**KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3

**Exercise 2.3 Class 10 Maths Triangles Question 1.**

State which pairs of triangles In the following 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 ∆ABC and ∆PQR,

∠A = ∠P = 60°

∠B=∠Q = 80°

∠C =∠R = 40°

∴ SimilarIty criterion of two triangles is A.A.A.

∴ ∆ABC ~ ∆PQR.

(ii)

In ∆ABC and PQR, we have

\(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}=\frac{1}{2}\)

i.e., sides of one triangle are proportional to the sides of the other triangle.

∴SImilarity criterion is S.S.S.

∴ ∆ABC ~ ∆PQR.

(iii)

∆LMP and ∆DEF are not similar triangles because one pair of sides are not proportional to each other.

\(\quad \frac{\mathrm{MP}}{\mathrm{DE}}=\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{2}{4}=\frac{3}{6}=\frac{1}{2}\)

But, \(\frac{\mathrm{LM}}{\mathrm{EF}}=\frac{2.7}{5} \neq \frac{1}{2}\)

∴ These are not similar triangles

(iv)

In ∆MNL and ∆PQR, we have

∠M = ∠Q = 70°

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

∴ These are not similar triangles

(v)

In ∆ABC and ∆DEF, we have

\(\frac{A B}{D F}=\frac{B C}{E F}=\frac{1}{2}\)

∠A = ∠F = 80°

∴ Similarly criterion for these triangles is S.A.S

∴ ∆ABC ~ ∆DEF

(vi)

Sum of three angles of is 180°, but in ∆DEF. sum of two angles. ∠F= 30°. In ∆PQR, ∠P = 70°.

In ∆DEF and ∆PQR. we have

∠D = ∠P = 70°

∠E =∠R = 80°

∠F = ∠R = 30°

∴ Three angles of these triangles are equal to each other.

∴ SimIlarity criterion here is A.A.A.

∴ ∆DEF ~ ∆PQR.

**KSEEB Solutions For Class 10 Maths Triangles Question 2.**

In the following figure. ∆OBA ~ ∆ODC, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:

∆OBA ~ ∆ODC (data)

In ∆OBA and ∆ODC, we have

∠ODC = ∠OBA = 70° (. Alternate angle)

∠AOB = 180°- 125° = 55° ( Adjacent angle)

∴ ∠AOB = ∠DOC = 55° (: VertIcally opposite angles)

∠OAB = ∠OCD = 55° (. Alternate angles)

∴ ∠DOC=55°

∠DCO = 55%

∠OAB= 55%.

**10th Maths Triangles Exercise 2.3 Solutions 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:

Data: Diagonals AC and BD of a trapezium ABCD with AB || DC. intersect each other at the point O.

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

In the trapezium ABCD AB || DC.

∴ ∠OCD = ∠OAB (Alternate angles)

∠ODC = ∠OBA (Alternate angles)

∠DOC = ∠AOB (Vertically opposite angles)

∴ These are equlangular triangles.

∴ These triangles are similar.

∴ ∆ODC ~ ∆OAB

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

∵ Corresponding sides are in proportional.

**Triangles Exercise 2.3 Question 4.**

In the following figure. \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:

Data: Here \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\)and ∠1 = ∠2

To prove: ∆PQS ~ ∆TQR.

In ∆PQS and ∆TQR, we have

\(\frac{\mathrm{Q} \mathrm{R}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\)

∠PQR = ∠PRQ (∵ ∠1 = ∠2)

Here, Similarity criterion used here is side, angle, side (SAS).

∴ ∆PQS ~ ∆TQR.

**Triangles Class 10 Exercise 2.3 Question 5.**

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

Solution:

Data: S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS.

To Prove: ∆RPQ ~ ∆RTS.

In ∆RPQ and ∆RTS,

∠P = ∠RTS (Data)

∠PRQ = ∠SRT (Common)

∴ 3rd angle ∠PRQ = ∠SRT

∴ These are equiangularangular triangles.

∴ Here A.A.A. similarity criterion.

∴ ∆RPQ ~ ∆RTS.

**Triangle Exercise 2.3 Question 6.**

In the given fIgure. ∆ABE ≅ ∆ACD. show that ∆ADE ~ ∆ABC.

Solution:

Data: ∆ABE ≅ ∆ACD.

To Prove: ∆ADE ~ ∆ABC.

∆ABE ≅ ∆ACD

AB = AC

AD = AE.

Then DC = BE

AB = AC

AD + DB = AE + EC

∴ DB = EC

(∵ DA = AE)

In ∆ADE and ∆ABC,

\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}\)

Here, corresponding sides are in proportion.

∴ Similarity criterion for ∆ is side, side, side

∴ ∆ADE ~ ∆ABC

**Triangles Class 10 Exercise 2.3 Solutions Question 7.**

In following figure. altitudes AD and CE of ∆ABC Intersect each other at the point P. Show that:

i) ∆AEP ~ ∆CDP

ii) ∆ABD ~ ∆CBE

iii) ∆AEP ~ ∆AÐB

iv) ∆PDC ~ ∆BEC.

Solution:

Data: altitudes AD and CE of ABC

Intersect each other at the point P.

To Prove: i) ∆AEP ~ ∆CDP

ii) ∆ABD ~ ∆CBE

iii) ∆AEP ~ ∆ADB

iv) ∆PDC ~ ∆BEC

i) In ∆AEP and ∆CDP.

∠AEP = ∠CDP = 90° (data)

∠APE = ∠CPD (Vertically opposite angle)

∴ ∠PAE = ∠PCD

These are equiangular triangles.

Similarity criterion for ∆ is A.A.A

∴ ∆AEP ~ ∆CDP

(ii) In ∆ABD and ∆CBE.

∠ADB = ∠CEB = 90° (data)

∠ABD = ∠CBE (common)

∴ ∠DAB = ∠BCE

These are equiangular triangles.

∴ Similarity criterion for is A.A.A.

∴ ∆ABD ~ ∆CBE

(iii) In ∆AEP and ∆ADB.

∠AEB = ∠ADB = 90° (data)

∠PAE = ∠DAB (common)

∴ ∠APE = ∠ABD.

∴ These are equiangular triangles.

∴ Similarity criterion for is A.A.A.

∴ ∆AEP ~ ∆ADB

(iv) In ∆PDC and ∆BEC.

∠PDC = ∠BEC = 90° (data)

∠PCD = ∠BCE (common)

∴ ∠CPD = ∠CBE

∴ These are equiangularangular triangles.

Similarity criterion for ∆ is A.A.A.

∴ ∆PDC ~ ∆BEC.

**Triangle Lesson Exercise 2.3 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:

Data: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.

To Prove: ∆ABE ~ ∆CFB

In \(\) ABCD Adjacent angles are equal.

Let ∠DAB = ∠BCD = 70°

∠DAB = ∠EAF = 70° (∵ Corresponding angle)

In ∆EDF, ∠DEF = 30° then,

∠EFD = 80°.

∠EFD = ∠BFC = 80° (vertIcally opposite angles)

In ∆FBC, ∠FBC = 30°. ,

Now in ∆ABE and ∆CFB,

∆EAB = ∆BCF = 70°

∆AEB = ∆FBC = 30°

∆ABE = ∆BFC = 80°

∴ Similarity criterion for ∆ is A.A.A.

∴ ∆ABE ~ ∆CFB

**Triangles Class 10 Solutions KSEEB Question 9.**

In the given figure. ∆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:

Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely.

To Prove: i) ∆ABC ~ ∆AMP

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

(i) In ∆ABC and ∆AMP,

∠ABC = ∠AMP = 90° (data)

∠CAB = ∠MAP (common)

∴ ∠ACB = ∠MPA

∴ These are equiangular triangles.

Similarity criterion for ∆ is A.A.A.

∴ ∆ABC ~ ∆AMP

(ii) ∆ABC ~ ∆AMP(Proved)

∴ Corresponding sides are in proportion.

LC and LP are corresponding angles.

∴ Adjacent sides are CA, PA.

Similarly BC and MP are adjacent sides.

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

**KSEEB Solutions For Class 10 Maths Question 10.**

CD and OH 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 ~ ∆EFG, show that

i) \(\frac{C D}{G H}=\frac{A C}{F G}\)

ii) ∆DCB ~ ∆HGE

iii) ∆DCA ~ ∆HGF

Solution:

Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively.

∆ABC ~ ∆EFG

To Prove: i) \(\frac{C D}{G H}=\frac{A C}{F G}\)

ii) ∆DCB ~ ∆HGE

iii) ∆DCA ~ ∆HGF

Proof: ∆ABC ~ ∆EFG (data given)

∴ Their corresponding sides are in proportion.

\(\quad \frac{\mathrm{AB}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{FG}}=\frac{\mathrm{AC}}{\mathrm{EG}}\)

∠B = ∠F, ∠A = ∠E, ∠C = ∠G.

(i) In ∆ADC and ∆EHG,

∠A = ∠E .

∠ACD = ∠EGH

∴ Their sides are in proportion.

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

(ii) In ∆DCB and ∆HGE.

\(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{BC}}{\mathrm{GF}}\)

∴ ∆DCB ~ ∆HGE

(iii) In ∆DCA and ∆HGE,

\(\frac{\mathrm{DC}}{\mathrm{GH}}=\frac{\mathrm{AD}}{\mathrm{EH}}=\frac{\mathrm{AC}}{\mathrm{EG}}\)

∴ ∆DCA ~ ∆HGE

**KSEEB Solutions Maths Question 11.**

In the following figure. 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:

Data: E Is a point on side CB produced of an isosceles ∆ABC with AB = AC. AD ⊥ BC and EF ⊥ AC,

To Prove: ∆ABD ~ ∆ECF

In ∆ABD and ∆ECF.

∠ADB = ∠EFC = 90° (data)

∠ABD = ∠FCE (∵ ∠B = ∠C)

∠BAD = ∠FEC

∴ Equiangular triangles.

∴ Similarity criterion for triangles is A.A.A.

∴ ∆ABD ~ ∆ECF.

**Triangles Class 10 KSEEB Solutions Question 12.**

Sides AB and BC and median AD of a ¿ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR. (Sec figure gIven below) Show that ∆ABC ~ ∆PQR

Solution:

Data: Sides AB and BC and median AD of a ∆ABC are respectively proportional to sides P9 and QR and medIan PM of ∆PQR.

To Prove: ∆ABC ~ ∆PQR

In ∆ABC and ∆PQR,

Side \(\frac{\mathrm{AB}}{\mathrm{PG}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

D is the mid-point of BC

∴ BD = DC

M is the mid-point of QR.

∴ QM = MR.

In ∆ADC and ∆PMR,

\(\frac{\mathrm{AD}}{\mathrm{PM}}=\frac{\mathrm{DC}}{\mathrm{MR}}=\frac{\mathrm{AC}}{\mathrm{PR}} \quad \left(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}\right)\)

Similarity criterion is S.S.S.

∴ ∆AÐC ~ ∆PMR

Now, In ∆ABC and ∆PQR.

\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)

∴ SimilarIty criterion is S.S.S.

∴ ∆ABC ~ ∆PQR.

**KSEEB Maths Solutions For Class 10 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:

Data: D is a point on the side BC of a triangle ∆ABC such that ∠ADC = ∠BAC.

To Prove: CA^{2} = CB × CD

Let ∠ADC = ∠BAC = 100°

In ∆ABC, If ∠B = 50°,then ∠C = 30°

In ∆ADC, If ∠C = 30°. then ∠DAC = 50°

In ∆BCP, ∠A= 100°, ∠B= 50°, ∠C= 30°

In ∆ADC, ∠ADC = 100. ∠DAC = 50°.

∠ACD = 30°

Similarity criterion of ∆ is A.A.A.

∴ In ∆ABC and ∆ADC,

\(\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{DC}}{\mathrm{CA}}\)

∴ CA × CA = BC × DC

∴ CA^{2} = BC × DC.

**KSEEB Solutions For Class 10 Maths Kannada Medium Question 14.**

Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and mediam PM of another triangle ∆PQR. Show that ∆ABC ~ ∆PQR.

Solution:

Data: 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.

To Prove: ∆ABC ~ ∆PQR

In ∆ABC and ∆PQR

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

Corresponding sides are proportional.

Similarity criterion for ∆ is S.S.S

∴ ∆ABC ~ ∆PQR.

**10th KSEEB Maths Solutions 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:

Height of PQ =?

In ∆ABC and ∆PQR,

∠B = ∠Q = 90°

∴ Hypotenuse AC is proportional to PQ

\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{MR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\)

\(\frac{6}{\mathrm{PQ}}=\frac{4}{28}\)

∴ 4 × PQ = 6 × 28

\(\quad \mathrm{PQ}=\frac{6 \times 28}{4} \quad \quad \mathrm{PQ}=42 \mathrm{m}\)

**Maths Solutions Class 10 KSEEB 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:

Data : AD and PM are medians of triangles ∆ABC and ∆PQR respectively. ∆ABC ~ ∆PQR.

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

∆ABC ~ ∆PQR.(data)

∴ we have \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)

D is the mid-point of BC. BD = DC.

M Is the mid-point of QR. QM = MR

\(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)

\(\quad \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)

In ∆ABD and ∆PQM, we have

\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

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

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