#### Chapter 10 Congruent Triangles R.D. Sharma Solutions for Class 9th MCQ's

**Multiple Choice Questions**

Q1. If âˆ†ABC â‰…âˆ†LKM, then side of âˆ†LKM equal to side AC of âˆ†ABC is

(a) LK

(b) KM

(c) LM

(d) None of these

**Solution**

Corresponding sides of congruent triangles are equal. The similarity is written accordingly. That means If âˆ†ABC â‰…âˆ†LKM, then corresponding equal sides will be

AB = LK

BC = KM

And, hence,Â

AC= LM

So, LM is correct answer, optionÂ ( c) is correct.

Q2. If âˆ†ABCâ‰… âˆ†ACB, then âˆ†ABC is isosceles withÂ

(a) AB=AC

(b) AB=BC

(c) AC=BC

(d) None of these

**Solution**

Corresponding sides of congruent triangles are equal. The similarity is written accordingly.Â

That means, If âˆ†ABC â‰…âˆ†ACB, then corresponding equal side of AB is AC

AB=AC

So, option (a) is correct.Â

Q3. If âˆ†ABCâ‰… âˆ†PQR andÂ âˆ†ABC is not congruent to âˆ†RPQ, which of the following is not true

(a) BC=PQ

(b) AC=PR

(c) AB=PQ

(d) QR=RC

**Solution**

Since,

âˆ†ABCâ‰… âˆ†PQR

So,

Only corresponding parts are needed to be equal

So,

AB=PQ

AC=PR

BC=RQ

So, here, Option (c) AB=PQ is true

(b) AC=PR

(d) QR=BC are true.Â

(b) AC=PR

(d) QR=BC are true.Â

Also, it is given thatÂ âˆ† ABC is not congruent to âˆ†RPQ.

So,Â BCâ‰ PQ

So,Â BCâ‰ PQ

So, Option (a) is correct answer.Â BC=PQÂ is not true.Â

Q4. In triangles ABC and PQR three equality relations between some parts are as follows.

AB=PQ, âˆ B= âˆ P and BC=PR

State whichÂ congruence condition applies

(a) SAS

(b) ASA

(c) SSS

(d) RHS

**Solution**

Here are two triangles ABC and PQR

It is given

AB=PQ

âˆ B= âˆ PÂ

BC=PR

So, two sides are equal, and angle between these two equal sides are equal, therefore, both the triangles are concurrent by Side Angle Side (SAS) Criteria. Here, option (A) is correct.Â

Q5. In triangle ABC and PQR, if âˆ A=âˆ R, âˆ B=âˆ P and AB=RP. Then which of the following congruence condition applies:Â

(a)SASÂ

(b) ASAÂ

(c) SSSÂ

(d) RHS

(b) ASAÂ

(c) SSSÂ

(d) RHS

**Solution**

âˆ A=âˆ R, âˆ B=âˆ P and AB=RP

So, Two corresponding angles of both triangles are equal, and side between these two triangles is also equal, therefore, both triangles are concurrent and condition is Angle side Angle (ASA).Â Thus, option (b) is correct.Â

Q6. If âˆ†PQRâ‰…âˆ†EFD, then ED=

(a) PQÂ

(b) QRÂ

(c) PRÂ

(d) None of these

(b) QRÂ

(c) PRÂ

(d) None of these

**Solution**

Since,

âˆ†PQRâ‰…âˆ†EFD,

Corresponding sides will be equal.Â

PQ=EF

QR= FD

PR=ED

So, ED=PR as corresponding sides of concurrent triangles are equal

So, here, option (c) is true.Â

Q7. If âˆ†PQRâ‰…âˆ†EFD, then âˆ E=

(a) âˆ PÂ

(b) âˆ QÂ Â

(c) âˆ RÂ Â

(d) None of these

(b) âˆ QÂ Â

(c) âˆ RÂ Â

(d) None of these

**Solution**

Since,Â

âˆ†PQRâ‰…âˆ†EFD

Corresponding Angles are equal

âˆ P=âˆ E

Â Since, Angle E is corresponding part of angle P.Â

So, here, option (a) is correct

Q8.Â In a âˆ† ABC, If AB=AC and BC is produced to D such that âˆ ACD=100Â° , then âˆ A=Â

(a) 20Â°

(b) 40Â°Â

(c) 60Â°

(d) 80Â°

(b) 40Â°Â

(c) 60Â°

(d) 80Â°

**Solution**

In âˆ†ABC, If AB=AC

Then, angles opposite to these sides will also be equal.

âˆ B=âˆ C = x (Let)

The Side BC of the triangle ABC is extended to D.Â

BCD is a straight line. So, âˆ ACB+âˆ ACD=180Â°

So, x+100Â°=180Â° ( From the diagram) (It is given ACD=100Â° and we have assumed âˆ ACB=100Â°.Â

x=180Â°-100Â°

â‡’x=80Â°

That means âˆ B=âˆ C = x= 80Â°

Since, Sum of all angles of triangle is 180Â°.Â

Therefore,Â

âˆ A+âˆ B+âˆ C=180Â°

â‡’ âˆ A+80Â°+80Â°=180Â°

â‡’âˆ A+160Â°=180Â°

â‡’âˆ A=180Â°-160Â°

â‡’âˆ A=20Â°

So, âˆ A=20Â° , hence, option (a) is correct.Â

Q9. In an Isosceles triangle, If the vertex angle is twice the sum of the base angles, then the vertex measure of the vertex angle of the triangle .

(a) 100Â°

(b) 120Â°

(c) 110Â°

(d) 130Â°

(b) 120Â°

(c) 110Â°

(d) 130Â°

**Solution**

Let the bases angles be x, then according to the question,Â the vertex of angle according to the question is twice the sum of base angles

So, vertex angle is 2(x+x) = 2Ã—2x = 4x

Sum of all angles will be 180Â°

So, x+x+4x=180Â°

â‡’ 6x=180Â°

â‡’ x=180Â°/6

â‡’ x=30Â°

VertexÂ Angle 4x = 4Ã—30Â°=120Â°

Q10. D, E, F are the mid-points of the sides BC, CA and AB respectively of âˆ†ABC. Then âˆ†DEF is congruentÂ triangle.Â

(a) ABCÂ

(b) AEFÂ Â

(c) BFD, CDEÂ Â

(d) AEF, BFD, CDE

(b) AEFÂ Â

(c) BFD, CDEÂ Â

(d) AEF, BFD, CDE

**Solution**

All, AEF, BFD, CDE are concurrent to each other.Â Â

Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it.Â

DE = 1/2 ACÂ â‡’ DE = AF = CFÂ

EF = 1/2 ABÂ Â â‡’ EF = AD = BD

DF =Â Â Â 1/2 BCÂ â‡’ DF = BE = CEÂ

DE = 1/2 ACÂ â‡’ DE = AF = CFÂ

EF = 1/2 ABÂ Â â‡’ EF = AD = BD

DF =Â Â Â 1/2 BCÂ â‡’ DF = BE = CEÂ

InÂ â–³DEF andÂ â–³CFE , we haveÂ

DE = CF [from (i)]

DF = CE [from (ii)]

EF = FE [common]Â

â‡’Â â–³DEFÂ â‰…Â â–³CFE [by SSS criterion of congruence ]

DE = CF [from (i)]

DF = CE [from (ii)]

EF = FE [common]Â

â‡’Â â–³DEFÂ â‰…Â â–³CFE [by SSS criterion of congruence ]

Similarly, we haveÂ â–³DEFÂ â‰…Â Â â–³BDE andÂ â–³DEFÂ â‰…Â â–³AFDÂ

Thus, â–³DEFÂ â‰…Â Â â–³CFEÂ â‰…Â â–³BDEÂ Â â‰…Â â–³AFDÂ Â

Hence, â–³ABC is divided into four congruent triangles

Thus, â–³DEFÂ â‰…Â Â â–³CFEÂ â‰…Â â–³BDEÂ Â â‰…Â â–³AFDÂ Â

Hence, â–³ABC is divided into four congruent triangles

Q11.Â Which of the following is not a criterion for congruence of triangles?

(a) SASÂ

(b) SSA

(c) ASA

(d) SSS

(b) SSA

(c) ASA

(d) SSS

**SolutionÂ**

SAS, ASA and SSS are criterion for congruence of triangle. SSA is not a criterion. It is the criterion, but in special condition, that is the RHS criteria, SSA can be the condition. In general it is not. So, option (b) is the correct Answer.

Q12. In fig 10.135, the measure of âˆ Bâ€™Aâ€™Câ€™ is

(a) 50Â°

(b) 60Â°

(c) 70Â°

(d) 80Â°

(b) 60Â°

(c) 70Â°

(d) 80Â°

**Solution**

From the figure

AB=Aâ€™Bâ€™=6cm

BC=Bâ€™Câ€™ =5cm

âˆ B=âˆ Bâ€™=60Â°

So, By SAS Criteria,Â

âˆ†ABCâ‰…âˆ†A'B'C'

So,

âˆ A=âˆ Aâ€™

â‡’3x=2x+20

â‡’3x-2x=20

â‡’x=20

We have to findÂ

âˆ Aâ€™Bâ€™Câ€™=2x+20

=2Ã—20+20=60

So, âˆ Aâ€™Bâ€™Câ€™=60Â°

Hence, here, option (b) is correct.

Q13. If ABC and DEF are two triangles such thatÂ

âˆ†ABCâ‰…âˆ†FDE and AB=5cm, âˆ B=40Â° and âˆ A=80Â°. Then which of the following is true?

DF=5cm, âˆ F=60Â°

DE=5cm, âˆ E=60Â°

DF=5cm, âˆ E=60Â°

DE=5cm, âˆ D=40Â°

**Solution**

The above diagram, is according to question.Â

Since, Sum of all angles is 180Â°

âˆ A+âˆ B+âˆ C=180Â°

â‡’80Â°+40Â°+âˆ C=180Â°

â‡’120Â°+âˆ C=180Â°

â‡’âˆ C=180Â°-120Â°=60Â°

Now, Since, âˆ†ABCâ‰…âˆ†FDE

Corresponding sides and angles will be equalÂ

That means

DF = AB = 5cm

âˆ F=âˆ A=80Â°

âˆ E=âˆ C=60Â°

So, DF= 5cm, and âˆ E=60Â°

That is option (c) here. So, option (c)Â DF=5cm, âˆ E=60Â° isÂ correct.Â

Q14. In figure, 10. 136, AB âŸ‚BE and FE âŸ‚BE.Â If BC=DE and AB=EF, then âˆ†ABD is congruent toÂ

(a) âˆ†EFC

(b) âˆ†ECF

(c) âˆ† CEFÂ

(d) âˆ†FEC

(b) âˆ†ECF

(c) âˆ† CEFÂ

(d) âˆ†FEC

**Solution**Â

From the diagramÂ

BC=DE

BC+CD=DE+CDÂ

BD=EC

Also, it is given AB âŸ‚BE and FE âŸ‚BE

That makesÂ

âˆ ABD=90Â°

âˆ FEC=90Â°

Now consider triangles ABD and FCEÂ

AB=FE (given in question)

âˆ ABD=âˆ FEC=90Â°

BD=EC

So, by SAS criteria, âˆ†ABDâ‰…âˆ†FEC

So, here, âˆ†ABD is congruent with âˆ†FEC.

So, here Option (D) is correct.Â

Q15.Â ABC is an Isosceles triangle Such that AB=AC and AD is median to base BC. Then âˆ BAD=

(a) 55Â°

(b) 70Â°

(c) 35Â°

(d) 110Â°

(b) 70Â°

(c) 35Â°

(d) 110Â°

**Solution**

For Isosceles triangles, the median of unequal side, is perpendicular to the side.Â

So, AD is perpendicular to side AC

So,Â

âˆ ADB=âˆ 90Â°

In triangle ABD,Â

âˆ ADB+âˆ ABD+âˆ BAD=180Â°

â‡’90Â°+35Â°+âˆ BAD=180Â°

â‡’125Â°+âˆ BAD=180Â°

â‡’âˆ BAD=180Â°-125Â°=55Â°

So, âˆ BAD=55Â°, hence, here option (a) is correct.Â

Q16. In fig 10.138, if AE II DC, and AB=AC, the value of âˆ ABD is

(a) 70Â°Â

(b)Â 110Â°Â

(c) 120Â°Â

(d) 130Â°

(b)Â 110Â°Â

(c) 120Â°Â

(d) 130Â°

**Solution**

Since, AEâˆ¥DC

We can say,

âˆ ACB=âˆ 1=70Â° (corresponding Angle)

Since,

AB=AC

Triangle ABC is isosceles triangle.

So, angles opposite to equal side will be equalÂ

âˆ ABC=âˆ ACB=70Â°

Since, CD is straight line, âˆ ABC and âˆ ABD are linear pair

Therefore,

âˆ ABC+âˆ ABD=180Â°

70Â°+âˆ ABD=180Â°

âˆ ABD=180Â°-70Â°=110Â°

So, âˆ ABD=110Â° , hence, here option B is correct

Q 17. In fig 10.139, ABC is isosceles triangle whose sides AC is produced to E. Through C, CD is drawn parallel to BA. The value of x

**Solution**

Since, opposite angles of equal sides of the isosceles triangle are equal,Â therefore

âˆ ACB=âˆ ABC=52Â°

Â âˆ ACB+âˆ ABC+âˆ BAC=180Â°

â‡’ 52Â°+52Â°+âˆ BAC=180Â°

â‡’ 104Â°+âˆ BAC=180Â°

âˆ BAC=180Â°-104Â°

âˆ BAC=76Â°

Since, AB II CD

âˆ ACD=âˆ BAC=76Â°

Since, âˆ ACD and âˆ CDE are linear pair.

âˆ ACD+âˆ CDE=180Â°Â

76Â°+xÂ°=180Â°Â (from diagram, âˆ CDE=xÂ°)

â‡’ x =180Â°-76Â°=104Â°

So, x=104, hence, here option (d) is correct.Â

Q18. In figure, 10.140, X is a point in the interior of the square ABCD. AXYZ is also a square. If DY=3cm, and AZ=2cm, then BY=

(a) 5cm

(b) 6cm

(c) 7cm

(d) 8cm

(b) 6cm

(c) 7cm

(d) 8cm

**Solution**

Since, sides of square are equalÂ

ZY=YX=XA=AZ=2cm

Consider âˆ†DAZ and âˆ†BAX

AD = BA (sides of Square âˆ†BCD)

AZ=AX (Sides of Square âˆ†XYZ)

âˆ DZA=âˆ BXA=90Â° (Angles of square âˆ†XYZ)

So, by RHS Criteria, âˆ†DAZâ‰…âˆ†BAX

So, BX=DZÂ (c.p.c.t)

Also,

From the diagram

BX=DY+YZ

â‡’ BX=3cm+2cm = 5cm

â‡’ BX=5cm

Q19. In fig 10.141, ABC is triangle in which âˆ B=2âˆ C. D is a point on Side BC such that AD bisects âˆ BAC and AB=CD. BE is bisector of âˆ B. The measure of âˆ BAC isÂ

(a) 72Â°

(b) 73Â°

(c) 74Â°

(d) 95Â°

(b) 73Â°

(c) 74Â°

(d) 95Â°

**Solution**

âˆ B=2âˆ C

â‡’âˆ B/2 = âˆ C

Also,

BE bisects âˆ ABC

So, âˆ ABC /2 = âˆ ABE = âˆ EBDÂ

So, finally

âˆ ABE = âˆ EBC = âˆ C =âˆ ABC/2 = x (Let)

Or

âˆ ABC = 2x

Also,

AD bisects âˆ BAC

âˆ BAD= âˆ DAC = âˆ BAC/2Â =y (Let)

OrÂ

âˆ A =2y

Considering âˆ†BEC

âˆ EBC = âˆ ECB

So, BE= EC

Also, It is given AB=CDÂ

So, considering âˆ†ABE and âˆ†DCE,

BE=EC

AB=CD

âˆ ABE = âˆ DCE =x (from diagram)

So,

âˆ†ABEâ‰… âˆ†DCE (SAS Criteria)

So,Â

âˆ EDC = âˆ BAE = 2xÂ (c.p.c.t)

AE = DE (c.p.c.t)

In âˆ†AED

Since, AE = DE

âˆ EDA= âˆ EAD = y (angles equal to opposite side)

From diagramÂ

âˆ ADC = âˆ ADE+âˆ EDC

Â â‡’âˆ ADC = y+2y = 3y

Now, in triangle ADC

âˆ DAC+âˆ ADC+âˆ ACD =180Â°

â‡’ y+3y+x=180Â°

â‡’ x+4y=180Â° --- (i)

Also, in triangle ABC

âˆ A+âˆ B+âˆ C = 180Â°

â‡’ 2y+2x+x =180Â°

â‡’ 2y+3x=180Â° --- (ii)

So, we have two equations

x+4y =180Â° (i)

2y+3x=180Â° (ii)

Multiplying eq (i) by 3 and subtracting (ii) from that

3x+12y= 3(180Â°)

â‡’ -3x-2y=-180Â°

â‡’ 10y=2Ã—180Â°

â‡’ y=(2Ã—180Â°)/10 = 36Â°

We have to find angle âˆ BAC=2y = 2Ã—36Â°=72Â°

So, here, option (a) is correct.

Q20. In fig 10.142, If AC is bisector of âˆ BAD, Such that AB=3cm and AC=5cm, then CD=

(a) 2cmÂ

(b) 3cmÂ

(c) 4cmÂ

(d) 5cm

(b) 3cmÂ

(c) 4cmÂ

(d) 5cm

**Solution**

Since,Â

AC bisects âˆ BAD, thenÂ

âˆ CBA = âˆ CADÂ

Also,Â

âˆ ABC = âˆ ADC=90Â°

So, Considering âˆ† ABC and âˆ†ADC

âˆ CBA = âˆ CADÂ

âˆ ABC = âˆ ADC

AC (common Side)

âˆ†ABC â‰… âˆ†ADC

AB = AD = 3cm (c.p.c.t)

Now, In âˆ†ADC

âˆ ADC=90Â°

AD = 3cm

AB=5cm

We can use Pythagoras Theorem to findÂ Side AC

AC

^{2}Â = AB^{2}-AD^{2}
â‡’ AC = âˆš(AB

^{2}-AD^{2})
â‡’ AC=âˆš(5

^{2}-3^{2}) cm
â‡’ AC =âˆš(25-9)cm

â‡’ AC=âˆš16cm = 4cm

So, here, option (c) is correct