RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

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 :


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,

∴ △ABC ~ △QRP
(By SSS similarity criterion)
i. e., both △ are similar.

(iii) In △LMPand △DEF,

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.

Question 2.
In Fig., △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

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.

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)

Question 4.
In Fig. \(\frac{\mathrm{QR}}{\mathrm{QS}}\) = \(\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that △PQS ~ △TQR.

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)

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

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)

Question 6.
In Fig., if △ABE = △ACD, show that △ADE ~ △ABC.

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)

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 :

(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)

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

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) \(\frac{\mathrm{CA}}{\mathrm{PA}}\) = \(\frac{\mathrm{BC}}{\mathrm{MP}}\)

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)

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:

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)

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.

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.

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)

∴ By SAS similarity criterion we have
△ABC ~ △PQR (Hence Proved)

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)

∴ \(\frac{\mathrm{AC}}{\mathrm{DC}}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
(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 \(\frac{\mathrm{AB}}{\mathrm{PQ}}\) = \(\frac{\mathrm{AC}}{\mathrm{PR}}\) = \(\frac{\mathrm{AD}}{\mathrm{PM}}\)

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)

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)

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:

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.

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)

(If two triangles are similar, then their corresponding sides are proportional.)
(Hence Proved)