Class- XI-CBSE-Mathematics

Triangles Practice

1. In fig, if AD =BC and BAD = ABC, then ACB is equal to

(A) ABD

(B) BAD

(C) BAC

(D) BDA

Ans. (D) BDA

2. IN fig, if ABCD is a quadrilateral in which AD= CB, AB=CD, and D= B,

then CAB is equal to

(A) ACD

(B) CAD

(C) ACD

(D) BAD

Ans. (C) ACD

3. If O is the mid – point of AB and BQO = APO, then OAP is equal to

(A) QPA

(B) OQB

(C) QBO

(D) BOQ

Ans. (C) QBO

4. IF AB BC and A = c, then the true statement is

(A) AB AC

(B) AB=BC

(C) AB=AD

(D) AB=AC

Ans. (B) AB=BC

5. If ABC is an isosceles triangle and B= , find x.

(a)

(b)

(c)

(d) none of these

Ans. (c)

6. If AB=AC and ACD= , find A

(a)
(b)

(c)

(d) none of these

Ans. (b)

7. What is the sum of the angles of a quadrilateral:

(a)

(b)

(c)

(d)

Ans. (b)

8. The sum of the angles of a triangle will be:

(a)

(b)

(c)

(d)

Ans. (c)

9. An angle is more than its complement. Find its measure.

(A) 42

(B) 32
(C) 52

(D) 62

Ans. (C) 52

10. An angle is 4 time its complement. Find measure.

(A) 62

(B) 72

(C) 52

(D) 42

Ans. (B) 72

2 Marks Quetions

1. In quadrilateral ABCD (See figure). AC = AD and AB bisects A. Show that

ABC ABD. What can you say about BC and BD?

Ans. Given: In quadrilateral ABCD, AC = AD and AB bisects A.

To prove: ABC ABD
Proof: In ABC and ABD,
AC = AD [Given]
BAC = BAD [ AB bisects A]
AB = AB [Common]
ABC ABD [By SAS congruency]
Thus BC = BD [By C.P.C.T.]
2. ABCD is a quadrilateral in which AD = BC and DAB = CBA. (See figure).
Prove that:

(i) ABD BAC

(ii) BD = AC

(iii) ABD = BAC

Ans. (i) In ABC and ABD,

BC = AD [Given]

DAB = CBA [Given]

AB = AB [Common]

ABC ABD [By SAS congruency]

Thus AC = BD [By C.P.C.T.]

(ii) Since ABC ABD

AC = BD [By C.P.C.T.]

(iii) Since ABC ABD

ABD = BAC [By C.P.C.T.]

3. and are two parallel lines intersected by another pair of parallel

lines and (See figure). Show that ABC CDA.
Ans. AC being a transversal. [Given]

Therefore DAC = ACB [Alternate angles]

Now [Given]

And AC being a transversal. [Given]

Therefore BAC = ACD [Alternate angles]

Now In ABC and ADC,

ACB = DAC [Proved above]

BAC = ACD [Proved above]

AC = AC [Common]

ABC CDA [By ASA congruency]

4. Line is the bisector of the angle A and B is any point on BP and BQ are
perpendiculars from B to the arms of A. Show that:

(i) APB AQB

(ii) BP = BQ or P is equidistant from the arms of A (See figure).

Ans. Given: Line bisects A.

BAP = BAQ
(i) In ABP and ABQ,

BAP = BAQ [Given]

BPA = BQA = [Given]

AB = AB [Common]

APB AQB [By ASA congruency]

(ii) Since APB AQB

BP = BQ [By C.P.C.T.]

B is equidistant from the arms of A.

5. In figure, AC = AB, AB = AD and BAD = EAC. Show that BC = DE.

Ans. Given that BAD = EAC

Adding DAC on both sides, we get

BAD + DAC = EAC + DAC

BAC = EAD ……….(i)

Now in ABC and AED,

AB = AD [Given]

AC = AE [Given]

BAC = DAE [From eq. (i)]

ABC ADE [By SAS congruency]

BC = DE [By C.P.C.T.]
6. AB is a line segment and P is the mid-point. D and E are points on the same
side of AB such that BAD = ABE and EPA = DPB. Show that:

(i) DAF FBPE D

(ii) AD = BE (See figure)

Ans. Given that EPA = DPB

Adding EPD on both sides, we get
EPA + EPD = DPB + EPD
APD = BPE ……….(i)
Now in APD and BPE,

PAD = PBE [ BAD = ABE (given), PAD = PBE]

AP = PB [P is the mid-point of AB]
APD = BPE [From eq. (i)]
DPA EBP [By ASA congruency]
AD = BE [ ByC.P.C.T.]

7. In an isosceles triangle ABC, with AB = AC, the bisectors of B and C

intersect each other at O. Join A to O. Show that:

(i) OB = OC

(ii) AO bisects A.

Ans. (i) ABC is an isosceles triangle in which AB = AC.

C= B [Angles opposite to equal sides]

 OCA + OCB = OBA + OBC

OB bisects B and OC bisects C

OBA = OBC and OCA = OCB

OCB + OCB = OBC + OBC

2 OCB = 2 OBC

OCB = OBC

Now in OBC,

OCB = OBC [Prove above]

OB = OC [Sides opposite to equal sides]

(ii) In AOB and AOC,

AB = AC [Given]

OBA = OCA [Given]

And B= C

B= C

OBA = OCA

OB = OC [Prove above]

AOB AOC [By SAS congruency]

OAB = OAC [By C.P.C.T.]

Hence AO bisects A.

8. In ABC, AD is the perpendicular bisector of BC (See figure). Show that ABC

is an isosceles triangle in which AB = AC.
Ans. In AOB and AOC,

BD = CD [AD bisects BC]

ADB = ADC = [AD BC]

AD = AD [Common]

ABD ACD [By SAS congruency]

AB = AC [By C.P.C.T.]

Therefore, ABC is an isosceles triangle.

9. ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC

and AB respectively (See figure). Show that these altitudes are equal.

Ans. In ABE and ACF,

A= A [Common]

AEB = AFC = [Given]

AB = AC [Given]

ABE ACF [By ASA congruency]

BE = CF [By C.P.C.T.]

Altitudes are equal.

10. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal
(See figure). Show that:

(i) ABE ACF

(ii) AB = AC or ABC is an isosceles triangle.

Ans. (i) In ABE and ACF,

A= A [Common]

AEB = AFC = [Given]

BE = CF [Given]

ABE ACF [By ASA congruency]

(ii) Since ABE ACF

BE = CF [By C.P.C.T.]

ABC is an isosceles triangle.

11. ABC and DBC are two isosceles triangles on the same base BC (See figure).
Show that ABD = ACD.

Ans. In isosceles triangle ABC,

AB = AC [Given]

ACB = ABC …….(i) *Angles opposite to equal sides+

Also in Isosceles triangle BCD.

BD = DC
 BCD = CBD ……….(ii) *Angles opposite to equal sides+

Adding eq. (i) and (ii),

ACB + BCD = ABC + CBD

ACD = ABD

Or ABD = ACD

12. ABC is a right angled triangle in which A= and AB = AC. Find B and
C.

Ans. ABC is a right triangle in which,

A= And AB = AC

In ABC,

AB = AC C= B ……….(i)

We know that, in ABC, A+ B+ C= [Angle sum property]

B+ B= [ A= (given) and B= C (from eq. (i)]

2 B=

B=

Also C= [ B= C]

13. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:

(i) AD bisects BC.

(ii) AD bisects A.
Ans. In ABD and ACD,

AB = AC [Given]

ADB = ADC = [AD BC]

AD = AD [Common]

ABD ACD [RHS rule of congruency]

BD = DC [By C.P.C.T.]

AD bisects BC

Also BAD = CAD [By C.P.C.T.]

AD bisects A.

14. Show that in a right angles triangle, the hypotenuse is the longest side.

Ans. Given: Let ABC be a right angled triangle, right angled at B.

To prove: Hypotenuse AC is the longest side.

Proof: In right angled triangle ABC,

A+ B+ C=

A+ + C= [ B= ]
 A+ C=

And B=

B > C and B> A

Since the greater angle has a longer side opposite to it.

AC > AB and AC > AB

Therefore B being the greatest angle has the longest opposite side AC, i.e.
hypotenuse.

15. In figure, sides AB and AC of ABC are extended to points P and Q

respectively. Also PBC< QCB. Show that AC > AB.

Ans. Given: In ABC, PBC< QCB

To prove: AC > AB

Proof: In ABC,

4 > 2 [Given]

Now 1+ 2= 3+ 4= [Linear pair]

1> 3[ 4 > 2]

AC > AB [Side opposite to greater angle is longer]

3 Marks Quetions
1. Prove that in a right triangle, hypotenuse is the longest (or largest) side.

Ans. Given a right angled triangle ABC in which

Now, since

Hence, the side opposite to is the hypotenuse and the longest side of the
triangle.

2. Show that the angles of an equilateral triangle are 60o each.

Ans. Let ABC be an equilateral triangle.

AB = BC = AC AB = BC

C= A……….(i)
Similarly, AB = AC

C= B……….(ii)

From eq. (i) and (ii),

A= B= C……….(iii)

Now in ABC

A+ B+ C= ……….(iv)

A+ A+ A= 3 A=

A=

Since A= B= C[From eq. (iii)]

A= B= C=

Hence each angle of equilateral triangle is

3. ABC and DBC are two isosceles triangles on the same base BC and
vertices A and D are on the same side of BC (See figure). If AD is extended to
intersect BC at P, show that:

(i) ABD ACD

(ii) ABP ACP

(iii) AP bisects A as well as D.

(iv) AP is the perpendicular bisector of BC.

Ans. i) ABC is an isosceles triangle.

 AB = AC

DBC is an isosceles triangle.

BD = CD

Now in ABD and ACD,

AB = AC[Given]

BD = CD[Given]

AD = AD[Common]

ABD ACD[By SSS congruency]

BAD = CAD*By C.P.C.T.+……….(i)

(ii)Now in ABP and ACP,

AB = AC[Given]

BAD = CAD[From eq. (i)]

AP = AP

ABP ACP[By SAS congruency]

(iii)Since ABP ACP[From part (ii)]

BAP = CAP[By C.P.C.T.]

AP bisects A.

Since ABD ACD[From part (i)]

ADB = ADC*By C.P.C.T.+……….(ii)

Now ADB + BDP = *Linear pair+……….(iii)

And ADC + CDP = *Linear pair+……….(iv)

From eq. (iii) and (iv),

ADB + BDP = ADC + CDP

ADB + BDP = ADB + CDP[Using (ii)]

BDP = CDP

DP bisects DorAP bisects D.

(iv) Since ABP ACP[From part (ii)]

BP = PC*By C.P.C.T.+……….(v)

And APB = APC*By C.P.C.T.+……….(vi)

Now APB + APC = [Linear pair]

APB + APC = [Using eq. (vi)]

2 APB =

APB =

AP BC……….(vii)

From eq. (v), we have BP PC and from (vii), we have proved AP B. So,
collectively AP is perpendicular bisector of BC.
4. Two sides AB and BC and median AM of the triangle ABC are respectively

equal to side PQ and QR and median PN of PQR (See figure). Show that:

(i) ABM PQN

(ii) ABC PQR

Ans. AM is the median of ABC.

BM = MC = BC……….(i)

PN is the median of PQR.

QN = NR = QR……….(ii)

Now BC = QR[Given] BC = QR

BM = QN……….(iii)

(i)Now in ABM and PQN,

AB = PQ[Given]

AM = PN[Given]

BM = QN[From eq. (iii)]

ABM PQN[By SSS congruency]

B= Q*By C.P.C.T.+……….(iv)

(ii)In ABC and PQR,

AB = PQ[Given]

B= Q[Prove above]
BC = QR[Given]

ABC PQR[By SAS congruency]

5. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence
rule, prove that the triangle ABC is isosceles.

Ans. In BEC and CFB,

BEC = CFB[Each ]

BC = BC[Common]

BE = CF[Given]

BEC CFB[RHS congruency]

EC = FB*By C.P.C.T.+…..(i)

Now In AEB and AFC

AEB = AFC [Each ]

A= A[Common]

BE = CF[Given]

AEB AFC[ASA congruency]

AE = AF*By C.P.C.T.+…………(ii)

Adding eq. (i) and (ii), we get,

EC + AE = FB + AF AB = AC

ABC is an isosceles triangle.