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.