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.

ML Aggarwal Solution Class 9 Chapter 12 Pythagoras Theorem Exercise 12

  Exercise 12

---

Q1 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 1

Lengths of sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse:

(i) 3 cm, 8 cm, 6 cm

(ii) 13 cm, 12 cm, 5 cm

(iii) 1.4 cm, 4.8 cm, 5 cm

Sol :

We use Pythagoras Theorem's converse :

(i) Sides of a triangle are 3cm, 8cm, 6cm

⇒3²+6²=9+36=45

and 8²=64

∵45≠64

∴It is not a right triangle

(ii) Sides are 13 cm, 12 cm and 5 cm
⇒12²+5²=144+25=169
and 13²=169
∵12²+5²=13²

∴It is a right angled triangle


(iii) 1.4 cm, 4.8 cm, 5 cm
and (1.4)²+(5)²=1.96+25=26.96

and (4.8)²=23.04

∵(1.4)²+(5)²≠4.8²

∴It is not a right angled triangle

---

Q2 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 2

Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Sol :

Let Ab be wall and AC be the ladder

Ladder AC=10 cm

BC=6 cm

Let height of wall AB=h

By Pythagoras Theorem

⇒AC²=BC²+AB²

⇒10²=6²+h²
⇒100=36+h²
⇒h²=100-36=64=(8)²
∵h=8

∴Height of wall=8 cm

---

Q3 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 3

A guy attached a wire 24 m long to a vertical pole of height 18 m and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be tight?

Sol :

Let AB be the pole and AC be the wire attached

AB=18 m and AC=24 m

In right ΔABC,

⇒AC²=BC²+AB² (Pythagoeas Theorem)

⇒24=BC²+18² 
⇒BC²=24²-18²
⇒BC=√576-324=√252

⇒√4×9×7=2×3√7=6√7 m

---

Q4 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 4

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Sol :

To poles AB and CD are 12m apart

AB=6 m , CD=11 m

From A, draw AE||BD

Then AE=BD=12 m

CE=CD-ED=CD-AB

=11-6=5 m

Now in right ΔACE

⇒AC²=AE²+CE² (Pythagoras Theorem)

⇒12²+5²=144+25=169=(13)²

∴AC=13 m

∴Distance between their tops=13 m

---

Q5 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 5

In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is 4:3, find the sides.

Sol :

In right angled triangle hypotenuse=20 cm

ratio of other two sides=4 : 3

Let first side=4x

then Second side=3x

By Pythagoras theorem'

⇒(Hypotenuse)²=(First side)²+(Second side)² 

∴(20)²=(4x)²+(3x)² 

⇒(20)²=16x²+9x² 

⇒400=25x²

⇒x²$=\frac{400}{25}$

⇒x²=16

⇒x=√16=4

∴First side=4x=4×4 cm=16 cm
Second side=3x=3×4 cm=12 cm

Hence , other two sides of right angled triangle=16 cm and 12 cm

---

Q6 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 6

If the sides of a triangle are in the ratio 3:4:5, prove that it is right-angled triangle.

Sol :

Let three sides of given triangle ABC is AB

BC and CA=3 : 4 : 5 

Let AB=3x , BC=4x and CA=5x

Here (AB)²+(BC)²

=(3x)²+(4x)²

=9x²+16x²=25x²

Also, (CA)²=(5x)²=25x²

i.e. (AB)²+(BC)²=(CA)²

Hence ,ABC is right angled triangle

---

Q7 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 7

For going to a city B from city A, there is route via city C such that AC ⊥ CB, AC = 2x km and CB=2(x+ 7) km. It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of highway.

Sol :

In right ΔABC, ∠C=90°

⇒(2x)²+[2(x+7)]²=26²

⇒4x²+4(x²+14x+49)=676

⇒4x²+4x²+56x+196-676=0

⇒8x²+56x-480=0

⇒x²+7x-60=0  (Dividing by 8)

⇒x(x+12)-5(x+12)=0

⇒(x+12)(x-5)=0

Either x+12=0, then x=-12 which is not possible being negative 

or x-5=0, then x=5

Now distance between AC=2x

=2×5=10km

and between BC=2(x+7)=2(5+7)

=2×12=24

∴Distance from A to C and B to C=10+24=34 km

∴Distance saved=34-26=8km

---

Q8 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 8

The hypotenuse of right triangle is 6m more than twice the shortest side. If the third side is 2m less than the hypotenuse, find the sides of the triangle.

Sol :

Let the shortest side of right angled triangle= x m

Hypotenuse=(2x+6) m

Third side=[(2x+6)-2] m

By Pythagoras theorem,

⇒(2x+6)²=x²+[(2x+6)-2]²

⇒4x²+36+24x=x²+(2x+4)²

⇒4x²+36+24x=x²+4x²+16+16x

⇒36+24x=x²+16+16x

⇒0=x²+16+16x-36-24x

⇒0=x²-8x-20

⇒x²-8x-20=0

⇒x-10x+2x-20=0

⇒x(x-10)+2(x-10)=0

Either x+2=0 or x-10=0

x=-2 (Which is not possible)

or x=10

Hence , shortest=x=10 m

Hypotenuse=(2x+6)=(2×10+6)=26 m

Third side=(2x+6)-m=26m-24m=24 m

---

Q9 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 9

ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Sol :

ΔABC is an isosceles right triangle, right angle at C, AC=BC

To prove : AB²=AC²

Proof : In right ΔABC

⇒∠C=90°

⇒AB²=AC²+BC² (Pythagoras Theorem)

=AC²+BC² (∵BC=AC)

=2AC²

---

Q10 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 10

In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².

Sol :

In ΔABC, AD⟂BC

To prove : AB²+CD²=AC²+BD²

Proof : In ΔABC , AD⊥BC

∴ΔABD and ΔACD are right triangle

In right ΔADB

⇒AB²=AD²+BC² (Pythagoras theorem)

⇒AD²=AB²-BD²...(i)

Similarly in right ΔADB

⇒AC²=AD²+CD²

⇒AD²=AC²-CD²..(ii)

From (i) and (ii)

⇒AB²-BD²=AC²-CD²

⇒AB²+CD²=AC²+BD²

---

Q11 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 11

In ∆PQR, PD ⊥ QR, such that D lies on QR. If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a – b) = (c + d) (c – d).

Sol :

In ΔPQR, PQ⟂QR

PQ=a , PR=b, QD=c, DE=d

To prove : (a+b)(a-b)=(c+d)(c-d)

Proof : In ΔPQR, PQ⟂QR

Now in right ΔPQD

⇒PQ²=PD²+QD² (Pythagoras theorem)

⇒PD²=PQ²-QD²=a²-c²..(i)

Similarly in right ΔPDR

⇒PR²=PD²+DR²

⇒PD²=PR²-DR²

⇒b²-d²...(ii)

From (i) and (ii)

⇒a²-c²=b²-d²

⇒a²-b²=c²-d²

⇒(a+b)(a-b)=(c+d)(c-d)

---

Q12 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 12

ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. Find the altitude on BC and Hence, calculate its area.

Sol :

To find : Altitude on BC i.e. value of AD 

In isosceles triangle perpendicular from vertex bisects the base

∴BD=DC

∴BD$=\frac{1}{2}\times 8$=4cm

In right angled triangle ABD

By Pythagoras theorem

⇒AD²+BD²=AB²

⇒AD²+(4)²=(12)²

⇒AD²+16=144

⇒AD²=128

⇒AD=√128=√64×2=8√2

∴Altitude of ΔABC$=\frac{1}{2}\times (base)\times (altitude)$

$=\frac{1}{2}\times 8\times 8\sqrt{2}cm^2$

=4×8√2 cm²

=32√2 cm²

---

Q13 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 13

Find the area and the perimeter of a square whose diagonal is 10 cm long.

Sol :

Let ABCD be a square whose diagonal AC=10 cm

Let length of sides of squared=x cm

In ΔABC

By Pythagoras theorem

⇒AC²+AB²+BC²

⇒(10)²=x²+x²

⇒2x²=100

⇒x²$=\frac{100}{2}=50$

⇒x=√50

⇒x=√25×2

⇒x=5√2

Area of square=side×side

=5√2×5√2

=25×2 cm²

Perimeter of square=4×side

=4×5√2 cm

=20√2 cm

---

Q14 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 14

(a) In fig. (i) given below, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, ∠ ABD = ∠BCD = 90°. Calculate the length of AB.

(b) In fig. (ii) given below, ABCD is a quadrilateral in which AB = AD, ∠A = 90° =∠C, BC = 8 cm and CD = 6 cm. Find AB and calculate the area of ∆ ABD

Sol :

(a) Given : ABCD is a quadrilateral in which AD=13 cm, DC=12 cm, BC=3 cm, and ∠ABD=∠BCD=90°

To calculate : The length of AB

In right angled triangle BCD

By Pythagoras theorem

⇒BD²=BC²+DC²

⇒BD²=(3)²+(12)²

⇒BD²=9+144

⇒BD²=153

Now, In right angled ΔABD,

By Pythagoras theorem

⇒AD²=AB²+BD²

⇒AB²=AD²-BD²

⇒(13)²-153  (∵BD²=153)

⇒169-153=16

⇒AB=√16=4

Hence, Length of AB=4 cm

(b) In right angled triangle BCD

By Pythagoras theorem

⇒BD²=BC²+CD²

⇒(8)²+(6)²

⇒64+36=100

⇒BD=√100=10 cm

∴BD=10 cm

In right angled triangle ABD,

⇒BD²=AB²+AD²

⇒BD²=AB²+AB² (∵AB=AD (given))

⇒(10)²=2AB²

⇒2AB²=100

⇒AB²$=\frac{100}{2}=50$

⇒AB=√50

⇒√25×2=5√2

∴AB=5√2 cm

Area of ΔABD$=\frac{1}{2}\times AB\times AD$

$=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}~cm^2$  (∵AB=AD)

$=\frac{25\times 2}{2}$

=25 cm²

---

Q15 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 15

(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm.Calculate the length of BD.

(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.

Sol :

(a) Here AB=12 cm, AC=13 cm , CE=10 cm and DE=6 cm

To calculate the length of BD

In right angled ΔABC

By Pythagoras theorem,

⇒AC²=AB²+BC²

⇒(13)²=(12)²+BC²

⇒BC²=(13)²-(12)²

⇒BC²=169-144

⇒BC²=25

⇒BC=√25=5

∴BC=5 cm...(1)

In right angled ΔCED

By Pythagoras theorem

⇒CE²=CD²+DE²

⇒(10)²=CD²+(6)²

⇒CD²=100-36

⇒CD²=64

⇒CD=√64

⇒CD=8...(2)

∴CD=8 cm

Hence, length of BD=BC+CD

=5 cm+8 cm  [Putting from (1) and (2)]

=13 cm

(b) Here ∠PSR=90°

PQ=10 cm, QS=6 cm and RQ=9 cm

To calculate the length of PR

In right angled ΔPQS

By Pythagoras theorem

⇒PQ²=PS²+QS²

⇒(10)²=PS²+(6)²

⇒(10)²-(6)²=PS²

⇒100-36=PS²

⇒PS²=64

⇒PS=√64=8

∴PS=8 cm

Now, in right angled ΔPSR

By Pythagoras theorem

⇒PR²=PS²+RS²

⇒PR²=(8)²+(15)²  (RS=RQ+QS)

⇒PR²=64+225

=(9+6)cm=15cm

⇒PR²=289

⇒PR=√289=17

∴PR=17 cm

(c) Here ∠D=90°

⇒AB=16 cm, BC=12 cm and CA=6 cm

To find CD

Let the value of CD= x cm

By Pythagoras theorem

⇒AB²=AD²+BD²

⇒(16)²=AD²+(BC+CD)²

⇒(16)²=AD²+(12+x)²

⇒AD²=(16)²-(12+x)²...(1)

Now, in right angle ΔACD

By Pythagoras theorem

⇒AC²=AD²+CD²

⇒(6)²=[(16)²-(12+x)]²+x² (∵Find (1) putting the value of AD)

⇒36=256-(144+x²+24x)+x²

⇒36=256-144-x²-24x+x²

⇒36=256-144-24x

⇒24x=256-144-36

⇒24x=76

⇒$x=\frac{76}{24}=\frac{19}{6}=3\frac{1}{6}$

Hence , CD$=3\frac{1}{6}$ cm

---

Q16 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 16

(a) In figure (i) given below, BC = 5 cm,

∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.

(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove that AB² = 4AD² – 3AC².

Sol :

(a) Here BC=5 cm , ∠B=90° , AB=5AE , CD=2AE, AC=ED

To calculate the lengths of EA, CD , AB and AC

In right angled ΔABC

By Pythagoras theorem

⇒AC²=AB²+BC²...(i)

Also, in right angled ΔBED

⇒ED²,in right angled ΔBED

⇒ED²=BE²+BD²...(ii)

But AC=ED

⇒AC²=ED²...(iii)

From (i) , (ii) and (iii)

⇒AB²+BC²=BE²+BD²

⇒(5EA)²+(5)²=(4EA)²+(BC+CD)²

(∵BE=AB-EA=5EA-EA=4EA)

⇒25EA²+25=16EA²+(5+2EA)²

(∵CD=2EA)

⇒25EA²+25-16EA²=25+4EA²+20EA

⇒25x²+25-16x²=25+4x²+20x (Let EA=x cm)

⇒9x²-4x²=20x

⇒5x²=20x

⇒x=4 cm  (∵x≠0)

∴EA=4 cm

CD=2AE=2×4 cm=8 cm

AB=5AE=5×4 cm=20 cm

In right angled ΔABC

By Pythagoras theorem

⇒AC²=AB²+BC²

⇒AC²=(20)²+(5)²

⇒AC²=400+25=425

⇒AC=√425=√25×17=5√17

Hence , AC=5√17

(b) In right ΔABC, ∠C=90°

D is mid point of BC

To prove : AB²=4AD²-3AC²

Proof : In right ΔABC,∠C=90° 

⇒AB²=AC²+BC²...(i)

(Pythagoras theorem)

But in right ΔADC

⇒AD²=AC²+DC²

⇒AC²=AD²-DC²...(ii)

From (i) and (ii)

⇒AC²=AD²$-\left(\frac{BC}{2}\right)^2$

(∵D is mid point of BC)

⇒AC²=AD²$-\frac{BC^2}{4}$

⇒4AC²=4AD²-BC²

⇒AC²+3AC²=4AD²-BC²

⇒AC²+BC²=4AD²-3AC²

But BC²+AC²=AB²  [from (i)]

∴AB²=4AD²-3AC²

---

Q17 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 17

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

Sol :

Given : In ∆ABC, AB=AC=x, BC=10 cm and area of ∆ABC=60 cm²

In isosceles triangle ABC

⇒BD$=\frac{1}{2}\times$ BC

⇒BD$=\frac{1}{2}\times 10$ cm=5 cm

In
right angled ABD

By Pythagoras theorem

⇒AB²=BD²+AD²

⇒x²=(5)²+AD²

⇒AD²=x²-25²

⇒AD$=\sqrt{x^2-25}$

Area of ΔABC$=\frac{1}{2}\times base \times height$

⇒$60=\frac{1}{2} \times 10 \times \sqrt{x^{2}-25}$

⇒$\frac{60 \times 2}{10}=\sqrt{x^{2}-25}$

⇒$12=\sqrt{x^{2}-25}$

Squaring both sides , we get

⇒$(12)^{2}=\left(\sqrt{x^{2}-25}\right)^{2}$

⇒144=x²-25

⇒144+25=x²

⇒x²=169

⇒x=√169=13

∴Hence , x=13 cm

---

Q18 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 18

In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.

Sol :

Given : AC=30 cm and BD=40 cm where AC and BD are diagonals of rhombus ABCD

Required : Side of rhombus 

We know that in rhombus diagonals are bisect each other also perpendicular to each other

∴AO$=\frac{1}{2}$AC$=\frac{1}{2}\times 30$ cm=15 cm

and BO$=\frac{1}{2}BD$ $=\frac{1}{2}\times 40$=20 cm

In right angled ΔAOB

By Pythagoras theorem

⇒AB²=AO²+BO²

⇒(15)²+(20)²

⇒225+400=625

⇒AB=√625=25

Side of rhombus (a)=25 cm

Perimeter of rhombus=4a=4×25=100 cm

---

Q19 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 19

(a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the height of the trapezium ABCD.

(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.

(c) In figure (iii) given below, ABCD is a square of side 7 cm. if

AE = FC = CG = HA = 3 cm,

(i) prove that EFGH is a rectangle.

(ii) find the area and perimeter of EFGH.

Sol :

(a) Given : AB||DC , BC=AD=13 cm, AB=22 cm and DC=12 cm

Required : Height of trapezium ABCD

Here CD=MN=12 cm

Also, AM=BN

∴AB=AM+MN+BN

⇒22=AM+12+AM

⇒22-12=2AM

⇒10=2AM

⇒AM$=\frac{10}{2}=5$

∴AM=5 cm

In right angled ΔAMD

⇒AD²=AM²+DM²

⇒(13)²=(5)²+DM²

⇒DM²=(13)²-(5)²

⇒DM²=169-25=144

⇒DM²=√144=12 cm

Hence , height of trapezium=12 cm

(b) Given : AB||DC ,∠A=90°, DC=7 cm

AB=7 cm and AC=25 cm

Required : BC

In right angled triangle

⇒AC²=AD²+CD² (By Pythagoras theorem)

⇒(25)²=AD²+(7)² 

⇒AD²=625-49=576 

⇒AD=√576=24

∴AD=24 cm

Also, DM=MC=24 cm  (∵AB||DC)

Also, Am=DC=7 cm

i.e. AM=7 cm

∴BM=AB-AM=10 cm

In right angled triangle

⇒BC²=MC²+BM² (By Pythagoras theorem)

⇒(24)²+(10)²

⇒576+100=676=(26)²

⇒BC=26

∴BC=26 cm

(c) Given : ABCD is a square of side=7 cm

AE=FC=CG=HA=3 cm

To prove : (i) EFGH is a rectangle

(ii) To find the area and perimeter of EFGH

Proof : BE=BF=DG=DH=7-3=4 cm

In right angled ΔAEH

⇒HE²=HA²+AE² (By Pythagoras theorem)

⇒HE²=(3)²+(3)² 

⇒9+9=18

⇒HE=√18=3√2 cm

∴HE=GF=3√2 cm

Again In right angled ΔEBF

⇒EF²=EB²+BF²

⇒(4)²+(4)²

⇒16+16=32

EF=√32=√16×2=4√2 cm

∴EF=HG=4√2 cm

Join EG

In ΔEFG

⇒EF²+GF²

=(3√2)²+(4√2)²

=18+32=50

Also, ⇒EH²+HG²=(3√2)²+(4√2)²

=18+32=50

∴EF²+GF²=EH²+HG²

i.e. EG²=HF²

i.e. EG=HF

i.e.Diagonals of quadrilaterals are equal

∴EFGH is a rectangle 

Area of rectangle EFGH=HE×EF

=3√2×4√2 cm²

=24 cm²

Perimeter of rectangle EFGH=2(EF+HE)

=2(4√2+3√2)

=2×7√2 cm

=14√2 cm

---

Q20 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 20

AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².

Sol :

Given : ABC is an equilateral triangle and AD⟂BC

AD⟂BC

To prove : 4AD² = 3AB².

Proof : Since ABC is an equilateral triangle 

∴AB=BC=CA

In right angled triangle ABD

⇒AB²=BD²+AD² (by Pythagoras theorem)

⇒AB²=$\left(\frac{BC}{2}\right)^2$+AD²  $\left[\because \mathrm{BD}=\frac{\mathrm{BC}}{2}\right]$

⇒$\mathrm{AB}^{2}=\frac{(\mathrm{AB})^{2}}{4}+\mathrm{AD}^{2}$  [∵AB=BC]

⇒$A B^{2}-\frac{A B^{2}}{4}=A D^{2}$

⇒$\frac{4 \mathrm{AB}^{2}-\mathrm{AB}^{2}}{4}=\mathrm{AD}^{2}$

⇒$\frac{3 \mathrm{AB}^{2}}{4}=\mathrm{AD}^{2}$

⇒$3 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$

⇒$4 \mathrm{AD}^{2}=3 \mathrm{AB}^{2}$

Hence ,the result is proved

---

Q21 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 21

In figure (i) given below, D and E are mid-points of the sides BC and CA respectively of a ΔABC, right angled at C.

Sol :

Prove that :

(i) 4AD²=4AC²+BC²

(ii) 4BE²=4BC²+AC²

(iii) 4(AD²+BE²)=5AB²

(a) Given : In ΔABC , right angled at C. D and E are mid points of the sides BC and CA respectively

To prove : (i) 4AD²=4AC²+BC²

(ii) 4BE²=4BC²+AC²

(iii) 4(AD²+BE²)=5AB²

Proof : In right angle ΔACD,

⇒AD²=AC²+CD² (by Pythagoras theorem)

⇒4AD²=4AC²+4BD²

(Multiplying both sides by 4)

⇒4AD²=4AC²+(2BD)²

⇒4AD²=4AC²+BC²...(1)

(∵2BD=BC ∴D is mid points of BC)

(ii) In right angled ΔBCE

⇒BE²=BC²+CE²  (By Pythagoras theorem)

⇒4BE²=4BC²+4CE² (Multiplying both sides by 4)

⇒4BE²=4BC²+(2CE)²

⇒4BE²=4BC²+AC²...(1)

(∵2CE=AC ∴E is mid points of AC)

Adding (1) and (2) , we get

⇒4AD²+4BE²=4AC²+BC²+AC²

⇒4(AD²+BE²)=5AC²+5BC²

⇒5(AC²+BC²)

⇒5(AB²)

(∵ In right angled ΔABC, AC²+BC²=AB²)

Hence , 4(AD²+BE²)=5AB²

---

Q22 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 22

If AD, BE and CF are medians of ΕABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Sol :

Given : AD, BE and CF are medians of ΔABC.

To prove : 3(AB²+BC²+CA²)=4(AD²+BE²+CF²)

Construction : Draw AP⟂BC

Proof : In right angled ΔAPB

⇒AB²=AP²+BP²

⇒AP²+(BD-PD)²

⇒AP²+BD²+PD²-2BD.PD

⇒(AP²+PD²)+BD²-2BD.PD

⇒$\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{BC}\right)^{2}-2 \times\left(\frac{1}{2} \mathrm{BC}\right) \cdot \mathrm{PD}$

(∵AP²+PD²=AD² and $\mathrm{BD}=\frac{1}{2} \mathrm{BC}$)

⇒$A D^{2}+\frac{1}{4} B C^{2}-B C . P D$...(1)

Now , in ΔAPC

⇒AC²=AP²+PC² (By Pythagoras theorem)

⇒AP²+(PD+DC)²

⇒AP²+PD²+DC²+2PD.DC

⇒$\left(\mathrm{AP}^{2}+\mathrm{PD}^{2}\right)+\left(\frac{1}{2} \mathrm{BC}\right)^{2}+2 \mathrm{PD} \times\left(\frac{1}{2} \mathrm{BC}\right)$ $\left(\because \mathrm{DC}=\frac{1}{2} \mathrm{BC}\right)$

⇒$\mathrm{AD}^{2}+\frac{1}{4} \mathrm{BC}^{2}+\mathrm{PD} \cdot \mathrm{BC}$...(2)

Adding (1) and (2) 

∴$A B^{2}+A C^{2}=2 A D^{2}+\frac{1}{2} B C^{2}$...(3)

Similarly . Draw the perpendicular from B and C on AC and AB respectively, we get

⇒$\mathrm{BC}^{2}+\mathrm{CA}^{2}=2 \mathrm{CF}^{2}+\frac{1}{2} \mathrm{AB}^{2}$...(4)

⇒$\mathrm{AB}^{2}+\mathrm{BC}^{2}=2 \mathrm{BE}^{2}+\frac{1}{2} \mathrm{AC}^{2}$...(5)

Adding (3), (4) and (5), we get

⇒2(AB²+BC²+CA²)

⇒$2\left(\mathrm{AD}^{2}+\mathrm{BE}^{2}+\mathrm{CF}^{2}\right)+\frac{1}{2}\left(\mathrm{BC}^{2}+\mathrm{AB}^{2}+\mathrm{AC}^{2}\right)$

⇒$2\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)-\frac{1}{2}\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\right.$$\left.\mathrm{CA}^{2}\right)=2\left(\mathrm{AD}^{2}+\mathrm{BE}^{2}+\mathrm{CF}^{2}\right)$

⇒$\frac{3}{2}\left(\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CA}^{2}\right)=2\left(\mathrm{AD}^{2}+\mathrm{BE}^{2}+\mathrm{CF}^{2}\right)$

∴3(AB²+BC²+CA²)=4(AD²+BE²+CF²)

Hence proved

---

Q23 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 23

(a) In fig. (i) given below, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that

AB² + CD² = AD² + BC².

(b) In figure (ii) given below, OD⊥BC, OE ⊥CA and OF ⊥ AB. Prove that :

(i) OA² + OB² + OC² = AF² + BD² + CE² + OD² + OE² + OF².

(ii) OAF² + BD² + CE² = FB² + DC² + EA².

Sol :

Given : In quadrilateral ABCD the diagonals AC and BD intersects at O at right angles

To prove : AB²+CD²=AD²+BC²

Proof : In right angled ΔAOB

⇒AB²=AO²+OB²...(1)

(By Pythagoras theorem)

In right angled ΔCOD

⇒CD²=OD²+OC²...(2)

Adding (1) and (2) 

⇒AB²+CD²=(AO+OB)²+(OD²+OC²)

⇒AB²+CD²=(OA²+OD²)+(OB²+OC²)..(3)

Now , in right angled triangle AOD and BOC

By Pythagoras theorem

⇒OA²+OD²=AD²...(4)

⇒OB²+OC²=BC²...(4)

From (3),(4) and (5), we get

⇒AB²+CD²=AD²+BC²

Hence , the result

(b) Given : OD⊥BC, OE⊥CA and OF⊥AB

To prove : 

(i) OA²+OB²+OC²=AF²+BD²+CE²+OD²+OE²+OF²

(ii) AF²+BD²+CE²=FB²=DC²+EA²

Proof : 
In right angled ΔAOF

⇒OA²=AF²+OF²...(1)
In right angled ΔBOD 

⇒OB²=BD²+OD²...(2)

In right angled ΔCOE

⇒OC²=CE²+OE²...(3)

Adding (1),(2) and (3) , we get
⇒OA²+OB²+OC²=AF²+BD²+CE²+OD²+OE²+OF²...(proved (i) part)

(ii) Also, OA²+OB²+OC²
=AF²+BD²+CE²=OA²+OB²+OC²-OD²-OE²-OF²...(4)

Again in ΔBOF, ΔCOD, ΔAOE, 
⇒BF²=OB²-OF²
⇒DC²=OC²-OD²
and EA²=OA²-OE²

Adding above , we get

⇒BF²+DC²+EA²=OB²-OF²+OC²-OD²+OA²-OF²

⇒BF²+DC²+EA²=OA²+OB²+OC²-OD²-OE²-OF²...(5)

From (4) and (5)

⇒AF²+BD²+CE²=BF²+DC²+EA²

Hence , the result

---

Q24 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 24

In a quadrilateral, ABCD ,∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².

Sol :

Given : In quadrilateral ABCD , ∠B = 90° and ∠D = 90°

To prove : 2AC² – BC2 = AB² + AD² + DC².

Construction : Join AC

Proof : In right angled ΔACD

⇒AC²=AB²+BC²...(1)

(by Pythagoras theorem)

In right angled ΔACD

⇒AC²=AD²+DC²...(2)

(by Pythagoras theorem)

Adding (1) and (2) , we get

⇒AC²+AC²=AB²+BC²+AD²+DC²

⇒2AC²=AB²+BC²+AD²+DC²

⇒2AC²-BC²=AB²+AD²+DC²

Hence , the result

---

Q25 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 25

In a ∆ABC, ∠A = 90°, CA = AB and D is a point on AB produced. Prove that :

DC² – BD² = 2AB. AD.

Sol :

Given : ∆ABC in which ∠A = 90° , CA=AB and D is point on AD produced

To prove : DC²-BD²=2AB.AD

Proof : In right angled ΔACD

⇒DC²=AC²+AD²
⇒DC²=AC²+(AB+BD)²

⇒DC²=AC²+AB²+BD²+2AB.BD

⇒DC²-BD²=AC²+AB²+2AB.BD

But AC=AB (given)

⇒DC²-BD²=AB²+AB²+2AB.BD

⇒DC²-BD²=2AB²+2AB.BD

⇒DC²-BD²=2AB(AB+BD)

⇒DC²-BD²=2AB.AD

Hence , the result

---

Q26 | Ex-12 | ML Aggarwal | Class 9 | Pythagoras Theorem | Chapter 12 | myhelper

OPEN IN YOUTUBE

Question 26

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

Sol :

Given : Isosceles ΔABC such that AB=AC.

D is mid point on BC produced

To prove : AD²=AC²+BD.CD

Construction : Draw AP⊥BC

Proof : In right angled ΔAPD

⇒AD²=AP²+PD²

⇒AD²=AP²+(PC+CD)²

⇒AD²=AP²+PC²+CD²+2PC.CD

In right angled ΔAPC

⇒AC²=AP²+PC²

∴AD²=AC²+CD²+2PC.CD

But ΔABC is isosceles triangle and AP⟂BC
∴$\mathrm{PC}=\frac{1}{2} \mathrm{BC}$
∴AD²=AC²+CD²+$2\times \frac{1}{2}$BC.CD
⇒AD²=AC²+CD²+BC.CD

⇒AD²=AC²+CD.BD

i.e. AD²=AC²+BD.CD

Hence, the result

Question P.Q.

(a) In figure (i) given below, PQR is a right angled triangle, right angled at Q. XY is parallel to QR. PQ = 6 cm, PY = 4 cm and PX : QX = 1:2. Calculate the length of PR and QR.

(b) In figure (ii) given below, ABC is a right angled triangle, right angled at B.DE || BC.AB = 12 cm, AE = 5 cm and AD : DB = 1: 2. Calculate the perimeter of A ABC.

(c)In figure (iii) given below. ABCD is a rectangle, AB = 12 cm, BC – 8 cm and E is a point on BC such that CE = 5 cm. DE when produced meets AB produced at F.

(i) Calculate the length DE.

(ii) Prove that ∆ DEC ~ AEBF and Hence, compute EF and BF.

Sol :

(a) Given : In right angled ΔPQR, XY||QR , PQ=6 cm , PY=4 cm and PX : QX=1 : 2

Required : The length of PR and QR

PX : QX=1: 2

Let PX=x cm
then QX=2x cm

∴PQ=PX+QX

⇒6=x+2x

⇒3x=6
⇒$x=\frac{6}{3}=2$

∴PX=2 cm and QX=2×2cm=4 cm

In right angled ΔPXY

⇒PY²=PX²+XY² (by Pythagoras theorem)

⇒(4)²=(2)²+XY² 
⇒XY² =(4)²-4

⇒XY² =12

⇒XY=√12=2√3

Also ,XY||QR

⇒$\frac{P X}{P Q}=\frac{X Y}{Q R}$

⇒$\frac{2}{6}=\frac{2 \sqrt{3}}{\mathrm{QR}}$

⇒6√3

Also , $\frac{P X}{P Q}=\frac{P Y}{P R}$

⇒$\frac{2}{6}=\frac{4}{P R}$

⇒PR$=\frac{6 \times 4}{2}=\frac{24}{2}$=12 cm

Hence, PR=12 cm and QR=6√3 cm

(b) Given : In right angled ΔABC

∠B=90° , DE||BC , AB=12 cm, AE=5 cm and 

AD : DB = 1 : 2

Required : The perimeter of ΔABC

AD :  DB= 1 : 2

Let AD=x cm

then DB=2x cm

∴AB=AD+DB

⇒12=x+2x

⇒3x=12

⇒$x=\frac{12}{3}=4$

∴AD=x=4 cm and DB=2x=2×4=8cm

In right angled ΔADE

⇒AE²=AD²+DE²  (by Pythagoras theorem)

⇒(5)²=(4)²+DE² 

⇒25=16+DE² 

⇒DE²=25-16 

⇒DE²=9 

⇒DE=√9=3 cm

Now, DE||BC (given)

∴$\frac{A D}{A B}=\frac{D E}{B C}$

⇒$\frac{4}{12}=\frac{3}{\mathrm{BC}}$

⇒$\mathrm{BC}=\frac{12 \times 3}{4}=$

⇒3×3=9 cm

Also , $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$

⇒$\frac{4}{12}=\frac{5}{\mathrm{AC}}$

⇒$\mathrm{AC}=\frac{12 \times 5}{4}$

⇒3×5=15

Perimeter of ΔABC=AB+BC+AC

=12cm+9cm+15cm=36cm

(c) Given : ABCD is rectangle , AB=12cm , BC=8 cm, and E is a point on BC such that CE=5 cm

Required : (i) The length of DE

(ii) To prove : ΔDEC~ΔEBF and Hence, find EF and BF

(i) In right angled ΔCDE

DE²=CD²+CE²

DE²=AB²+CE² [CD=AB]

DE²=(12)²+(5)²

DE²=144+25=169

DE=√169=13 cm

(ii) In ΔDEC and ΔEBF

∠DEC=∠BEF (vertically opposite angles)

∠DCE=∠EBF (each 90°)

∴ΔDCE ~ ΔEBF (by A.A axiom of similarity)

∴$\frac{C E}{B E}=\frac{D E}{E F}$

⇒$\frac{5}{3}=\frac{13}{\mathrm{EF}}$  (∵BE=8 cm-5 cm=3 cm)

⇒5×EF=13×3

⇒EF$=\frac{13 \times 3}{5}=\frac{39}{5}$=7.8 cm

Also ,$\frac{C E}{B E}=\frac{D E}{B F}$

⇒$\frac{5}{3}=\frac{12}{\mathrm{BF}}$

(∵BF=8.5 cm-5 cm=3 cm also CD=AB=12 cm)

⇒BF×5=12×3

⇒BF$=\frac{12 \times 3}{5}=\frac{36}{5}$=7.2 cm

Hence , DE=13 cm, EF=7.8 cm and BF=7.2 cm