# PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

## PSEB 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.

State which pairs of triangles in Fig. 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:

Solution:

(i) In ∆ABC and ∆PQR,

∠A = ∠P (each 60°)

∠B = ∠Q (each 80°)

∠C = ∠R (each 40°)

∴ ∆ABC ~ PQR [AAA Similarity criterion]

(vi) In ΔDEF, ∠D = 70°, ∠E = 80°

∠D + ∠E + ∠F = 180°

70° + 80° + ∠F = 180° [Angle Sum Propertyl

∠F= 180° – 70° – 80°

∠F = 30°

In ΔPQR,

∠Q = 80°, ∠R = 30°

∠P + ∠Q + ∠R = 180°

(Sum of angles of triangle)

∠P + 80° + 30° = 180°

∠P = 180° – 80° – 30°

∠P = 70°

In ΔDEF and ΔPQR,

∠D = ∠P (70° each)

∠E = ∠Q (80° each)

∠F = ∠R (30° each)

∴ ΔDEF ~ ΔPQR (AAA similarity criterion).

Question 2.

In Fig., ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. FInd ∠DOC, ∠DCO and ∠OAB.

Solution:

Given that: ∠BOC = 125°

∠CDO = 70°

DOB is a straight line

∴ ∠DOC + ∠COB = 180°

[Linear pair Axiom]

∠DOC + 125° = 180°

∠DOC = 180°- 125°

∠DOC = 55°

∠DOC = ∠AOB = 55°

[Vertically opposite angle]

But ΔODC ~ ΔOBA

∠D = ∠B = 70°

In ΔDOC, ∠D + ∠O + ∠C = 180°

70° + 55° + ∠C = 180°

∠C= 180° – 70° – 55°

∠C = 55°

∠C = ∠A = 55°

Hence ∠DOC = 55°

∠DCO = 55°

∴ ∠OAB = 55°.

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.

Solution:

Question 4.

In Fig., and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Question 5.

S and T are points on skies PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.

Solution:

S and T are the points on side PR and QR such that ∠P = ∠RTS.

To Prove. ∆RPQ ~ ∆RTS

Proof: In ∆RPQ and ∆RTS

∠RPQ = ∠RTS (given)

∠R = ∠R [common angle]

∴ RPQ ~ ARTS

[By AA similarity critierion which is the required result.]

Question 6.

In figure ∆ABE ≅ ∆ACD show that ∆ADE ~ ∆ABC.

Question 7.

In Fig., altitudes AD and CE of ∆ABC intersect each other at the point P.

Show that:

(i) ∆AEP ~ ∆CDP

(ii) ∆ABD ~ ∆CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆PDC ~ ∆BEC

Solution:

Given. ∆ABC, AD ⊥ BC CE⊥AB,

To Prove. (i) ∆AEP ~ ∆CDP

(ii) ∆ABD ~ ∆CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆PDC ~ ∆BEC

Proof:

(i) In ∆AEP and ∆CDP,

∠E = ∠D (each 90°)

∠APE = ∠CPD (vertically opposite angle)

∴ ∆AEP ~ ∆CDP [By AA similarity criterion].

(ii) In ∆ABD and ∆CBE,

∠D = ∠E (each 90°)

∠B = ∠B (common)

∴ ∆ABD ~ ∆CBE [AA Similarity criterion]

(iii) In ∆AEP and ∆ADB.

∠E = ∠D (each 90°)

∠A = ∠A (common)

∴ ∆AEP ~ ∆ADB [AA similarity criterion].

(vi) In ∆PDC and ∆BEC,

∠C = ∠C

∠D = ∠E

∴ ∆SPDC ~ ∆BEC [AA similarity criterion].

Question 8.

E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that AABE – &CFB.

Solution:

Given. Parallelogram ABCD. Side AD is produced to E, BE intersects DC at F.

To Prove. ∆ABE ~ ∆CFB

Proof. In ∆ABE and ∆CFB.

∠A = ∠C (opposite angle of || gm)

∠ABE = ∠CFB (alternate angle)

∴ ∆ABE ~ ∆CFB (AA similarity criterion)

Question 9.

In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

Q. 10.

CD and GH are respectively the vectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and

∆EFG respectively. If ∆ABC ~ ∆FEG, show

(ii) In ∆DCB and ∆HGE,

∠B = ∠E [Proved above]

∠1 = ∠3 [Proved above]

∴ ∆DCB ~ ∆HGE [∵ AA similarity criterion]

(iii) In ∆DCA and ∆HGF

∠A = ∠F [Proved above]

∠2 = ∠4 [Proved above]

∴ ∆DCA ~ ∆HGF [∵ AA similarity criterion].

Question 11.

In Fig., E is a point on side CB produced of an Isosceles triangle ABC with AB = AC. IfAD ⊥BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Solution:

Given. ∆ABC, isosceles triangle with AB = AC AD ⊥ BC, side BC is produced to E. EF ⊥ AC

To Prove. ∆ABD ~ ∆ECF

Proof. ∆ABC is isosceles (given)

AB = AC

∴ ∠B = ∠C [Equal sides have equal angles opposite to it)

In ∆ABD and ∆ECF,

∠ABD = ∠ECF (Proved above)

∠ADB = ∠EFC (each 90°)

∴ ∠ABD – ∠ECF [AA similarity).

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 ∆PQR (see Fig.). Show that ∆ABC ~ ∆PQR.

Question 13.

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

Solution:

Given. ∆ABC, D is a point on side BC such that ∠ADC = ∠BAC

To Prove. CA^{2} = BC × CD

Proof. In ABC and ADC,

Question 14.

Sides AB and AC and median AD of a triangle ABC are proportional to sides PQ and PR and median PM of another

triangle PQR. Prove that ∆ABC ~ ∆PQR.

Solution:

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.

Solution:

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