## In this post, you will find the NCERT solutions for class 10 maths ex 6.3. These solutions are based on the latest syllabus of NCERT Maths class 10.

**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 [

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 Prove**: OA/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

⇒ AB/AD = AC/AE

⇒ AB/AC = AD/AE ………. (i)

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**

**(iii) **∆**AEP **~ ∆**ADB**

**(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

∠ADB = ∠CEB = 90° [Since AD
⊥ BC, CE ⊥ AB]

And
∠ABD = ∠CBE
[Common]

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

**(iii)** In ∆AEP
and ∆ADB, we have

∠AEP = ∠ADB = 90° [Since AD
⊥ BC, CE ⊥ AB]

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]

∠ADB = ∠EFC = 90° [Since AD
⊥ BC and EF ⊥ AC]

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

AB/PQ
= BC/QR = AD/PM

**To prove**: ∆ABC
~ ∆PQR

**Proof**: BD = ½ BC [Given]

And
QM = ½ QR [Given]

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

⇒ AB/PQ = 2BD/2QM = AD/PM

⇒ AB/PQ = BD/QM = AD/PM

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 CA ^{2} = CB.CD.**

**Solution: **

In
∆ABC and ∆DAC,
we have

∠ADC = ∠BAC [Given]

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

⇒ CA^{2}
= 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

AB/PQ
= AC/PR = AD/PM

**To prove**: ∆ABC ~ ∆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]

⇒ 2AE/2PS = 2DE/2SM = AD/PM

⇒ AE/PS = DE/SM = AD/PM

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.

**To prove**: AB/PQ
= AD/PM

**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.

⇒ AB/PQ = BD/QM = AD/PM

⇒ AB/PQ = AD/PM Hence proved.