TOPICS

Exercise - 6.3

Triangles

**Question-1 :-** State which pairs of triangles in Figure 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 :

(i) In Δ ABC & Δ PQR ∠A = ∠P = 60^{o}∠B = ∠Q = 80^{o}∠C = ∠R = 40^{o}Therefore, Δ ABC ~ Δ PQR (By AAA similarity rule) Hence, AB/QR = BC/RP = CA/PQ

(ii) In Δ ABC & Δ PQR AB = 2, QR = 4 BC = 2.5, PR = 5 AC = 3, PQ = 6 By using proportional rule, we get QR/AB = PR/BC = PQ/AC 4/2 = 5/2.5 = 6/3 = 2 Therefore, Δ ABC ~ Δ PQR (By SSS similarity rule)

(iii) Both triangles are not similar as the corresponding sides are not proportional.

(iv) In Δ LMN & Δ PQR PQ = 5, QR = 10 MN = 2.5, ML = 5 ∠M = ∠Q = 70^{o}By using Proportional rule, we get PQ/MN = QR/ML 5/2.5 = 10/5 = 2 Therefore, Δ LMN ~ Δ PQR (By SAS similarity rule)

(v) Both triangles are not similar as the corresponding sides are not proportional.

(vi) In Δ DEF, ∠D +∠E +∠F = 180^{o}(Sum of the measures of the angles of a triangle is 180^{o}.) 70^{o}+ 80^{o}+ ∠F = 180^{o}∠F = 30^{o}Similarly, in Δ PQR, ∠P +∠Q + ∠R = 180^{o}(Sum of the measures of the angles of a triangle is 180^{o}.) ∠P + 80^{o}+ 30^{o}= 180^{o}∠P = 70^{o}In Δ DEF and Δ PQR, ∠D = ∠P (Each 70^{o}) ∠E = ∠Q (Each 80^{o}) ∠F = ∠R (Each 30^{o}) Therefore, ΔDEF ∼ ΔPQR (By AAA similarity criterion)

**Question-2 :-** In Figure, Δ ODC ~ Δ OBA, ∠ BOC = 125^{o} and ∠ CDO = 70^{o}. Find ∠ DOC, ∠ DCO and ∠ OAB.

Here, DOB is a straight line. So, ∠ DOB = 180^{o}, ∠ BOC = 125^{o}(Given) Now, ∠ DOB = ∠ DOC + ∠ BOC 180^{o}= ∠ DOC + 125^{o}∠ DOC = 180^{o}- 125^{o}∠ DOC = 55^{o}In Δ DOC, Given that ∠ CDO = 70^{o}∠ DOC + ∠ DCO + ∠ CDO = 180^{o}(Sum of the measures of the angles of a triangle is 180^{o}) 55^{o}+ ∠ DCO + 70^{o}= 180^{o}∠ DCO = 180^{o}- 125^{o}∠ DCO = 55^{o}In figure, Given that Δ ODC ~ Δ OBA So, ∠OAB = ∠OCD (Corresponding angles are equal in similar triangles) ∠OAB = 55^{o}

**Question-3 :-** Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

Given that : Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Prove that : OA/OC = OB/OD Proof : In Δ DOC and Δ BOA, ∠CDO = ∠ABO (Alternate interior angles as AB || CD) ∠DCO = ∠BAO (Alternate interior angles as AB || CD) ∠DOC = ∠BOA (Vertically opposite angles) By AAA similarity criterion Rule Therefore Δ DOC ∼ Δ BOA Hence, OD/OB = OC/OA (Corresponding sides are Proportional) OA/OC = OB/OD

**Question-4 :-** In Figure, QR/QS = QT/PR and ∠1 = ∠2. Show that Δ PQS ~ Δ TQR.

Given that : QR/QS = QT/PR and ∠1 = ∠2. Prove that : Δ PQS ~ Δ TQR Proof : In Δ PQR, ∠1 = ∠2 or ∠PQR = ∠PRQ (Given) Therefore, PQ = PR .......(i) Also, given that QR/QS = QT/PR By using (i), we get QR/QS = QT/QP ......(ii) In Δ PQS and Δ TQR QR/QS = QT/QP By using (ii), we get ∠1 = ∠1 By SAS similarity rule, we get Δ PQS ~ Δ TQR

**Question-5 :-** S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS. Show that Δ RPQ ~ Δ RTS.

Given that : S and T are points on sides PR and QR of Δ PQR such that ∠ P = ∠ RTS. Proove that : Δ RPQ ~ Δ RTS Proof : In ΔRPQ and ΔRST, ∠RTS = ∠QPS (Given) ∠PRQ = ∠TRS (Common angle) Therefore, By AA similarity rule ΔRPQ ∼ ΔRTS

**Question-6 :-** In Figure, if Δ ABE ≅ Δ ACD, show that Δ ADE ~ Δ ABC.

Given that : Δ ABE ≅ Δ ACD Prove that : Δ ADE ~ Δ ABC Proof : Δ ABE ≅ Δ ACD (Given) Therefore, AB = AC .....(i) (By CPCT Rule) And, AD = AE ......(ii) (By CPCT Rule) In ΔADE and ΔABC, AD/AB = AE/EC (By Dividing equation (ii) by (i)) ∠QBAC = ∠QDAE (Common angle) Therefore, By SAS similarity Rule, Δ ADE ~ Δ ABC

**Question-7 :-** In Figure, altitudes AD and CE of Δ ABC intersect each other at the point P. Show that:

(i) Δ AEP ~ Δ CDP

(ii) Δ ABD ~ Δ CBE

(iii) Δ AEP ~ Δ ADB

(iv) Δ PDC ~ Δ BEC

Given that : Altitudes AD and CE of Δ ABC intersect each other at the point P. Prove that : (i) Δ AEP ~ Δ CDP (ii) Δ ABD ~ Δ CBE (iii) Δ AEP ~ Δ ADB (iv) Δ PDC ~ Δ BEC Proof : (i) In ΔAEP and ΔCDP, ∠AEP = ∠CDP (Each 90^{o}) ∠APE = ∠CPD (Vertically opposite angles) Therefore, By using AA similarity rule, ΔAEP ∼ ΔCDP

Proof : (ii) In ΔABD and ΔCBE, ∠ADB = ∠CEB (Each 90^{o}) ∠ABD = ∠CBE (Common angle) Therefore, By using AA similarity rule, ΔABD ∼ ΔCBE

Proof: (iii) In ΔAEP and ΔADB, ∠AEP = ∠ADB (Each 90^{o}) ∠PAE = ∠DAB (Common angle) Therefore, By using AA similarity rule, ΔAEP ∼ ΔADB

Proof : (iv) In ΔPDC and ΔBEC, ∠PDC = ∠BEC (Each 90^{o}) ∠PCD = ∠BCE (Common angle) Therefore, By using AA similarity rule, ΔPDC ∼ ΔBEC

**Question-8 :-** E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB.

Given that : E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Prove that : Δ ABE ~ Δ CFB Proof : In ΔABE and ΔCFB, ∠BAE = ∠BCF (Opposite angles of a parallelogram) ∠AEB = ∠CBF (Alternate interior angles as AE || BC) Therefore, By AA similarity rule, ΔABE ∼ ΔCFB

**Question-9 :-** In Figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) Δ ABC ~ Δ AMP

(ii) CA/PA = BC/MP

Given that : ABC and AMP are two right triangles, right angled at B and M respectively. Prove that : (i) Δ ABC ~ Δ AMP (ii) CA/PA = BC/MP Proof : (i) In ΔABC and ΔAMP, ∠ABC = ∠AMP (Each 90^{o}) ∠BAC = ∠MAP (Common angle) Therefore, By AA similarity rule ΔABC ∼ ΔAMP

Proof : (ii) ΔABC ∼ ΔAMP, So CA/PA = BC/MP (Corresponding sides of similar triangles are proportional)

**Question-10 :-** CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:

(i) CD/GH = AC/FG

(ii) Δ DCB ~ Δ HGE

(iii) Δ DCA ~ Δ HGF

Given that : CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. Δ ABC ~ Δ FEG Prove that : (i) CD/GH = AC/FG (ii) Δ DCB ~ Δ HGE (iii) Δ DCA ~ Δ HGF Proof : (i) Δ ABC ~ Δ FEG (Given) Therefore, ∠BAC = ∠EFG, ∠ABC = ∠FEG and ∠ACB = ∠FGE ∠ACD = ∠FGH (Angle Bisectors) ∠DCB = ∠HGE (Angle Bisectors) In Δ ACD and Δ FGH, ∠BAC = ∠FEG (Δ ABC ~ Δ FEG) ∠ACD = ∠FGH (Above Proved) CD/GH = AC/FG

Proof : (ii) In Δ DCB and Δ HGE, ∠DCB = ∠HGE (Proved above) ∠ABC = ∠FEG (Proved above) Δ DCB ∼ Δ HGE (By AA similarity rule)

Proof : (iii) In Δ DCA and Δ HGF, ∠ACD = ∠FGH (Proved above) ∠BAC = ∠EFG (Proved above) Δ DCA ∼ Δ HGF (By AA similarity rule)

**Question-11 :-** In Figue, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.

Given that : E is a point on side CB produced of an isosceles triangle ABC with AB = AC. Also, AD ⊥ BC and EF ⊥ AC Prove that : Δ ABD ~ Δ ECF Proof : Δ ABC is an isosceles triangle. (Given) Therefore, AB = AC, ∠ABD = ∠ECF In Δ ABD and Δ ECF, ∠ADB = ∠EFC (Each 90^{o}) ∠BAD = ∠CEF (Proved above) Δ ABD ∼ Δ ECF (By using AA similarity criterion)

**Question-12 :-** Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR (see Figure). Show that Δ ABC ~ Δ PQR.

Given that : Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Δ PQR . Prove that : Δ ABC ~ Δ PQR Proof : Sides AB and BC and median AD of a Δ ABC. Median divides the opposite side. Therefore, BD = BC/2 and QM = QR/2 Also given that AB/PQ = BC/QR = AD/PM AB/PQ = (BC/2)/(QR/2) = AD/PM AB/PQ = BD/QM = AD/PM In Δ ABD and Δ PQM, AB/PQ = BD/QM = AD/PM (Proved above) Δ ABD ∼ ΔPQM (By SSS similarity rule) ∠ABD = ∠PQM (Corresponding angles of similar triangles) In Δ ABC and Δ PQR, ∠ABD = ∠PQM (Proved above) AB/PQ = BC/QR Δ ABC ∼ Δ PQR (By SAS similarity rule)

**Question-13 :-** D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA^{2} = CB . CD.

Given that : D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Prove that : CA^{2}= CB . CD Proof : In Δ ADC and Δ BAC, ∠ADC = ∠BAC (Given) ∠ACD = ∠BCA (Common angle) Δ ADC ∼ Δ BAC (By AA similarity rule) We know that corresponding sides of similar triangles are in proportion. Therefore, CA/CB = CD/CA CA^{2}= CB . CD

**Question-14 :-** Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Δ ABC ~ Δ PQR.

Given that : AB/PQ = AC/PR = AD/PM Prove that : Δ ABC ~ Δ PQR Construction : Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L. Proof : We know that medians divide opposite sides. Therefore, BD = DC and QM = MR Also, AD = DE (By construction) And, PM = ML (By construction) In quadrilateral ABEC, diagonals AE and BC bisect each other at point D. Therefore, quadrilateral ABEC is a parallelogram. So, AC = BE and AB = EC (Opposite sides of a parallelogram are equal) Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR It was given that AB/PQ = AC/PR = AD/PM AB/PQ = BE/QL = 2AD/2PM AB/PQ = BE/QL = AE/PL Therefore, Δ ABE ∼ Δ PQL (By SSS similarity rule) We know that corresponding angles of similar triangles are equal. Therefore, ∠BAE = ∠QPL ....... (i) Similarly, it can be proved that Δ AEC ∼ Δ PLR and ∠CAE = ∠RPL ...... (ii) Adding equation (i) and (ii), we get ∠BAE + ∠CAE = ∠QPL + ∠RPL ∠CAB = ∠RPQ .......... (iii) In Δ ABC and Δ PQR, AB/PQ = AC/PR (Given) ∠CAB = ∠RPQ (Using equation (iii)) Δ ABC ∼ Δ PQR (By SAS similarity rule)

**Question-15 :-** A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Let AB and CD be a tower and a pole respectively. Let the shadow of BE and DF be the shadow of AB and CD respectively. At the same time, the light rays from the sun will fall on the tower and the pole at the same angle. Therefore, ∠DCF = ∠BAE And, ∠DFC = ∠BEA ∠CDF = ∠ABE (Tower and pole are vertical to the ground) ΔABE ∼ ΔCDF (AAA similarity criterion) AB/CD = BE/DF AB/6 = 28/4 AB = 42 m Therefore, the height of the tower will be 42 metres.

**Question-16 :-** If AD and PM are medians of triangles ABC and PQR, respectively where Δ ABC ~ Δ PQR, prove that AB/PQ = AD/PM.

Given that : AD and PM are medians of triangles ABC and PQR and Δ ABC ∼ Δ PQR. Prove that : AB/PQ = AD/PM Proof : Δ ABC ∼ Δ PQR (Given) We know that the corresponding sides of similar triangles are in proportion. Therefore, AB/PQ = AC/PR = BC/QR ....... (i) Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ....... (ii) Since AD and PM are medians, they will divide their opposite sides. Therefore, BD = BC/2, QM = QR/2 .......(iii) From equations (i) and (iii), we get AB/PQ = BD/QM ......... (iv) In Δ ABD and Δ PQM, ∠B = ∠Q (Using equation (ii)) AB/PQ = BD/QM (Using equation(iv)) Δ ABD ∼ Δ PQM (By SAS similarity rule) AB/PQ = BD/QM = AD/PM

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