# ICSE Solutions for Selina Concise Chapter 13 Pythagoras Theorem Class 9 Maths

### Exercise 13(A)

1. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

The pictorial representation of the given problem is given below, Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, AB is the hypotenuse. Therefore applying the Pythagoras theorem we get,

132 = 52 + AC2

⇒ 169 = 25 + AC2

⇒ AC2 = 169 – 25

⇒ AC2 = 144

⇒ AC = 12

Therefore, the distance of the other end of the ladder from the ground is 12m.

2. A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.

Here, we need to measure the distance AB as shown in the figure below,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Therefore, in this case

AB2 = BC2 + CA2

⇒ AB2 = 502 + 402

⇒ AB2 = 2500 + 1600

⇒ AB= 4100

⇒ AB = 64.03

Therefore, the required distance is 64.03 m.

3. In the figure: PSQ = 90o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

PQ2 = PS2 + QS2
⇒ 102 = PS2 + 62
⇒ PS2 = 100 – 36
⇒ PS = 8
Now, we consider the ∆PRS and applying Pythagoras theorem we get,
PR2 = RS2 + PS2
⇒ PR2 = 152 + 82
⇒ PR = 17
The length of PR 17Cm
First, we consider the ΔPQS and applying Pythagoras theorem we get,

4. The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ABD = BCD = 90o. Calculate the length of AB.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔBCD and applying Pythagoras theorem we get,

DB2 = DC2 + BC2
⇒ DB 2 = 122 + 32
⇒ DB 2 = 144 + 9
⇒ DB2 = 153
Now, we consider the ∆ABD and applying Pythagoras theorem we get,
DA 2 = DB 2 + BA2
⇒ 132 = 153 + BA2
⇒ BA = 4
The length of AB is 4 cm.

5. AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of same measure and the perpendicular AD will divide BC in two equal parts..

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, we consider the and applying Pythagoras theorem we get

AB2 = AD2 + BD2
⇒ AD2 = 102 - 52 [Given, BC = 10 CM = AB, BD = (1/2)BC]
⇒ AD2 = 100 – 25
⇒ AD2 = 75
⇒ AD = 8.7
Therefore, the length of AD is 8.7 cm

6. In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC= 3 cm. Calculate the length of OC.

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the and applying Pythagoras theorem we get,

AB2 = AO2 + OB2
⇒ AO2 = AB2 – OB2
⇒ AO2 = AB2 – (BC + OC) [Let, OC = x]
⇒ AO2 = AB2 – (BC + x)2 …(i)
First, we consider the ∆ACO, and applying pythagoras theorem we get,
AC2 = AO2+ x2
⇒ AO2 = AC2 – x2 …(ii)
Now, from (i) and (ii),
AB2 – (BC + x)2 = AC2 – x2
⇒ 82 – (6+x)2 = 32 – x2 [Given, AB = 8cm, BC = 8cm and AC = 3cm]
⇒ x = 1(7/12)cm
Therefore, the length of OC will be 1(7/12)cm

7. In triangle ABC,

AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm2. Find x.

Here, the diagram will be,

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since, ABC is an isosceles triangle, therefore perpendicular from vertex will cut the base in two equal

First, we consider the ΔABD, and applying Pythagoras theorem we get,

AB2 = AD2 + BD2
⇒ AD2 = x2 - 52
⇒ AD2 = x2 – 25
⇒ AD = √(x2 – 25) …(i)
Now,
Area = 60
⇒ (1/2)×10× √(x2 – 25) = 60
⇒ x = 13
Therefore, x is 13cm.

8. If the sides of triangle are in the ratio 1:1, show that is a right-angled triangle.

Let, the sides of the triangle be, x, √2 x and x
Now, x2 + x2 = 2x2 = (√2x)2 …(i)

Here, in (i) it is shown that, square of one side of the given triangle is equal to the addition of square of other two sides. This is nothing but Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore, the given triangle is a right angled triangle.

9. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.

The diagram of the given problem is given below, We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Here, 11 - 6 = 5 m  (Since DC is perpendicular to BC)
Base = 12m
Applying Pythagoras theorem we get,
Hypotenuse2 = 52 + 122
⇒ h2 = 25 + 144
⇒ h2 = 169
⇒ h = 13
Therefore, the distance between the tips will be 13m.

10. In the given figure, AB||CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.

Take M be the point on CD such that AB = DM.

So DM = 7cm and MC = 10 cm

Join points B and M to form the line segment BM.

So BM || AD also BM = AD. In right - angled ΔBAD
BD2 = AD2 +BA2
⇒ (25)2 = AD2 + (7)2
⇒ AD2 = (25)2  - (7)2
⇒ AD2  = 576
⇒ AD = 24
In right - angled △CMB
CB2  = CM2 + MB2
⇒ CB2  = (10)2  + (24)2  [MB = AD]
⇒ CB2  = 100 + 576
⇒ CB2  = 676
⇒ CB = 26CM

11. In the given figure, ∠B = 90°, XY || BC, AB = 12cm, AY = 8cm and AX: XB = 1: 2 = AY: YC. Find the lengths of AC and BC.

Given that AX:XB = 1:2 = AY:YC.

Let x be the common multiple for which this proportion gets satisfied.

So, AX = 1x and XB = 2x

AX + XB = 1x + 2x = 3x

⇒ AB = 3x

⇒ 12 = 3x

⇒ x = 4

AX = 1x = 4 and XB = 2x = 2×4 = 8

Similarly,

AY = 1y and YC = 2y

⇒ AY = 8 (given)

⇒ 8 = y

∴ YC = 2y = 2×8 = 16

∴ AC = AY + YC = 8 + 16 = 24 cm

∆ABC is a right angled triangle. (Given)

∴ By Pythagoras Theorem, we get

⇒ AB2 + BC2 = AC2

⇒ BC= AC2 – AB2

⇒ BC= (24)2 – (12)2

⇒ BC= 576 – 144

⇒ BC= 432

⇒ BC = 12√3 cm
∴AC = 24cm and BC = 12√3 cm

12. In ΔABC, ∠B = 90°. Find the sides of the triangle, if:

(i) AB = (x – 3) cm, BC = (x + 4) cm and AC = (x + 6) cm

(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm (i) In right - angled ΔABC
AC2 = AB2 + BC2
⇒ (x+6)2  = (x-3)2 +(x+4)2
⇒ (x2 + 12x +36) = (x2 - 6x +9) + (x2 + 8x + 16)
⇒ x2 - 10x - 11 = 0
⇒ (x - 11)(x+1) = 0
⇒ x = 11 or x = -1
But length of the side of a triangle can not be negative.
⇒ x = 11cm
∴ AB = (x -3) = (11-3) = 8cm
BC = (x+4) = (11+4) = 15cm
AC = (x+6) = (11+6) = 17cm

(ii) In right - angled ΔABC
AC2 = AB2 + BC2
⇒ (4x +5)2 = (x)2 + (4x+4)2
⇒ (16x2 + 40x + 25) = (x2) + (16x2 + 32x + 16)
⇒ x2 -8x-9 = 0
⇒ (x - 9)(x+1) = 0
⇒ x = 9 or x = -1
But length of the side of a triangle can not be negative.
x = 9 cm
∴ AB = x = 9 cm
BC = (4x +4) = (36+4) = 40cm
AC = (4x + 5) = (36+5) = 41cm

### Exercise 13(B)

1. In the figure, given below, AD BC. Prove that: c2 = a2 + b2 – 2ax.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the ΔACD and applying Pythagoras theorem we get,
AB2 = AD2 + BD2
⇒ c2 = h2 +(a-x)2
⇒ h2 = c2 - (a-x)2  ...(i)
First, we consider the ΔACD and applying Pythagoras theorem we get,
AC2  = AD2 + CD2
⇒ b2 = h2 + x2
⇒ h2 = b2 -x2 ...(ii)
From (i) and (ii) we get,
c2 - (a-x)2  = b2 - x2
⇒ c2 -a2 -x2 +2ax = b2 - x2
⇒ c2 = a2 + b2 -2ax

2. In equilateral Δ ABC, AD…. BC and BC = x cm. Find, in terms of x, the length of AD.

In right - angled ΔADC
AC2 = AD2 + DC2
⇒ (x)2 = AD2 + (x/2)2
⇒ AD2 = (x)2 - (x/2)2
⇒ AD2 = (x/2)2
⇒ AD = (x/2)cm

3. ABC is a triangle, right-angled at B. M is a point on BC. Prove that:

AM2 + BC2 = AC2 + BM2.

The pictorial form of the given problem is as follows,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
First, we consider the ΔABM and applying Pythagoras theorem we get,
AM2 = AB2 + BM2
⇒ AB2 = AM2  - BM2  ...(i)
Now, we consider the ΔABC and applying Pythagoras theorem we get,
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2  ...(ii)
From (i) and (ii) we get,
AM2 - BM2  = AC2  - BC2
⇒ AM2 + BC2 = AC2 + BM2
Hence, Proved

4. M and N are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at . Prove that:

(i) PM2 + RN2 = 5 MN2

(ii) 4 PM2 = 4 PQ2 + QR2

(iii) 4 RN2 = PQ2 + 4 QR2

(iv) 4 (PM2 + RN2) = 5 PR2

We draw, PM,MN,NR

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since, M and N are the mid-points of the sides QR and PQ respectively, therefore, PN=NQ, QM=RM

(i) First, we consider the ΔPQM, and applying Pythagoras theorem we get,
PM2 = PQ2 + MQ2
= (PN + NQ)2 + MQ2
= PN2 + NQ2 + 2PN.NQ + MQ2
= MN2 +PN2 + 2PN.NQ  [From, ΔMNQ, MN2 = NQ2 + MQ2 ] ...(i)
Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN2 = NQ2 + RQ2
= NQ2 + (QM+RM)2
= NQ2 + QM2 + RM2 + 2QM.RM
= MN2 + RM2 + 2QM.RM  ...(ii)
Adding (i) and (ii) we get,
PM2 + RN2 = MN2 + PN2 +2PN.NQ +MN2 +RM2 + 2QM.RM
⇒ PM2 +RM2 = 2MN2 +PN2 + RM2 + 2PN.NQ + 2QM.RM
⇒ PM2 +RN2 = 2MN2 + NQ2 + QM2 + 2(QN2) +2(QM2)
⇒ PM2 +RN2 = 2MN2 + MN2 +2MN2
⇒ PM2 +RN2 = 5MN2
Hence, Proved

(ii) We consider the ΔPQM, and applying Pythagoras theorem we get,
PM2 = PQ2 +MQ2
⇒ 4PM2 = 4PQ2 + 4MQ2  [Multiplying both sides by 4]
⇒ 4PM2 = 4PQ2 + 4[(1/2)QR]2  [MQ = (1/2)QR]
⇒ 4PM2 = 4PQ2 + 4.(1/4)QR2
⇒ 4PM2 = 4PM2 + QR2
Hence, Proved.

(iii) We consider the ΔRQN, and applying Pythagoras theorem we get,
RN2 = NQ2 + RQ2
⇒ 4RN2 = 4NQ2 + 4QR2  [Multiplying both sides by 4]
⇒ 4RN2 = 4QR2 + 4[(1/2)PQ]2  [NQ = (1/2)PQ]
⇒ 4RN2 = 4QR2 + 4. (1/4)PQ2
⇒ 4RN2 = PQ2 + 4QR2
Hence, Proved

(iv) First, we consider the ΔPQM, and applying Pythagoras theorem we get
PM2 = PQ2 + MQ2
= (PN+ NQ)2 + MQ2
= PN2 + NQ2 + 2PN. NQ + MQ2
= MN2 + PN2 + 2PN.NQ  [From, ΔMNQ, MN2 = NQ2 + MQ2 ] ...(i)
Now, we consider the ΔRNQ, and applying Pythagoras theorem we get,
RN2 = NQ2 + RQ2
= NQ2 +(QM+RM)2
= NQ2 + QM2 + RM2 + 2QM. RM
= MN2 + RM2 + 2QM.RM ...(ii)
Adding (i) and (ii) we get,
PM2 + RN2 = MN2 + PN2 + 2PN.NQ+MN2 + RM2 +2QM.RM
⇒ PM2 + RN2 = 2MN2 +PN2 +RM2 + 2PN.NQ+2QM.RM
⇒ PM2 + RN2 = 2MN2  NQ2 + QM2 + 2(QN2)+2(QM2)
⇒ PM2 + RN2 = 2MN2 + MN2 + 2MN2
⇒ PM2 + RN2 = 5MN2
⇒ 4(PM2 + RN2 )= 4.5 (NQ2 + MQ2 )

5. In triangle ABC, ∠B = 90o and D is the mid-point of BC. Prove that: AC2 = AD2 + 3CD2 Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
In triangle ABC, ∠B = 90° and D is the mid-point of BC. Join AD . Therefore, BD = DC
First, we consider the ΔADB, and applying Pythagoras theorem we get,
AD2 = AB2 + BD2
⇒ AB2 = AD2 - BC2  ...(i)
Similarly, we get rom rt. angle triangles ABC we get,
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2  ...(ii)
From (i) and (ii),
AC2 - BC2 = AD2 - BD2
⇒ AC2 =AD2 - BD2 + BC2
⇒ AC2 = AD2 - CD2 + 4CD2  [BD = CD = (1/2)BC]
⇒ AC2 = AD2 + 3CD2
Hence, proved

6. In a rectangle ABCD, prove that:

AC2 + BD2 = AB2 + BC2 + CD2 + DA2. Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
Since, ABCD is a rectangle angles A,B,C and D are rt. angles.
First, we consider the ΔACD, and applying Pythagoras theorem we get,
AC2 = DA2 + CD2  ...(i)
Similarly, we get from rt. angle triangle BDC we get,
BD2 = BC2 + CD2
= BC2 + AB2  [In a rectangle, opposite sides are equal, CD = AB] ...(ii)
Adding (i) and (ii),
AC2 + BD2 = AB2 + BC2 + CD2 + DA2
Hence, proved.

7. In a quadrilateral ABCD, ∠B = 900 and D = 900. Prove that: 2AC2 – AB2 = BC2 + CD2 + DA2

In quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
So, ΔABC and ΔADC are right-angled triangles.
In ΔABC using Pythagoras theorem,
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2 ...(i)
In ΔADC, using Pythagoras theorem,
AC2 = AD2 + DC2
LHS = 2AC2 - AB2
= 2AC2 - (AC2 - BC2) [from(i)]
= 2AC2 - AC2 + BC2
= AC2 + BC2
= AD2 + DC2 + BC2 [from (ii)]
= RHS

8. O is any point inside a rectangle ABCD. Prove that: OB2 + OD2 = OC2 + OA2.

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Using Pythagorean theorem we have from the above diagram:

OA= AH+ OH= AH+ AE2

⇒ OC= CG+ OG= EB+ HD2

⇒ OB= EO+ BE= AH+ BE2

⇒ OD= HD+ OH= HD+ AE2

Adding these equalities we get:

OA+ OC= AH+ HD+ AE+ EB2

⇒ OB+ OD= AH+ HD+ AE+ EB2

From which we prove that for any point within the rectangle there is the relation

OA+ OC= OB+ OD2

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 + BR2

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the ΔARO and applying Pythagoras theorem we get,
AO2 = AR2 + OR2
⇒ AR2 = AO2 - OR2 ...(i)
similarly, from triangles, BPO, COQ, AOQ, CPO and BRO we get the following results,
BP2 = BO2 - OP2 ...(ii)
CQ2 = OC2 - OQ2 ...(iii)
AQ2 = AO2 - OQ2 ...(iv)
CP2 = OC2 - OP2 ...(v)
BR2 = OB2 - OR2 ...(vi)
Adding (i), (ii) and (iii), we get
AR2 + BP2 + CQ2 = AO2 - OR2 + BO2 - OP2 + OC2 - OQ2 ...(vii)
Adding (iv), (v) and (vi), we get,
AQ2 +CP2 + BR2 = AO2 - OR2 + BO2 - OP2 + OC2 - OQ2  ...(viii)
From (vii) and (viii), we get,
AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2
Hence, proved.

10. Diagonals of rhombus ABCD intersect each other at point O. Prove that:

OA2 + OC2 = 2AD2 Diagonals of the rhombus are perpendicular to each other.
In quadrilateral ABCD, ∠AOD = ∠COD = 90°,
So, ΔAOD and ΔCOD are right - angled triangles.
In ΔAOD using Pythagoras theorem,
AD2 = OA2 + OD2
⇒ OA2 =  AD2 - OD2 ...(i)
In ΔCOD using Pythagoras theorem,
CD2 = OC2 + OD2
⇒ OC2 = CD2 - OD2 ...(ii)
LHS = OA2 + OC2
= AD2 - OD2 + CD2 - OD2 [from (i) and (ii)]
= AD2 + CD2 - 2OD2
= AD2 + AD2 - 2(BD/2)2  [AD = CD and OD = BD/2]
= 2AD2 - BD2/2
= RHS

11. In the figure AB = BC and AD is perpendicular to CD. Prove that:

AC2 = 2BC. DC.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the ΔACD and applying Pythagoras theorem we get,

AC2 = AD2 +DC2
= (AB2 - DB2) + (DB+BC)2
=BC2 - DB2 + DB2 + BC2 + 2DB.BC (Given, AB = BC)
= 2BC2 + 2DB.BC
= 2BC(BC+DB)
= 2BC.DC
Hence, Proved.

12. In an isosceles triangle ABC; AB = AC and D is point on BC produced. Prove that:

AD2 = AC2 + BD.CD.

Construct AE perpendicular BC.
Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
We consider the rt. angled ΔAED and applying Pythagoras theorem we get,
AD2 = AE2 + ED2
⇒ AD2 = AE2 +(EC+CD)2 ...(i) [∵ ED = EC + CD]
Similarly, in △AEC,
AC2 = AE2 + EC2
⇒ AE2 = AC2 - EC2 ...(ii)
putting AE2 = AC2 - EC2  in (i), we get,
AD2 = AC2 - EC2 + (EC + CD)2
= AC2 + CD(CD+2EC)
AD2 = AC2 + BD.CD  [∵ 2EC + CD = BD]
Hence, Proved

13. In triangle ABC, angle A = 90o, CA = AB and D is point on AB produced. Prove that DC2 -BD2 = 2AB.AD.

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled and applying Pythagoras theorem we get,
CD2 = AC2 + AD2
⇒ CD2 = AC2 + (AB+ BD)2  [∵ AD = AB + BD]
⇒ CD2 = AC2 + AB2 + BD2 + 2AB. BD ...(i)
Similarly, in ΔABC,
BC2 = AC2 + AB2
⇒ BC2 = 2AB [AB= AC]
⇒ AB2 = (1/2)BC2 ...(ii)
Putting, AB2  from (ii) in (i) we get,
CD2 = AC2 + (1/2)BC2  +  BD2 + 2AB.BD
⇒ CD2 - BD2 = AB2 + AB2 + 2AB.(AD-AB)
⇒ CD2 - BD2 = AB2 + AB2 +2AB.AD - 2AB2
⇒ CD2 - BD2 = 2AB. AD
⇒ DC2 - BD2 = 2AB.AD
Hence, Proved.

14. In triangle ABC, =AB = AC and BD is perpendicular to AC. Prove that: BD2 – CD2 = 2CD × AD.

⇒ AC2 = AD2 + DC2 +2AD×DC
⇒ AC2 = AB2 - BD2 + DC2 + 2AD×DC  [from (i)]
⇒ AC2 = AC2 - BD2 + DC2 + 2AD × DC  [AB = AC]
⇒ BD2 - DC2 = 2AD×DC

15. In the following figure, AD is perpendicular to BC and D divides BC in the ratio 1: 3.

Prove that : 2AC2 = 2AB2 + BC2