Tuesday, July 29, 2014

CBSE IX Congruence of Triangle Solved Questions

CBSE Exam  Congruence of  Triangle Solved Questions
Q. 1. Prove that Sum of Two Sides of a triangle is greater than twice the length of median drawn to third side.
Given: Δ ABC in which AD is a median.
To prove: AB + AC > 2AD.
Construction: Produce AD to E, such that AD = DE. Join EC.
Proof: In ΔADB and ΔEDC,
AD = DE              (Construction)
BD = BD             (D is the mid point of BC)
ADB = EDC       (Vertically opposite angles)
ΔADB      ΔEDC   (SAS congruence criterion)
AB = ED               (CPCT)
In ΔAEC,
AC + ED > AE           (Sum of any two sides of a triangles is greater than the third side)
AC + AB > 2AD      (AE = AD + DE = AD + AD = 2AD & ED = AB)

Q. 2. ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that BCD is a right angle.
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)
In ΔACD,      AC = AD
⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)
In ΔBCD,
∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)
⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º
⇒ 2(∠ACB + ∠ACD) = 180º         
⇒ 2(∠BCD) = 180º            
⇒ ∠BCD = 90º
  
Q.3.Given: two triangles ABC and PQR in which AB=PQ, BC=QR , median AM =median PN prove that triangle ABC is congruent to triangle PQR.

In ∆ ABM  and ∆ PQN 
AB   =  PQ                           ( Given )
AM  =  PN                           ( Given )
And  BM   =  QN   (  As M and N are the midpoint of sides BC and QR  respectively and given BC=  QR ) ∆ ABM 
 ∆ PQN             ( By SSS rule )
SO,
 ABM   =   PQN             ( by  CPCT )
Now  In ∆ ABC  and ∆ PQR
AB   =  PQ                           ( Given )
BC   =  QR                           ( Given )
And
 ABC   =   PQR              ( As we proved ) 
 ∆ ABC    ∆ PQR            ( By SAS  rule )                                       ( Hence proved )

Q.4. The vertex angle of an isosceles triangle is twice the sum of its base angles. Find the measure of all the angles.
Let ABC be an isosceles ∆.Let the measure of each of the base angles = x
Let B = C = x
Now, vertex angle = A = 2x
Now,A + B + C = 180°   [angle sum property]
2x + x + x = 180°4x = 180x = 180/4=450
So, measure of each of the base angles = 45°
Now, measure of the vertex angle = 90°

Q. 5. Prove that the triangle formed by joining the midpoints of the sides of an equilateral triangle is also equilateral.
Let DEF be the midpoints of sides of a triangle ABC( with D on BC, E on AB and F on AC ).
 Now, considering triangles AEF and ABC, angles
EAF = BAC and AE / AB = 1/2 and AF/AC = 1/2. 
Hence, both triangles are similar by the SAS ( Side - Angle - Side ) criterion and correspondingly as AE/AB=AF/AC=EF/BC ( similar triangle properties ), EF =BC/2.
The cases DF=AC/2 and DE=AB/2 can be proved in the same way.
So, AB=BC=AC (from the given data)
2DF=2EF=2DE
DE=EF=DF
So triangle DEF is also Equilateral Triangle
The triangle formed by joining the mid-points of the equilateral triangle is also an equilateral triangle

Q. 6. In triangle PQR, PQ> PR. QS and RS are the bisectors of angle Q and angle R. Prove that SQ> SR
In PQR, we have,       
PQ > PR               [given]
 PRQ > PQR    [angle opposite to longer side of a  is greater]
12PRQ > 12PQR     ........(1)
Since, SR bisects R, thenSRQ = 1/2PRQ      ........(2)
Since SQ bisects P, thenSQR = 1/2PQR   .......(3)
Now, from (1), we have     1/2PRQ > 1/2PQR
⇒∠SRQ > SQR     [using (2) and (3)]
Now, in SQR, we have    SRQ > SQR       [proved above]
 SQ > SR           [side opposite to greater angle of a  is longer

Q.7. In triangle ABC (A at the top) , D is any point on the side BC. Prove that AB+BC+CA 2AD
In triangle ABD,
AB+BD >AD (Sum of two sides of a triangle is greater than the third side) ... (1)
In triangle ACD,
AC+CD>AD (Sum of two sides of a triangle is greater than the third side)  ...(2)
Adding eq. (1) and (2)
AB+(BD+CD)+AC> AD+AD
AB+BC+AC> 2AD

Q.8. In triangle ABC, if AB is the greatest side, then prove that angle c is greater than 60 degrees
It is given that, AB is the longest side of the ∆ABC.
 AB > BC   and  AB > AC.Now,    AB > BC⇒∠C > A    (angle opposite to longer side is greater)  ....(1)
Also,AB > AC⇒∠C > B    (angle opposite to longer side is greater)   ....(2)
adding (1) and (2) , 
we getC + C > A + B
2C > A + B2C + C > A + B + C3C > 180°⇒∠C > 60°

Q.9. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that A > C and B > D.
Let us join AC.
In Δ ABC,
AB < BC (AB is the smallest side of quadrilateral ABCD)
∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) ... (1)
In ΔADC,
AD < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) ... (2)
On adding equations (1) and (2), we obtain
∠2 + ∠4 < ∠1 + ∠3
⇒ ∠C < ∠A
⇒ ∠A > ∠C

Let us join BD.
In ΔABD,
AB < AD (AB is the smallest side of quadrilateral ABCD)
∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) ... (3)
In ΔBDC,
BC < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠7 < ∠6 (Angle opposite to the smaller side is smaller) ... (4)
On adding equations (3) and (4), we obtain
∠8 + ∠7 < ∠5 + ∠6
⇒ ∠D < ∠B            ⇒ ∠B > ∠D

Q.10.  If S. is any point on the side QR of triangle PQR, prove that PQ+QR+RP> 2PS
 In   ΔPQS,
PQ + QS > PS   (i) ……………..(Sum of two sides of a triangle is greater than the third side)
In   ΔPSR,
PR + SR > PS  ……(ii)… Sum of two sides of a triangle is greater than the third side)
Adding (i) and (ii), we get
PQ + QS + PR + SR > 2PS
PQ + QR + PR > 2PS  (QS + SR = QR) Hence proved.

Q.11. Prove that the difference of any two sides of a triangle is less than the third side.
Construction: Take a Point D on AB such that AD = AC and join CD
Prove that : AB – AC < BC , AB – BC < AC and BC-AC <AB
Proof: In Δ ACD, Ext <4 > <2
but ,  AD = AC => <1 =  <2
So , < 4  > < 1 ----------------(i)
Now , In Δ BCD, ext <1 > <3 -------------(ii)
Then from (i)  and  (ii) 
< 4  > <3      =>       BC > BD
But, BD = AB – AD and AD = AC         => BD = AB – AC
So, BC > AB – AC

Q.12. that Sum of any two sides of  triangle is greater than third side .
Solution:.
Construction: Extend BA to D Such that AD = AC
Proof : In Δ DACD,  DA=CA.
Therefore, ADC=ACD [ isosceles triangle have two equal angles]
ADC + <1  > ACD 
Thus, BCD >BDC [by Euclid's fifth common notion.]
In  DCB 
BCD >  BDC, So, BD>BC.
But  BD=BA+AD, and AD=AC.
Thus,  BA+AC>BC.
A similar argument shows that AC+BC>BA and BA+BC>AC.

OR, Another way to prove
Draw a triangle,  ABC and line perpendicular to AC passing through vertex B.
Prove that BA + BC > AC

From the diagram, AM is the shortest distance from vertex A to BM. and CM is the shortest distance from vertex C to BM.
i.e. AM < BA and CM < BC
By adding these inequalities, we have
AM + CM < BA + BC
=> AC < BA + BC (
 AM + CM = AC)
BA + BC > AC (Hence Proved)

Q.13. if one acute angle in a right angled triangle is double the other then prove that the hypotenuse is double the shortest side
Given: In Δ ABC , <B = 900 and <ACB = 2 <CAB
Prove that AC = 2BC
Construction: Produce CB to D such that BC  = BD Join  to AD
Proof :  In Δ ABD, and ABC
BD = BC ; AB = AB and <B = <B = 900
By SAS congruency ,    D  ABD ≅ ABC
By CPCT, AD = AC
<DAB = <BAC = X0
So, < DAC =  2X0  
=> <ACB = <ACD
Now in Triangle Δ ADC, <DAC = <ACD= 2X0
So, AD = DC
=> AC = DC = 2BC Proved

Q. 14. Prove that in a triangle the side opposite to the largest angle is the longest.
Solution:
Given , in Δ ABC,  <ABC < <ACB
There is a triangle ABC, with angle ABC > ACB.      
Assume line AB = AC
Then angle ABC = ACB, This is a contradiction       
Assume line AB > AC
Then angle ABC < ACB, This also contradiction our hypothesis
So we are left with only one possibility ,AC> AB, which must be true
Hence proved:  AB < AC       

Q. 15. Prove that in a triangle the angle opposite to the longer side is the longest.
Solution:
Given, in Δ ABC,  AC > AB.
Construction: Take a point D on AC such that AB = AD
Proof: Angle ADB > DCB      
< ADB = <ABD          
So < ABD > <DCB (or ACB) 
< ABC >  <ABD, so < ABC > <ACB 

Q. 16.In a Δ ABC ,<B = 2<C. D is a point on BXC such that AD bisect < BAC and AB = CD. Prove that < BAC = 72 degree
In ΔABC, we have
∠B = 2∠C or, ∠B = 2y, where ∠C =  y
AD is the bisector of ∠BAC. So, let ∠BAD = ∠CAD =  x
Let BP be the bisector of ∠ABC. Join PD.
In ΔBPC, we have
∠CBP = ∠BCP =  y  ⇒ BP = PC ... (1)
Now, in ΔABP and ΔDCP, we have
∠ABP = ∠DCP =  y
AB = DC  [Given]
and, BP = PC  [Using (1)]
So, by SAS congruence criterion, we have
Δ ABP  Δ DCP
<BAP = < CPD and AP = DP
<CDP = 2x  then <ADP = < DAP = x    [<A = 2x]
In ΔABD, we have
∠ADC = ∠ABD + BAD ⇒  x  + 2x   = 2y  +  x  ⇒  x  =  y
In ΔABC, we have
∠A + ∠B + ∠C = 180°
⇒ 2x  + 2y  +  y  = 180°
⇒ 5x  = 180°
⇒  x  = 36°
Hence, ∠BAC = 2x  = 72°

You may also use this way:

Q.17,  If o is any point in the interior of triangle ABC .Prove that  
(a)  AB + AC > OB + OC
(b) AB + BC + CA > OA + OB + OC
(c )OA +OB+OC>1/2(AB+BC+CA)
Construction: Produce BO to meet AC at D
In D ABD, AB + AD > BD => AB + AD > OB + OD   ------(i)
In D OCD, OD + DC > OC    ------(ii)
Adding (i) and (ii) we get,
AB + AD + OD + DC  > OB + OD + OC    
=> AB + AC > OB + OC    --------- (iii)                   Hence prove (a)
Similarly we get ,
BC + BA > OA + OC                ---------(iv)
and , CA + CB > OA + OB       ---------(v)
Adding (iii),(iv)and (v) we get,
2(AB + BC + CA) > 2(OA + OB + OC)
AB + BC + CA > OA + OB + OC                       Hence prove (b)
In D  OAB , D OBC and D OCA
[OA + OB > AB ] + [OB + OC>BC] + [ OC + AO > AC]
2[OA + OB + OC]  > AB + BC + CA
[OA + OB + OC]  > ½ [AB + BC + CA]          
Hence prove (c)
Check more stuff on CBSE IX  Congruence of  Triangle
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CBSE IX Congruence of Triangle Solved Questions          Download File

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