NCERT Solutions for Class 10 Maths Exercise 6.3

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NCERT Solutions for Class 10 Maths Exercise 6.3 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Maths Triangles Download as PDF

NCERT Solutions for Class 10 Maths Exercise 6.3

NCERT Solutions for Class 10 Maths Triangles

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:

P

Ans. (i)In s ABC and PQR, we observe that,

By AAA criterion of similarity,

(ii) In s ABC and PQR, we observe that,

By SSS criterion of similarity,

(iii) In s LMP and DEF, we observe that, the ratio of the sides of these triangles is not equal.

Therefore, these two triangles are not similar.

(iv) In s MNL and QPR, we observe that,

But,

These two triangles are not similar as they do not satisfy SAS criterion of similarity.

(v) In s ABC and FDE, we have,

But, [ AC is not given]

These two triangles are not similar as they do not satisfy SAS criterion of similarity.

(vi) In s DEF and PQR, we have,

[ ]

And

By AAA criterion of similarity,


NCERT Solutions for Class 10 Maths Exercise 6.3

2. In figure, ODC OBA, BOC = and CDO = Find DOC, DCO and OAB.

Ans. Since BD is a line and OC is a ray on it.

In CDO, we have

It is given that ODC OBA

HenceDOC = DCO = and OAB =


NCERT Solutions for Class 10 Maths Exercise 6.3

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

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

To Prove:

Proof: In s OAB and OCD, we have,

5 = 6 [Vertically opposite angles]

1 = 2 [Alternate angles]

And 3 = 4 [Alternate angles]

By AAA criterion of similarity,OAB ODC

Hence,


NCERT Solutions for Class 10 Maths Exercise 6.3

4. In figure, and 1 = 2. Show that PQS TQR.

Ans. We have,

……….(1)

Also, 1 = 2 [Given]

PR = PQ ……….(2) [Sides opposite to equal s are equal]

From eq.(1) and (2), we get

In s PQS and TQR, we have,

and PQS = TQR = Q

By SAS criterion of similarity,PQS TQR


NCERT Solutions for Class 10 Maths Exercise 6.3

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

Ans. In s RPQ and RTS, we have

RPQ = RTS [Given]

PRQ = TRS [Common]

By AA-criterion of similarity,

RPQ RTS


NCERT Solutions for Class 10 Maths Exercise 6.3

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

Ans. It is given that ABE ACD

AB = AC and AE = AD

……….(1)

In s ADE and ABC, we have,

[from eq.(1)]

And BAC = DAE [Common]

Thus, by SAS criterion of similarity, ADE ABC


NCERT Solutions for Class 10 Maths Exercise 6.3

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

(i) AEP CDP

(ii)ABD CBE

(iii) AEP ADB

(iv) PDC BEC

Ans. (i) In s AEP and CDP, we have,

AEP = CDP = [ CEAB, ADBC]

AndAPE = CPD[ Vertically opposite]

By AA-criterion of similarity, AEP CDP

(ii) In s ABD and CBE, we have,

ADB = CEB =

AndABD = CBE[ Common]

By AA-criterion of similarity, ABD CBE

(iii) In s AEP and ADB, we have,

AEP = ADB = [ ADBC, CEAB]

AndPAE = DAB[ Common]

By AA-criterion of similarity, AEP ADB

(iv) In s PDC and BEC, we have,

PDC = BEC = [ CEAB, ADBC]

AndPCD = BEC[ Common]

By AA-criterion of similarity, PDC BEC


NCERT Solutions for Class 10 Maths Exercise 6.3

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE CFB.

Ans. In s ABE and CFB, we have,

AEB = CBF[Alt. s]

A = C [opp. s of a gm]

By AA-criterion of similarity, we have

ABE CFB


NCERT Solutions for Class 10 Maths Exercise 6.3

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

(i) ABC AMP

(ii)

Ans. (i) In s ABC and AMP, we have,

ABC = AMP = [Given]

BAC = MAP [Common angles]

By AA-criterion of similarity, we have

ABC AMP

(ii) We have ABC AMP [As prove above]


NCERT Solutions for Class 10 Maths Exercise 6.3

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

(i)

(ii) DCB HE

(iii) DCA HGF

Ans. We have, ABC FEG

A = F………(1)

And C = G

C = G

1 = 3 and 2 = 4 ……….(2)

[CD and GH are bisectors of C and G

respectively]

In s DCA and HGF, we have

A = F[From eq.(1)]

2 = 4[From eq.(2)]

By AA-criterion of similarity, we have

DCA HGF

Which proves the (iii) part

We have,DCA HGF

Which proves the (i) part

In s DCA and HGF, we have

1 = 3[From eq.(2)]

B = E[ DCB HE]

Which proves the (ii) part


NCERT Solutions for Class 10 Maths Exercise 6.3

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.

Ans. Here ABC is isosceles with AB = AC

B = C

In s ABD and ECF, we have

ABD = ECF[ B = C]

ABD = ECF = [ ADBC and EFAC]

By AA-criterion of similarity, we have

ABD ECF


NCERT Solutions for Class 10 Maths Exercise 6.3

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.

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

To prove: ABC PQR

Proof: BD = BC [Given]

And QM = QR [Given]

Also [Given]

ABD PQM[By SSS-criterion of similarity]

B = Q[Similar triangles have corresponding angles equal]

And [Given]

By SAS-criterion of similarity, we have

ABC PQR


NCERT Solutions for Class 10 Maths Exercise 6.3

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

ANS. In triangles ABC and DAC,

ADC = BAC [Given]

and C = C[Common]

By AA-similarity criterion,

ABC DAC

= CB.CD


NCERT Solutions for Class 10 Maths Exercise 6.3

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.

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

To prove: ABC PQR

Proof: BD = BC [Given]

And QM = QR and [Given]

ABD PQM[By SSS-criterion of similarity]

B = Q[Similar triangles have corresponding angles equal]

And [Given]

By SAS-criterion of similarity, we have

ABC PQR


NCERT Solutions for Class 10 Maths Exercise 6.3

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.

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

Let DE = meters

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

In the triangles ABC and DEF,

A = D =

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

By AA-similarity criterion,

ABC DEF

= 42 m

Hence, the height of the tower is 42 meters.


NCERT Solutions for Class 10 Maths Exercise 6.3

16. If AD and PM are medians of triangles ABC and PQR respectively, where ABC PQR, prove that

Ans. Given: AD and PM are the medians of triangles

ABC and PQR respectively, where

ABC PQR

To prove:

Proof: In triangles ABD and PQM,

B = Q [Given]

And [ AD and PM are the medians of BC and QR respectively]

By SAS-criterion of similarity,

ABD PQM

=

=

NCERT Solutions for Class 10 Maths Exercise 6.3

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62 thoughts on “NCERT Solutions for Class 10 Maths Exercise 6.3”

  1. The main question in this exercise Q14 is wrongly done.The solution does not match that question but the site has just put up solution of Q12 again.

  2. I’d like to let u know that the solution of Q14 ex-6.3 class 10 maths ncert is absolutely wrong.. Please….. Reread the Q we are ratio of AC/PR instead of ratio of bases that u’ve given

  3. The mistake made by u has totally changed, simplified, but mistakened& wronged the very solution of the word problem.. Plz understand my suggestion so as to help other students avoid any deduction of valuable marks in boards

  4. Hey in Q14 what you did is wrong…please read the question again n then post the answer it is completely different from Q12

  5. There are about 20 – 30 comments complaining about question number 14.
    I want to ask you a question that do you even read the comments???
    Such act was not admirable from a educational site. You shoul correct it.

  6. Guyzz the q14 is right
    This is the property og side and median
    For info. About this property contact me
    At facebook page – Mukul Bhavesh

  7. I didn’t find any mistake in question number 14.Guys first see and understand the question then make a comment.

  8. Question no 10 is also wrong what are you doing ABC ~ TO HE what is this some more mistakes is there in this exercise you did not have any sense of doing questions bhai sahi kar sari mistakes

  9. 14Q is wrong they given AC is proportional to PR but u take BC is proportional to QR which is not mentioned in question

  10. This was the website i trusted the most….Nd now its very disappointing to see the mistake…students are here so that they can understand the sums…nd if here only mistake occurs, wat will the students learn ???

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