Pioneer Education {The Best Way To Success} IIT JEE ... AD is the bisector of A ? 1 = 2 ..(i) Since exterior angle of a triangle is greater than each of interior opposite

Pioneer Education {The Best Way To Success} IIT JEE /AIPMT/NTSE/Olympiads Classes

www.pioneermathematics.com S.C.O. - 326, Sector 40D, CHD. Phone: 9815527721, 4617721 3

Revision Question Bank Triangles 1. In the given figure, find the values of x and y.

Solution.

Since, AB = AC

C =B

[angles opposite to the equal sides are equal]

x = 50

Also, the sum of all angles of a triangle is 180.

50 +x + y=180

50 + 50 + y = 180 [put x = 50]

y= 180 100 = 80

Hence, x = 50 and y = 80

2. In the given figure, PQ > PR, QS and RS are the bisectors of Q and R, respectively.

Then, prove that SQ > SR

Solution.

Given, PQ > PR

PRQ > PQR

[angle opposite to the longer angle is longer]

1

2 PRO >

1

2PQR

[dividing both sides by 2]

SRQ > SQR

SQ > SR [side opposite to the longer angle is longer]



3. In the given figure, BA AC,DE DF such that BA=DE and BF = EC. Show that

ABC DEF

Solution.

In ABC andDEF,

BA =DE [given]

BF = EC [given]

BF + FC = EC + CF

[adding both sides by FC]

BC = EF

Also, CAB = FDF [each 90]

ABC DEF [by RHS congruence rule]

4. In figure, AB = DE, BC=EF and median AP = median DQ. Prove that B = E.

Solution.

Given ABC and DEF in which AB = DE, BC = EF and median AP = median DQ.

To prove B = E

Proof BC = EF [given]

1 1BC EF

2 2 [dividing both sides by 2]

BP = EQ

[ AP and DQ are medians]

In ABP and DEQ,

AB = DE [given]

AP=DQ [given]

and BP = EQ [proved above]



ABP DEQ

[by SSS congruence rule]

B =E [by CPCT]

Hence proved.

5. Show that a median of a triangle divides it into two triangles of equal areas.

Solution.

Given ABC in which AD is the median i.e., BD = CD

To prove ar(ABD) = ar(ACD)

Construction Draw AE BC

Proof ar(ABD)1

2 BD AE ...(i)

and ar (ACD) = 1

2 CD AE

= 1

2 BD AE ...(ii)

[ CD = BD, given]

From Eqs. (i) and (ii), we get

ar(ABD) = ar(ACD)

Hence proved.

6. In the given figure, XYZ and PYZ are two isosceles triangles on the base YZ with

XY = XZ and PY= PZ. If P=120 and XYP=40, then find YXZ.

Solution.

Given, in PXZ, PY = PZ

PZY = PYZ [opposite angles of equal sides are equal]



Let PYZ = PZY =x

P + PYZ + PZY = 180

[ the sum of all angles of a triangle is 180]

120+x + x = 180

2x = 180 120

2x = 60

x = 30

PYZ = PZY = 300

Now, XYZ = XYP + PYZ

= 40 + 30 = 70

In XYZ, YXZ + XYZ + XZY = 180

[ sum of all angles of a triangle is 180]

YXZ + 70 + 70 = 180

YXZ = 180 140 = 40

7. In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is

double the smallest side.

Solution.

We have, AABC in which B is right angle and C is double A.

Produce CB upto D such that BC = BD and join AD. In ABD and ABC,

BD = BC [by construction]

ABD = ABC [each 90]

and AB = AB [common side]

ABD ABC [ by SAS Congruence rule]

AD = AC [by CPCT]

and DAB = CAB = x [CPCT]



Now, DAC = DAB +CAB

DAC = x + x = 2x

DAC = ACB [ ACB = 2x]

DC = AD

[ sides opposite to equal angles are equal]

2BC = AD = AC [ BC = DB]

Hence proved.

8. In the given figure, RS = QT and QS=RT. Prove that PQ =PR.

Solution.

Given In PQR, RS = QT and QS = RT

To prove PQ = PR

Proof In RESQ and QTR,

RS = QT .. (i) [given]

SQ = RT ..(ii) [given]

and RQ = RQ ...(iii) [common side]

RSQ QTR

[by SSS congruence rule]

QRS = RQT ...(i)

[by CPCT]

and QRT = RQS [by CPCT] ...(ii)

On subtracting Eq. (ii) from Eq. (i), we get

QRS QRT = RQT RQS

TRS = SQT

Also, RSQ =QTR

On multiplying both sides by -1 and then adding 180, we get

180 RSQ =180 QTR

QSP = RTP

RPT QPS [by ASA congruence rule]

PQ = PR Hence proved.

9. ABC is a triangle in which B = 2C. D is a point on side BC such that AD bisects BAC

and AB = CD. Prove that BAC=72.

Solution.



Given B = 2C

AD is the bisector of A and AB = DC

To prove BAC =72

Construction Draw BE as angle bisector of B.

Join DE. If C = x, then

ABE = EBC = x

Proof Since, AD is the bisector of A.

BAD = CAD = y [let]

Now, EBC is. an isosceles triangle.

So, EBC = ECB= x EB = EC

Consider ABE DCE [by SAS congruence rule]

EDC = 2y [by CPCT] ...(i)

Consider isosceles

AED as EA = ED

EA = ED [by CPCT]

LDA = EAD = y ... (ii)

From Eqs. (i) and (ii), we get

ADC = y + 2y =3y ... (iii)

or ADC = 2x + y ... (iv)

[exterior angles of AABD ]

From Eqs. (iii) and (iv), we get

3y = 2x + y

x = y ...(v)

Now, in ABC,

A + B +C=180

[ sum of all angles of a triangle is 180)

2y + 2x + x =180

2y + 2y + y = 180 [from Eq. (v)]

5y =180

0

0180y 365



A = 2y = 2 36 = 72

Hence, BAC = 72 Hence proved.

10. In the given figure, if E >A, C > D, then AD > EC. Is it true?

Solution.

True, In AEB, E > A

AB > BE ...(i)

In BDC, C > D

BD > BC ...(ii)

On adding Eqs. (i) and (ii), we get

AB + BD > BE + BC AD > EC

Chapter Test {Triangles}

M: Marks: 40 M: Time: 40

1. BE and CF are two equal altitudes of a ABC . Using RHS congruence rule, prove that

ABC is an isosceles triangle. [4]

Solution.

Given In a ABC, BE and CF are two equal altitudes of a ABC.

To prove ABC is an isosceles triangle.



Proof In ABC and CFB,

BE=CF

[given, equal altitudes]

BEC = CFB [each 90]

and BC = BC [common side]

BEC CFB [by RHS congruence rule]

ECB = FBC [by CPCT]

or ACB = ABC Hence proved.

2. In figure, S is any point on the side QR of PQR. Prove that PQ + QR+ RP>2PS. [4]

Solution.

We know that, sum of the two sides of a triangle is greater than the third side.

In PQS,

PQ + QS>PS ...(i)

and in PRS,

PR + SR>PS ...(ii)


(PQ + QS) + (PR + SR)> 2PS

PQ + (QS + SR) +PR > 2PS

PQ + QR + RP>2PS

Hence proved.

3. In figure, AB = AC, D is a point in the interior of ABC such that DBC = DCB.

Prove that AD bisects BAC of ABC. [4]

Solution.



In BDC, we have

DBC = DCB [Given]

DC = DB ..(i) Sides opposite to equal

angles of DBC are equal

Now, in ABD and ACD, we have

AB = AC [Given]

BD = CD [From (i)]

and, AD = AD [Common side]

So, by SSS congruency criterion, we have

ABD ACD

BAD = CAD [c.p.c.t.]

Hence, AD is the bisector of BAC.

4. In a ABC, AB = AC and AD is bisector of BAC.

Anju has to prove that:ABD ACD, which as follows:

Here, AB = AC [given]

BAD = CAD [given]

and B = C [corresponding angles of a equal side are equal)

ABD ACD

[by ASA congruence rule]

Further Anju shows this proof to his classmate Amita and she find that there is some

errors in the proof.

(i) Write the correct proof. (ii) What is the mistake in Anju's proof?

(iii) Which value is depicted from this action? [4]

Solution.

(i) Given In ABC, AB = AC and AD is bisector of BAC.

To prove ABD ACD

Proof In ABD and ACD

AB = AC [given]

BAD = CAD [given]

and AD = AD [common side]

ABD ACD [by SAS congruence rule]

(ii) Anju has used the result B = C for proving ABD = ACD, which is

wrong.

Because for proving B = C, firstly we prove that ABD ACD.

(iii) The value depicted from this action, it is cooperative and learning among



students without any religion bias.

5. In the adjoining figure, ABC is a triangle and D is any point in its interior. Show that

BD + DC < AB + AC. [4]

Produced BD to meet AC in E.

In ABE, AB + AE>BE

[ sum of two sides is greater than the third side]

AB+AE > BD + DE ...(i)

[ BE =BD + DE]

In CDE, DE + EC > DC ...(ii)

[ sum of two sides is greater than the third side]


AB + AE +DE +EC > BD + DE + DC

AB + (AE +EC)>BD + DC

AB + AC>BD + DC

Hence, BD + DC < AB + AC

6. In ABC, if AD is the bisector of A, then show that AB > BD and AC > DC. [4]

Solution :

In ABC, AD is the bisector of A

1 = 2 ..(i)

Since exterior angle of a triangle is greater than each of interior opposite angle.

Therefore, in ADC, we have

Ext. ADC > 2

3 > 2

3 > 1 [Using (i)]

Thus, in ABD, we have

3 > 1

AB >BD [Side opp, to greater angle is larger]



Hence, AB > BD.

Similarly, we can prove that AC < CD.

7. In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that

(i) PT = PS (ii) PSR = 15 [4]

Solution

Given : QRST is a square and PQR is an equilateral triangle.

To prove : (i) PS = PS

(ii) PSR = 150

Proof : Since QRST is a square and PQR is an equilateral triangle.

TQR = 900 and PQR = 600

TQR + PQR = 900 + 600

TQP = 1500

Thus, in TQP and SRP, we have

TQ = RS

TQP = SRP = 1500

and QP = RP

TQP SRP (by SAS congruence criteria)

PT = PS (by cpct)

(ii) In PSR

SR = RP ( QR = SR)

RPS = PSR = x (say)

Now, RPS + PSR + SRP = 1800 (by angle sum property)

x + x + 1500 = 1800

2x + 1500 = 1800

2x = 1800 1500



2x = 300

x = 150

PSR = 150

8. A plot is in the form of AEF. Owner of this plot wants to build old age home,

health centre and dispensary for elderly people as shown in figure. In which ABCD is a

parallelogram and AB = BE and AD = DF.

(i) Prove that BEC DCF . (ii) What values are depicted here? [4]

Solution :

Given : ABCD is a parallelogram

AB = BE and AD = DF

To prove : BEC DCF

Proof :

Since ABCD is a parallelogram

AB = DC

But AB = BE (given)

DC = BE ..(i)

Also, AD = BC

but AD = DF (given)

BC = DF ..(ii)

As ABCD is a parallelogram ,



AD BC and AE as a transversal,

1 = 2 (corresponding angles) ..(iii)

Also, AB DC and AF as a transversal,

1 = 3 (corresponding angles) ..(iv)

from (iii) and (iv), we get

2 = 3 ..(v)

Therefore, in BEC and DCF, we have

BE = DC (from (i)

BC = DF (from (ii)

2 = 3 (from (v)

BEC DCF (by SAS congruence rule)

(ii) Kindness and respectful for elder people.

9. In the figure, ABCD is a quadrilateral in which AD = BC and DAB = CBA. [4]

Prove that

(i) ABD BAC (ii) BD = AC

Solution :

Given : AD = BC and DAB = CBA

To prove : (i) ABD ABC

(ii)BD = AC

Proof : In ABD and BAC, we have :

AD = BC (Given)

DAB = CBA (Given)

AB = AB (common)

Therefore, ABD BAC (by SAS congruence rule)

(ii) BD = AC (by cpct)

10. ABCD is a square. P and Q are points on DC and BC respectively, such that

AP = DQ, prove that

(i) ADP DCQ (ii) DMP = 90 [4]



Solution :

Given : ABCD is a square, AP = AQ

To prove : (i) ADP DCQ

(ii) DMP = 900

Proof :

(i) In ADP and DCQ

ADP = DCQ = 900

AP = DQ (Given)

AD = CD ( ABCD is a square)

ADP DCQ A (by RHS congruence rule)

(ii) Since ADP DCQ (by (i))

DAP + DAP + DPA = 1800

(by angle sum property)

900 + DAP + DPA = 1800

DAP + DPA = 900

QDC + DPA = 900 (by (i))

MDP + DPM = 900 .. (ii)

In MDP,

MDP + DPM + PMD = 180 (by angle sum property by ii)

900 + PMD = 1800

PMD = 1800 900

PMD = 900

Documents

Pioneer Education {The Best Way To Success} IIT JEE ... AD is the bisector of A ? 1 = 2 ..(i) Since exterior angle of a triangle is greater than each of interior opposite