Segments & Midpoints
B is between A and C. AB = 4x − 7, BC = 2x + 9, and AC = 5x + 20. Find x and all three lengths. Show your equation.
x = 18; AB = 65; BC = 45; AC = 110.
AB + BC = AC
(4x − 7) + (2x + 9) = 5x + 20
6x + 2 = 5x + 20 → x = 18
AB = 65; BC = 45; AC = 110. Check: 65 + 45 = 110.
Find the distance between A(−7, 4) and B(5, −1). Show the coordinate differences and substitution into the distance formula.
AB = 13 units.
Horizontal change: 5 − (−7) = 12. Vertical change: −1 − 4 = −5.
AB = √[12² + (−5)²] = √(144 + 25) = √169 = 13.
Two angles are complementary. The larger angle is 12° more than twice the smaller angle. Find both measures. Define a variable and show your equation.
The angles are 26° and 64°.
Let s be the smaller measure; the larger is 2s + 12.
s + (2s + 12) = 90 → 3s = 78 → s = 26
Larger = 2(26) + 12 = 64°. Check: 26 + 64 = 90.
Two parallel lines are cut by a transversal. Alternate exterior angles measure (5x + 11)° and (8x − 25)°. Find x, their common measure, and the measure of an angle forming a linear pair with either one.
x = 12; common measure = 71°; linear-pair partner = 109°.
Alternate exterior angles are congruent: 5x + 11 = 8x − 25.
36 = 3x → x = 12
5(12) + 11 = 71°; 180 − 71 = 109°.
A regular dodecagon has 12 congruent sides. Each side is (2x + 1) cm and its perimeter is (30x − 6) cm. Find x, one side length, and the perimeter. Show your equation.
x = 3; side = 7 cm; perimeter = 84 cm.
12(2x + 1) = 30x − 6
24x + 12 = 30x − 6 → 18 = 6x → x = 3
Side = 2(3) + 1 = 7 cm; P = 12(7) = 84 cm.
M is the midpoint of PQ. PM = 5x − 9 and PQ = 14x − 42. Find x, MQ, and PQ. Show how the midpoint changes your equation.
x = 6; MQ = 21; PQ = 42.
Since M is the midpoint, 2(PM) = PQ.
2(5x − 9) = 14x − 42
10x − 18 = 14x − 42 → 24 = 4x → x = 6
PM = MQ = 5(6) − 9 = 21; PQ = 2(21) = 42.
A segment has endpoints P(−6, 8) and Q(9, −2). Find its midpoint and its exact length in simplest radical form. Show both formulas.
Midpoint M(3/2, 3); PQ = 5√13 units.
M = ((−6 + 9)/2, (8 − 2)/2) = (3/2, 3).
PQ = √[(9 + 6)² + (−2 − 8)²]
PQ = √(225 + 100) = √325 = 5√13.
Two angles form a linear pair. Their measures are (2x + 6)° and (5x − 8)°. Find x, both angle measures, and the measure of an angle vertical to the smaller angle. Show work.
x = 26; angles 58° and 122°; the vertical angle is 58°.
(2x + 6) + (5x − 8) = 180
7x − 2 = 180 → x = 26
2(26) + 6 = 58°; 5(26) − 8 = 122°.
Vertical angles are congruent, so the requested angle is 58°.
Two parallel lines are cut by a transversal. Same-side interior angles measure (3x + 8)° and (7x − 18)°. Find x, both measures, and the measure of an angle corresponding to the smaller one.
x = 19; angles 65° and 115°; corresponding angle = 65°.
Same-side interior angles are supplementary.
(3x + 8) + (7x − 18) = 180 → 10x − 10 = 180 → x = 19
3(19) + 8 = 65°; 7(19) − 18 = 115°.
Corresponding angles are congruent, so the requested angle is 65°.
A rectangular garden has length (3x − 1) m, width (x + 4) m, and perimeter 70 m. Find x and both dimensions. Then find the fencing needed if a 4 m gate is left open along the boundary.
x = 8; length = 23 m; width = 12 m; fencing = 66 m.
2(3x − 1) + 2(x + 4) = 70
8x + 6 = 70 → x = 8
Length = 23 m; width = 12 m.
Fencing = perimeter − gate = 70 − 4 = 66 m.
A(3a − 4, 2b + 1) and B(a + 8, 4b − 7) have midpoint M(10, 12). Find a, b, and both endpoints. Show work for each coordinate.
a = 4; b = 5; A(8, 11); B(12, 13).
(3a − 4 + a + 8)/2 = 10 → 4a + 4 = 20 → a = 4
(2b + 1 + 4b − 7)/2 = 12 → 6b − 6 = 24 → b = 5
A = (8, 11); B = (12, 13).
Check: ((8 + 12)/2, (11 + 13)/2) = (10, 12).
A(−2, 3) and B(x, −5) are 17 units apart. Find ALL possible values of x, and explain why there are two. Show your equation.
x = 13 or x = −17.
17² = (x + 2)² + (−5 − 3)²
289 = (x + 2)² + 64 → (x + 2)² = 225
x + 2 = ±15 → x = 13 or x = −17.
B can lie 15 units to either side of A horizontally; both have vertical separation 8.
Rays OA, OB, OC, and OD occur in that order inside a 150° angle AOD. m∠AOB = (3x − 5)°, m∠BOC = (2x + 10)°, and m∠COD = (4x + 1)°. Find x, m∠AOC, and m∠BOD. Show your equation.
x = 16; m∠AOC = 85°; m∠BOD = 107°.
(3x − 5) + (2x + 10) + (4x + 1) = 150
9x + 6 = 150 → x = 16
m∠AOB = 43°; m∠BOC = 42°; m∠COD = 65°.
m∠AOC = 43 + 42 = 85°; m∠BOD = 42 + 65 = 107°.
Two parallel lines are cut by a transversal. Corresponding angles A and B measure (4x + 8)° and (6x − 28)°. Angle C forms a linear pair with B and measures (3y + 4)°. Find x, y, and all three angle measures.
x = 18; y = 32; A = B = 80°; C = 100°.
Corresponding angles: 4x + 8 = 6x − 28 → 36 = 2x → x = 18.
A = B = 4(18) + 8 = 80°.
B + C = 180° → 80 + (3y + 4) = 180
3y = 96 → y = 32; C = 100°.
A triangle has perimeter 58 cm and sides (x + 3) cm, (2x − 1) cm, and (3x − 4) cm. Find x and all side lengths. Then find the new perimeter if every side length is doubled.
x = 10; sides 13, 19, 26 cm; new perimeter = 116 cm.
(x + 3) + (2x − 1) + (3x − 4) = 58
6x − 2 = 58 → x = 10
Side lengths: 13, 19, and 26 cm.
Doubling each side doubles perimeter: 2(58) = 116 cm.
A, B, C, and D lie on a line in that order. B is the midpoint of AC, and C is the midpoint of BD. AB = 2x + 4 and AD = 7x + 2. Find x, BC, and BD. Show your reasoning.
x = 10; BC = 24; BD = 48.
B midpoint of AC gives AB = BC. C midpoint of BD gives BC = CD.
Thus AD = AB + BC + CD = 3(AB).
7x + 2 = 3(2x + 4) → 7x + 2 = 6x + 12 → x = 10
AB = BC = CD = 24; BD = 24 + 24 = 48.
Triangle ABC has A(−4, −1), B(8, 4), and C(3, 16). Find all three exact side lengths. Classify the triangle by its sides AND angles, using calculations to justify both.
AB = BC = 13; AC = 13√2 units. It is a right isosceles triangle.
AB = √(12² + 5²) = 13; BC = √[(−5)² + 12²] = 13.
AC = √(7² + 17²) = √338 = 13√2.
AB = BC, so it is isosceles.
13² + 13² = 338 = (13√2)², so ∠B is a right angle.
Rays OA, OB, OC, and OD occur in that order inside ∠AOD. OB bisects ∠AOC, and OC bisects ∠BOD. m∠AOD = 132° and m∠AOB = (5x − 6)°. Find x, m∠BOC, and m∠BOD. Explain the equal parts.
x = 10; m∠BOC = 44°; m∠BOD = 88°.
OB bisects AOC, so m∠AOB = m∠BOC.
OC bisects BOD, so m∠BOC = m∠COD.
The three equal parts total 132°, so each is 44°.
5x − 6 = 44 → x = 10; m∠BOD = 44 + 44 = 88°.
Two parallel lines are cut by a transversal. Same-side interior angles A and B satisfy m∠A = 2(m∠B) + 18°. Angle C is alternate exterior to an angle vertical to B. Find A, B, C, and the ratio A:B in simplest form.
A = 126°; B = 54°; C = 54°; A:B = 7:3.
A + B = 180 and A = 2B + 18.
3B + 18 = 180 → B = 54°; A = 126°.
An angle vertical to B is 54°. Its alternate exterior angle C is also 54°.
126:54 = 7:3.
Rectangle ABCD has A(−1, 1), B(3, 4), C(9, −4), and D(5, −7). Find its perimeter using distances. Then find the perimeter when all side lengths are multiplied by 3/2. Show work.
Original perimeter = 30 units; new perimeter = 45 units.
AB = √(4² + 3²) = 5; BC = √(6² + (−8)²) = 10.
CD = √[(−4)² + (−3)²] = 5; DA = √[(−6)² + 8²] = 10.
P = 5 + 10 + 5 + 10 = 30.
New P = (3/2)(30) = 45 units.
M is the midpoint of a nonzero segment AB. AM = x² − 9 and MB = 2x − 1. Find every geometrically valid value of x and AB. Explain why any algebraic solution must be rejected.
Only x = 4 is valid; AB = 14.
AM = MB → x² − 9 = 2x − 1
x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0
x = 4 or x = −2.
At x = 4, AM = MB = 7, so AB = 14.
Reject x = −2: both proposed lengths are −5, which is impossible.
Point P has coordinates (t, t). P is equally distant from A(−1, 0) and B(6, 7). Find t, the coordinates of P, and the common distance. Set up and solve an equation.
t = 3; P(3, 3); PA = PB = 5 units.
Equate squared distances: (t + 1)² + t² = (t − 6)² + (t − 7)².
2t² + 2t + 1 = 2t² − 26t + 85
28t = 84 → t = 3.
PA = √(4² + 3²) = 5; PB = √[(−3)² + (−4)²] = 5.
An acute angle has the following property: its supplement is 15° more than four times its complement. Find the angle, its complement, and its supplement. Show the equation and check your answer.
Angle = 65°; complement = 25°; supplement = 115°.
Let θ be the angle. Complement = 90 − θ; supplement = 180 − θ.
180 − θ = 4(90 − θ) + 15
180 − θ = 375 − 4θ → 3θ = 195 → θ = 65°.
90 − 65 = 25°; 180 − 65 = 115°. Check: 4(25) + 15 = 115.
Two parallel lines are cut by a transversal. Same-side interior angles measure (x² + 10)° and (12x + 10)°. Find every geometrically valid x and both angle measures. Explain why any algebraic solution is rejected.
Only x = 8 is valid; the angles are 74° and 106°.
(x² + 10) + (12x + 10) = 180
x² + 12x − 160 = 0 → (x + 20)(x − 8) = 0
x = −20 or x = 8.
At x = 8: angles are 74° and 106°.
Reject x = −20: it gives 410° and −230°, impossible for these interior angles.
A net has 6 congruent squares: four in a horizontal row, with one attached above and one below the second square. Each square side is (2x − 1) cm. The outside perimeter is (20x + 34) cm. Find x, the side length, and perimeter. Explain your edge count.
x = 6; square side = 11 cm; outside perimeter = 154 cm.
Six separate squares have 24 edges. There are 5 shared edges.
Exposed edges = 24 − 2(5) = 14.
14(2x − 1) = 20x + 34 → 28x − 14 = 20x + 34
8x = 48 → x = 6; side = 2(6) − 1 = 11 cm.
P = 14(11) = 154 cm (also 20(6) + 34).