166-NCERT-10-6-Triangles - Ex- 6.3

NCERT

10th Mathematics

Exercise 6.3

Topic: 6 Triangles

Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.2

EXERCISE 6.3

Q1. State which pairs of triangles in the following 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)

1) In ∆ ABC and ∆ PQR,

a) < A = < P = 60⁰

b) < B = < Q = 80⁰
c) < C = < R = 40⁰

2) So by AAA similarity rule, ∆ ABC ~ ∆ PQR.

 

(ii)

1) In ∆ ABC and ∆ QRP,

Note: For ∆ ABC and ∆ QRP, points A ↔ Q, B ↔ R, C ↔ P. That is 

i) point A is associated with poin Q, 

ii) point B is associated with poin R,
iii) point B is associated with poin R,

a) (AB)/(QR) = 2/4 = 1/2 -------- equation 1

b) (BC)/(RP) = 2.5/5 = 1/2 -------- equation 2

c) (AC)/(QP) = 3/6 = 1/2 -------- equation 3

2) So from equations 1, 2, and 3, the corresponding ratios are proportional, so by SSS

test, ∆ ABC ~ ∆ QRP.

(iii)

1) In ∆ LMP and ∆ FED,

Note: For ∆ LMP and ∆ FED, points L ↔ F, M ↔ E, and P ↔ D. That is 

i) point L is associated with poin F, 

ii) point M is associated with poin E,
iii) point P is associated with poin D,

a) (LM)/(FE) = 2.7/5 = 27/50 -------- equation 1

b) (MP)/(ED) = 2/4 = 1/2 -------- equation 2

c) (LP)/(FD) = 3/6 = 1/2 -------- equation 3

2) So from equations 1, 2, 3, (LM)/(FE) ≠ (MP)/(ED) = (LP)/FD), so 

∆ ABC ≁ ∆ QRP.

(iv)

1) In ∆ LMN and ∆ RQP,

Note: For ∆ LMN and ∆ RQP, points L ↔ R, M ↔ Q, and N ↔ P. That is 

i) point L is associated with poin R, 

ii) point M is associated with poin Q,
iii) point N is associated with poin P,

a) (LM)/(RQ) = 5/10 = 1/2 -------- equation 1

b) < NML = < PQR = 70⁰-------- equation 2

c) (MN)/(QP) = 2.5/5 = 1/2 -------- equation 3

2) So from equations 1, 2, 3, (LM)/(RQ) = (MN)/(QP) = 1/2, so by SAS property,

∆ LMN ~ ∆ RQP.

(v)

1) In ∆ ABC and ∆ DFE,

Note: For ∆ ABC and ∆ DFE, points A ↔ D, B ↔ F, and C ↔ E. That is 

i) point A is associated with poin D, 

ii) point B is associated with poin F,
iii) point C is associated with poin E,

a) (AB)/(DF) = 2.5/5 = 1/2 -------- equation 1

b) < ABC ≠ < DFE -------- equation 2

c) (BC)/(FE) = 3/6 = 1/2 -------- equation 3

2) So from equations 1, 2, 3, (AB)/(DF) = (BC)/(FE) = 1/2, but < ABC ≠ < DFE so,

∆ ABC ≁ ∆ DFE.

(vi)

1) In ∆ DEF and ∆ PQR,

Note: For ∆ DEF and ∆ PQR, points D ↔ P, E ↔ Q, and F ↔ R. That is 

i) point D is associated with poin P, 

ii) point E is associated with poin Q,
iii) point F is associated with poin R,

a) in ∆ DEF,

< D + < E + < F = 180

70 + 80 + < F = 180

150 + < F = 180

< F = 180 - 150

< F = 30

< DFE = 30

< DFE = < PRQ = 30 -------- equation 1

b) in ∆ PQR,

< P + < Q + < R = 180
< P + 80 + 30 = 180
< P + 110 = 180
< P = 180 - 110
< P = 70
< QPR = 70
< QPR = < EDF = 70 -------- equation 2 

c) < DEF = < PQR  = 80 -------- equation 3

2) From equations 1, 2, and 3, by AAA similarity rule, ∆ DEF ~ ∆ PQR.

Q2. In the following Fig., ∆ ODC ~ ∆ OBA, < BOC = 125° and < CDO = 70°. 

Find < DOC, < DCO and < OAB.

Solution:

1) In the above figure,

a) < DOC and < COB are angles in linear pairs, so

< DOC + < COB = 180

< DOC + 125 = 180

< DOC = 180 - 125

< DOC = 55 -------- equation 1

b) So < DOC = 55

2) In ∆ PQR,

< DOC + < DCO + < CDO = 180

< 55 + < DCO + 70 = 180

< DCO + 125 = 180

< DCO = 180 - 125

< DCO = 55 -------- equation 2

So < DCO = 55

3) It's given that, ∆ ODC ~ ∆ OBA,

< OAB = < OCD -------- equation 3

4) From equations 2 and 3, we have, 

< OAB = 55

Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at point O. Using a similarity criterion for two triangles, show that 

(OA)/(OB) = (OC)/(OD).

Solution:

1) In trapezium  ABCD, AB ‖ CD, and diagonals AC and BD intersect at O.

2) In ∆ AOB and ∆ COD,

a) < AOB = < COD (vertically opposite angles are equal)

b) < ABO = < CDO (alternate interior angles are equal)

c) < BAO = < DCO (alternate interior angles are equal)

3) So, by AAA similarity test, ∆ AOB ~ ∆ COD

3) So, as corresponding sides are proportional, we have,

(OA)/(OB) = (OC)/(OD). Hence proved.

Q4. In the following Fig., (QR)/(QS) = (QT)/(PR) and < 1 = < 2. Show that

∆ PQS ~ ∆ TQR.

Solution:

1) In ∆ PQR,

< 1 = < 2, so

(QP) = (PR) -------- equation 1

2) It is given that,

(QR)/(QS) = (QT)/(PR) -------- equation 2

3) From equations 1 and 2, we have,

(QR)/(QS) = (QT)/(QP) -------- equation 3

4) In ∆ PQS and ∆ TQR,

< PQS = < TQR (same angle) -------- equation 4

5) From equations 3 and 4 and by SAS property of similarity we have,

∆ PQS ~ ∆ TQR. Hence proved.

Q5. S and T are points on sides PR and QR of ∆ PQR such that < P = < RTS. Show that ∆ RPQ ~ ∆ RTS.

Solution:

1) In ∆ PQR, and ∆ TSR,

Note: For ∆ PQR and ∆ TSR, points P ↔ T, Q ↔ S, and R ↔ R. That is 

i) point P is associated with poin T,

ii) point Q is associated with poin S,

        iii) point R is associated with point R,

a) < QPR = < STR       (given)
b) < PRQ = < TRS       (same angles)

2) By AA similarity test, ∆ PQR ~ ∆ TSR. Hence proved.

Q6. In the following fig., if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC.

Solution:

1) It is given that ∆ ABE ≅ ∆ ACD, corresponding sides of congruent triangles are

    congruent, so,

a) AD = AE ------- equation 1

b) AB = AC ------- equation 2

2) In ∆ DAE and ∆ BAC, and from equations 1 and 2, we have,

a) (AD)/(AB) = (AE)/(AC) 

b) < DAE = <BAC     (same angle)

3) ∆ DAE ~ ∆ BAC. Hence proved.

Q7. In the following fig., altitudes AD and CE of ∆ ABC intersect each other at point P. Show that:

Solution:

(i) ∆ AEP ~ ∆ CDP

1) In ∆ AEP and ∆ CDP

Note: For ∆ AEP and ∆ CDP, points A ↔ C, E ↔ D, and P ↔ P. That is 

i) point A is associated with poin C,

ii) point E is associated with poin D,

iii) point P is associated with poin P, 

2) In ∆ AEP and ∆ CDP

a) < AEP = < CDP = 90     (given) -------- equation 1

b) < APE = < CPD     (Vertically oposite angles) -------- equation 2

3) From equations 1 and 2, by AA similarity test, we have,

a) ∆ AEP ~ ∆ CDP, hence proved.

(ii) ∆ ABD ~ ∆ CBE

1) In ∆ ABD and ∆ CBE

Note: For ∆ ABD and ∆ CBE, points A ↔ C, B ↔ B, and D ↔ E. That is 

i) point A is associated with poin C,

ii) point B is associated with poin B,

iii) point D is associated with poin E, 

2) In ∆ ABD and ∆ CBE

a) < ADB = < CEB = 90     (given) -------- equation 1

b) < ABD = < CBE     (common angles) -------- equation 2

3) From equations 1 and 2, by AA similarity test, we have,

a) ∆ ABD ~ ∆ CBE, hence proved.

(iii) ∆ AEP ~ ∆ ADB

1) In ∆ AEP and ∆ ADB

Note: For ∆ AEP and ∆ ADB, points A ↔ A, E ↔ D, and P ↔ B. That is 

i) point A is associated with poin A,

ii) point E is associated with poin D,

iii) point P is associated with poin B, 

2) In ∆ AEP and ∆ ADB

a) < AEP = < ADB = 90     (given) -------- equation 1

b) < PAE = < BAD     (common angles) -------- equation 2

3) From equations 1 and 2, by AA similarity test, we have,

a) ∆ AEP ~ ∆ ADB, hence proved.

(iv) ∆ PDC ~ ∆ BEC

1) In ∆ PDC and ∆ BEC

Note: For ∆ PDC and ∆ BEC, points P ↔ B, D ↔ E, and C ↔ C. That is 

i) point P is associated with poin B,

ii) point D is associated with poin E,

iii) point C is associated with poin C, 

2) In ∆ PDC and ∆ BEC

a) < PDC = < BEC = 90     (given) -------- equation 1

b) < PCD = < BCE     (common angles) -------- equation 2

3) From equations 1 and 2, by AA similarity test, we have,

a) ∆ PDC ~ ∆ BCE, hence proved.

Q8. E is a point on the side AD produced of a parallelogram ABCD and BE

intersect CD at F. Show that ∆ ABE ~ ∆ CFB.

Solution:

1) In ∆ ABE and ∆ CFB

Note: For ∆ ABE and ∆ CFB, points A ↔ C, B ↔ F, and E ↔ B. That is 

i) point A is associated with poin C,

ii) point B is associated with poin F,

iii) point E is associated with poin B, 

2) In ∆ ABE and ∆ CFB

a) < BAE = < FCB     (Opposite angles of a parallelogram) -------- equation 1

b) < AEB = < CBF     (Alternate interior angles as AE || BC) -------- equation 2

3) From equations 1 and 2, by AA similarity test, we have,

a) ∆ ABE ~ ∆ CFB, hence proved.

Q9. In the following fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: 

(i) ∆ ABC ~ ∆ AMP     (ii) (CA)/(PA) = (BC)/(MP) 

Solution:

(i) ∆ ABC ~ ∆ AMP

1) In ∆ ABC and ∆ AMP

Note: For ∆ ABC and ∆ AMP, points A ↔ A, B ↔ M, and C ↔ P. That is 

i) point A is associated with poin A,

ii) point B is associated with poin M,

iii) point C is associated with poin P, 

2) In ∆ ABC and ∆ AMP

a) < ABC = < AMP = 90 -------- equation 1

b) < BAC = < MAP   (Common angle) -------- equation 2

3) From equations 1 and 2, by AA similarity test, we have,

a) ∆ ABC ~ ∆ AMP, hence proved.

 (ii) (CA)/(PA) = (BC)/(MP)

1) In ∆ ABC and ∆ AMP

a) ∆ ABC ~ ∆ AMP (we proved this)

b) (CA)/(PA) = (BC)/(MP) (By basic propotinality theorem)

Q10. 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) (CD)/(GH) = (AC)/(FG) 

(ii) ∆ DCB ~ ∆ HGE 

(iii) ∆ DCA ~ ∆ HGF

Solution:

(i) (CD)/(GH) = (AC)/(FG)

1) ∆ ABC ~ ∆ FEG   (given), so we have,

a) < ACB = < FGE --------- equation 1

b) CD is bisector of < ACB and GH is the bisector of < FGE 

c) (< ACB)/2 = (< FGE)/2 --------- equation 2

2) From equations 1 and 2, and in ∆ DCA and ∆ HGF we have

< DCA = < HGF --------- equation 3

< CAD = < GFH --------- equation 4

3) From equations 3 and 4, and the AA similarity test, we have,

∆ DCA ~ ∆ HGF --------- equation 5

4) So, (CD)/(GH) = (AC)/(FG) hence proved. 

(ii) ∆ DCB ~ ∆ HGE

1) In ∆ DCB and ∆ HGE (CD is bisector of < ACB and GH is the bisector of < FGE)

a) < DCB = < HGE --------- equation 1

b) < DBC = < HEG (Common angle) --------- equation 2

2) From equations 1 and 2, we have,

∆ DCB ~ ∆ HGE, hence proved.

(iii) ∆ DCA ~ ∆ HGF

It is already proved in example: (i) above. (See equation 5 for example (i)).

Q11. In the following 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:

1) ∆ ABC is an isosceles triangle with AB = AC, so we have,

a) < ABC = < ACB --------- equation 1

2) In ∆ ABD and ∆ ECF and from equation 1

a) < ABD = < ECF (same angle) --------- equation 2 

c) < ADB = < EFC (each angle is 90⁰) --------- equation 3

3) From equations 2 and 3, and the AA similarity test, we have,

a) ∆ ABD ~ ∆ ECF, hence proved.

Q12. 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 the following fig.). Show that ∆ ABC ~ ∆ PQR.

Solution:

1) It is given that:

a) (AB)/(PQ) = (BC)/(QR) = (AD)/(PM) ---------- equation 1.

b) (AB)/(PQ) = [(BC)/2]/[(QR)/2] = (AD)/(PM)

c) (AB)/(PQ) = (BD)/(QM) = (AD)/(PM) ---------- equation 2.

2) From equation 2 and the basic proportionality theorem, we have,

∆ ABD ~ ∆ PQM ---------- equation 3.

3) In ∆ ABC and ∆ PQR,

a) (AB)/(PQ) = (BC)/(QR) ---------- equation 4.

4) As ∆ ABD ~ ∆ PQM, we have,

a) < ABC = < PQR ---------- equation 5.

5) From equations 4 and 5 and by SAS test, we have,

∆ ABC ~ ∆ PQR, hence proved.

Q13. D is a point on the side BC of a triangle ABC such that < ADC = < BAC. Show that CA² = CB.CD.

Solution:

1) In ∆ ADC and ∆ BAC,

Note: For ∆ ADC and ∆ BAC, points A ↔ B, D ↔ A, and C ↔ C. That is 

i) point A is associated with poin B,

ii) point D is associated with poin A,

        iii) point C is associated with point C,

a) < ADC = < BAC       (given)
b) < ACD = < BCA       (common angles)

2) By AA similarity test, ∆ ADC ~ ∆ BAC.

3) By the basic proportionality theorem, we have,

(CA)/(CB) = (CD)/(CA) 

(CA) x (CA) = (CD) x (CB)

(CA)² = (CB) x (CD), hence proved.

Q14. 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:

1) Extend AD to E such that AD = ED.

2) Extend PM to N such that PM = NM.

3) In ∆ ABC and ∆ PQR,

(AB/PQ) = (AC/PR) = (AD/PM) ---------- equation 1.

4) In ∆ ABD and ∆ ECD,

a) (AD) = (ED)        (construction)

b) < ADB = < EDC  (vertically opposite angles)

c) (BD) = (CD)        (AD is median of ∆ ABC)

5) So by SAS test ∆ ABD ≅ ∆ ECD.

6) So here, (AB) = (EC) ---------- equation 2

7) In ∆ PQM and ∆ NRM,

a) (PM) = (NM)        (construction)

b) < PMQ = < NMR  (vertically opposite angles)

c) (QM) = (RM)        (PM is median of ∆ PQR)

8) So by SAS test ∆ PQM ≅ ∆ NRM.

9) So here, (PQ) = (NR) ---------- equation 3

10) From equations 1, 2, and 3, we have,

a) (AB/PQ) = (AC/PR) = (AD/PM)

b) (EC/NR) = (AC/PR) = 2(AD)/2(PM)

c) (EC/NR) = (AC/PR) = (AE)/(PN) -------  (As 2AD = AE, and 2PM = PN)

11) So, by SSS similarity test ∆ ACE ~ ∆ PRN,

12) So we have,

a) < CAD = < RPM,

b) [2(< CAD)] = [2(< RPM)] (as AD and PM are medians of ∆ ABC and ∆ PQR) 

c) < CAB = < RPQ ---------- equation 4

13) From equations 1 and 4, we have,

a) (AB/PQ) = (AC/PR) and < CAB = < RPQ by SAS similarity test, we have,

b) ∆ ABC ~ ∆ PQR hence proved.

Q15. 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:

1) In ∆ ABC and ∆ PQR,

a) < BCA = < QRP        (angle of elevation)

b) < ABC = < PQR = 90  (angle between pole or tower and their shadows)

c) ∆ ABC ~ ∆ PQR by AA similarity test,

2) So by basic proportionality theorem.

(AB)/(PQ) = (BC)/(QR)

(6)/(PQ) = (4)/(28)

(6)/(PQ) = (1)/(7)

6 x 7 = (PQ)

(PQ) = 42 m

3) So, the height of the tower is 42 m.

Q16. If AD and PM are medians of triangles ABC and PQR, respectively where

∆ ABC ~ ∆ PQR, prove that AB/PQ = AD/PM.

Solution:

1) As ∆ ABC ~ ∆ PQR,

a) (AB)/(PQ) = (AC)/(PR) = (BC)/(QR)

b) (AB)/(PQ) = (AC)/(PR) = [(BC)/2]/[(QR)/2] 

2) As D and M are midpoints of BC and QR respectively, we have,

a) (AB)/(PQ) = (AC)/(PR) = (BD)/(QM) ---------- equation 1

3) As AD and PM are the medians of ∆ ABC and ∆ PQR,

a) < BAD = < QPM ---------- equation 2

4) In ∆ ABD and ∆ PQM, and from equations 1 and 2, we have,

a) (AB)/(PQ) = (AD)/(PM) = (BD)/(QM)

b) < BAD = < QPM

5) By the SAS similarity test, we have ∆ ABD ~ ∆ PQM

6) So by basic proportionality theorem.

a) (AB)/(PQ) = (AD)/(PM), hence proved.

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Click here for ⇨ NCERT-10-6-Triangles - Ex- 6.4

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