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Exam practice

Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.

15 questions

Unit 14 · Quadrilateral

Question 1

a.
2xx
Figure 14.15
Answer

The angles 2x and x are neighbors, so they add up to 180°: 3x=180∘ 3x = 180^\circ , x=60∘ x = 60^\circ (the angles are 120° and 60°).

b.
yx118°
Figure 14.16
Answer

On the straight base, the angle inside the triangle at the foot of the line is 180∘−118∘=62∘ 180^\circ - 118^\circ = 62^\circ . In the triangle, the left side and the line are equal, so the angles facing them are equal: y faces the line, and the 62° faces the left side, so y=62∘ y = 62^\circ . Then x=180∘−62∘−62∘=56∘ x = 180^\circ - 62^\circ - 62^\circ = 56^\circ .

c.
(y − 10)°(5x − 10)°(12x − 14)°
Figure 14.17
Answer

The angles at the two right-hand corners are neighbors: (5x − 10) + (12x − 14) = 180, so 17x = 204 and x = 12. The angle at the bottom right is then 12×12−14=130∘ 12 \times 12 - 14 = 130^\circ , and the top-left angle is opposite it: y − 10 = 130, so y = 140. (The angles are 130°, 50°, 130°, 50°.)

d.
8x3x + 5
Figure 14.18
Answer

The diagonals of a parallelogram bisect each other: 8x = 3x + 5, so 5x = 5 and x = 1 (each half is 8).

e.
3x − 11x + 5
Figure 14.19
Answer

One diagonal is already bisected. The diagonals of a parallelogram bisect each other (Theorem 6), so the other must be bisected too, and then it is a parallelogram (Theorem 7): 3x − 11 = x + 5, so 2x = 16 and x = 8 (each half is 13).

f.
2x − 1x + 5
Figure 14.20
Answer

2x − 1 = x + 5, so x = 6 (each half is 11).

Question 2

a.

ABCD is a parallelogram, and BC is produced to E.

120°xABCDE
Figure 14.21
Answer

∠BCD=∠BAD=120∘ \angle BCD = \angle BAD = 120^\circ (opposite angles), and BCE is a straight line, so x=180∘−120∘=60∘ x = 180^\circ - 120^\circ = 60^\circ .

b.
54°68°yxABCDE
Figure 14.22
Answer

ABDE is a parallelogram (AB∥ED AB \parallel ED , AE∥BD AE \parallel BD ), so its opposite angles are equal: y=∠EDB=∠EAB=54∘ y = \angle EDB = \angle EAB = 54^\circ .
In △BED, ∠EBD=180∘−68∘−54∘=58∘ \angle EBD = 180^\circ - 68^\circ - 54^\circ = 58^\circ . EB∥DC EB \parallel DC , so x = ∠BDC and ∠EBD are alternate angles: x=58∘ x = 58^\circ .
(Or: BCDE is a parallelogram, so ∠CDE=180∘−68∘=112∘ \angle CDE = 180^\circ - 68^\circ = 112^\circ and x=112∘−54∘=58∘ x = 112^\circ - 54^\circ = 58^\circ .)

Question 3

3.

ABCD is a rhombus and ∠DAC=34∘ \angle DAC = 34^\circ . Find ∠DAB, ∠ABC, ∠BCA and ∠CDA.

34°ABCD
Figure 14.23
Answer

The diagonal of a rhombus bisects its angles, so ∠CAB=∠DAC=34∘ \angle CAB = \angle DAC = 34^\circ and ∠DAB=68∘ \angle DAB = 68^\circ .
Neighboring angles add up to 180°: ∠ABC=180∘−68∘=112∘ \angle ABC = 180^\circ - 68^\circ = 112^\circ .
In △ABC, AB = BC, so ∠BCA=∠BAC=34∘ \angle BCA = \angle BAC = 34^\circ .
∠CDA is opposite ∠ABC: ∠CDA=112∘ \angle CDA = 112^\circ .
Some books give 112°, 112°, 68°, 68° here. ∠BCA is only the half of the angle at C on one side of the diagonal, so it is 34°.

Question 4

4.

Two opposite angles of a parallelogram are (63−3x)∘ (63 - 3x)^\circ and (4x−7)∘ (4x - 7)^\circ . Find all the angles of the parallelogram.

Answer

Opposite angles are equal: 63 − 3x = 4x − 7, so 7x = 70 and x = 10. Each of these angles is 63−30=33∘ 63 - 30 = 33^\circ , and the other two are 180∘−33∘=147∘ 180^\circ - 33^\circ = 147^\circ . The angles are 33°, 147°, 33° and 147°.

Question 5

5.

In the quadrilateral MNOP, MN = OP and PM = ON. Prove that MNOP is a parallelogram.

MNOP
Figure 14.24
Answer

Join M and O.

Statements Reasons
1. In △MNO and △OPM: i. MN = OP (S) ii. NO = PM (S) iii. MO = OM (S) i.given ii. given iii. common side
2. △MNO≅△OPM \triangle MNO \cong \triangle OPM by S.S.S.
3. ∠NMO = ∠POM and ∠NOM = ∠PMO corresponding angles of congruent triangles
4. MN∥PO MN \parallel PO and NO∥MP NO \parallel MP from 3: the alternate angles are equal
5. MNOP is a parallelogram both pairs of opposite sides are parallel

Proved.

Question 6

6.

PQRS is a parallelogram, and X and Y are the midpoints of PQ and RS. Prove that PXRY is a parallelogram.

PQRSXY
Figure 14.25
Answer
Statements Reasons
1. PQ = SR and PQ∥SR PQ \parallel SR opposite sides of a parallelogram
2. PX=12PQ PX = \frac12 PQ and YR=12SR YR = \frac12 SR X and Y are midpoints (given)
3. PX = YR from 1 and 2: halves of equal lengths
4. PX∥YR PX \parallel YR parts of the parallel lines PQ and SR
5. PY = XR and PY∥XR PY \parallel XR from 3 and 4, by Theorem 1 (joining the ends on the same side)
6. PXRY is a parallelogram from 4 and 5: both pairs of opposite sides are parallel

Proved.

Question 7

7.

In the figure, AD = EF, AD∥EF AD \parallel EF , EF = BC and EF∥BC EF \parallel BC . Prove that ABCD is a parallelogram.

ABCDEF
Figure 14.26
Answer
Statements Reasons
1. AD = EF and EF = BC given
2. AD = BC from 1: both equal EF
3. AD∥EF AD \parallel EF and EF∥BC EF \parallel BC given
4. AD∥BC AD \parallel BC from 3: lines parallel to the same line are parallel to each other
5. AB = DC and AB∥DC AB \parallel DC from 2 and 4, by Theorem 1 (AB and DC join the ends on the same side)
6. ABCD is a parallelogram from 4 and 5

Proved.

Question 8

8.

In the quadrilateral SLOW, SL = LO = OW = SW. Prove that SLOW is a parallelogram.

Answer
SLOW
Figure 14.27
Statements Reasons
1. SL = OW and LO = WS given (all four sides are equal)
2. SLOW is a parallelogram a quadrilateral whose opposite sides are equal is a parallelogram (Theorem 4)

Proved. (With all four sides equal, it is in fact a rhombus.)

Question 9

9.

MOAT is a quadrilateral whose diagonals MA and OT intersect at R, with MR = RA and TR = OR. Prove that MOAT is a parallelogram.

Answer
MOATR
Figure 14.28
Statements Reasons
1. In △MRT and △ARO: i. MR = AR (S) ii. ∠MRT = ∠ARO (A) iii. RT = RO (S) i.given ii. vertically opposite angles iii. given
2. △MRT≅△ARO \triangle MRT \cong \triangle ARO by S.A.S.
3. MT = AO and ∠RMT = ∠RAO corresponding parts of congruent triangles
4. MT∥AO MT \parallel AO from 3: the alternate angles are equal
5. MO = TA and MO∥TA MO \parallel TA from 3 and 4, by Theorem 1
6. MOAT is a parallelogram its opposite sides are equal and parallel

Proved. (This is Theorem 7 for the quadrilateral MOAT.)