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## S Chand Class 9 ICSE Maths Solutions Chapter 10 Pythagoras Theorem Ex 10(B)

Question 1.

ABCD is a square, prove that AC² = 2AB².

Solution:

Given : ABCD is a square in which AC is its diagonal

To prove : AC² = 2Ab²

Proof: In ∆ABC, ∠B = 90°

∴ AC² = AB² + BC² (Pythagoras Theorem)

= (AB)² + (AB)²

{∵ AB = BC sides of the square}

= 2Ab²

Hence AC² = 2AB²

Question 2.

In the figure, AB = BC = CA = 2a and segment AD ⊥ side BC. Draw that

(i) AD = a\(\sqrt{3}\)

(ii) area of ∆ABC = a²\(\sqrt{3}\)

Solution:

Given : In ∆ABC,

AB = BC = CA = 2a

AD ⊥ BC

To prove :

(i) AD = a\(\sqrt{3}\)

(ii) area of ∆ABC = a²\(\sqrt{3}\)

Proof:

(i) ∵ The sides of ∆ABC are equal (each = 2d)

∴ It is an equilateral triangle

∴ AD bisects BC at D

i.e. BD = DC = a

Now in right ∆ABD

AB² = AD² + BD² (Pythagoras Theorem)

⇒ (2a)² = AD² + (a)²

⇒ 4a² = AD² + a² ⇒ AD² = 4a² – a²

⇒ AD² = 3a² ⇒ AD = \(\sqrt{3}\)

Hence AD = a

(ii) Now area of ∆ABC = \(\frac { 1 }{ 2 }\) base x altitude

= \(\frac { 1 }{ 2 }\) BC x AD

= \(\frac { 1 }{ 2 }\) (2a) (\(\sqrt{3}\)a)

= \(\sqrt{3}\) a² = a²\(\sqrt{3}\)

Hence proved.

Question 3.

In the figure, prove that Ab² – AD² = CD² – CB².

Solution:

Given : In quadrilateral ABCD,

∠B = 90° and ∠D = 90°

To prove : Ab² – AD² = CD² – CB²

Construction : Join AC

Proof: In right ∆ABC, ∠B = 90°

∴ AC² = AB² + BC² … (i)

(Pythagoras Theorem) Similarly in right ∆ADC,

AC² = AD² + CD² … (ii)

From (i) and (ii),

AB² + BC² = AD² + CD²

⇒ AB² – AD² = CD² – BC²

⇒ AB² – AD² = CD² – CB²

Hence proved.

Question 4.

In a ∆ABC, AD ⊥ BC. Prove that AB² + CD² = AC² + BD².

Solution:

Given : In ∆ABC, AD ⊥ BC

To prove : AB² + CD² = AC² + BD²

Proof: In right ∆ABD (∵ AD ⊥ BC)

AB² = BD² + AD² (Pythagoras Theorem)

⇒ AD² = AB² – BD² … (i)

Similarly in right ∆ACD

AC² = AD² + CD²

⇒ AD² = AC² – CD² … (ii)

From (i) and (ii),

AB² – BD² = AC² – CD²

∴ AB² + CD² = AC² + BD²

Hence proved.

Question 5.

In a quadrilateral ABCD, the diagonals AC, BD intersect at right angles. Prove that Ab² + CD² = BC² + DA²

[No marks will be given if it is assumed that ABCD is either a rhombus or a square].

Solution:

Given: In quadrilateral ABCD, diagonals AC and BD intersect each other at right angles at O

To prove : AB² + CD² = BC² + DA²

Proof: In right ∆ABO,

AB² = AO² + BO² (Pythagoras Theorem)

Similarly in right ∆BOC,

BC² = BO² + CO²

In right ∆COD

CD² = CO² + DO²

and in right ∆AOD

DA² = AO² + DO²

Now AB² + CD² = AD² + BO² + CO² + DO² … (i)

and BC² + DA² = BO² + CO² + AO² + DO²

= AO² + BO² + CO² + DO² … (ii)

From (i) and (ii),

AB² + CD² = BC² + DA²

Hence proved.

Question 6.

In ∆ABC, ∠B = 90° and D is mid-point of BC. Prove that

(i) AC² = AD² + 3CD²

(ii) BC² = 4 (AD² – AB²)

Solution:

Given : In ∆ABC, ∠B = 90°

D is mid-point of BC

AD is joined

To prove :

(i) AC² = AD² + 3CD²

(ii) BC² = 4 (AD² – AB²)

Proof:

In right ∆ABD,

AD² = AB² + BD² (Pythagoras Theorem)

⇒ AD² = AB² + (\(\frac { 1 }{ 2 }\)BC)² (∵ D is mid-point of BC)

⇒ AD² = AB² + \(\frac{B C^2}{4}\)

⇒ 4AD² = 4AB² + BC²

∴ BC² = 4AD² – 4AB² = 4 (AD² – AB²)

(i) and in right ∆ABC

(ii) AC² = AB² + BC²

= AB² + (2CD)² (∵D is mid-point of BC)

= AB² + 4CD²

= (AD² – BD²) + 4CD² {∵ AD² = AB² + BD²}

= AD² – CD² + 4CD² (∵ BD = CD)

= AD² + 3CD²

Hence proved.

Question 7.

The side BC of a square ABCD is produced to any point E. Prove that AE² = 2BC. BE + CE².

Solution:

Given : ABCD is a square whose side BC is produced to E

EA is joined

To prove : AE² = 2BC.BE + CE²

Proof: In right ∆ABE

AE² = AB² + BE² = AB² + (BC + CE)²

= AB² + BC² + CE² + 2BC.CE

= BC² + BC² + CE² + 2BC (BE – BC)

= 2BC² + CE² + 2BC.BE – 2BC²

= 2BC.BE + CE²

Hence proved.

Question 8.

ABCD is a rhombus. Prove that AC² + BD² = 4AB².

Solution:

Given : In rhombus ABCD, diagonals AC and BD bisect each other at right angle at O

In right ∆AOB, ∠AOB = 90°

∴ AB² = AO² + OB²

⇒ AB² = (\(\frac { 1 }{ 2 }\)AC)² + (\(\frac { 1 }{ 2 }\)²BD

⇒ AB² = \(\frac { 1 }{ 4 }\) AC² + \(\frac { 1 }{ 4 }\)BD²

⇒ 4AB² = AC² + BD²

Hence AC² + BD² = 4AB²

Hence proved.

Question 9.

In the figure, ∠B of ∆ABC is an acute angle and AD ⊥ BC. Prove that

AC² = AB² + BC² – 2BC.BD

Solution:

Given : In ∆ABC, ∠B is an acute angle AD ⊥ BC

To prove : AC² = AB² + BC² – 2BC.BD

Proof: In right ∆ABD

AB² = AD² + BD² (Pythagoras Theorem)

⇒ AD² = AB² – BD² … (i)

Similarly in ∆ADC

AC² = AD² + DC²

= AB² – BD² + (BC – BD)²

= AB² – BD² + BC² + BD² – 2BC.BD

= AB² + BC² – 2BC.BD

Hence proved.

Question 10.

In a quadrilateral ∆BCD, ∠B = 90° and AD² = AB² + BC² + CD². Prove that ∠ACD = 90°.

Solution:

Given : In quadrilateral ABCD, ∠B = 90° and AD² = AB² + BC² + CD²

To prove : ∠ACD = 90°

Proof: In right ∆ABC

AC² = AB² + Bc² … (i)

(Pythagoras Theorem)

∵ AD² = AB² + BC² + CD² (given)

AD² = AC² + CD² [From (i)]

∴ In ∆ACD,

∠ACD = 90°

(Converse of Pythagoras Theorem)

Hence proved.

Question 11.

ABC is a triangle right angled at A and p is the length of the perpendicular from A on BC. Show that

(i) pa = bc Hence deduce that

(ii) \(\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2}\)

Solution:

Given : In ∆ABC, ∠A = 90°

AD ⊥ BC

AD = p, AB = c, BC = a and AC = b

Solution:

(i) pa = bc

(ii) \(\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2}\)

Proof: In right ∆ABC, ∠A = 90°

∴ BC² = AB² + AC² (Pythagoras Theorem)

a² = c² + b² = b² + c² …. (i)

(i) Area of ∆ABC = \(\frac { 1 }{ 2 }\) base x altitude

= \(\frac { 1 }{ 2 }\) BC x AD

= \(\frac { 1 }{ 2 }\) ap

and also area of ∆ABC = \(\frac { 1 }{ 2 }\) AB x AC

= \(\frac { 1 }{ 2 }\) c.b

∴ \(\frac { 1 }{ 2 }\) pa = \(\frac { 1 }{ 2 }\) bc ⇒ Pa = bc

(ii) ∵ pa = bc

Squaring both sides,

p²a² = b²c²

p² (b² + c²) = b²c² [from (i)]

p² = \(\frac{b^2 c^2}{b^2+c^2} \Rightarrow \frac{1}{p^2}=\frac{b^2+c^2}{b^2 c^2}\)

⇒ \(\frac{1}{p^2}=\frac{b^2}{b^2 c^2}+\frac{c^2}{b^2 c^2}\)

⇒ \(\frac{1}{p^2}=\frac{1}{c^2}+\frac{1}{b^2}\)

Hence proved.

Question 12.

In the given figure, ∠B is acute and segment

AD ⊥ side BC. Show that

(i) b² = h² + a² + x² – 2ax

(ii) b² = a² + c² – 2ax

Solution:

Given : In ∆ABC, ∠B acute angle

AD ⊥ BC

To prove :

(i) b² – h² + a² + x² – 2ax

(ii) b² = a² + c² – 2ax

Proof: In right ∆ABD

AB² = AD² + BD² (Pythagoras Theorem)

⇒ c² = h² + x² … (i)

(ii) b² = a² + h² + x² – 2ax

= a² + c² – 2ax [From (i)]

∴ b² = a² + c² – 2ax

Hence proved.