SELINA Solution Class 9 Chapter 13 Pythagoras Theorem [Proof and simple applications with converse] Exercise 13B

Question 1

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

Sol:

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 ΔABD 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
Hence Proved.

Question 2

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

Sol:

In equilateral Δ ABC, AD ⊥ BC.
Therefore, BC = x cm.

Area of equilateral ΔABC = ${\displaystyle \frac{\sqrt{3}}{4}\times {\text{side}}^2 =\frac{1}{2}\times \text{base}\times \text{height}}$

= ${\displaystyle \frac{\sqrt{3}}{4}\times x^2=\frac{1}{2}\times x\times \text{AD}}$

AD = ${\displaystyle \frac{\sqrt{3}}{2}x}$

Question 3

ABC is a triangle, right-angled at B. M is a point on BC.
Prove that: AM² + BC² = AC² + BM².

Sol:

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 = AB  + 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.

Question 4

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

Sol:

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
= PN + 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 
PM ² + RN²  = 5MN²                 Hence Proved.

(ii) We consider the ΔPQM, and applying Pythagoras theorem we get,
PM² = PQ² + MQ²
4PM² = 4PQ² + 4MQ²             ...[ Multiply both sides by 4]
4PM² = 4PQ² + 4.( 1/2 QR )² ...[ MQ = ${\displaystyle \frac{1}{2}}$ QR ]
4PM² = 4PQ² + 4PQ + 4 . ${\displaystyle \frac{1}{2}}$ QR²
4PM² = 4PQ²  + 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 = ${\displaystyle \frac{1}{2}}$ PQ ]
4RN² = 4QR² + 4 .${\displaystyle \frac{1}{4}}$ 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 
PM ² + RN²  = 5MN² 
4( PM² + RN² ) = 4.5. (NQ² + MQ²)
4( PM² + RN² ) = 4.5. ${\displaystyle [{\left(\frac{1}{2}PQ\right)}^2+{\left(\frac{1}{2}RQ\right)}^2] ....[\because NQ=\frac{1}{2}PQ,MQ=\frac{1}{2}QR]}$
4 ( PM² + RN² ) = 5PR²
Hence Proved.

 

Question 5

In triangle ABC, ∠B = 90^o and D is the mid-point of BC.
Prove that: AC² = AD² + 3CD².

Sol:

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^o 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 - BD²                 ....(i)

Similarly, we get from 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 = ${\displaystyle \frac{1}{2}}$ BC ]
AC² = AD² + 3CD² 
Hence proved.

Question 6

In a rectangle ABCD,
prove that: AC² + BD² = AB² + BC² + CD² + DA².

Sol:

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.

Question 7

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC² - AB² = BC² + CD² + DA²

Sol:

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                   ....(ii)

LHS = 2AC² - AB²
= 2AC² - ( AC² - BC² )           .....[ From(i) ]
= 2AC² - AC² + BC²
= AC² + BC²
= AD² + DC² + BC²            ....[ From(ii) ]
= RHS

Question 8

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

Sol:

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:
OA² = AH² + OH² = AH² + AE²
OC² = CG² + OG² = EB² + HD²
OB² = EO² + BE² = AH² + BE²
OD² = HD² + OH² = HD² + AE²

Adding these equalities we get:
OA² + OC² = AH² + HD² + AE² + EB²
OB² + OD² = AH² + HD² + AE² + EB²

From which we prove that for any point within the rectangle there is the relation
OA² + OC² = OB² + OD²
Hence Proved.

Question 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

Sol:

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,

AQ² = 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.

Question 10

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA² + OC² = 2AD² - ${\displaystyle \frac{{\text{BD}}^2}{2}}$

SOl:

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${\displaystyle {\left(\frac{\text{BD}}{2}\right)}^2}$ ...[ AD = CD and OD = ${\displaystyle \frac{\text{BD}}{2}}$]

= 2AD² - ${\displaystyle \frac{{\left(\text{BD}\right)}^2}{2}}$

= RHS.

Question 11

In figure AB = BC and AD is perpendicular to CD.
Prove that: AC² = 2BC. DC.

Sol:

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.

Question 12

In an isosceles triangle ABC; AB = AC and D is the point on BC produced.
Prove that: AD² = AC² + BD.CD.

Sol:

In an isosceles triangle ABC; AB = AC and
D is the point on BC produced.
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 ΔAEC,
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.

Question 13

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

Sol:

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  ΔACD 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 = 2AB²                                        ...[ AB = AC ]
AB² = ${\displaystyle \frac{1}{2}}$BC²                                   ...(ii)

Putting, AB2 from (ii) in (i), We get,
CD² = AC² + ${\displaystyle \frac{1}{2}}$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.

Question 14

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

Sol:

In right-angled ΔADB,
AB² = AD² + BD²
⇒ AD² = AB² - BD²                                      .....(i)

AC = AD + DC
⇒ AC² = ( AD + DC )²
⇒  AC² = AD² + DC² + 2AD x DC
⇒ AC² = AB² - BD² + DC² + 2AD x DC      ...[ From(i) ]
⇒ AC² = AC² - BD² + DC² + 2AD x DC     ...[ AB = AC ]
⇒ BD² - DC² = 2AD x DC.

Question 15

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

Prove that : 2AC² = 2AB² + BC²

Sol:

Here,
BD : DC = 1 : 3.
⇒ BD = ${\displaystyle \frac{1}{4}\text{BC}\text{and}CD=\frac{3}{4}}$BC

AC² = AD² + CD² and AB² = AD² + BD² 
Therefore,
AC² - AB² = CD² - BD²

= ${\displaystyle {\left(\frac{3}{4}\text{BC}\right)}^2-{\left(\frac{1}{4}\text{BC}\right)}^2}$

= ${\displaystyle \frac{9}{16}{\text{BC}}^2- \frac{1}{16}{\text{BC}}^2}$

= ${\displaystyle \frac{1}{2}{\text{BC}}^2}$

∴ 2AC² - 2AB² = BC²
2AC² = 2AB² + BC²
Hence proved.

SELINA Solution Class 9 Chapter 13 Pythagoras Theorem [Proof and simple applications with converse]Exercise 13A

Question 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.

Sol:

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,
AB²  = BC² + CA² 
13² = 5² + CA² 
CA² = 13² -  5² 
CA²  = 144
CA = 12 m
Therefore, the distance of the other end of the ladder from the ground is 12m.

Question 2

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

Sol:

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
AB² = 4100
AB = 64.03
Therefore the required distance is 64.03 m.

Question 3

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

Sol:

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 ΔPQS and applying Pythagoras theorem we get,
PQ²  = PS² + QS² 
10²  = PS² + 6² 
PS² = 100 - 36
PR  = 8
Now, we consider the ΔPRS and applying Pythagoras theorem we get,
PR²  = RS² + PS² 
PR²  = 15² + 8² 
PR = 17
The length of PR 17 cm.

Question 4

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

Sol:

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 ΔBDC and applying Pythagoras theorem we get,
DB²  = DC²  + BC² 
DB²  = 12²  + 3² 
DB²  = 144  + 9 
DB²  = 153
Now, we consider the ΔABD and applying Pythagoras theorem we get,
DA²  = DB²  + BA² 
13² = 153  + BA²  
BA²  = 169 - 153 
BA = 4
The length of AB is 4 cm.

Question 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.

Sol:

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of the same measure and the perpendicular AD will divide BC into 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 ΔABD and applying Pythagoras theorem we get,
AB²  = AD²  + BD²  
AD²  = 100²  - 5²    ......[ Given, BC = 10 cm = AB, BC = ${\displaystyle \frac{1}{2}}$ BC ]
AD²  = 100 - 25
AD²  = 75
AD = 8.7
Therefore, the length of AD is 8.7 cm

Question 6

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

Sol:

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 ΔABD, and applying Pythagoras theorem we get,
AB²  = AO²  + OB²  
AO²  = AB²  -  OB²
AO²  = AB²  - ( BD+ OC )²          .....[ Let, OC = x ]
AO²   = AB²   - ( BC+ x )²           ......(i)
First, we consider the ΔACO, and applying Pythagoras theorem we get,
AC²  = AO²  -  x ²
AO² = AC²  -  x ²                        ......(ii)

Now, from (i) and (ii),
AB²  - ( BC+ x )² = AC²  - x ² 
8² - ( 6+ x )² = 3²  - x ²    ...[ Given, AB = 8 cm, BC = 8 cm and AC = 3 cm ]
x = ${\displaystyle 1\frac{7}{12}}$ cm
Therefore , the length of OC will be ${\displaystyle 1\frac{7}{12}}$ cm.

Question 7

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

Sol:

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 segments.

First, we consider the ΔABD, and applying Pythagoras theorem we get,
AB² = AD² + BD²
AD² = x² - 5²
AD² = x² - 25
AD = ${\displaystyle \sqrt{x^2-25}}$                .....(i)
Now,
Area = 60
${\displaystyle \frac{1}{2}\times 10\times \text{AD}}$ = 60
${\displaystyle \frac{1}{2}\times 10\times \sqrt{x^2-25}}$ = 60
x = 13.
Therefore, x is 13 cm.

Question 8

If the sides of the triangle are in the ratio 1: ${\displaystyle \sqrt{2}}$: 1, show that is a right-angled triangle.

Sol:

Let, the sides of the triangle be, x: √2x and x.
Now,
x² + x² = 2x² = ${\displaystyle {\left(\sqrt{2}x\right)}^2}$              ....(i)

Here, in (i) it is shown that a square of one side of the given triangle is equal to the addition of square of the 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.

Question 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.

Sol:

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 = 5m            ...( Since DC is perpendicular to BC )
base = 12 cm

Applying Pythagoras theorem we get,
hypotenuse² = 5² + 12²
h² = 25 + 144
h² = 169
h = 13

Therefore, the distance between the tips will be 13m.

Question 10

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

Sol:

Take M to 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 ΔCMB,
CB² = CM² + MB²
CB² = (10)² + (24)²            ...[ MB = AD ]
CB² = 100 + 576
CB² = 676
CB = 26 cm

Question 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.

Sol:

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                                        .….(A - X - B)
⇒ 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
⇒ BC² = AC² - AB²
⇒ BC² = (24)² - (12)²
⇒ BC² = 576 - 144
⇒ BC² = 432
⇒ BC = 12√3 cm
∴ AC = 24 cm and BC = 12√3 cm. 

Question 12

In ΔABC,  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

Sol:

(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 = 0             or         x = - 1
But length of the side of a triangle can not be negative.
⇒ x = 11 cm
∴ AB = ( x - 3 ) = ( 11 - 3 ) = 8 cm
BC = ( x + 4 ) = ( 11 + 4 ) = 15 cm
AC = ( x + 6 ) = ( 11 + 6 ) = 17 cm.

(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 ) = 40 cm
AC = ( 4x + 5 ) = ( 36 + 5 ) = 41 cm.