Each solution shows the postulate or definition used, the equation, the algebra and a check. Compare every step with your own work, not only the final answer, and redraw any figure that did not match.
Problem 1: Naming points and rays
- P, Q and R lie on one line: P, Q, R.
- S is off the line, so S with any two of the others works: P, Q, S.
- Endpoint first: ray QR.
Problem 2: Defining a segment
A line segment is the part of a line that consists of two endpoints and all the points of the line between them.
Problem 3: Segment addition: find the whole
Segment Addition Postulate: \(AC=AB+BC=7+12=19\). AC = 19 in.
Problem 4: Segment addition: find a part
Subtract: \(BC=AC-AB=31-18=13\). BC = 13 cm. Check: \(18+13=31\).
Problem 5: Segment addition with algebra
\((3x+2)+(2x+7)=34\), so \(5x+9=34\) and \(5x=25\). x = 5.
\(AB=3(5)+2=17\) and \(BC=2(5)+7=17\). AB = 17 and BC = 17. Check: \(17+17=34\). B is the midpoint of \(\overline{AC}\).
Problem 6: Midpoint with algebra
A midpoint makes the halves equal: \(4x-3=2x+9\). Then \(2x=12\) and x = 6.
\(PM=4(6)-3=21\) and \(MQ=2(6)+9=21\). \(PQ=21+21=42\). PQ = 42.
Problem 7: Midpoint on a number line
\(AB=|10-(-4)|=14\). The midpoint is halfway: \(\frac{-4+10}{2}=\frac{6}{2}=3\). AB = 14; midpoint at 3. Check: \(3-(-4)=7\) and \(10-3=7\).
Problem 8: Angle addition: find the whole
\(m\angle ABC\)\(=\)\(m\angle ABD\)\(+\)\(m\angle DBC\)\(=38^{\circ}+47^{\circ}=85^{\circ}\). 85°.
Problem 9: Angle addition: find a part
\(m\angle BOC=112^{\circ}-65^{\circ}=47^{\circ}\). 47°. Check: \(65+47=112\).
Problem 10: Angle addition with algebra
\((4x+8)+(2x+4)=96\), so \(6x+12=96\) and \(6x=84\). x = 14.
\(m\angle PQS\)\(=4(14)+8=64^{\circ}\) and \(m\angle SQR\)\(=2(14)+4=32^{\circ}\). Check: \(64+32=96\). 64° and 32°.
Problem 11: Angle bisector
A bisector makes two equal angles: \(3x+5=5x-9\). Then \(14=2x\) and x = 7.
\(m\angle ABD\)\(=3(7)+5=26^{\circ}\) and \(m\angle DBC\)\(=5(7)-9=26^{\circ}\). \(m\angle ABC\)\(=26+26=52^{\circ}\). 52°.
Problem 12: Complement of an angle
\(90-63=27\). 27°.
Problem 13: Supplement of an angle
\(180-127=53\). 53°.
Problem 14: Complementary angles with algebra
\((2x+3)+(3x+7)=90\), so \(5x+10=90\) and \(5x=80\). x = 16.
The angles are \(2(16)+3=35^{\circ}\) and \(3(16)+7=55^{\circ}\). Check: \(35+55=90\). 35° and 55°.
Problem 15: Supplementary angles with algebra
\((4x-10)+(6x+20)=180\), so \(10x+10=180\) and \(10x=170\). x = 17.
The angles are \(4(17)-10=58^{\circ}\) and \(6(17)+20=122^{\circ}\). Check: \(58+122=180\). 58° and 122°.
Problem 16: Angles at an intersection
Angle a forms a linear pair with the \(48^{\circ}\) angle: \(a=180-48=132^{\circ}\).
Angle b is vertical to the \(48^{\circ}\) angle, so \(b=48^{\circ}\). Angle c is vertical to a, so \(c=132^{\circ}\). a = 132°, b = 48°, c = 132°.
Problem 17: Vertical angles with algebra
Vertical angles are congruent: \(7x-12=4x+18\). Then \(3x=30\) and x = 10.
\(m\angle 1\)\(=7(10)-12=58^{\circ}\) and \(m\angle 3\)\(=4(10)+18=58^{\circ}\). Angle 2 forms a linear pair with angle 1: \(m\angle 2\)\(=180-58=122^{\circ}\). 58° and 122°.
Problem 18: Perpendicular lines
Perpendicular lines form right angles, so \(3x+6=90\). Then \(3x=84\) and x = 28. Check: \(3(28)+6=90\).
Problem 19: Points and a circle
A circle is the set of all points in the plane at the same distance from the center. Only D is 6 cm from C. E is closer than 6 cm (inside) and F is farther (outside). Only point D is on the circle.
Problem 20: Road trip on a straight highway
Distance: \(74-18=56\) miles. Half of that is \(56: 2=28\) miles.
Rest stop: \(18+28=46\). 56 miles apart; mile marker 46. Check: \(74-46=28\).
Problem 21: Ladder against a wall
The wall and the ground are perpendicular, so the ladder splits a right angle into two complementary angles.
\(90-75=15\). 15°.
Problem 22: Laptop lid
The new angle plus the \(40^{\circ}\) rotation makes the original angle (Angle Addition Postulate): \(115-40=75\). 75°.
Problem 23: Four angles at an intersection
A linear pair is supplementary: \((2x+15)+(3x-5)=180\), so \(5x+10=180\) and \(5x=170\). x = 34.
\(m\angle 1\)\(=2(34)+15=83^{\circ}\) and \(m\angle 2\)\(=3(34)-5=97^{\circ}\). Angle 3 is vertical to angle 1 and angle 4 is vertical to angle 2.
83°, 97°, 83°, 97°. Check: \(83+97=180\).
See every chapter on the Geometry math page.