Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13B Solutions
EXERCISE – 13B
(1) In the figure, given below, AD parallel to BC. Prove that c2 = a2 + b2 – 2ax.
=>
To prove :
c2 = a2 + b2 – 2ax
Proof:
In △ ADB,
by Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AB)2 = (BD)2 + (AD)2
c2 = (a – x)2 + h2
c2 – (a – x)2 = h2 ……(i)
In △ ADC,
by Pythagoras theorem,
(AC)2 = (DC)2 + (AD)2
b2 = x2 + h2
b2 – x2 = h2 …..(i)
Equating equations (i) and (ii),
c2 – (a – x)2 = b2 – x2
c2 – (a2 + x2 – 2ax) = b2 – x2
c2 – a2 – x2 + 2ax = b2 – x2
c2 – a2 = b2 – 2ax
c2 = a2 + b2 – 2ax Hence proved.
(2) In equilateral △ ABC, AD parallel to BC and BC = x cm find, in terms of x, the length of AD.
=>
Given :
In equilateral △ ABC, AD parallel to BC and BC = x cm.
In △ ADB,
by Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(AB)2 = (BD)2 + (AD)2
x2 = (x/2)2 + (AD)2
x2 = x2/4 + AD2
x2 – x2/4 = AD2
4x2 – x2/4 = AD2
3x2/4 = AD
Taking square root on both sides,
√3/4 x2 = √AD2
√3/2 x = AD
∴ AD = √3/2 x
(3) ABC is a triangle right angled at B. M is a point on BC. Prove that ; AM2 + BC2 = AC2 + BM2.
=> Given,
ABC is a triangle, right-angled at B . M is a point on BC.
To prove :
AM2 + BC2 = AC2 + BM2
Proof:
In △ ABM,
by Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
AM2 = BM2 + AB2 …..(i)
In △ ABC,
by Pythagoras theorem,
AC2 = BC2 + AB2
AC2 – AB2 = BC2
BC2 = AC2 – AB2 ……(ii)
Adding equation (i) and (ii)
AM2 + BC2 = BM2 + AB2 + AC2 – AB2
AM2 + BC2 = BM2 + AC2
AM2 + BC2 = AC2 + BM2 Hence proved.
(4) M and N are the mid-points of the sides QR and PQ respectively of a triangle PQR, right-angled at Q. Prove that:
(i) PM2 + RN2 = 5MN2
(ii) 4 PM2 = 4PQ2 + QR2
(iii) 4RN2 = PQ2 + 4 QR2
(iv) 4(PM2 + RN2) = 5 PR2
=> Given,
M and N are the mid-points of the sides QR and PQ respectively.
To prove :
(i) PM2 + RN2 = 5MN2
(ii) 4 PM2 = 4PQ2 + QR2
(iii) 4RN2 = PQ2 + 4 QR2
(iv) 4(PM2 + RN2) = 5 PR2
Proof: In △ PQM,
by Pythagoras theorem,
(i) PM2 = PQ2 + QM2 ……. (i)
In △ NQR,
RN2 = QN2 + QR2 …..(ii)
Adding equation (i) and (ii),
PM2 + RN2 = PQ2 + QM2 + QN2 + QR2
PM2 + RN2 = (2QN)2 + QM2 + QN2 + (2QM)2
PM2 + RN2 = 4QN2 + QM2 + QN2 + 4QM2
PM2 + RN2 = 5QN2 + 5QM2
PM2 + RN2 = 5 (QN2 + QM2)
PM2 + RN2 = 5 (MN)2
(ii) In △ PQM2,
by Pythagoras theorem,
PM2 = PQ2 + QM2
PM2 = PQ2 + (QR/2)2
PM2 = PQ2 + QR2/4
PM2 = 4PQ2 + QR2/4
4PM2 = 4PQ2 + QR2 ….. (i)
(iii) 4RN2 = PQ2 + 4PQ2
In △ NQR,
by Pythagoras theorem,
RN2 = QN2 + QR2
RN2 = (PQ/2)2 + QR2
RN2 = PQ2/4 + QR2
RN2 = PQ2 + 4QR2/4
4RN2 = PQ2 + 4QR2 ….. (ii)
(iv) 4(PM2 + RN2) = 5PR2
Adding equation (i) and (ii),
4PM2 + 4RN2 = 4PQ2 + QR2 + PQ2 + 4QR2
4 (PM2 + RN2) = 4PQ2 + QR2 + PQ2 + 4QR2
4 (PM2 + RN2) = 5PQ2 + 5QR2
4 (PM2 + RN2) = 5 (PQ2 + QR2)
4 (PM2 + RN2) = 5PR2 (∵ PR2 = PQ2 + QR2)
Hence proved.
(5) In triangle ABC, ∠B = 90° and D is the mid-point of BC prove that;
AC2 = AD2 + 3CD2
=> Given :
In triangle ABC, ∠90° and D is the mid-point of BC
To prove :
AC2 = AD2 + 3CD2
Proof : In △ ABC,
AC2 = AB2 + BC2 …..(i)
In △ ABD,
AD2 = AB2 + BD2
AD2 – BD2 = AB2
from equation (i),
AC2 = AD2 – BD2 + BC2
AC2 = AD2 – CD2 + (2CD)2
AC2 = AD2 – CD2 + 4CD2
AC2 = AD2 + 3CD2
Hence proved.
(6) In a rectangle ABCD, prove that;
AC2 + BD2 = AB2 + BC2 + CD2 + DA2
=>
To prove :
ABCD is a rectangle, AC2 + BD2 = AB2 + BC2 + CD2 + DA2
Proof :
In △ ABC,
by Pythagoras theorem,
AC2 = AB2 + BC2 ….. (i)
In △ BAD,
BD2 = AB2 + AD2 …… (ii)
Adding equation (i) and (ii),
AC2 + BD2 = AB2 + BC2 + AB2 + AD2
AC2 + BD2 = AB2 + BC2 + CD2 + AD2
(AB = CD)
Hence proved.
(7) In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°. Prove that : 2AC2 – AB2 = BC2 + CD2 + DA2
=> Given,
ABCD is a quadrilateral,
∠B = 90° and ∠D = 90°
To prove:
2AC2 – AB2 = BC2 + CD2 + DA2
Proof:
In △ ABC,
by Pythagoras Theorem,
AC2 = AB2 + BC2 …. (i)
In △ ADC,
AC2 = AD2 + CD2 …… (ii)
Adding equation (i) and (ii),
AC2 + AC2 = AB2 + BC2 + AD2 + CD2
2AC2 – AB2 = BC2 + CD2 + AD2 Hence proved.
(8) O is any point inside a rectangle ABCD. Prove that : OB2 + OD2 = OC2 + OA2
=> Given,
O is any point inside a rectangle ABCD.
To prove :
OB2 + OD2 = OC2 + OA2
Proof: In △ OMB,
by Pythagoras Theorem,
OB2 = OM2 + MB2 …… (i)
In △ OPD,
by Pythagoras theorem,
OD2 = OP2 + DP2 …… (ii)
Adding equation (i) and (ii),
OB2 + OD2 = OM2 + MB2 + DP2 + DP2
OB2 + OD2 = OM2 + NC2 + OP2 + ON2
OB2 + OD2 = OM2 + OP2 + (NC2 + ON2)
OB2 + OD2 = (OM2 + AM2) + (NC2 + ON2)
OB2 + OD2 = OA2 + OC2 Hence proved.
(9) In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.
Prove that : AR2 + BP2 + CQ2 = AQ2 + CP2 + BP2
=>
To prove :
AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2
Proof :
In △ ARO,
by Pythagoras Theorem,
AO2 = AR2 + RO2
AO2 – RO2 = AR2 ….. (i)
In △ OPB,
by Pythagoras Theorem,
OB2 = OP2 + BP2
OB2 – OP2 = BP2 ……. (ii)
In △ OQC,
by Pythagoras Theorem,
OC2 = QC2 + CQ2
OC2 – QC2 = CQ2 …… (iii)
Adding equations (i), (ii) and (iii),
AO2 – RO2 + OB2 – OP2 + OC2 – QC2 = AR2 + BP2 + CQ2
(AO2 – OQ2) + (OC2 – OP2) + (OB2 – OP2) = AR2 + BP2 + CQ2
AQ2 + PC2 + BP2 = AR2 + BP2 + CQ2 Hence proved.
(10) Diagonals of rhombus ABCD intersect each other at point o prove that:
OA2 + OC2 = 2AD2 – BD2/2
=>
Given,
Diagonals of rhombus ABCD intersect each other at point O.
To prove :
OA2 + OC2 = 2AD2 – BD2/2
Proof :
In △ AOD,
by Pythagoras theorem,
AD2 = AO2 + DO2
AD2 – DO2 = AO2
AO2 = AD2 – DO2 ….. (i)
In △ COD,
by Pythagoras theorem,
BC2 = OC2 + OB2
BC2 – OB2 = OC2
OC2 = BC2 – OB2 …… (ii)
Adding equations (i) and (ii),
AO2 + OC2 = AD2 – DO2 + BC2 – OB2
OA2 + OC2 = AD2 + AD2 – (BD/2)2 – (BD/2)
OA2 + OC2 = 2AD2 – 2(BD/2)2
= 2AD2 – 2 × BD2/4
OA2 + OC2 = 2AD2 – BD2/2 Hence proved.
Here is your solution of Selina Concise Class 9 Maths Chapter 13 Pythagoras Theorem Exercise 13B
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.