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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.

31 questions

Unit 10 · Triangle and Quadrilaterals

Question 1

a.
4 cm12 cmABC
Figure 10.22
Answer

Area = ½ × 12 × 4 = 24 cm².

b.
4 cm7 cmABDC
Figure 10.23
Answer

Area = ½ × base × height = ½ × 7 × 4 = 14 cm².

c.
12 cmPQR
Figure 10.24
Answer

An equilateral triangle of side a has area 34a2 \dfrac{\sqrt3}{4}a^2 : 34×144 \dfrac{\sqrt3}{4} \times 144 = 363 36\sqrt3 cm², about 62.35 cm².

d.
6 cm9 cm5 cmADBEC
Figure 10.25
Answer

Area of a trapezium = ½ × (sum of the parallel sides) × height = ½ × (6 + 9) × 5 = 37.5 cm².

e.
6 in.7 in.QRPSM
Figure 10.26
Answer

Area = base × height = 6 × 7 = 42 sq. in.

f.
5 cm6 cmFGEH
Figure 10.27
Hint

EG is a diagonal, and △EFG is isosceles: its height from E meets FG at its midpoint.

Answer

In △EFG, EF = EG = 5 cm, so the height from E meets FG at its midpoint, 3 cm from F: height = 52−32 \sqrt{5^2 - 3^2} = 4 cm.
Area of △EFG = ½ × 6 × 4 = 12 cm², and the diagonal halves the parallelogram, so area of EFGH = 2 × 12 = 24 cm².

Question 2

a.
2√2 cmPWZXY
Figure 10.28

WXYZ is a square with diagonal WY=22 WY = 2\sqrt2 cm. Find the area of parallelogram PXYW.

Answer

A square’s diagonal is side × 2 \sqrt2 , so the side is 2 cm and the square’s area is 4 cm².
Parallelogram PXYW and square WXYZ stand on the same base XY between XY∥PZ XY \parallel PZ , so area of PXYW = 4 cm².

b.
10 cmAEDBFC
Figure 10.29

In parallelogram ABCD, E is on AD with BE = EC = 10 cm, EF⊥BC EF \perp BC and AD = 16 cm. Find the area of the parallelogram.

Answer

BC = AD = 16 cm. In the isosceles △BEC, EF meets BC at its midpoint, so BF = 8 cm and EF=102−82 EF = \sqrt{10^2 - 8^2} = 6 cm.
EF is the distance between AD and BC: area = 16 × 6 = 96 cm².

c.
8 cmPQRSTA
Figure 10.30

PQST is a parallelogram and R is on TS produced. SA⊥TP SA \perp TP , SA = 8 cm and the area of △PQR is 64 cm². Find the length of TP.

Answer

△PQR stands on the base PQ between PQ∥TR PQ \parallel TR , so area of PQST = 2 × 64 = 128 cm².
Taking TP as the base, the height is SA: TP × 8 = 128, so TP = 16 cm.

d.
PQRSOT
Figure 10.31

PQRS is a parallelogram and O is any point on QR. PO and SR are produced to meet at T, and Q is joined to T. Prove that area of △QOT = area of △ROS.

Answer
Statements Reasons
1. area of △PQT = ½ area of parallelogram PQRS same base PQ, between PQ∥ST PQ \parallel ST
2. area of △POS = ½ area of parallelogram PQRS same base PS, between PS∥QR PS \parallel QR
3. area of △PQO + area of △ROS = ½ area of parallelogram PQRS the rest of the parallelogram, from 2
4. area of △PQO + area of △QOT = area of △PQO + area of △ROS from 1 and 3, as △PQT is △PQO and △QOT together
5. area of △QOT = area of △ROS taking △PQO from both sides

Proved.

e.
ABCDE
Figure 10.32

AC∥DE AC \parallel DE and BC = EC. If the area of △ACE is 24 cm², find the area of quadrilateral ABCD.

Answer

△ABC and △ACE have equal bases BC = CE on one line and the same corner A, so area of △ABC = 24 cm².
△ACD and △ACE stand on the base AC between AC∥DE AC \parallel DE , so area of △ACD = 24 cm².
Area of ABCD = 24 + 24 = 48 cm².

Question 3

a.
18 cm13 cm23 cmPSQR
Figure 10.33

In trapezium PQRS, PS∥QR PS \parallel QR , PQ = 13 cm, PS = 18 cm, QR = 23 cm and SR⊥QR SR \perp QR . Find the area of △PSR.

Hint

Drop a perpendicular from P to QR: it cuts off 23 − 18 = 5 cm.

Answer

The perpendicular from P to QR meets it 23 − 18 = 5 cm from Q, so the height of the trapezium is 132−52 \sqrt{13^2 - 5^2} = 12 cm, the same as SR.
Area of △PSR = ½ × PS × SR = ½ × 18 × 12 = 108 cm².

b.
PQSTR
Figure 10.34

PQ∥SR PQ \parallel SR and PS∥QT PS \parallel QT . The area of trapezium PQRS is 95 cm² and the area of △QRT is 15 cm². Find the area of △RPQ.

Answer

PQTS is a parallelogram (both pairs of sides parallel), and it is the trapezium without △QRT: 95 − 15 = 80 cm².
△RPQ stands on the base PQ between PQ∥SR PQ \parallel SR , so it is half of PQTS: 40 cm².

c.
ABCDEF
Figure 10.35

DE∥AB DE \parallel AB , AD∥BC AD \parallel BC and F is the midpoint of BE. If the area of △AFE is 12 cm², find the area of parallelogram ABCD.

Answer

AF is a median of △ABE, so area of △ABE = 2 × 12 = 24 cm².
△ABE stands on the base AB with E on the line DC∥AB DC \parallel AB , so area of ABCD = 2 × 24 = 48 cm².

d.
6 cm9 cmABCDE
Figure 10.36

In parallelogram ABCD, E is on AD, ∠BEC=90° \angle BEC = 90° , BE = 6 cm and CE = 9 cm. Find the area of ABCD.

Answer

Area of the right-angled △BEC = ½ × 6 × 9 = 27 cm². It stands on the base BC with E on AD∥BC AD \parallel BC , so area of ABCD = 2 × 27 = 54 cm².

Question 4

a.
QRXYPS
Figure 10.37

PQRS is a parallelogram and QRXY is a square. Prove that their areas are equal.

Answer
Statements Reasons
1. parallelogram PQRS and square QRXY stand on the same base QR between QR∥PX QR \parallel PX given figure
2. area of PQRS = area of QRXY a square is a parallelogram; Theorem 1

Proved.

b.
ABCDE
Figure 10.38

ABCD is a quadrilateral, and DE is drawn parallel to AC, meeting BC produced at E. Prove that area of quadrilateral ABCD = area of △ABE.

Answer
Statements Reasons
1. area of △ACD = area of △ACE same base AC, between AC∥DE AC \parallel DE
2. area of △ABC + area of △ACD = area of △ABC + area of △ACE adding △ABC to both sides
3. area of ABCD = area of △ABE from 2, the whole is the sum of its parts

Proved. This is how a quadrilateral is turned into a triangle of the same area in Unit 11.

c.
ABCDEFG
Figure 10.39

ABCD and EFGC are parallelograms, with E on AD and B on FG. Prove that their areas are equal.

Hint

Join EB. △EBC is half of each parallelogram.

Answer

Join E and B.

Statements Reasons
1. area of △EBC = ½ area of parallelogram ABCD same base BC, between BC∥AD BC \parallel AD
2. area of △EBC = ½ area of parallelogram EFGC same base EC, between EC∥FG EC \parallel FG
3. area of ABCD = area of EFGC from 1 and 2

Proved.

d.
PQRSTO
Figure 10.40

PQRS is a parallelogram, SR is produced to T, and PT meets QR at O. Prove that area of △POR = area of △QOT.

Answer
Statements Reasons
1. area of △PQR = area of △PQT same base PQ, between PQ∥ST PQ \parallel ST
2. area of △PQR − area of △PQO = area of △PQT − area of △PQO taking △PQO from both sides
3. area of △POR = area of △QOT from 2, what is left of each

Proved.

e.
PQRSTO
Figure 10.41

PQRS is a parallelogram, SR is produced to T, PT meets QR at O, and S is joined to O and Q to T. Prove that area of △QOT = area of △ROS.

Answer
Statements Reasons
1. area of △POS = ½ area of parallelogram PQRS same base PS, between PS∥QR PS \parallel QR
2. area of △PQO + area of △ROS = ½ area of parallelogram PQRS the rest of the parallelogram, from 1
3. area of △PQT = ½ area of parallelogram PQRS same base PQ, between PQ∥ST PQ \parallel ST
4. area of △PQO + area of △QOT = area of △PQO + area of △ROS from 2 and 3
5. area of △QOT = area of △ROS taking △PQO from both sides

Proved.

f.
ABCDXY
Figure 10.42

ABCD is a parallelogram, X is any point on AD and Y any point on CD; triangles XBC and YAB are drawn. Prove that area of △ABY = area of △ABX + area of △CDX.

Answer
Statements Reasons
1. area of △ABY = ½ area of parallelogram ABCD same base AB, between AB∥DC AB \parallel DC
2. area of △XBC = ½ area of parallelogram ABCD same base BC, between BC∥AD BC \parallel AD
3. area of △ABX + area of △CDX = ½ area of parallelogram ABCD the rest of the parallelogram, from 2
4. area of △ABY = area of △ABX + area of △CDX from 1 and 3

Proved.

g.
ABCDEF
Figure 10.43

ABCD is a parallelogram and E is any point on BC. DE and AB are produced to meet at F, and C is joined to F. Prove that area of △CEF = area of △ABE.

Answer
Statements Reasons
1. area of △DCF = area of △DCB same base DC, between DC∥AF DC \parallel AF
2. area of △DCF − area of △DCE = area of △DCB − area of △DCE taking △DCE from both sides
3. area of △CEF = area of △DEB from 2, what is left of each
4. area of △DEB = area of △AEB same base EB, between BC∥AD BC \parallel AD
5. area of △CEF = area of △ABE from 3 and 4

Proved.

h.
ABCDPQ
Figure 10.44

ABCD is a trapezium with AD∥PQ∥BC AD \parallel PQ \parallel BC , P on AB and Q on DC. Prove that area of △AQB = area of △DPC.

Hint

PQ cuts each of the two triangles into two pieces.

Answer
Statements Reasons
1. area of △APQ = area of △DPQ same base PQ, between PQ∥AD PQ \parallel AD
2. area of △BPQ = area of △CPQ same base PQ, between PQ∥BC PQ \parallel BC
3. area of △APQ + area of △BPQ = area of △DPQ + area of △CPQ adding 1 and 2
4. area of △AQB = area of △DPC from 3: P is on AB and Q on DC

Proved.

i.
PQRST
Figure 10.45

QT∥RS QT \parallel RS , PR∥TS PR \parallel TS and PS∥QR PS \parallel QR . Prove that area of △PQR = area of △PTS.

Answer
Statements Reasons
1. area of △PQR = area of △SQR same base QR, between QR∥PS QR \parallel PS
2. area of △SQR = area of △STR same base RS, between RS∥QT RS \parallel QT
3. area of △STR = area of △STP same base TS, between TS∥PR TS \parallel PR
4. area of △PQR = area of △PTS from 1, 2 and 3

Proved.

j.
ABCDXY
Figure 10.46

ABCD is a trapezium, Y is a point on AC and X a point on BD, with AD∥XY∥BC AD \parallel XY \parallel BC . Prove that area of △AXC = area of △BYD.

Answer
Statements Reasons
1. area of △AXY = area of △DXY same base XY, between XY∥AD XY \parallel AD
2. area of △CXY = area of △BXY same base XY, between XY∥BC XY \parallel BC
3. area of △AXY + area of △CXY = area of △DXY + area of △BXY adding 1 and 2
4. area of △AXC = area of △BYD from 3: Y is on AC and X on BD

Proved.

k.
ABCDE
Figure 10.47

ABCD is a quadrilateral with DB∥CE DB \parallel CE , E on AB produced. Prove that area of △ADE = area of quadrilateral ABCD.

Answer
Statements Reasons
1. area of △BDE = area of △BDC same base BD, between BD∥CE BD \parallel CE
2. area of △ABD + area of △BDE = area of △ABD + area of △BDC adding △ABD to both sides
3. area of △ADE = area of ABCD from 2, the whole is the sum of its parts

Proved.

l.
AFBEPCD
Figure 10.48

ABCDEF is a hexagon with AF∥BE∥CD AF \parallel BE \parallel CD and AB∥CF∥DE AB \parallel CF \parallel DE , and BE and CF meet at P. If parallelograms ABPF and CDEP have equal areas, prove that EF∥BC EF \parallel BC .

Hint

Join BF and CE. Show that △BCF and △BCE have equal areas, then use the converse of Theorem 3.

Answer

Join B to F and C to E.

Statements Reasons
1. area of △BPF = ½ area of ABPF the diagonal BF halves parallelogram ABPF
2. area of △CPE = ½ area of CDEP the diagonal CE halves parallelogram CDEP
3. area of △BPF = area of △CPE from 1 and 2, as ABPF and CDEP are equal
4. area of △BPF + area of △BPC = area of △CPE + area of △BPC adding △BPC to both sides
5. area of △BCF = area of △BCE from 4
6. EF∥BC EF \parallel BC equal triangles on the same base BC, on the same side of it, lie between the same parallels

Proved.

m.
PQRSM
Figure 10.49

The diagonal QS of quadrilateral PQRS bisects the other diagonal PR at M. Prove that area of △PQS = ½ area of quadrilateral PQRS.

Answer
Statements Reasons
1. area of △PQM = area of △RQM QM is a median of △PQR (PM = MR)
2. area of △PSM = area of △RSM SM is a median of △PSR
3. area of △PQS = area of △RQS adding 1 and 2
4. area of △PQS = ½ area of PQRS the two triangles make up PQRS and are equal

Proved.

n.
ABCPQR
Figure 10.50

In △ABC, P is the midpoint of AB and Q is any point on BC. R is a point on AB with RC∥PQ RC \parallel PQ . Prove that area of △BQR = ½ area of △ABC.

Hint

Join PC: it is a median of △ABC.

Answer

Join P and C, and Q and R.

Statements Reasons
1. area of △PQR = area of △PQC same base PQ, between PQ∥RC PQ \parallel RC
2. area of △BPQ + area of △PQR = area of △BPQ + area of △PQC adding △BPQ to both sides
3. area of △BQR = area of △BPC from 2, the whole is the sum of its parts
4. area of △BPC = ½ area of △ABC CP is a median of △ABC
5. area of △BQR = ½ area of △ABC from 3 and 4

Proved.

o.
ABCDEF
Figure 10.51

In quadrilateral ABCD, AD∥BC AD \parallel BC , E is a point on AD and F a point on CD. If area of △ABE = area of △ACF, prove that EF∥AC EF \parallel AC .

Answer

Join C and E.

Statements Reasons
1. area of △ABE = area of △ACE same base AE, between AD∥BC AD \parallel BC
2. area of △ABE = area of △ACF given
3. area of △ACE = area of △ACF from 1 and 2
4. EF∥AC EF \parallel AC equal triangles on the same base AC, on the same side of it, lie between the same parallels

Proved.

Questions to discuss

a.

Draw a triangle and a quadrilateral on the same base and between the same parallels so that the triangle’s area is equal to, half of, a quarter of or a third of the quadrilateral’s. What measurements did you choose, how and why? Justify each one.

Answer

Between the same parallels every figure has the same height, so the areas compare as half the triangle’s base to the parallelogram’s base. Some choices for a parallelogram on a 6 cm base:

  • Equal: the triangle on a base of 12 cm (twice as long).
  • Half: the triangle on the same 6 cm base (Theorem 2).
  • A quarter: the triangle on 3 cm, half the base.
  • A third: the triangle on 4 cm, two thirds of the base: ½ × 4 = 2, a third of 6.