Haryana State Board HBSE 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 6 Triangles Ex 6.3

Question 1.

State which pairs of triangles in below figures 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 âˆ†ABC and âˆ†PQR

âˆ A = âˆ P = 60Â°

âˆ B = âˆ Q = 80Â°

âˆ C = âˆ R = 40Â°

By AAA similarity criterion.

âˆ†ABC ~ âˆ†PQR.

(ii) In âˆ†ABC and âˆ†QRP

\(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{2}{4}=\frac{1}{2}\)

\(\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{2 \cdot 5}{5}=\frac{1}{2}\)

\(\frac{\mathrm{AC}}{\mathrm{PQ}}=\frac{3}{6}=\frac{1}{2}\)

âˆµ \(\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{AC}}{\mathrm{PQ}}\)

âˆ´ By SSS similarity criterion

âˆ†ABC ~ âˆ†QRP

(iii) In âˆ†LMP and âˆ†DEF

\(\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2 \cdot 7}{4}\) = 0.675

\(\frac{\mathrm{MP}}{\mathrm{EF}}=\frac{2}{5}\) = 0.4

\(\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{3}{6}\) = 0.5

âˆµ \(\frac{\mathrm{LM}}{\mathrm{DF}} \neq \frac{\mathrm{MP}}{\mathrm{EF}} \neq \frac{\mathrm{LP}}{\mathrm{DF}}\)

âˆ´ These two triangles are not similar because they do not satisfy the SSS similarity criterion.

(iv) In âˆ†MNL and âˆ†QPR

âˆ M = âˆ Q = 70Â°

\(\frac{\mathrm{MN}}{\mathrm{PQ}}=\frac{2 \cdot 5}{6}\) (approx)

\(\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}\) = 0.5

âˆµ \(\frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}\)

âˆ´ These two triangles are not similar because they do not satisfy the SAS similarity criterion.

(v) In âˆ†ABC and âˆ†FDE

âˆ A = âˆ F = 80Â°

\(\frac{\mathrm{AB}}{\mathrm{DF}} \neq \frac{\mathrm{AC}}{\mathrm{EF}}\) [âˆµ AC is not given]

âˆ´ These two triangles are not similar because they do not satisfy the SAS similarity criterion.

(vi) In âˆ†DEF and âˆ†PQR

âˆ P = 180Â° (80Â° + 30Â°)

âˆ P = 180Â° – 110Â°

âˆ P = 70Â°

Now, âˆ D = âˆ P = 70Â°

âˆ E = âˆ Q = 80Â°

By AA similarity criterion âˆ†DEF ~ âˆ†PQR.

Question 2.

In given figure âˆ†ODC – âˆ†OBA, âˆ BOC = 125Â° and âˆ CDO = 70Â°. Find âˆ DOC, âˆ DCO and âˆ OAB.

Solution:

We have, âˆ CDO = 70Â° and âˆ COB = 125Â°

âˆµ BD is a line and CO ray on it

âˆ DOC + âˆ COB = 180Â°

âˆ DOC + 125Â° = 1800

âˆ DOC = 180Â° – 125Â°

âˆ DOC = 55Â°

In âˆ†COD, we have

âˆ DOC + âˆ DCO + âˆ CDO = 180Â° [Sum of âˆ s of a triangle is 180Â°]

55Â° + âˆ DCO + 70Â° = 180Â°

125Â° + âˆ DCO = 180Â°

âˆ DCO = 180Â° – 125Â°

âˆ DCO = 55Â°

Now âˆ†ODC ~ âˆ†OBA (given)

âˆ´ âˆ OAB = âˆ DCO

(Corresponding âˆ s of similar âˆ†s)

âˆ OAB = 55Â°

Hence, âˆ DOC = 55Â°, âˆ DCO = 55Â°, âˆ OAB = 55Â°.

Question 3.

Diagonals AC and BD of a trapezium ABCD with AB DC ntersect 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 : ABCD is a trapezium in which AB || DC. Its diagonals AC and BD meet at O.

To Prove: \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

âˆ OAB = âˆ OCD [âˆµ AB || DC]

âˆ´ Alternate âˆ s are equal.

âˆ AOB = âˆ COD (vertically opposite âˆ s)

âˆ´ âˆ†OAB ~ âˆ†OCD (By AA similarity criterion)

âˆ´ \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

Hence Proved.

Question 4.

In given figure \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and âˆ 1 = âˆ 2. Show that âˆ†PQS ~ âˆ†TQR.

Solution:

Given: \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and âˆ 1 = âˆ 2.

To Prove: âˆ†PQS ~ âˆ†TQR

Proof: We have

âˆ´ âˆ†PQS ~ âˆ†TQR [By SAS similarity criterion]

Hence Proved.

Question 5.

S and T are points on sides PR and QR of âˆ†PQR such that âˆ P = âˆ RTS. Show that âˆ†RPQ – âˆ†RTS.

Solution :

Given : S and T are points on sides PR and QR of âˆ†PQR such that âˆ P = âˆ RTS

To Prove: âˆ†RPQ ~ âˆ†RTS.

Proof: In âˆ†RPQ and âˆ†RTS.

âˆ P = âˆ RTS (given)

âˆ R = âˆ R (common)

âˆ´ âˆ†RPQ ~ âˆ†RTS (By AA similarity criterion)

Hence Proved.

Question 6.

In given figure, If âˆ†ABE â‰… âˆ†ACD, show that âˆ†ADE ~ âˆ†ABC.

Solution:

Given: âˆ†ABE â‰… âˆ†ACD

ToProve: âˆ†ADE ~ âˆ†ABC

Proof: âˆ†ABE ~ âˆ†ACD

âˆ´ AB = AC

and AD = AE [CPCT]

â‡’ \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = 1 and \(\frac{\mathrm{AD}}{\mathrm{AE}}\) = 1

â‡’ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)

Now, In âˆ†ADE and âˆ†ABC, we have

â‡’ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)

â‡’ \(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}\)

and âˆ DAE = âˆ BAC (each = âˆ A)

âˆ†ADE ~âˆ†ABC. [By AA similarity criterion]

Hence Proved .

Question 7.

In given figure altitudes AD and CE of AABC intersect each other at the point P. Show that :

(i) âˆ†AEP ~ âˆ†CDP

(ii) âˆ†ABD ~ âˆ†CBE

(iii) âˆ†AEP ~ âˆ†ADB

(iv) âˆ†PDC ~ âˆ†BEC.

Solution :

Given : CE âŠ¥ AB and AD âŠ¥ BC.

To Prove : (i) âˆ†AEP ~ âˆ†CDP

(ii) âˆ†ABD ~ âˆ†CBE

(iii) âˆ†AEP ~ âˆ†ADB

(iv) âˆ†PDC ~ âˆ†BEC.

Proof : (i) In âˆ†AEP and âˆ†CDP

âˆ AEP = âˆ CDP (each = 90Â°)

âˆ APE = âˆ CPD (vertically opposite âˆ s)

âˆ´ âˆ†AEP ~ âˆ†CDP

(By AA similarity criterion) Hence Proved.

(ii) In âˆ†ABD and âˆ†CBE

âˆ ADB =âˆ CEB (each = 90Â°)

âˆ B = âˆ B (common)

âˆ´ âˆ†ABD ~ âˆ†CBE

(By AA similarity criterion) Hence Proved.

(iii) In âˆ†AEP and âˆ†ADB

âˆ AEP = âˆ ADB (each = 90Â°)

âˆ PAE = âˆ DAB (same angle)

âˆ´ âˆ†AEP ~ âˆ†ADB

(By AA similarity criterion) Hence Proved.

(iv) In âˆ†PDC and âˆ†BEC

âˆ PDC = âˆ BEC (each = 90Â°)

âˆ PCD = âˆ BCE (same angle)

âˆ†PDC ~ âˆ†BEC

(By 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 âˆ†ABE ~ âˆ†CFB.

Solution :

Given : E is a point on side AD produced of a ||gm ABCD and BE intersects CD at F.

To Prove : âˆ†ABE ~ âˆ†CFB.

Proof : In âˆ†ABE and âˆ†CFB.

âˆ AEB = âˆ CBF (alternate interior âˆ s)

âˆ A = âˆ C (opposite âˆ s of a ||gm )

âˆ´ âˆ†ABE ~ âˆ†CFB

(By AA similarity criterion) Hence Proved.

Question 9.

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

(i) âˆ†ABC ~ âˆ†AMP.

(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\).

Solution:

Given: âˆ†ABC and âˆ†AMP are two right angled triangles right angles at B and M respectively.

To Prove :(i) âˆ†ABC ~ âˆ†AMP.

(ii) \(\frac{C A}{P A}=\frac{B C}{M P}\).

Proof : (i) In âˆ†ABC and âˆ†AMP.

âˆ ABC = âˆ AMP (each = 90Â°)

âˆ A = âˆ A (common)

âˆ´ âˆ†ABC ~ âˆ†AMP (By AA similarity criterion)

Hence Proved.

(ii) âˆµ âˆ†ABC ~ âˆ†AMP

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

(corresponding sides of similar triangles)

Hence Proved.

Question 10.

CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on sides AB and FE of âˆ†ABC and âˆ†EFG respectively. If âˆ†ABC ~ âˆ†FEG, show that:

(i) \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)

(ii) âˆ†DCB ~ âˆ†HGE

(iii) âˆ†DCA ~ âˆ†HGF

Solution:

Given : CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on the sides AB and EF of âˆ†ABC and âˆ†EFG and âˆ†ABC ~ âˆ†FEG.

To Prove :

(i) âˆ†DCB ~ âˆ†HGE

(ii) âˆ†DCA ~ âˆ†HGF

(iii) \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)

Proof : (i) âˆ†ABC ~ âˆ†FEG (given)

âˆ ACB = âˆ EGF (corresponding âˆ s of similar âˆ†s)

\(\frac{1}{2}\) âˆ ACB = \(\frac{1}{2}\) âˆ EGF

[CD and GH are bisectors of âˆ C and âˆ G.]

âˆ DCB = âˆ HGE ………………..(1)

and âˆ B = âˆ E

(corresponding âˆ s of similar âˆ†s)

â‡’ âˆ DBC = âˆ HEG ……………….(2)

From (1) and (2), we have

âˆ†DCB ~ âˆ†HGE

Hence Proved.

(ii) In âˆ†DCA and âˆ†HGF

âˆ C = âˆ G

(corresponding âˆ s of similar âˆ†s)

âˆ ACD = âˆ HGF [CD and GH are bisectors of âˆ C and âˆ G] ………………..(1)

âˆ A = âˆ F

(corresponding âˆ s of similar âˆ†s) ……………….(2)

âˆ ADC = âˆ FHG

(third âˆ s of two triangles) …………………(3)

âˆ´ âˆ†DCA ~ âˆ†HGF. (By AAA similarity criterion)

Hence Proved.

(iii) âˆµ âˆ†DCA ~ âˆ†HGF.

âˆ´ \(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)

(corresponding sides of similar âˆ†s)

Hence Proved.

Question 11.

In given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC If AD âŠ¥ BC and EF âŠ¥ AC, prove that âˆ†ABD ~ âˆ†ECF.

Solution :

Given : AB = AC, AD âŠ¥ BC and EF âŠ¥ AC.

To Prove : âˆ†ABD ~ âˆ†ECF

Proof : AB = AC (given)

âˆ B = âˆ C [âˆµ âˆ s opposite to equal side are equal]

Now, In âˆ†ABD and âˆ†ECF,

âˆ ABD = âˆ ECF [âˆµ âˆ B = âˆ C]

âˆ ADB = âˆ EFC = 90Â°

[âˆµ AD âŠ¥ BC and EF âŠ¥ AC]

âˆ´ âˆ ABD ~ âˆ†ECF

(By AA similarity criterion)

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 âˆ†PQR (see figure below). Show that âˆ†ABC ~ âˆ†PQR.

Solution:

(By SSS similarity criterion)

âˆ B = âˆ Q

(Corresponding âˆ s of Similar As)

Now, in âˆ†ABC and âˆ†PQR.

\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) (given)

âˆ B = âˆ Q (Proved above)

âˆ´ âˆ†ABC ~ âˆ†PQR [By SAS similarity criterion]

Hence Proved.

Question 13.

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

Solution:

Given : âˆ ADC = âˆ BAC

To prove : CA^{2} = CB.CD

Proof : In âˆ†ABC and âˆ†DAC

âˆ ADC = âˆ BAC (given)

âˆ C = âˆ C (common)

âˆ†ABC ~ âˆ†DAC (By AA similarity criterion)

â‡’ CA^{2} = CB Ã— CD.

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 âˆ†ABC ~ âˆ†PQR.

Solution :

Given : In âˆ†ABC and âˆ†PQR

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

To Prove : âˆ†ABC ~ âˆ†PQR

Construction : Produce AD to E such that AD = DE and PM to N such that PM = MN Join CE and NR.

Proof : In âˆ†ADB and âˆ†EDC

AD = ED (By construction)

BD = CD (AD is median)

âˆ ADB = âˆ CDE (vertically opposite âˆ s)

âˆ†ADB â‰… âˆ†EDC (By SAS Congruence criterion)

â‡’ AB = EC (By CPCT) …………….(1)

Similarly, âˆ†PMQ â‰… âˆ†NMR (By SAS congruence criterion)

â‡’ PQ = NR (By CPCT) …………..(2)

Now, \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)(given)

â‡’ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

[From (1) and (2), AB = EC and PQ = NR]

â‡’ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \mathrm{AD}}{2 \mathrm{PM}}\)

â‡’ \(\frac{\mathrm{EC}}{\mathrm{NR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AE}}{\mathrm{PN}}\)

[By construction AD = DE and PM = MN)

âˆ´ âˆ†ACE ~ âˆ†PRN (By SSS similarity criterion)

âˆ 1 = âˆ 3 ……………….(3)

(Corresponding âˆ s of similar triangles)

Similarly âˆ 2 = âˆ 4 ………………(4)

Adding (3) and (4) we get

âˆ 1 + âˆ 2 = âˆ 3 + âˆ 4.

â‡’ âˆ A = âˆ P.

Now âˆ†ABC and âˆ†PQR.

âˆ A = âˆ P (Proved above)

\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}\) (given)

âˆ´ âˆ†ABC ~ âˆ†PQR.

Question 15.

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

Solution:

Let AB be the vertical pole of length 6m casts a shadow BC and PQ be the tower casts a shadow QR. Let height of tower berm. Join AC and PR.

In âˆ†ABC and âˆ†PQR

âˆ B = âˆ Q (each = 90Â°)

âˆ C = âˆ R (Angle of elevation of the Sun)

âˆ´ âˆ†ABC ~ âˆ†PQR (By AA similarity criterion)

âˆ´ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)

[Corresponding sides of similar triangles]

â‡’ \(\frac{6}{x}=\frac{4}{28}\)

â‡’ x = \(\frac{6 \times 28}{4}\)

x = 42.

Hence, height of tower = 42 m.

Question 16.

If AD and PM are medians of triangles ABC and PQR, respectively where âˆ†ABC ~ âˆ†PQR, prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\).

Solution :

Given : AD and PM are medians of âˆ†ABC and âˆ†PQR respectively and âˆ†ABC ~ âˆ†PQR.

To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

Proof : âˆ†ABC ~ âˆ†PQR

âˆ´ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}\) ………………(1)

(corresponding sides of similar triangles)

â‡’ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)

â‡’ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}\)

[âˆµ AD and PM are medians

âˆ´ BD = CD and QM = MR]

â‡’ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)

and âˆ B = âˆ Q

[Corresponding âˆ s of similar âˆ†s ABC and PQR]

âˆ´ âˆ†ADB ~ âˆ†PMQ [By SAS similarity criterion]

\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\) (Corresponding sides of similar âˆ†s.)

Hence Proved.