NCERT Solutions for Class 10 Maths Exercise 6.3 Chapter 6 Triangles – FREE PDF Download

NCERT Class 10 Maths Ch 6 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
6 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Triangles solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 6- Triangles

1. State which pairs of triangles in the given 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:

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,

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 =

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,

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

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

6. In the given 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

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

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

9. In the given 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]

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

11. In the given 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

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

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

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

$\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M} . . . . . . . . (1)$

To prove: ABC PQR

Proof:

Let us extend AD to point D such that AD = DE and PM up to point L such that PM = ML

Join B to E. C to E,and Q to L, and R to L
We know that medians is the bisector of opposite side
Hence
BD = DC
Also, AD = DE (by construction)

Hence in quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
AC = BE
AB = EC (opposite sides of ||gm are equal ) ………. (2)

Similarly, we can prove that PQLR is a parallelogram
PR = QL
PQ = LR opposite sides of ||gm are equal ) ………. (3)

Given that

$\frac{A B}{P Q} = \frac{A C}{P R} = \frac{A D}{P M}$
$\frac{A B}{P Q} = \frac{B E}{Q L} = \frac{A D}{P M}$ [from (2) (3) ]
$\frac{A B}{P Q} = \frac{B E}{Q L} = \frac{2 A D}{2 P M}$

$\frac{A B}{P Q} = \frac{B E}{Q L} = \frac{A E}{P L}$ $[a s A D = D E, A E = A D + D E = 2 A D P M = M L . P L = P M + M L = 2 P M$

$△ A B E \sim △ P Q L$ ( By SSS Similiarity Criteria)

We know that corresponding angles of similar triangles are equal.
$∠ B A E = ∠ Q P L$ (4)
Similarly, we can prove that $△ A E C \sim △ P L R$ .

We know that corresponding angles of similar triangles are equal.
$∠ C A E = ∠ R P L$ (5)
Adding (4) and (5),
$∠ B A E + ∠ C A E = ∠ Q P L + ∠ R P L$
$∠ C A B = ∠ R P Q I n △ A B C a n d △ P Q R,$

$\frac{A B}{P Q} = \frac{A C}{P R}$
$∠ C A B = ∠ R P Q$