=-\frac{1}{3}(x-1)\).
At \(x=3\): \(y=3-\frac{1}{3}(2)=\frac{7}{3}\). Check: the squared distance to A is \(3^{2}+(\frac{7}{3})^{2}=\frac{130}{9}\), and to C is \(1^{2}+(\frac{11}{3})^{2}=\frac{130}{9}\). Answer: \((3,\frac{7}{3})\).
Problem 19: Incenter and inscribed circle
- Area \(=\frac{1}{2}(5)(12)=30\text{ ft}^{2}\). Semiperimeter \(s=\frac{5+12+13}{2}=15\text{ ft}\).
- \(30=r\cdot15\), so \(r=2\text{ ft}\).
- The incenter is equidistant from all three sides, so each distance is 2 ft.
Problem 20: Angles at the incenter
I lies on the angle bisectors of \(\angle A\) and \(\angle B\). So \(m\angle BAI=\frac{50}{2}=25^{\circ}\) and \(m\angle ABI=\frac{70}{2}=35^{\circ}\).
In \(\triangle ABI\): \(m\angle AIB=180-25-35=120^{\circ}\). Answer: \(25^{\circ}\), \(35^{\circ}\), \(120^{\circ}\).
Problem 21: Where to put the fountain
The fountain is the circumcenter. The perpendicular bisector of the Oak-Elm segment is \(x=6\).
The Oak-Pine segment has midpoint \((2,4)\) and slope \(2\), so its perpendicular bisector has slope \(-\frac{1}{2}\): \(y-4=-\frac{1}{2}(x-2)\). At \(x=6\): \(y=4-2=2\). The fountain is at \((6,2)\).
Distance to the Oak: \(\sqrt{6^{2}+2^{2}}=\sqrt{40}\), which is about 6.3 yd. Pine: \(\sqrt{(4-6)^{2}+(8-2)^{2}}=\sqrt{40}\), the same. Answer: \((6,2)\), about 6.3 yd from each tree.
Problem 22: Largest circle in a triangle
The largest circle is the inscribed circle. Since \(9^{2}+12^{2}=225=15^{2}\), the triangle is a right triangle with area \(\frac{1}{2}(9)(12)=54\text{ ft}^{2}\).
\(s=\frac{9+12+15}{2}=18\), and \(54=r\cdot18\) gives \(r=3\text{ ft}\). The diameter is \(2\times 3=6\). Answer: 6 ft.
Problem 23: Balance point of a triangle
A uniform triangle balances at its centroid, the average of the vertices.
\(x=\frac{0+12+3}{3}=5\) and \(y=\frac{0+0+9}{3}=3\). Answer: \((5,3)\), 5 inches right and 3 inches up from the lower left corner.
Problem 24: Roof gable angles
- \(180-110=70\), shared by two congruent base angles: \(70: 2=35^{\circ}\).
- Exterior angle at B: \(180-35=145^{\circ}\). Check: remote angles are \(110\) and \(35\), and \(110+35=145\). The exterior angle is \(145^{\circ}\).
Problem 25: Midsegment triangle perimeter
Each side of \(\triangle DEF\) is a midsegment, half as long as the parallel side of \(\triangle ABC\): \(DE=\frac{1}{2}(18)=9\) (parallel to \(\overline{CA}\)), \(EF=\frac{1}{2}(14)=7\) (parallel to \(\overline{AB}\)), and \(DF=\frac{1}{2}(22)=11\) (parallel to \(\overline{BC}\)).
Perimeter of \(\triangle DEF\): \(9+7+11=27\text{ in}\). Perimeter of \(\triangle ABC\): \(14+22+18=54\text{ in}\). Answer: 27 in, exactly half of 54 in.
Problem 26: Orthocenter on a grid
The altitude from C to the horizontal side \(\overline{AB}\) is vertical: \(x=2\).
\(\overline{BC}\) has slope \(\frac{6-0}{2-8}=-1\), so the altitude from A has slope \(1\): \(y=x\). At \(x=2\), \(y=2\). The orthocenter is \((2,2)\).
Check: the altitude from B should be perpendicular to \(\overline{AC}\), which has slope \(3\). The slope from B to H is \(\frac{2-0}{2-8}=-\frac{1}{3}\), and \(3\times (-\frac{1}{3})=-1\). Answer: \((2,2)\).
See every chapter on the Geometry math page.