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RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

April 16, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.3

Question 1.
State which pairs of triangles in Fig. are similar. Write the similarity criterion used by you for answering the qpestion and also write the pairs of similar triangles in the symbolic form :
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 1
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 2
Solution:
(i) In △ABC and APQR,
∠A = ∠P (Each 60°)
∠B = ∠Q (Each 80°)
∠C = ∠R (Each 40°)
∴ △ABC ~ △PQR
i. e., both △ are similar.
(By AAA similarity criterion)

(ii) In △ABC and △PQR,
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 3
∴ △ABC ~ △QRP
(By SSS similarity criterion)
i. e., both △ are similar.

(iii) In △LMPand △DEF,
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 4
i. e., both the triangles are not similar.

(iv) In △MNL and △QPR,
\(\frac{\mathrm{ML}}{\mathrm{QR}}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{MN}}{\mathrm{PQ}}\) = \(\frac{2.5}{5}\) = \(\frac{1}{2}\)
∠M = ∠Q = 70° (Each 70°)
By SAS similarity criterion,
△MNL ~ △QPR

(v) In △ABC and △DFE,
\(\frac{\mathrm{AB}}{\mathrm{DF}}\) = \(\frac{2.5}{5}\) = \(\frac{1}{2}\)
\(\frac{\mathrm{BC}}{\mathrm{EF}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∠B ≠ ∠F
i. e., △ABC and △DEF are not similar.

(vi) In △DEF,
∠D = 70°, ∠E = 80°
∵ ∠D + ∠E + ∠F = 180° or
70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
∴ ∠F = 30°
In △PQR,
∠Q = 80°, ∠R = 30°
∵ ∠P + ∠Q + ∠R = 180°
or ∠P + 80° + 30° =180°
or ∠P = 180°- 80°- 30°
Hence ∠P = 70°
In △DEF and △PQR,
∠D = ∠P (Each angle 70°)
∠E = ∠Q (Each angle 80°)
∠F = ∠R (Each angle 30°)
∴ △DEF ~ △PQR
(AAA similarity criterion)
i.e., both the triangles are similar.

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 2.
In Fig., △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 5
Solution:
∠BOC = 125°
∠CDO = 70°
DOB is a straight line
∴ ∠DOC + ∠COB = 180°
or ∠DOC + 125° = 180°
or ∠DOC = 180° – 125°
∠DOC = 55°
∠DOC = ∠AOB = 55° (Vertically opposite angles)
∴ △ODC ~ △OBA
∴ ∠D = ∠B = 70°
In △DOC,
∠D + ∠O + ∠C = 180°
∴ 70° + 55° + ∠C = 180°
or ∠C = 180° – 70° – 55°
∠C = 55°
∠C = ∠A = 55°
+DOC = 55°
∴ +DQO = 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 = \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\).
Solution:
Given : A trapezium ABCD in which AB || CD and the diagonals AC and BD intersect each other at the point O.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 6
To prove : \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\)
Proof: ∵ AB || DC
∴ In △DOC and △BOA,
∠1 = ∠2 (Alternate angles)
∠5 = ∠6 (Vertically opposite angles)
∠3 = ∠4 (Alternate angles
∴ △DOC ~ △BOA (AAA similarity criterion)
∴ \(\frac{\mathrm{OB}}{\mathrm{OD}}\) = \(\frac{\mathrm{OC}}{\mathrm{OA}}\)
(If two triangles are similar, then their corresponding sides are proportional)
⇒ \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{OB}}{\mathrm{OD}}\) (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 4.
In Fig. \(\frac{\mathrm{QR}}{\mathrm{QS}}\) = \(\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that △PQS ~ △TQR.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 7
Solution:
Given : △TQR is a triangle in which
\(\frac{\mathrm{QR}}{\mathrm{QS}}\) = \(\frac{\mathrm{QT}}{\mathrm{PR}}\)
∴ ∠1 = ∠2
To prove : △PQS ~ △TQR
Proof: In △PQR,
∠1 =∠2 (Given)
∴ PR = PQ (i)
(Sides opposite to equal angles are equal)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 8

Question 5.
S and T are points on sides PR and QR of △PQR such that ∠P = ∠RTS. Show that △RPQ ~ △RTS.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 9
Solution:
Given : Points S and T are on sides PR and QR respectively of △PQR such that ∠P = ∠RTS.
To prove : △RPQ ~ △RTS
Proof: In △RPQ and △RTS,
∠RPQ = ∠RTS (Given)
∠PRQ = ∠TRS (Common angle)
∴ △RPQ ~ △RTS
(By AA similarity criterion)
(Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 6.
In Fig., if △ABE = △ACD, show that △ADE ~ △ABC.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 10
Solution:
Given : △ABE ≅ △ACD
To prove : △ADE – △ABC
Proof: △ABE ≅ △ACD (Given)
△RPQ ~ △RTS
∴ AB = AC
(Corresponding sides of congruent triangles)
and AE = AD
(Corresponding sides of congruent triangles)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 11
In △ADE and △ABC. \(\frac{\mathrm{AD}}{\mathrm{AE}}\) = \(\frac{\mathrm{AD}}{\mathrm{AC}}\)
∠A = ∠S
(Common angle)
∴ △ADE ~ △ABC
(By SAS similarity criterion)
(Hence Proved)

Question 7.
In Fig., altitudes AD and CE of △ABC intersect each other at the point P. Show that :
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3. 24
(i) △AEP ~ △CDP
(ii) △ABD ~ △CBE
(iii) △AEP ~ △ADB
(iv) △PDC ~ △BEC
Solution: Given : In △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 angles)
∴ △AED ~ △CDP
(AA similarity criterion) (Hence Proved)

(ii) In △ABD and △CBE,
∠D = ∠E (Each 90°)
∠B = ∠B (Common angle)
∴ △AABD ~ △CBE
(AA similarity criterion) (Hence Proved)

(iii) In △AEP and △ADB,
∠E = ∠D (Each 90°)
∠A = ∠A (Common angle)
∴ △AEP ~ △ADB
(AA similarity criterion) (Hence Proved)

(iv) In △PDC and △BEC,
∠C = ∠C (Common angle)
∠D =∠E (Each 90°)
△PDC ~ △BEC
(AA similarity criterion) (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ABE ~ △CFB.
Solution:
Given : E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
To prove : △ABE ~ △CFB
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 12
Proof: In △ABE and △CFB,
∠A = ∠C (Opposite angle of || gm)
∠ABE = ∠CFB (Alternate angles)
∴ △ABE ~ △CFB (AA similarity criterion)
(Hence Proved)

Question 9.
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) △ABC ~ △AMP
(ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 13
Solution:
Given : △ABC and △AMP are two right triangles, right angled at B and M respectively.
To prove : (i) △ABC ~ △AMP
(ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)
Proof: (i) In △ABC and △AMP,
∠A = ∠A (Common angle)
∠B = ∠M (Each 90°)
∴ △ABC ~ △AMP
(By AA similarity criterion) (Hence Proved)

(ii) ∵ △ABC ~ △AMP [From (i)]
∴ \(\frac{\mathrm{AC}}{\mathrm{AP}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)
(If two triangles are similar, then their corresponding sides are proportional.)
So \(\frac{\mathrm{CA}}{\mathrm{PA}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\) (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

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 △ABC and △EFG respectively. If △ABC ~ △FEG, show that:
(i) \(\frac{\mathrm{CD}}{\mathrm{GH}}\) = \(\frac{\mathrm{AC}}{\mathrm{FG}}\)
(ii) △DCB ~ △HGE
(iii) △DCA ~ △HGF
Solution:
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 22
Given : In △ABC and △EFG, CD and GH are respectively the bisectors of ∠ACB and ∠EGF i.e., ∠1 = ∠2 and ∠3 = ∠4 and △ABC ~ △FEG
To prove: (i) \(\frac{\mathrm{CD}}{\mathrm{GH}}\) = \(\frac{\mathrm{AC}}{\mathrm{FG}}\)
(ii) △DCB ~ △HGE
(iii) △DCA ~ △HFG
Proof: (i) △ABC ~ △FEG (Given)
∠C = ∠G
(If two triangles are similar, then their corresponding angles are equal.)
⇒ \(\frac{1}{2}\)∠C = \(\frac{1}{2}\)∠G
or ∠1 = ∠3
or ∠2 = ∠4
i.e., ∠1 = ∠4
∠2 =∠3
Now, in △ACD and △FGH,
∠A = ∠F (∵ △ABC ~ △FEG)
∠2 = ∠4 (Proved above)
∴ △ACD ~ △FGH
(By AA similarity criterion)
So, \(\frac{\mathrm{CD}}{\mathrm{GH}}\) = \(\frac{\mathrm{AC}}{\mathrm{FG}}\)
(∵ Corresponding sides of similar triangles are proportional.) (Hence Proved)

(ii) Now, in △DCB and △HGE,
∠B = ∠E (∵ △ABC ~ △FEG)
∠1 = ∠4 (Proved above)
∴ △DCB ~ △HGE
(By AA similarity criterion) (Hence Proved)

(iii) Now, in △DCA and △HGF,
∠A = ∠F (∵ △ABC ~ △FEG)
∠2 = ∠3 (Proved above)
△DCA ~ △HGF
(By AA similarity criterion) (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 11.
In Fig., E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that △ABD ~ △ECF.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 14
Solution:
Given : E is a point on side CB produced of an isosceles triangle ABC with AB = AC. AD ⊥ BC and EF ⊥ AC.
To prove : △ABD ~ △ECF
Proof: △ABC is an isosceles triangle. (Given)
∴ AB = AC
(Since in a triangle angles opposite to equal sides are equal)
∴ ∠B = ∠C (Equal angles)
In △ABD and △ECF,
∠ABD = ∠ECF (Proved above)
∠ADB = ∠EFC (Each 90°)
∴ △ABD ~ △ECF (AA similarity) (Hence Proved)

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.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 15
Solution:
Given : Median AD of △ABC and median PM of △PQR are such that
\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)
To prove : △ABC ~ △PQR
Proof: BD = \(\frac{1}{2}\) BC (Given)
and QM = \(\frac{1}{2}\) QR (Given)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3. 23
∴ By SAS similarity criterion we have
△ABC ~ △PQR (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

Question 13.
D is a point on the sideBC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
Solution:
Given : D is a point on the side BC of a triangle ABC such that
∠ADC = ∠BAC
To prove: CA2 = BC.CD
Proof: In △ABC and △ADC,
∠C = ∠C (Common angle)
∠BAC = ∠ADC (Given)
∴ △ABC ~ △DAC (By AA similarity criterion)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 16
∴ \(\frac{\mathrm{AC}}{\mathrm{DC}}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
(If two triangles are similar, then their corresponding sides are proportional.)
⇒ AC2 = BC.DC (Hence Proved)

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 △ABC ~ △PQR.
Solution:
Given : In two triangles ABC and PQR, D is the mid-point of BC and M is the mid-point of QR.
and \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AC}}{\mathrm{PR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3. 19
To prove : △ABC ~ △PQR
Construction : Produce AD to E such that AD = DE. Join BE and CE.
Produce PM to N such that PM = MN
Join QN and NR.
Proof : Since the diagonals AE and BC of the quadrilateral ABEC bisect each other at D, therefore the quadrilateral ABEC is a parallelogram.
Similarly it can be shown that the quadrilateral PQNR is a parallelogram.
Since ABEC is a parallelogram
∴ BE = AC …..(ii)
Similarly, since PQNR is a parallelogram
∴ QN = PR ……..(iii)
Dividing (ii) by (iii)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 18
So, From △ABE and △PQN
\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BE}}{\mathrm{QN}}\) = \(\frac{\mathrm{AE}}{\mathrm{PN}}\)
∴ △ABC ~ △PQN
∴ ∠BAE = ∠QPN ……(vi)
Similarly, it can be proved that
△AEC ~ △PNR
∴ ∠EAC = ∠NPR ……(vii)
Now. adding (vi) and (vii)
∠BAE + ∠EAC = ∠QPN + ∠NPR
or ∠BAC = ∠QPR
Now, in △ABC and △PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AC}}{\mathrm{PR}}\)
∠A = ∠P
∴ △ABC ~ △PQR
(By SAS similarity criterion) (Hence Proved)

RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

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:
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 17
Length of the vertical pole = 6 m
Length of the shadow of the pillar = 4 nr
Let the height of the tower = H m
Length of the shadow of the tower = 28 m
In △ABC and △PMN,
∠C = ∠N
(Length of the shadow of the tower)
∠B = ∠M (Each 90°)
∴ △ABC ~ △PMN
∴ \(\frac{\mathrm{AB}}{\mathrm{PM}}\) = \(\frac{\mathrm{BC}}{\mathrm{MN}}\)
(If two triangles are similar, then their corresponding sides are proportional.)
∴ \(\frac{6}{\mathrm{H}}=\frac{4}{28}\) ⇒ H = \(\frac{6 \times 28}{4}\)
H = 6 × 7
H =42m
Hence, the height of the tower = 42 m

Question 16.
If AD and PM are medians of triangles ABC and PQR, respectively where △ABC ~ △PQR, prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\).
Solution:
Given : AD and PM are the medians of △ABC and △PQR respectivelv and △ABC ~ △PQR.
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3. 20
To prove : \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)
Proof: △ABC ~ △PQR (Given)
∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{BC}}{\mathrm{QR}}\) = \(\frac{\mathrm{AC}}{\mathrm{PR}}\) ….(i)
(If two triangles are similar, then their corresponding sides are proportional.)
∠A = ∠P
(If two triangles are similar, then their corresponding angles are equal.)
∠B = ∠Q, ∠C = ∠R
∵ D is the mid-point of BC
∴ BD = DC = \(\frac{1}{2}\)BC ….(ii)
∵ M is die mid-point of QR
∴ QM = MR = \(\frac{1}{2}\)QR ……(iii)
RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3. 21
(If two triangles are similar, then their corresponding sides are proportional.)
(Hence Proved)

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