Important Questions for CBSE Class 9 Maths Chapter 7 - Triangles
CBSE Class 9 Maths Chapter-7 Important Questions - Free PDF Download
1 Marks Quetions
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) ABAC
(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
11. Find the measure of angles which is equal to its supplementary.
(A)
(B)
(C)
(D)
Ans. (D)
12. Which of the following pairs of angle are supplementary?
(A)
(B)
(C)
(D) None of these.
Ans. (B)
13. Find the measure of each exterior angle of an equilateral triangle.
(A)
(B)
(C)
(D)
Ans. (C)
14. In an isosceles ABC, if AB=AC and
, Find
B.
(A)
(B)
(C)
(D)
Ans. (A)
15. In a ABC, if
B=
C=
, Which is the longest side.
(A) BC
(B) AC
(C) CA
(D) None of these.
Ans. (A) BC
16. In a ABC, if AB=AC and
B=
, Find
A.
(A)
(B)
(C)
(D)
Ans. (A)
17. Determine the shortest sides of the triangles.
(a) AC
(b) BC
(c) CA
(d) none of these
Ans. (b) BC
18. In an ABC, if
=
and
=
, determine the longest sides of the triangle.
(a) AC
(b) CA
(c) BC
(d) none of these
Ans. (a) AC
19. The sum of two angles of a triangle is equal to its third angle. Find the third angles.
(a)
(b)
(c)
(d)
Ans. (a)
20. Two angles of triangles are respectively. Find third angles.
(a)
(b)
(c)
(d)
Ans. (d)
21. ABC is an isosceles triangle with AB=AC and
=
, find
.
Ans.
[angle opposite to equal sides are equal]
But,
And,
22. 1. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB (See figure)
Ans. In BOC and
AOD,
OBC =
OAD =
[Given]
BOC =
AOD [Vertically Opposite angles]
BC = AD [Given]
BOC
AOD [By ASA congruency]
OB = OA and OC = OD [By C.P.C.T.]
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
ThereforeB 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]
16. In figure, B <
A and
C <
D. Show that AD < BC.
Ans. In AOB,
B <
A [Given]
OA < OB ……….(i) [Side opposite to greater angle is longer]
In COD,
C <
D [Given]
OD < OC ……….(ii) [Side opposite to greater angle is longer]
Adding eq. (i) and (ii),
OA + OD < OB + OC
AD < BC
17. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Ans. Let ABC be a triangle.
Draw bisectors of B and
C.
Let these angle bisectors intersect each other at point I.
Draw IK BC
Also draw IJ AB and IL
AC.
Join AI.
In BIK and
BIJ,
IKB =
IJB =
[By construction]
IBK =
IBJ
[ BI is the bisector of
B (By construction)]
BI = BI [Common]
BIK
BIJ [ASA criteria of congruency]
IK = IJ [By C.P.C.T.] ……….(i)
Similarly, CIK
CIL
IK = IL [By C.P.C.T.] ……….(ii)
From eq (i) and (ii),
IK = IJ = IL
Hence, I is the point of intersection of angle bisectors of any two angles of ABC equidistant from its sides.
18. In quadrilateral ACBD, AB=AD and AC bisects A. show
ABC
ACD?
Ans. IN ABC and
ACD,
AD=AB……… (Given)
BAC=
CAD……. (AC bisects
A)
And AC= AC ………….. (Common)
ABC
ACD ……….. (SAS axiom)
19. If DA and CB are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Ans. In AOD and
BOC,
AD=BC ……….. (Given)
And (vert opp. Angles)
(AAS rule)
(CPCT)
Hence CD bisects AB.
20. l and m, two parallel lines, are intersected by Another pair of parallel lines p and C. show that ABC
CDA.
Ans. cuts them – (Given)
(alternate angles)
(Given)
(Alternate angles)
AC=CA (common)
(ASA rule)
21. In fig, the bisector AD of ABC is
to the opposite side BC at D. show that
ABC is isosceles?
Ans. In ABD and
ACD
22. If AE=AD and BD=CE. Prove that AEB
ADC
Ans. We have,
AE=AD and CE=BD
AE+CE=AD+BD
AC=AB(i)
Now, in AEB and
ADC,
AE=AD [given]
EAB=
DAC [common]
AB=AC [from (i)]
AEB
ADC [by SAS]
23. In quadrilateral ACBD, AC=AD and AB bisects A. show that
ABC
ABD. What can you say about BC and BD?
Ans. In ABC and
ABD,
AC=AD [given]
CAB=
DAB[AB bisects
A]
AB=AB [common]
ABC
ABD[SAS criterion]
BC=BD [CPCT]
24. In ABC, the median AD is
to BC. Prove that
ABC is an isosceles triangle.
Ans.
BD =CD [D is mid-point of BC]
AD=AD [Common]
[each
]
[By SAS]
[CPCT]
Hence, triangle ABC is an isosceles triangle.
25. Prove that ABC is isosceles if altitude AD bisects
BAC.
Ans.
AD=AD [common]
[By AAs]
[CPCT]
Thus, is an isosceles triangle.
26. ABC is An isosceles triangle in which altitudes BE and CF are drawn to side AC and AB respectively. Show that these altitudes are equals.
Ans.
AB=AC [given]
[AAS rule]
[CPCT]
27. If AC= AE, AB=AD and =
show that BC =DE.
Ans.
AB=AD [given]
AC=AE [given]
Also, [given]
[SAS criterior]
[CPCT]
28. Line is the bisector of an angle
A and B is any point on line l. BP and BQ are
from B to the arms of
A show that :
(i) APB
AQB
(ii) BP = BQ or B is A equidistant from the arms of
Ans.
[common]
AB=AB [Common]
[AAS rule]
[CPCT]
29. In the given figure, ABC is an isosceles triangle and
B =
, find x.
Ans.
AB=AC
[Angles opposite to equal sides are equal]
But
So,
30. If E>
A and
C>
D. prove that AD>EC.
Ans.
[given]
[Side opposite to greater angle is larger] …..(i)
Similarly, in
Adding (i) and (ii)
31. If PQ= PR and S is any point on side PR. Prove that RS<QS.
Ans.
PQ=PR [given]
[angle opposite to equal side are equal]
Now,
[side opposite to smaller angle in
]
32. Prove that MN+NO +OP+PM>2MO.
Ans.
MN+NO>MO [Sum of any two side of is greater than third sides] …(i)
Similarly in
OP+PM>MO ….(ii)
Hence from (i) and (ii)
Or MN+NO+OP+PM>2MO
33. Prove that MN+NO+OP>PM.
Ans.
MN+NO>MO [Sum of any two side of is greater than third sides] …(i)
Similarly in
MO+OP>PM ….(ii)
Hence from (i) and (ii)
Or MN+NO+OP+MO>MO+PM
Or MN+NO+OP>PM
34. ABC is an equilateral triangle and
=
, find
.
Ans.
AB=AC
[angles opposite to equal sides are equal]
But
So,
35. In the figure, AB = AC and.
Ans.
[angles opposite to equal sides are equal]
Also, [Linear pair]
and,
36. In the given figure, find
Ans.
[sum of three angles of a]
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 =
SinceA =
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)SinceABP
ACP[From part (ii)]
BAP =
CAP[By C.P.C.T.]
AP bisects
A.
SinceABD
ACD[From part (i)]
ADB =
ADC[By C.P.C.T.]……….(ii)
NowADB +
BDP =
[Linear pair]……….(iii)
AndADC +
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) SinceABP
ACP[From part (ii)]
BP = PC[By C.P.C.T.]……….(v)
AndAPB =
APC[By C.P.C.T.]……….(vi)
NowAPB +
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 + AFAB = AC
ABC is an isosceles triangle.
6. ABC is an isosceles triangles with AB = AC. Draw AP BC and show that
B =
C.
Ans. Given: ABC is an isosceles triangle in which AB = AC
To prove:B =
C
Construction: Draw AP BC
Proof: In ABP and
ACP
APB =
APC =
[By construction]
AB = AC[Given]
AP = AP[Common]
ABP
ACP[RHS congruency]
B =
C[By C.P.C.T.]
7. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (See figure). Show that A >
C and
B >
D.
Ans. Given: ABCD is a quadrilateral with AB as smallest and CD as longest side.
To prove:(i)A >
C(ii)
B >
D
Construction: Join AC and BD.
Proof:(i)In ABC, AB is the smallest side.
4 <
2 ……….(i)
[Angle opposite to smaller side is smaller]
In ADC, DC is the longest side.
3 <
1 ……….(ii)
[Angle opposite to longer side is longer]
Adding eq. (i) and (ii),
4 +
3 <
1 +
2
C <
A
(ii) In ABD, AB is the smallest side.
5 <
8 ……….(iii)
[Angle opposite to smaller side is smaller]
In BDC, DC is the longest side.
6 <
7 ……….(iv)
[Angle opposite to longer side is longer]
Adding eq. (iii) and (iv),
5 +
6 <
7 +
8
D <
B
8. In figure, PR > PQ and PS bisects QPR. Prove that
PSR>
PSQ.
Ans. In PQR,PR > PQ[Given]
PQR>
PRQ…..(i)[Angle opposite to longer side is greater]
Again1 =
2…..(ii)[
PS is the bisector of
P]
PQR +
1 >
PRQ +
2……….(iii)
ButPQS +
1 +
PSQ =
PRS +
2 +
PSR =
[Angle sum property]
PQR +
1 +
PSQ =
PRQ +
2 +
PSR………(iv)
[PRS =
PRQ and
PQS =
PQR]
From eq. (iii) and (iv),
PSQ<
PSR
OrPSR>
PSQ
9. Show that all the line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Ans. Given: is a line and P is point not lying on
PM
N is any point on other than M.
To prove: PM <PN
Proof: In PMN,
M is the right angle.
N is an acute angle. (Angle sum property of
)
M >
N
PN> PM[Side opposite greater angle]
PM <PN
Hence of all line segments drawn from a given point not on it, the perpendicular is the shortest.
10. ABC is a triangle. Locate a point in the interior of ABC which is equidistant from all the vertices of
ABC.
Ans. Let ABC be a triangle.
Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC. Let PQ bisects AB at M and RS bisects BC at point N.
Let PQ and RS intersect at point O.
Join OA, OB and OC.
Now in AOM and
BOM,
AM = MB[By construction]
AMO =
BMO =
[By construction]
OM = OM[Common]
AOM
BOM[By SAS congruency]
OA = OB[By C.P.C.T.]…..(i)
SimilarlyBON
CON
OB = OC[By C.P.C.T.]…..(ii)
From eq. (i) and (ii),
OA = OB = OC
Hence O, the point of intersection of perpendicular bisectors of any two sides of ABC equidistant from its vertices.
11. In a huge park, people are concentrated at three points (See figure).
A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
Where should an ice cream parlour be set up so that maximum number of persons can approach it?
Ans. The parlour should be equidistant from A, B and C.
For this let we draw perpendicular bisector say of line joining points B and C also draw perpendicular bisector say
of line joining points A and C.
Let and
intersect each other at point O.
Now point O is equidistant from points A, B and C.
Join OA, OB and OC.
Proof: In BOP and
COP,
OP = OP[Common]
OPB =
OPC =
BP = PC[P is the mid-point of BC]
BOP
COP[By SAS congruency]
OB = OC[By C.P.C.T.]…..(i)
Similarly, AOQ
COQ
OA = OC[By C.P.C.T.]…..(ii)
From eq. (i) and (ii),
OA = OB = OC
Therefore, ice cream parlour should be set up at point O, the point of intersection of perpendicular bisectors of any two sides out of three formed by joining these points.
12. If ABC, the bisector of
ABC and
BCA intersect each other at the point O prove that
BOC =
Ans.
Substituting this value of in (1)
So,
13. Prove that if one angle of a triangle is equal to the sum of the other two angles, the triangle is right angled:
Ans.Sum of three angles of triangle is
] …..(1)
Given that:
From (1) and (2)
Hence ABC is right angled.
14. IF fig, if PQ PS, PQ
SR,
SQR =
and
QRT =
, then find the values of X and Y.
Ans. is the transversal,
[pair of alternate angles]
Or
or
Also in
Or
15. If in fig, AD= AE and D and E are point on BC such that BD=EC prove that AB=AC.
Ans.
[Given]
[angles opposite to equal side are equal]
Now, [linear pair]
Also, [linear pair]
But,
Now in,
BD=CE
AD=AE
[By SAS]
[CPCT]
16. In the given figure, AC=BC, DCA=
ECB and
DBC=
EAC. Prove that
DBC and
EAC are congruent and hence DC=EC.
Ans. We have,
[Given]
[adding ECD on both sides]
...(i)
[From (i)
Now, in
[given]
[given]
[By SAS]
[CPCT]
17. From the following figure, prove that BAD=3
ADB.
Ans.
Exterior = 2Q
Hence
18. O is the mid-point of AB and CD. Prove that AC=BD and ACBD.
Ans.
[O is the mid – point of AB]
[vertically opposite angles]
[O is the mid-point of CD]
[By SAS]
[CPCT]
[CPCT]
Now, AC and BD are two lines inter sected by a transversal AB such that i.e. alternate angle are equal.
19. ABCD is a quadrilateral in which AD=BC and DAB=
CBA. Prove that.
(i) ABD
BAC
(ii) BA=AC
(iii) ABD=
BAC
Ans.
[given]
[given]
[common]
[SAS criterion]
[CPCT]
[CPCT]
20. AB is a line segment. AX and BY are equal two equal line segments drawn on opposite side of line AB such that AX BY. If AB and XY intersect each other at P. prove that
(i) APX
BPY,
(ii) AB and XY bisect each other at P.
Ans.
[alternate angle]
[vertically opposite angle]
AX=BY[given]
[By AAS]
[CPCT]
AB and XY bisects each other at P.
21. In an isosceles ABC, with AB =AC, the bisector of
B and
C intersect each other at o, join A to o. show that:
(i) OB=OC
(ii) AO bisects A.
Ans. (i)
[given]
[angles opposite to equal side]
[side opposite to equal angle]
(ii)
[given]
[Halves of equals]
OB=OC [proved]
[SAS rule]
[CPCT]
i.e. AO bisects
22. Two side AB and BC and median AM of a triangle ABC are respectively equal to side PQ and QR and median PN of PQR, show that
(i)
(ii)
Ans. (i)
AB=PQ [Given]
BM=QN [Halves of equal]
AP=PN[Given]
[SSS rules]
(ii)
Now, in
AB=PQ [Given]
BC=QR [Given]
[SAS rule]
23. In the given figure, ABC and DBC are two triangles on the same base BC such that AB=AC and DB=DC. Prove that ABD =
ACD,
Ans.
AB=AC[Given]
[angles opposite to equal side are equals]
Similarly in, [Given]…..(1)
……(2)
Adding (1) and (2)
24. Prove that the Angle opposite to the greatest side of a triangle is greater than two- third of a right angle i.e. greater than
Ans.
[angle opposite to large side is greater]….(i)
Similarly,
AB>AC
Adding (i) and (ii)
Adding to both sides,
[Sum of three angles of
]
25. AD is the bisector of A of
ABC, where D lies on BC. Prove that AB>BD and AC>CD.
Ans.
[Exterior angles of
is greater than each of the interior opposite angles]
But [Ad bisects
]
[Side opposite to greater angle is larger]
[Exterior angles of
is greater than each of the interior opposite angles]
But,
[Side opposite to greater angle is larger].
26. In the given figure, AB and CD are respectively the smallest and the largest side of a quadrilateral ABCD. Prove that A>
C and
B>
D.
Ans. Join AC.
[AB is the smallest sides of quadrilateral ABCD]
[Angle opposite to larger side is greater]…(i)
[CD is the largest side of quadrilateral ABCD]
[angle opposite to larger side is greater]…..(ii)
Adding (i) and (ii)
Similarly, by joining BD, we can show that
27. If the bisector of a vertical angle of a triangle also bisects the opposite side; prove that the triangle is an isosceles triangle.
Ans.
DC=DB [Given]
AD=ED [By construction]
[vertically opposite angle]
[By SAS]
[CPCT]
But, [
AD bisects
]
But BE=AC [Proved above]
28. ABC is an isosceles triangle with AB = AC. Draw APBC to show that
Ans.
AP=AP [common]
Hypotenuse AB = Hypotenuse AC [Given]
[RHS rule]
[CPCT]
29. AD is an altitude of an isosceles triangle ABC in which AB = AC. Prove that:
(i) AD bisects BC
(ii) AD bisects
Ans. (i) In right triangle ABD and ACD,
Side AD = Side AD[common]
Hypotenuse AB = Hypotenuse AC [Given]
[By RSH]
[CPCT]
Also, AD bisects BC
(ii) Also, [CPCT]
30. In the given figure, PQ>PR, QS and RS are the bisectors of the Q and
R respectively. Prove that SQ>SR.
Ans. Since PQ>PR
[angle opposite to larger side is larger]
[Side opposite to greater angle is larger]
4 Marks Quetions
1. In right triangle ABC, right angled