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Important Questions for CBSE Class 9 Maths Chapter 7 – Triangles

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. 

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 =  

BOC = AOD 

BC = AD 

 BOC AOD 

 OB = OA and OC = OD ....


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 
BAC = BAD [ AB bisects A]
AB = AB 
 ABC ABD 
Thus BC = BD ....


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 

DAB = CBA 

AB = AB 

 ABC ABD 

Thus AC = BD ....

(ii) Since ABC ABD

 AC = BD ....

(iii) Since ABC ABD

 ABD = BAC ....


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. 

Therefore DAC = ACB 

Now  

And AC being a transversal. 

Therefore BAC = ACD 

Now In ABC and ADC,

ACB = DAC 

BAC = ACD 

AC = AC 

 ABC CDA 


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 

BPA = BQA =  

AB = AB 

 APB AQB 

(ii) Since APB AQB

 BP = BQ ....

 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 

AC = AE 

BAC = DAE .()

 ABC ADE 

 BC = DE ....


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) DAFFBPE 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 
APD = BPE .()
 DPAEBP 
 AD = BE ....


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 = 

 OCA + OCB = OBA + OBC

 OB bisects B and OC bisects C

 OBA = OBC and OCA = OCB

 OCB + OCB = OBC + OBC

 2OCB = 2OBC

 OCB = OBC

Now in OBC,

OCB = OBC 

 OB = OC 

(ii) In AOB and AOC,

AB = AC 

OBA = OCA 

And B = C

 B = C

 OBA = OCA

 OB = OC 

 AOBAOC 

 OAB = OAC ....

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 

ADB = ADC = [AD  BC]

AD = AD 

 ABD ACD 

 AB = AC ....

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= 

AEB = AFC = 

AB = AC 

 ABE ACF 

 BE = CF ....

 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= 

AEB = AFC = 

BE = CF 

 ABE ACF 

(ii) Since ABE ACF

 BE = CF ....

 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 

ACB = ABC …….(i) 

Also in Isosceles triangle BCD.

BD = DC

 BCD = CBD ……….(ii) 

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 =  

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

 2B = 

 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 

ADB = ADC = [AD  BC]

AD = AD 

 ABD ACD 

 BD = DC ....

 AD bisects BC

Also BAD = CAD ....

 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 >

Now 1 + 2 = 3 + 4 = 

 1 >3 [4 >2]

 AC > AB 


16. In figure, B <A and C <D. Show that AD < BC.

Ans
. In AOB,

B <

 OA < OB ……….(i) 

In COD,

C <

 OD < OC ……….(ii) 

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 = 

IBK = IBJ

[ BI is the bisector of B (By construction)]

BI = BI 

 BIK BIJ 

 IK = IJ .... ……….(i)

Similarly, CIKCIL

 IK = IL .... ……….(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 bisectsA)

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 
EAB=DAC 
AB=AC ()
AEBADC 


23. In quadrilateral ACBD, AC=AD and AB bisects A. show that ABCABD. What can you say about BC and BD?

Ans. 
In ABC and ABD,
AC=AD 
CAB=DAB[AB bisects A]
AB=AB 
ABCABD
BC=BD 


24. In ABC, the median AD isto BC. Prove that ABC is an isosceles triangle.

Ans. 

BD =CD 
AD=AD 

[each  ]


Hence, triangle ABC is an isosceles triangle.


25. Prove that ABC is isosceles if altitude AD bisects BAC.

Ans. 


AD=AD 


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 


27. If AC= AE, AB=AD and = show that BC =DE.

Ans. 

AB=AD 
AC=AE 
Also, 




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. 



AB=AB 



29. In the given figure, ABC is an isosceles triangle and B =, find x.

Ans. 

AB=AC

But 

So,



30. If E>A and C>D. prove that AD>EC.

Ans. 


 …..(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 

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

But 
So, 


35. In the figure, AB = AC and.

Ans. 

Also, 

and, 


36. In the given figure, find

Ans. 




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 = ACAB = 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 = 3A = 

A = 

SinceA = B = C.()

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) ABPACP

(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

BD = CD

AD = AD

ABD ACD

BAD = CAD....……….(i)

(ii)Now in ABP and ACP,

AB = AC

BAD = CAD.()

AP = AP

ABPACP

(iii)SinceABPACP()

BAP = CAP....

AP bisects A.

SinceABD ACD()

ADB = ADC....……….(ii)

NowADB + BDP = ……….(iii)

AndADC + CDP = ……….(iv)

From eq. (iii) and (iv),

ADB + BDP = ADC + CDP

ADB + BDP = ADB + CDP()

BDP = CDP

DP bisects DorAP bisects D.

(iv) SinceABPACP()

BP = PC....……….(v)

AndAPB = APC....……….(vi)

NowAPB + APC = 

APB + APC = .()

2APB = 

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 BC = QR

BM = QN……….(iii)

(i)Now in ABM and PQN,

AB = PQ

AM = PN

BM = QN.()

ABM PQN

B = Q....……….(iv)

(ii)In ABC and PQR,

AB = PQ

B = Q

BC = QR

ABC PQR


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

BE = CF

BECCFB

EC = FB....…..(i)

Now In AEB and AFC

AEB = AFC [Each ]

A = A

BE = CF

AEBAFC

AE = AF....…………(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 = 

AB = AC

AP = AP

ABPACP

B = C....


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)

In ADC, DC is the longest side.

3 <1 ……….(ii)

Adding eq. (i) and (ii),

4 + 3 <1 + 2C <A

(ii) In ABD, AB is the smallest side.

5 <8 ……….(iii)

In BDC, DC is the longest side.

6 <7 ……….(iv)

Adding eq. (iii) and (iv),

5 + 6 <7 + 8D <B


8. In figure, PR > PQ and PS bisects QPR. Prove that PSR>PSQ.

Ans
. In PQR,PR > PQ

PQR>PRQ…..(i)

Again1 = 2…..(ii)[ PS is the bisector of P]

PQR + 1 >PRQ + 2……….(iii)

ButPQS + 1 + PSQ = PRS + 2 + PSR = 

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

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

AMO = BMO = 

OM = OM

AOMBOM

OA = OB....…..(i)

SimilarlyBON CON

OB = OC....…..(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

OPB = OPC = 

BP = PC

BOP COP

OB = OC....…..(i)

Similarly, AOQCOQ

OA = OC....…..(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, PQSR, SQR = and QRT =, then find the values of X and Y.

Ans. 
is the transversal,

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. 

Now, 

Also, 

But, 

Now in, 

BD=CE

AD=AE


16. In the given figure, AC=BC, DCA=ECB and DBC=EAC. Prove that DBC andEAC are congruent and hence DC=EC.

Ans. 
We have,

[adding ECD on both sides]

 ...(i)

[From (i)

Now, in 


17. From the following figure, prove that BAD=3ADB.

Ans.

Exterior  = 2Q


Hence 


18. O is the mid-point of AB and CD. Prove that AC=BD and ACBD.

Ans. 

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) ABDBAC

(ii) BA=AC

(iii) ABD=BAC

Ans. 


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.

 

AX=BY

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) 




(ii)


OB=OC 


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 
BM=QN 
AP=PN

(ii) 
Now, in 
AB=PQ 
BC=QR 


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

 

Similarly in, …..(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.

….(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 ]



[Exterior angles of  is greater than each of the interior opposite angles]
But, 


.


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.

…(i)

…..(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 

AD=ED 

But, [AD bisects ]


But BE=AC 


28. ABC is an isosceles triangle with AB = AC. Draw APBC to show that

Ans.

AP=AP 

Hypotenuse AB = Hypotenuse AC 


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

Hypotenuse AB = Hypotenuse AC 

Also, AD bisects BC

(ii) Also, 


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

  

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