RS Aggarwal Class 9 Chapter 5 – Congruence And Inequalities In Triangle Ex 5B (5.2) Solutions Free PDF
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Question 1:
In the given figure, if x = y and AB = CB, then prove that AE = CD.
Solution:
Given: AB = BC
and, xo = yo
To prove: AE = CD
Proof: In \(\Delta\)ABE, we have,
Exterior \(\angle\)AEB = \(\angle\)EBA + \(\angle\)BAE
\(\Rightarrow\) yo = \(\angle\)EBA + \(\angle\)BAE
Again, in \(\Delta\)BCD, we have
xo = \(\angle\)CBA + \(\angle\)BCD
Since, xo = yo [Given]
So, \(\angle\)EBA + \(\angle\)BAE = \(\angle\)CBA + \(\angle\)BCD
\(\Rightarrow \) \(\angle\)BAE = \(\angle\)BCD
Thus in \(\Delta\)BCD and \(\Delta\)BAE, we have
\(\angle\)B = \(\angle\)B [Common]
BC = AB [Given]
and, \) \(\angle\)BAE = \(\angle\)BCD [Proved above]
Thus by Angle-Side-Angle criterion of congruence, we have
\(\Delta\)BCD \(\cong\) \(\Delta\)BAE
The corresponding parts of the congruent triangles are equal.
So, CD = AE [Proved]
Question 2:
ABC is a triangle in which AB = AC. If the bisectors of \(\angle\)B and \(\angle\)C meet AC and AB in D and E respectively, prove that BD = CE.
Solution:
Given: A \(\Delta\)ABC in which AB = AC and BD and CE are the bisectors of \(\angle\)B and \(\angle\)C respectively.
To prove: BD = CE
Proof: In \(\Delta\)ABC and \(\Delta\)ACE
\(\angle\)ABD = ½ \(\angle\)B
and, \(\angle\)ACE = ½ \(\angle\)C
But \(\angle\)B = \(\angle\)C as AB = AC [In isosceles triangle, base angles are equal]
\(\Rightarrow \) \(\angle\)ABD = \(\angle\)ACE
AB = AC [Given]
\(\angle\)A = \(\angle\)A [Common]
Thus by Angle-Side-Angle criterion of congruence, we have
\(\Delta\)ABD \(\cong\) \(\Delta\)ACE
The corresponding parts of the congruent triangles are equal.
BD = CE [c.p.c.t]
Question 3:
In the adjoining figure, AD is a median of \(\Delta\)ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.
Solution:
Given: A triangle in which D is the midpoint of BC and BL \(\perp\) AD and CM \(\perp\) AD
To prove: BL = CM
Proof: In \(\Delta\)BLD and \(\Delta\)CMD
\(\angle\)BLD = \(\angle\)CMD = 90o [Given]
\(\angle\)BDL = \(\angle\)MDC [Vertically opposite angles]
BD = DC [Given]
Thus by Angle-Angle-Side criterion of congruence, we have
\(\Delta\)BLD \(\cong\) \(\Delta\)CMD
The corresponding parts of the congruent triangles are equal
So, BL = CM [c.p.c.t]
Question 4:
In \(\Delta\)ABC, D is the mid-point of BC. If DL \(\perp\) AB and DM \(\perp\) AC such that DL = DM, prove that AB = AC.
Solution:
Given: In a \(\Delta\)ABC, D is the midpoint of BC and DL \(\perp\) AB and DM \(\perp\) AC. Also, DL = DM
To prove: AB = AC
Proof: In right angled triangle \(\Delta\)BLD and \(\Delta\)CMD
\(\angle\)BLD = \(\angle\)CMD = 90o
Hypt. BD = Hypt. CD [Given]
DL = DM [Given]
Thus, by Right Angle-Hypotenuse_side criterion of congruence, we have
\(\Delta\)BLD \(\cong\) \(\Delta\)CMD
The corresponding parts of the congruent triangles are equal.
Therefore, \(\angle\)ABD = \(\angle\)ACD [c.p.c.t]
In \(\Delta\)ABC, we have
\(\angle\)ABD = \(\angle\)ACD
\(\Rightarrow \) AB = AC [Since opposite to equal angles are equal]
Question 5:
In \(\Delta\)ABC, AB = AC and the bisectors of \(\angle\)B and \(\angle\)C meet at a point O. Prove that BO = CO and the ray AO is the bisector of \(\angle\)A.
Solution:
Given: A \(\Delta\)ABC in which AB = AC, BO and CO are bisectors of \(\angle\)B and \(\angle\)C
To prove: In \(\Delta\)BOC, we have,
\(\angle\)OBC = ½ \(\angle\)B
and, \(\angle\)OCB = ½ \(\angle\)C
But, \(\angle\)B = \(\angle\)C [therefore AB = AC (given)]
So, \(\angle\)OBC = \(\angle\)OCB
Since base angles are equal, sides are equal
\(\Rightarrow \) OB = OC ——- (1)
Since OB and OC are the bisectors of angles,
\(\angle\)B and \(\angle\)C respectively, we have
\(\angle\)ABO = ½ \(\angle\)B
\(\angle\)ACO = ½ \(\angle\)C
\(\Rightarrow \) \(\angle\)ABO = \(\angle\)ACO Â ——— (2)
Now, in \(\Delta\)ABO and \(\Delta\)ACO
AB = AC [Given]
\(\angle\)ABO = \(\angle\)ACO [from (2)]
BO = OC [from(1)]
Thus, by Side-Angle-Side criterion of congruence, we have
\(\Delta\)ABO \(\cong\) \(\Delta\)ACO
The corresponding parts of the congruent triangles are equal
Therefore, \(\angle\)BAO = \(\angle\)CAO [by c.p.c.t]
i.e. , AO bisects \(\angle\)A.
Question 6:
In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that
(i) PT = PS
(ii) \(\angle\)PSR = 15o
Solution:
Given: PQR is an equilateral triangle and QRST is a square.
To prove: PT = PS
and \(\angle\)PSR = 15o
Proof: Since \(\Delta\) is an equilateral triangle,
\(\angle\) = 60o and \(\angle\)PRQ = 60o
Since QRTS is a square,
\(\angle\)RQT = 90o and \(\angle\)QRS = 90o
In \(\Delta\)PQT
\(\angle\)PQT = \(\angle\)PQR + \(\angle\)RQT
= 60o + 90o = 150o
In \(\Delta\)PRS
\(\angle\)PRS = \(\angle\)PRQ + \(\angle\)QRS
=60o + 90o = 150o ——- (1)
\(\Rightarrow \) \(\angle\)PQT = \(\angle\)PRS ——— (2)
Thus, in \(\Delta\)PQT and \(\Delta\)PRS
PQ = RS [sides of equilateral triangle PQR]
\(\angle\)PQT = \(\angle\)PRS ——– (2)
QT = RS [sides of square QRST]
Thus, by Side-Angle-Side criterion of congruence, we have
therefore, \(\Delta\)PQT \(\cong\) \(\Delta\)PRS
The corresponding parts of the congruent triangles are equal.
Therefore, PT = PS [c.p.c.t]
Now in \(\Delta\)PRS, we have
PR = RS
\(\Rightarrow \) \(\angle\) RPS = \(\angle\)PSR
But \(\angle\)PRS = 150o [from(1)]
So, by angle sum property in \(\Delta\)PRS
\(\angle\)PRS + \(\angle\)SPR + \(\angle\)PSR = 180o
\(\Rightarrow \) 150o + \(\angle\)PSR + \(\angle\)PSR = 180o
\(\Rightarrow \) 2\(\angle\)PSR = 180o – 150o
\(\Rightarrow \) \(\angle\)PSR = 30o/2 = 15o
Question 7:
In the given figure, ABC is a triangle, right angled at B. If BCDE is a square on side BC and ACFG is a square on AC, prove that AD = BF.
Solution:
Given: ABC is a triangle right angled at B. ACFG is a square and BCDE is a square.
To prove: AD = EF
Proof: Since BCDE is a square,
\(\angle\)BCD = 90o ——- (1)
In \(\Delta\)ACD,
\(\angle\)ACD = \(\angle\)ACB + \(\angle\)BCD
= \(\angle\)ACB + 90o ——- (2)
In \(\Delta\)BCF,
\(\angle\)BCF = \(\angle\)BCA + \(\angle\)ACF
Since ACFG is a square,
\(\angle\)ACF = 90o
Thus, we have
\(\angle\)BCF = \(\angle\)BCA + 90o ——- (3)
From (2) and (3), we have
\(\angle\)ACD = \(\angle\)BCF ——— (4)
Thus in \(\Delta\)ACD and \(\Delta\)BCF, we have
AC = CF [sides of a square]
\(\angle\)ACD = \(\angle\)BCF [from (4)]
CD = BC [sides of a square]
Thus, by Side-Angle-Side criterion of congruence, we have
Therefore, \(\Delta\)ACD \(\cong\) \(\Delta\)BCF
The corresponding parts of congruent triangles are equal.
So, AD = BF (c.p.c.t)
Question 8:
Prove that the median from the vertex of an isosceles triangle is the bisector of the vertical angle.
Solution:
Given: ABC is a isosceles triangle in which AB = AC and AD is the median through A.
To prove: \(\angle\)BAD = \(\angle\)DAC
Proof: In \(\Delta\)ABD and \(\Delta\)ADC
AB = AC [Given]
BD = DC [Given]
AD = AD [Common]
Thus by Side-Side-Side criterion of congruence, we have
\(\Delta\)ABD \(\cong\) \(\Delta\)ADC
The corresponding parts of the congruent triangles are equal.
Therefore, \(\angle\)BAD = \(\angle\)DAC (Proved)
Question 9:
In the given figure, ABCD is quadrilateral in which AB || DC and P is the mid-point of BC. On producing, AO and DC meet at Q. Prove that
(i) AB = CQ
(ii) DQ = DC + AB
Solution:
Given ABCD is a quadrilateral in which AB || DC
To prove:
i) AB = CQ
ii) DQ = DC + AB
Proof: In \(\Delta\)ABP and \(\Delta\)PCQ we have
\(\angle\)PAB = \(\angle\)PQC [alternate angles]
\(\angle\)APB = \(\angle\)CPQ [vertically opposite angles]
BP = PC [Given]
Thus by Angle-Angle-Side criterion of congruence, we have
\(\Delta\)ABP \(\cong\) \(\Delta\)PCQ
The corresponding parts of the congruent triangles are equal
Therefore, AB = CQ ——– (1)
Now, DQ = DC + CQ
= DC + AB [from (1)]
Question 10:
In the given figure, OA = OB and OP = OQ. Prove that
(i) PX = QX
(ii) AX = BX
Solution:
Given: OA = OB and OP = OQ
To prove:
(i) PX = QX
(ii) AX = BX
Proof: In \(\Delta\)OAQ and \(\Delta\)OPB, we have,
OA = OB [Given]
\(\angle\)O = \(\angle\)O [Common]
OQ = OP [Given]
Thus by Side-Angle-Side criterion of congruence, we have
\(\Delta\)OAQ \(\cong\) \(\Delta\)OPB
The corresponding parts of the congruent triangles are equal.
Therefore, \(\angle\)OBP = \(\angle\)OAQ ——— (1)
Thus, in \(\Delta\)BXQ and \(\Delta\)PXA, we have
BQ = OB – OQ
and, PA = OA – OP
But, OP = OQ
and OA =OB [Given]
Therefore, we have, BQ = PA ———- (2)
Now consider triangles \(\Delta\)BXQ and \(\Delta\)PXA.
\(\angle\)BXQ = \(\angle\)PXA [vertically opposite angles]
\(\angle\)OBP = \(\angle\)OAQ [from (1)]
BQ = PA [from (2)]
Thus by Angle-Angle-Side criterion of congruence, we have,
Therefore, \(\Delta\)BXQ \(\cong\) \(\Delta\)PPXA
PX = QX [c.p.c.t]
AX = BX [c.p.c.t]’
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