Question 1(i)
Complete: All circles are ____.
Steps only
Every circle has the same shape; only its radius may change.
Answer
Similar.
CLASS 10 · MATHEMATICS · CHAPTER 6
Triangles
Test similar figures, apply the Basic Proportionality Theorem and write clear similarity proofs using AA, SSS and SAS.
47 worked answersAll 3 exercise sets
What you will learn
- Recognise similar polygons and triangles.
- Apply the Basic Proportionality Theorem and its converse.
- Use triangle similarity to prove ratios and find lengths.
47 worked answers · short steps first, explanations when you need them
Exercise 6.1
Textbook pages 78
Question 1(ii)
Complete: All squares are ____.
Steps only
Every square has four right angles and equal sides.
Answer
Similar.
Question 1(iii)
Complete: All ____ triangles are similar.
Steps only
Every equilateral triangle has angles 60°, 60°, 60°.
Answer
Equilateral.
Question 1(iv)
Complete the conditions: corresponding angles are ____ and corresponding sides are ____.
Steps only
Corresponding angles must be equal.
Corresponding sides must be proportional.
Answer
Equal; proportional.
Question 2(i)
Give two pairs of similar figures.
Steps only
Pair 1: two circles of different radii
Pair 2: two equilateral triangles of different side lengths
Answer
Two circles; two equilateral triangles.
Question 2(ii)
Give two pairs of non-similar figures.
Steps only
Pair 1: a circle and a square
Pair 2: a square and a non-square rectangle
Answer
A circle with a square; a square with a non-square rectangle.
Question 3
A rhombus has side 1.5 cm and a square has side 3 cm. Are the quadrilaterals similar?
Steps only
Side ratios are all 1.5/3 = 1/2.
The square has four 90° angles, but the shown rhombus does not.
Answer
No, the quadrilaterals are not similar.
Exercise 6.2
Textbook pages 84–85
Question 1(i)
With DE ∥ BC, AD = 1.5 cm, DB = 3 cm and AE = 1 cm, find EC.
Steps only
AD/DB = AE/EC
1.5/3 = 1/EC
EC = 2 cm
Answer
EC = 2 cm.
Question 1(ii)
With DE ∥ BC, DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm, find AD.
Steps only
AD/7.2 = 1.8/5.4 = 1/3
AD = 7.2/3 = 2.4 cm
Answer
AD = 2.4 cm.
Question 2(i)
If PE = 3.9, EQ = 3, PF = 3.6 and FR = 2.4 cm, decide whether EF ∥ QR.
Steps only
PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 1.5
The ratios are unequal.
Answer
No, EF is not parallel to QR.
Question 2(ii)
If PE = 4, EQ = 4.5, PF = 8 and FR = 9 cm, decide whether EF ∥ QR.
Steps only
PE/EQ = 4/4.5 = 8/9
PF/FR = 8/9
The ratios are equal.
Answer
Yes, EF ∥ QR.
Question 2(iii)
If PQ = 1.28, PR = 2.56, PE = 0.18 and PF = 0.36 cm, decide whether EF ∥ QR.
Steps only
PE/PQ = 0.18/1.28 = 9/64
PF/PR = 0.36/2.56 = 9/64
The whole-side ratios are equal.
Answer
Yes, EF ∥ QR.
Question 3
In the given figure, LM ∥ CB and LN ∥ CD. Prove AM/AB = AN/AD.
Steps only
LM ∥ CB ⇒ AM/AB = AL/AC
LN ∥ CD ⇒ AN/AD = AL/AC
Therefore AM/AB = AN/AD.
Answer
Proved: AM/AB = AN/AD.
Question 4
In the given figure, DE ∥ AC and DF ∥ AE. Prove BF/FE = BE/EC.
Steps only
DE ∥ AC ⇒ BD/DA = BE/EC
DF ∥ AE ⇒ BD/DA = BF/FE
Therefore BF/FE = BE/EC.
Answer
Proved: BF/FE = BE/EC.
Question 5
In the given figure, DE ∥ OQ and DF ∥ OR. Show EF ∥ QR.
Steps only
DE ∥ OQ ⇒ PD/DO = PE/EQ
DF ∥ OR ⇒ PD/DO = PF/FR
PE/EQ = PF/FR ⇒ EF ∥ QR by converse BPT
Answer
Hence, EF ∥ QR.
Question 6
A, B and C lie on OP, OQ and OR. If AB ∥ PQ and AC ∥ PR, show BC ∥ QR.
Steps only
AB ∥ PQ ⇒ OA/AP = OB/BQ
AC ∥ PR ⇒ OA/AP = OC/CR
OB/BQ = OC/CR ⇒ BC ∥ QR
Answer
Hence, BC ∥ QR.
Question 7
Using BPT, prove that a line through the midpoint of one side and parallel to another side bisects the third side.
Steps only
Let D be midpoint of AB and DE ∥ BC.
AD/DB = AE/EC by BPT.
AD = DB, so AE = EC.
Answer
The parallel line bisects the third side.
Question 8
Using converse BPT, prove that the line joining the midpoints of two sides is parallel to the third side.
Steps only
Let AD = DB and AE = EC.
Then AD/DB = AE/EC = 1.
By converse BPT, DE ∥ BC.
Answer
The midpoint-joining line is parallel to the third side.
Question 9
ABCD is a trapezium with AB ∥ DC and diagonals meeting at O. Show AO/BO = CO/DO.
Steps only
∠AOB = ∠COD (vertical angles)
∠ABO = ∠CDO (alternate angles), so ΔAOB ~ ΔCOD
AO/CO = BO/DO ⇒ AO/BO = CO/DO
Answer
Proved: AO/BO = CO/DO.
Question 10
The diagonals of ABCD meet at O and AO/BO = CO/DO. Show ABCD is a trapezium.
Steps only
AO/BO = CO/DO ⇒ AO/CO = BO/DO
∠AOB = ∠COD, so ΔAOB ~ ΔCOD by SAS
∠ABO = ∠CDO, hence AB ∥ CD
Answer
ABCD is a trapezium because AB ∥ CD.
Exercise 6.3
Textbook pages 94–97
Question 1(i)
For pair (i), the angles of ABC and PQR are respectively 60°, 80° and 40°. Decide similarity.
Steps only
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
Therefore ΔABC ~ ΔPQR by AAA.
Answer
Similar: ΔABC ~ ΔPQR by AAA.
Question 1(ii)
For pair (ii), ABC has sides 2, 2.5, 3 and the matching sides QR, RP, PQ are 4, 5, 6. Decide similarity.
Steps only
AB/QR = 2/4 = 1/2
BC/RP = 2.5/5 = 1/2; CA/PQ = 3/6 = 1/2
Therefore ΔABC ~ ΔQRP by SSS.
Answer
Similar: ΔABC ~ ΔQRP by SSS.
Question 1(iii)
For pair (iii), compare LMP with sides 2.7, 2, 3 and DEF with sides 5, 4, 6.
Steps only
MP/DE = 2/4 = 1/2
LP/DF = 3/6 = 1/2
LM/EF = 2.7/5 ≠ 1/2
Answer
The triangles are not similar.
Question 1(iv)
For pair (iv), MN = 2.5, ML = 5, QP = 5, QR = 10 and ∠M = ∠Q = 70°. Decide similarity.
Steps only
MN/QP = 2.5/5 = 1/2
ML/QR = 5/10 = 1/2
Included angles are equal, so ΔMNL ~ ΔQPR by SAS.
Answer
Similar: ΔMNL ~ ΔQPR by SAS.
Question 1(v)
For pair (v), AB = 2.5, BC = 3, DF = 5, EF = 6 and one 80° angle is shown in each triangle. Decide similarity.
Steps only
AB/DF = BC/EF = 1/2
The equal 80° angles are not both included between these proportional side pairs.
Answer
The triangles are not proved similar; SAS does not apply.
Question 1(vi)
For pair (vi), DEF has angles D = 70°, E = 80°; PQR has Q = 80°, R = 30°. Decide similarity.
Steps only
∠F = 180° − 70° − 80° = 30°
∠P = 180° − 80° − 30° = 70°
D ↔ P, E ↔ Q, F ↔ R, so ΔDEF ~ ΔPQR by AAA.
Answer
Similar: ΔDEF ~ ΔPQR by AAA.
Question 2
Given ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°, find ∠DOC, ∠DCO and ∠OAB.
Steps only
∠DOC = 180° − 125° = 55°
∠DCO = 180° − 70° − 55° = 55°
∠OAB = ∠CDO = 70°
Answer
∠DOC = 55°, ∠DCO = 55° and ∠OAB = 70°.
Question 3
Diagonals AC and BD of trapezium ABCD, with AB ∥ DC, meet at O. Prove OA/OC = OB/OD.
Steps only
∠AOB = ∠COD (vertical angles)
∠ABO = ∠CDO (alternate angles)
ΔAOB ~ ΔCOD by AA, so OA/OC = OB/OD.
Answer
Proved: OA/OC = OB/OD.
Question 4
Given QR/QS = QT/PR and ∠1 = ∠2 in the figure, show ΔPQS ~ ΔTQR.
Steps only
In ΔPQR, ∠1 = ∠2 ⇒ PQ = PR
Thus QR/QS = QT/PQ
The included angle at Q is common, so ΔPQS ~ ΔTQR by SAS.
Answer
Proved: ΔPQS ~ ΔTQR.
Question 5
S and T lie on PR and QR of ΔPQR, and ∠P = ∠RTS. Show ΔRPQ ~ ΔRTS.
Steps only
∠PRQ = ∠SRT (same angle at R)
∠RPQ = ∠RTS (given)
Therefore ΔRPQ ~ ΔRTS by AA.
Answer
Proved: ΔRPQ ~ ΔRTS.
Question 6
In the figure, ΔABE ≅ ΔACD. Show ΔADE ~ ΔABC.
Steps only
Congruence gives AB = AC and AE = AD.
AD/AB = AE/AC and ∠DAE = ∠BAC.
Therefore ΔADE ~ ΔABC by SAS.
Answer
Proved: ΔADE ~ ΔABC.
Question 7(i)
Altitudes AD and CE meet at P. Show ΔAEP ~ ΔCDP.
Steps only
∠AEP = ∠CDP = 90°
∠APE = ∠CPD (vertical angles)
Therefore ΔAEP ~ ΔCDP by AA.
Answer
Proved: ΔAEP ~ ΔCDP.
Question 7(ii)
For the same figure, show ΔABD ~ ΔCBE.
Steps only
∠ADB = ∠CEB = 90°
∠ABD = ∠CBE (same angle at B)
Therefore ΔABD ~ ΔCBE by AA.
Answer
Proved: ΔABD ~ ΔCBE.
Question 7(iii)
For the same figure, show ΔAEP ~ ΔADB.
Steps only
∠AEP = ∠ADB = 90°
∠EAP = ∠DAB (same lines AB and AD)
Therefore ΔAEP ~ ΔADB by AA.
Answer
Proved: ΔAEP ~ ΔADB.
Question 7(iv)
For the same figure, show ΔPDC ~ ΔBEC.
Steps only
∠PDC = ∠BEC = 90°
∠PCD = ∠BCE (same lines BC and CE)
Therefore ΔPDC ~ ΔBEC by AA.
Answer
Proved: ΔPDC ~ ΔBEC.
Question 8
E lies on AD produced in parallelogram ABCD, and BE meets CD at F. Show ΔABE ~ ΔCFB.
Steps only
AB ∥ CD ⇒ ∠ABE = ∠CFB
AD ∥ BC ⇒ ∠AEB = ∠CBF
Therefore ΔABE ~ ΔCFB by AA.
Answer
Proved: ΔABE ~ ΔCFB.
Question 9(i)
ABC and AMP are right-angled at B and M. Prove ΔABC ~ ΔAMP.
Steps only
∠ABC = ∠AMP = 90°
∠BAC = ∠PAM (common angle at A)
Therefore ΔABC ~ ΔAMP by AA.
Answer
Proved: ΔABC ~ ΔAMP.
Question 9(ii)
Using ΔABC ~ ΔAMP, prove CA/PA = BC/MP.
Steps only
ABC corresponds to AMP.
AC corresponds to AP, and BC corresponds to MP.
Therefore CA/PA = BC/MP.
Answer
Proved: CA/PA = BC/MP.
Question 10(i)
ABC ~ FEG. CD and GH bisect angles C and G. Prove CD/GH = AC/FG.
Steps only
∠DCA = 1/2∠C = 1/2∠G = ∠HGF
∠DAC = ∠HFG, so ΔDCA ~ ΔHGF by AA
Corresponding sides give CD/GH = AC/FG.
Answer
Proved: CD/GH = AC/FG.
Question 10(ii)
Under the same conditions, show ΔDCB ~ ΔHGE.
Steps only
AC/FG = BC/EG from ABC ~ FEG
CD/GH = AC/FG, so CD/GH = BC/EG
∠DCB = ∠HGE; therefore ΔDCB ~ ΔHGE by SAS.
Answer
Proved: ΔDCB ~ ΔHGE.
Question 10(iii)
Under the same conditions, show ΔDCA ~ ΔHGF.
Steps only
∠DCA = ∠HGF (halves of equal angles)
∠DAC = ∠HFG (corresponding angles)
Therefore ΔDCA ~ ΔHGF by AA.
Answer
Proved: ΔDCA ~ ΔHGF.
Question 11
ABC is isosceles with AB = AC; E lies on CB produced, AD ⟂ BC and EF ⟂ AC. Prove ΔABD ~ ΔECF.
Steps only
∠ADB = ∠EFC = 90°
AB = AC ⇒ ∠ABC = ∠BCA, so ∠ABD = ∠ECF
Therefore ΔABD ~ ΔECF by AA.
Answer
Proved: ΔABD ~ ΔECF.
Question 12
AB, BC and median AD are proportional to PQ, QR and median PM. Show ΔABC ~ ΔPQR.
Steps only
BD = BC/2 and QM = QR/2, so BD/QM = BC/QR
AB/PQ = AD/PM = BD/QM, hence ΔABD ~ ΔPQM by SSS
∠ABC = ∠PQR and AB/PQ = BC/QR, so ΔABC ~ ΔPQR by SAS.
Answer
Proved: ΔABC ~ ΔPQR.
Question 13
D lies on BC and ∠ADC = ∠BAC. Show CA² = CB·CD.
Steps only
∠ADC = ∠BAC (given)
∠ACD = ∠BCA (same angle at C)
ΔADC ~ ΔBAC, so AC/BC = CD/AC ⇒ AC² = BC·CD.
Answer
Proved: CA² = CB·CD.
Question 14
AB, AC and median AD are proportional to PQ, PR and median PM. Show ΔABC ~ ΔPQR.
Steps only
Let AB/PQ = AC/PR = AD/PM = k.
Apollonius: AB² + AC² = 2(AD² + BD²) and PQ² + PR² = 2(PM² + QM²).
Substitution gives BD = kQM, hence BC = kQR; all three side pairs are proportional, so ΔABC ~ ΔPQR by SSS.
Answer
Proved: ΔABC ~ ΔPQR.
Question 15
A 6 m pole casts a 4 m shadow while a tower casts a 28 m shadow. Find the tower height.
Steps only
Height/shadow is the same at the same time.
6/4 = h/28
h = 28 × 6/4 = 42 m
Answer
The tower is 42 m high.
Question 16
If ΔABC ~ ΔPQR and AD, PM are corresponding medians, prove AB/PQ = AD/PM.
Steps only
AB/PQ = BC/QR from similarity
BD = BC/2 and QM = QR/2, so BD/QM = AB/PQ
∠ABD = ∠PQM; hence ΔABD ~ ΔPQM by SAS and AB/PQ = AD/PM.
Answer
Proved: AB/PQ = AD/PM.