Triangles | Exercise 6.3

Question 1

  1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Question diagram 1
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Solution
Understand the Question

Two triangles are similar if their corresponding angles are equal and their corresponding sides are proportional. The main similarity criteria are:

  • AAA / AA Criterion: If corresponding angles of two triangles are equal, the triangles are similar.
  • SSS Criterion: If the ratios of all three corresponding sides of two triangles are equal, the triangles are similar.
  • SAS Criterion: If two pairs of corresponding sides are in the same ratio and their included angles are equal, the triangles are similar.

When stating similarity symbolically, the order of vertices must match the corresponding equal angles and proportional sides.

(i) Check similarity for △ABC\triangle ABC with angles 60∘,80∘,40∘60^\circ, 80^\circ, 40^\circ and △PQR\triangle PQR with angles 60∘,80∘,40∘60^\circ, 80^\circ, 40^\circ.

Step 1 · Compare Corresponding Angles

Question diagram

In △ABC\triangle ABC and △PQR\triangle PQR:

∠A=∠P=60∘∠B=∠Q=80∘∠C=∠R=40∘\begin{aligned} \angle A &= \angle P = 60^\circ \\ \angle B &= \angle Q = 80^\circ \\ \angle C &= \angle R = 40^\circ \end{aligned}

Since all corresponding angles are equal, by AAA similarity criterion: △ABC∼△PQR\triangle ABC \sim \triangle PQR

Answer

(i) △ABC∼△PQR\triangle ABC \sim \triangle PQR (by AAA criterion)

(ii) Check similarity for △ABC\triangle ABC (AB=2,BC=2.5,AC=3AB = 2, BC = 2.5, AC = 3) and △PQR\triangle PQR (PQ=6,QR=4,PR=5PQ = 6, QR = 4, PR = 5).

Step 1 · Compare Ratios of Corresponding Sides

Compare the side lengths of △ABC\triangle ABC and △PQR\triangle PQR:

ABQR=24=12BCRP=2.55=12ACPQ=36=12\begin{aligned} \dfrac{AB}{QR} &= \dfrac{2}{4} = \dfrac{1}{2} \\[0.6em] \dfrac{BC}{RP} &= \dfrac{2.5}{5} = \dfrac{1}{2} \\[0.6em] \dfrac{AC}{PQ} &= \dfrac{3}{6} = \dfrac{1}{2} \end{aligned}

ABQR=BCRP=ACPQ=12\dfrac{AB}{QR} = \dfrac{BC}{RP} = \dfrac{AC}{PQ} = \dfrac{1}{2}

Since all three pairs of corresponding sides are proportional, by SSS similarity criterion: △ABC∼△QRP\triangle ABC \sim \triangle QRP

Answer

(ii) △ABC∼△QRP\triangle ABC \sim \triangle QRP (by SSS criterion)

(iii) Check similarity for △LMP\triangle LMP (LM=2.7,MP=2,LP=3LM = 2.7, MP = 2, LP = 3) and △DEF\triangle DEF (DE=4,EF=5,DF=6DE = 4, EF = 5, DF = 6).

Step 1 · Compare Ratios of Corresponding Sides

Comparing the ratios of corresponding sides:

LMDE=2.74=0.675MPEF=25=0.4LPDF=36=0.5\begin{aligned} \dfrac{LM}{DE} &= \dfrac{2.7}{4} = 0.675 \\[0.6em] \dfrac{MP}{EF} &= \dfrac{2}{5} = 0.4 \\[0.6em] \dfrac{LP}{DF} &= \dfrac{3}{6} = 0.5 \end{aligned}

LMDE≠MPEF≠LPDF\dfrac{LM}{DE} \ne \dfrac{MP}{EF} \ne \dfrac{LP}{DF}

Since the corresponding sides are not proportional, the triangles are not similar.

Answer

(iii) Not similar (corresponding sides are not proportional)

(iv) Check similarity for △NML\triangle NML (NM=2.5,ML=5,∠N=70∘NM = 2.5, ML = 5, \angle N = 70^\circ) and △PQR\triangle PQR (PQ=5,QR=10,∠Q=70∘PQ = 5, QR = 10, \angle Q = 70^\circ).

Step 1 · Check Side Ratios and Included Angle

Comparing side ratios:

NMPQ=2.55=12MLQR=510=12\begin{aligned} \dfrac{NM}{PQ} &= \dfrac{2.5}{5} = \dfrac{1}{2} \\[0.6em] \dfrac{ML}{QR} &= \dfrac{5}{10} = \dfrac{1}{2} \end{aligned}

Although ∠N=∠Q=70∘\angle N = \angle Q = 70^\circ, ∠N\angle N is not the included angle between sides NMNM and MLML (the included angle is ∠M\angle M).

For SAS similarity, the equal angle must be the included angle between the proportional sides. Therefore, the triangles are not similar.

Answer

(iv) Not similar (the equal angle is not the included angle)

(v) Check similarity for △ABC\triangle ABC (AB=2.5,AC=3,∠A=80∘AB = 2.5, AC = 3, \angle A = 80^\circ) and △DEF\triangle DEF (DF=5,EF=6,∠F=80∘DF = 5, EF = 6, \angle F = 80^\circ).

Step 1 · Check Side Ratios and Included Angle

In △ABC\triangle ABC, ∠A=80∘\angle A = 80^\circ is between sides ABAB and ACAC, but the given sides are not proportional to the corresponding sides of △DEF\triangle DEF with the included angle ∠F=80∘\angle F = 80^\circ.

Since the SAS criterion is not satisfied, the triangles are not similar.

Answer

(v) Not similar (corresponding sides and included angles do not satisfy SAS criterion)

(vi) Check similarity for △DEF\triangle DEF (∠D=70∘,∠E=80∘\angle D = 70^\circ, \angle E = 80^\circ) and △PQR\triangle PQR (∠Q=80∘,∠R=30∘\angle Q = 80^\circ, \angle R = 30^\circ).

Step 1 · Find the Missing Angles in Both Triangles

Using the angle sum property of a triangle:

In △DEF\triangle DEF:

∠D+∠E+∠F=180∘70∘+80∘+∠F=180∘150∘+∠F=180∘∠F=180∘−150∘=30∘\begin{aligned} \angle D + \angle E + \angle F &= 180^\circ \\ 70^\circ + 80^\circ + \angle F &= 180^\circ \\ 150^\circ + \angle F &= 180^\circ \\ \angle F &= 180^\circ - 150^\circ = 30^\circ \end{aligned}

In △PQR\triangle PQR:

∠P+∠Q+∠R=180∘∠P+80∘+30∘=180∘∠P+110∘=180∘∠P=180∘−110∘=70∘\begin{aligned} \angle P + \angle Q + \angle R &= 180^\circ \\ \angle P + 80^\circ + 30^\circ &= 180^\circ \\ \angle P + 110^\circ &= 180^\circ \\ \angle P &= 180^\circ - 110^\circ = 70^\circ \end{aligned}

Step 2 · Compare Corresponding Angles

Comparing angles of △DEF\triangle DEF and △PQR\triangle PQR:

∠D=∠P=70∘∠E=∠Q=80∘∠F=∠R=30∘\begin{aligned} \angle D &= \angle P = 70^\circ \\ \angle E &= \angle Q = 80^\circ \\ \angle F &= \angle R = 30^\circ \end{aligned}

By AAA similarity criterion: △DEF∼△PQR\triangle DEF \sim \triangle PQR

Answer

(vi) △DEF∼△PQR\triangle DEF \sim \triangle PQR (by AAA criterion)

Common Mistakes
  • Incorrect Vertex Order: Writing △ABC∼△PQR\triangle ABC \sim \triangle PQR in part (ii) instead of the correct corresponding order △ABC∼△QRP\triangle ABC \sim \triangle QRP.
  • Non-included Angle in SAS: Assuming two triangles are similar just because two sides are proportional and any one angle is equal. The angle must be between the two proportional sides (included angle).
  • Not Computing Third Angle: In part (vi), giving up because the two stated angles look different (70∘,80∘70^\circ, 80^\circ vs 80∘,30∘80^\circ, 30^\circ), without first computing the third angle using the angle sum property (180∘180^\circ).

More questions in Exercise 6.3

Q1
  1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Q2
  1. In Fig. 6.35, ΔODC∼ΔOBA\Delta \text{ODC} \sim \Delta \text{OBA}, ∠BOC=125∘\angle \text{BOC} = 125^\circ and ∠CDO=70∘\angle \text{CDO} = 70^\circ. Find ∠DOC\angle \text{DOC}, ∠DCO\angle \text{DCO} and ∠OAB\angle \text{OAB}.
Q3
  1. Diagonals AC\text{AC} and BD\text{BD} of a trapezium ABCD\text{ABCD} with AB∥DC\text{AB} \parallel \text{DC} intersect each other at the point O\text{O}. Using a similarity criterion for two triangles, show that OAOC=OBOD\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}.
Q4
  1. In Fig. 6.36, QRQS=QTPR\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}} and ∠1=∠2\angle 1 = \angle 2. Show that ΔPQS∼ΔTQR\Delta \text{PQS} \sim \Delta \text{TQR}.
Q5

S and T are points on sides PRPR and QRQR of ΔPQR\Delta \text{PQR} such that ∠P=∠RTS\angle \text{P} = \angle \text{RTS}. Show that ΔRPQ∼ΔRTS\Delta \text{RPQ} \sim \Delta \text{RTS}.

Q6
  1. In Fig. 6.37, if ΔABE≅ΔACD\Delta \text{ABE} \cong \Delta \text{ACD}, show that ΔADE∼ΔABC\Delta \text{ADE} \sim \Delta \text{ABC}.
Q7
  1. In Fig. 6.38, altitudes ADAD and CECE of ΔABC\Delta \text{ABC} intersect each other at the point PP. Show that:

(i) ΔAEP∼ΔCDP\Delta \text{AEP} \sim \Delta \text{CDP}

(ii) ΔABD∼ΔCBE\Delta \text{ABD} \sim \Delta \text{CBE}

(iii) ΔAEP∼ΔADB\Delta \text{AEP} \sim \Delta \text{ADB}

(iv) ΔPDC∼ΔBEC\Delta \text{PDC} \sim \Delta \text{BEC}

Q8
  1. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE∼ΔCFB\Delta \text{ABE} \sim \Delta \text{CFB}.
Q9
  1. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) ΔABC∼ΔAMP\Delta \text{ABC} \sim \Delta \text{AMP}

(ii) CAPA=BCMP\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}

Q10
  1. CD and GH are respectively the bisectors of ∠ACB\angle \text{ACB} and ∠EGF\angle \text{EGF} such that D and H lie on sides AB and FE of ΔABC\Delta \text{ABC} and ΔEFG\Delta \text{EFG} respectively. If ΔABC∼ΔFEG\Delta \text{ABC} \sim \Delta \text{FEG}, show that:

(i) CDGH=ACFG\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}

(ii) ΔDCB∼ΔHGE\Delta \text{DCB} \sim \Delta \text{HGE}

(iii) ΔDCA∼ΔHGF\Delta \text{DCA} \sim \Delta \text{HGF}

Q11

In Fig. 6.40, EE is a point on side CBCB produced of an isosceles triangle ABCABC with AB=ACAB = AC. If AD⊥BCAD \perp BC and EF⊥ACEF \perp AC, prove that ΔABD∼ΔECF\Delta ABD \sim \Delta ECF.

Q12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR\Delta PQR (see Fig. 6.41). Show that ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.

Q13

D is a point on the side BCBC of a triangle ABCABC such that ∠ADC=∠BAC\angle \text{ADC} = \angle \text{BAC}. Show that CA2=CB⋅CD\text{CA}^2 = \text{CB} \cdot \text{CD}.

Q14

Sides ABAB and ACAC and median ADAD of a triangle ABCABC are respectively proportional to sides PQPQ and PRPR and median PMPM of another triangle PQRPQR. Show that ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.

Q15

A vertical pole of length 6 m6\text{ m} casts a shadow 4 m4\text{ m} long on the ground and at the same time a tower casts a shadow 28 m28\text{ m} long. Find the height of the tower.

Q16

If ADAD and PMPM are medians of triangles ABCABC and PQRPQR, respectively where ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR, prove that

ABPQ=ADPM\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}

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