Triangles Exercise 6.3 – class 10

1. State which pairs of triangles in the following figure 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: 

(i)

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(ii)

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(iii)

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(iv)

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(v)

36.png

(vi)

37.png

Solution:

(i) ∠A = ∠P = 60°;

∠B = ∠Q = 80° ;

∠C = ∠R = 40°

Therefore, ∆ABC ∼ ∆PQR [By AAA similarity criterion]

 

AB/QR = BC/RP = CA/PQ

(ii) ∆ABC ∼ ∆PQR -by SSS similarity criterion]

(iii)The given triangles are not similar as the corresponding sides are not proportional.

(iv)The given triangles are not similar as the corresponding sides are not proportional.

(v)The given triangles are not similar as the corresponding sides are not proportional.

(vi) In ∆DEF, ∠D +∠E +∠F = 180º (Sum of the measures of the angles of a triangle is 180º.) 70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ∆PQR, ∠P +∠Q +∠R = 180º (Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ∆DEF and ∆PQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ∆DEF ∼ ∆PQR [By AAA similarity criterion]

 

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

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Solution:

DOB is a straight line.

∴ ∠DOC + ∠COB = 180°

⇒ ∠DOC = 180° − 125° = 55°

In ∆DOC, ∠DCO + ∠CDO + ∠DOC = 180° (Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ∆ODC ∼ ∆OBA.

∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°

 

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 AO/OC = OB/OD

Solution:

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In ∆DOC and ∆BOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ∆DOC ∼ ∆BOA [AAA similarity criterion]

DO/BO = CO/AO

⇒ AO/OC = OB/OD

 

4. In the following figure, QR/QS = QT/PR and ∠1 = ∠2

41.png

Show that ∆PQR ∼ ∆TQR

Solution:

42.png

In ∆PQR, ∠PQR = ∠PRQ

∴ PQ = PR ………………(i)

 

Given,

QR/QS = QT/PR

Using (i), we obtain

QR/QS = QT/QP —————(ii)

In ∆PQS and ∠TQR

QR/QS = QT/QP [Using (ii)]

∠Q = ∠Q

∴ ∆PQS ∼ ∠TQR [SAS similarity theorem]

 

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

Solution:

44.png

In ∆RPQ and ∆RST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ∆RPQ ∼ ∆RTS (By AA similarity criterion)

 

6. In the following figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ∼ ∆ABC.

45.png

Solution:

It is given that ∆ABE ≅ ∆ACD.

∴ AB = AC [By CPCT] …………………(1)

And, AD = AE [By CPCT] …………….(2)

In ∆ADE and ∆ABC,

AD/DB = AE/AC

[Dividing equation (2) by (1)]

∠A = ∠A [Common angle]

∴ ∆ADE ∼ ∆ABC [By SAS similarity criterion]

 

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

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(i) ∆AEP ∼ ∆CDP

(ii) ∆ABD ∼ ∆CBE

(iii) ∆AEP ∼ ∆ADB

(v) ∆PDC ∼ ∆BEC

Solution:

(i)

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In ∆AEP and ∆CDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion, ∆AEP ∼ ∆CDP

(ii)

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In ∆ABD and ∆CBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion, ∆ABD ∼ ∆CBE

(iii)

49.png

In ∆AEP and ∆ADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion, ∆AEP ∼ ∆ADB

(iv)

50.png

In ∆PDC and ∆BEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion, ∆PDC ∼ ∆BEC

 

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:

51.png

In ∆ABE and ∆CFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ∆ABE ∼ ∆CFB (By AA similarity criterion)

 

9. In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that: 

52.png

(i) ∆ABC ∼ ∆AMP

(ii) CA/PA = BC/MP

Solution:

In ∆ABC and ∆AMP, ∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ∆ABC ∼ ∆AMP (By AA similarity criterion)

CA/PA = BC/MP (corresponding sides of similar angles are proportional)

(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)CD/GH = AC/FG

(ii)∆DCB ∼ ∆HGE

(iii) ∆DCA ∼ ∆HGF

Solution:

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It is given that ∆ABC ∼ ∆FEG.

∴ ∠A = ∠F, ∠B = ∠E,

and ∠ACB = ∠FGE  ; ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = HGE (Angle bisector)

 

In ∆ACD and ∆FGH, ∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ ∆ACD ∼ ∆FGH (By AA similarity criterion)

CD/GH = AC/FG

 

In ∆DCB and ∆HGE, ∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ∆DCB ∼ ∆HGE (By AA similarity criterion)

 

In ∆DCA and ∆HGF, ∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ ∆DCA ∼ ∆HGF (By AA similarity criterion)

 

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 

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Solution:

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ∆ABD and ∆ECF, ∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Proved above)

∴ ∆ABD ∼ ∆ECF (By using AA similarity criterion)

 

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 the given figure). Show that ∆ABC ∼ ∆PQR. 

Solution:

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Median divides the opposite side.

∴ BD = BC/2 and QM = QR/2

Given that,

 

AB/PQ = BC/QR = AD/PM

⇒ AB/PQ = (BC/2)/(QR/2) = AD/PM

⇒ AB/PQ = BD/QM = AD/PM

In ∆ABD and ∆PQM,

AB/PQ = BD/QM = AD/PM [proved above]

∴ ∆ABD ∼ ∆PQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ∆ABC and ∆PQR, ∠ABD = ∠PQM (Proved above)

AB/PQ = BC/QR

∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)

 

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that 

CA² = CB.CD

Solution:

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In ∆ADC and ∆BAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ∆ADC ∼ ∆BAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion

CA/CB = CD/CA

⇒CA² = CB.CD

 

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 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:

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Given that,

AB/PQ = AC/PR = BC/QR

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

59.png

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that

 

AB/PQ = AC/PR = AD/PM

⇒ AB/PQ = BE/QL = 2AD/2PM

⇒ AB/PQ = BE/QL = AE/PL

∴ ∆ABE ∼ ∆PQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ∆AEC ∼ ∆PLR and ∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL ⇒ ∠CAB = ∠RPQ … (3)

In ∆ABC and ∆PQR,

AB/PQ = AC/PR [given]

∠CAB = ∠RPQ [Using equation (3)]

∴ ∆ABC ∼ ∆PQR (By SAS similarity criterion)

 

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

Solution:

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Let AB and CD be a tower and a pole respectively.

Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, ∠DCF = ∠BAE And, ∠DFC = ∠BEA ∠CDF = ∠ABE (Tower and pole are vertical to the ground)

∴ ∆ABE ∼ ∆CDF (AAA similarity criterion)

⇒AB/CD = BE/DF

⇒AB/6 = 28/4

⇒AB = 42m

Therefore, the height of the tower will be 42 metres.

 

16. If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ∼ ∆PQR provve thta, AB/PQ = AD/PM

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It is given that ∆ABC ∼ ∆PQR We know that the corresponding sides of similar triangles are in proportion.

∴AB/PQ = AC/PR = BC/QR————(1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ———— (2)

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = BC/2 and QM = QR/2 ————-(3)

From equations (1) and (3), we obtain

AB/PQ = BD/QM ————-(4)

In ∆ABD and ∆PQM,

∠B = ∠Q [Using equation (2)]

AB/PQ = BD/QM [using equation (4)]

∴ ∆ABD ∼ ∆PQM (By SAS similarity criterion)

⇒AB/PQ = BD/QM = AD/PM

 

 

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