NCERT Solutions for Class 10 Maths Ex 6.3

# NCERT Solutions for Class 10 Maths Ex 6.3

## NCERT Solutions for Class 10 Maths Ex 6.3

1. State which pairs of triangles in 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:

Solution: (i) In ∆ABC and ∆PQR, we observe that, A = P = 60°, B = Q = 80° and C = R = 40°

Therefore, by AAA criterion of similarity, ∆ABC ~ ∆PQR.

(ii) In ∆ABC and ∆PQR, we observe that, AB/QR = BC/RP = AC/PQ = ½

Therefore, by SSS criterion of similarity, ∆ABC ~ ∆PQR.

(iii) In ∆LMP and ∆DEF, we observe that, the ratios of the sides of these triangles are not equal. Therefore, these two triangles are not similar.

(iv) In ∆MNL and ∆QPR, we observe that, M = Q = 70°

And MN/PQ = ML/QR = 1/2

Therefore, by SAS criterion of similarity, ∆MNL ~ ∆QPR.

(v) In ∆ABC and ∆FDE, we observe that, A = F = 80°

But, AB/DF ≠ AC/EF        [  AC is not given]

Therefore, these two triangles are not similar as they do not satisfy SAS criterion of similarity.

(vi) In DEF and PQR, we have, D = P = 70°   [Since P = 180° 80° 30° = 70°]

And E = Q = 80°

Therefore, by AAA criterion of similarity, ∆DEF ~ ∆PQR.

2. In figure, ODC ~ OBA, BOC = 125° and CDO = 70°. Find DOC, DCO and OAB.

Solution: Since BD is a straight line and OC is a ray on it.

Therefore, DOC + BOC = 180°

DOC + 125° = 180°

DOC = 55°

In CDO, we have, CDO + DOC + DCO = 180°

70° + 55° + DCO = 180°

DCO = 55°

It is given that ODC ~ OBA.

Therefore, OBA = ODC, OAB = OCD

OBA = 70°, OAB = 55°

Hence, DOC = 55°, DCO = 55° and OAB = 55°

3. Diagonals AC and BD of a trapezium ABCD with AB CD intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD.

Solution: Given: ABCD is a trapezium in which AB DC.

To ProveOA/OC = OB/OD

Proof: In OAB and OCD, we have, 5 = 6   [Vertically opposite angles]

1 = 2      [Alternate angles]

And 3 = 4     [Alternate angles]

Therefore, by AAA criterion of similarity, OAB ~ OCD.

Hence, OA/OC = OB/OD

4. In figure, QR/QS = QT/PR and 1 = 2. Show that PQS ~ TQR.

Solution: We have, QR/QS = QT/PR

QT/QR = PR/QS  ………. (i)

Also, 1 = 2   [Given]

Therefore, PR = PQ   ………. (ii) [ Since sides opposite to equal angles are equal]

From equations (i) and (ii), we get

QT/QR = PR/QS  QT/QR = PQ/QS

PQ/QT = QS/QR

In ∆PQS and ∆TQR, we have,

PQ/QT = QS/QR and PQS = TQR = Q

Therefore, by SAS criterion of similarity, PQS ~ TQR.

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

Solution:

In ∆RPQ and ∆RTS, we have RPQ = RTS [Given]

PRQ = TRS [Common]

Therefore, by AAA criterion of similarity, RPQ ~ ∆RTS.

6. In figure, if ABE ACD, show that ADE ~ ∆ABC.

Solution: It is given that ABE ∆ACD

Therefore, AB = AC and AE = AD

Therefore, in ∆ADE and ∆ABC, we have

AB/AC = AD/AE         [From equation (i)]

And BAC = DAE     [Common]

Thus, by SAS criterion of similarity, ADE ~ ABC.

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

(i) AEP ~ CDP

(ii)ABD ~ CBE

(iv) PDC ~ BEC

Solution: (i) In AEP and CDP, we have

AEP = CDP = 90°  [Since CE AB and AD BC]

And APE = CPD    [Vertically opposite angles]

Therefore, by AAA criterion of similarity, AEP ~ CDP.

(ii) In ABD and CBE, we have

And ABD = CBE      [Common]

Therefore, by AAA criterion of similarity, ABD ~ CBE.

(iii) In AEP and ADB, we have

And PAE = DAB         [Common]

Therefore, by AAA criterion of similarity, AEP ~ ADB.

(iv) In ∆PDC and BEC, we have

PDC = BEC = 90°       [Since CE AB, AD BC]

And PCD = BCE         [Common]

Therefore, by AAA criterion of similarity, 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:

In ∆ABE and ∆CFB, we have

AEB = CBF        [Alternate angles]

A = C                [Opposite angles of a parallelogram]

Therefore, by AAA criterion of similarity, we have ABE ~ CFB.

9. In figure, 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) In ∆ABC and ∆AMP, we have

ABC = AMP = 90°        [Given]

BAC = MAP                 [Common angles]

Therefore, by AAA criterion of similarity, we have ABC ~ AMP.

(ii) We have ABC ~ AMP        [As proved above]

CA/PA = BC/MP

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:

(i) We have, ABC ~ FEG

A = F        ……… (i)

And C = G

½ C = ½ G

1 = 3 and 2 = 4       ………. (ii)

[Since CD and GH are bisectors of C and G respectively]

In ∆DCA and ∆HGF, we have

A = F       [From equation (i)]

2 = 4           [From equation (ii)]

Therefore, by AAA criterion of similarity, we have DCA ~ ∆HGF.

Therefore, CD/GH = AC/FG

(ii) In ∆DCB and ∆HGE, we have

B = E          [Since ABC ~ FEG]

1 = 3           [From equation (ii)]

Therefore, by AAA criterion of similarity, we have DCB ~ ∆HGE.

(iii) In part (i), we have proved that DCA ~ ∆HGF.

11. In 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: Here ABC is an isosceles triangle with AB = AC.

Therefore, B = C

In ∆ABD and ∆ECF, we have

ABD = ECF          [Since B = C]

Therefore, by AAA criterion of similarity, we have ABD ~ ∆ECF.

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of a PQR (see figure). Show that ABC ~ ∆PQR.

Solution: Given: AD is the median of  ∆ABC and PM is the median of ∆PQR such that

To proveABC ~ ∆PQR

Proof: BD = ½ BC    [Given]

And QM = ½ QR      [Given]

Also AB/PQ = BC/QR = AD/PM       [Given]

Therefore, ∆ABD ~ ∆PQM                   [By SSS criterion of similarity]

B = Q     [Similar triangles have corresponding angles equal]

And AB/PQ = BC/QR    [Given]

Therefore, by SAS criterion of similarity, we have ABC ~ ∆PQR.

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

Solution:

In ABC and DAC, we have

And C = C        [Common]

Therefore, by AAA criterion of similarity, ABC ~ ∆DAC

AB/DA = BC/AC = AC/DC

BC/AC = AC/DC

Or, CB/CA = CA/CD

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 ABC ~ ∆PQR.

Solution:

Given: AD is the median of ABC and PM is the median of PQR such that

To proveABC ~ PQR

Construction: Draw DE ∥ AC and MS ∥ PR

Proof: In ABC, D is the mid-point of BC      [Given]

Therefore, E is the mid-point of AB.              [By converse of mid-point theorem]

DE = ½ AC and SM = ½ PR

Similarly, AE = ½ AB and PS = ½ PQ

Now, AB/PQ = AC/PR = AD/PM                  [Given]

Therefore, ADE ~ ∆PMS           [By SSS criterion of similarity]

Thus, DAE = MPS

Similarly, DAC = MPR

DAE + DAC = MPS + MPR

BAC = QPR

Now, in ABC and ∆PQR,

AB/PQ = AC/PR and A = P

Therefore, ABC ~ ∆PQR           [By SAS criterion of similarity]

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:  Let AB be the vertical pole and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF.

Let DE = x metres

Here, AB = 6 m, AC = 4 m and DF = 28 m

In the ∆ABC and ∆DEF, A = D = 90°

And C = F       [Each is the angular elevation of the sun at the same time]

Therefore, ∆ABC ~ ∆DEF           [By AAA criterion of similarity]

AB/DE = AC/DF

6/x = 4/28

6/x = 1/7

x = 42 m

Hence, the height of the tower is 42 metres.

16. If AD and PM are medians of triangles ABC and PQR respectively, where ABC ~ PQR, prove that AB/PQ = AD/PM.

Solution:

Given: AD and PM are the medians of ABC and PQR respectively, where ABC and PQR.

Proof: In ABC and PQR,

B = Q [Given]

BD = ½ BC and QM = ½ QR

Therefore, AB/PQ = BC/QR          [Since ABC ~ PQR]

AB/PQ = 2BD/2QM

AB/PQ = BD/QM

Therefore, by SAS criterion of similarity, ABD ~ ∆PQM.