ML Aggarwal Pythagoras Theorem Chapter Test Class 9 ICSE Maths Solutions

ML Aggarwal Pythagoras Theorem Chapter Test Class 9 ICSE Maths APC Understanding Solutions. Solutions of  Chapter Test . This post is the Solutions of  ML Aggarwal  Chapter 12- Pythagoras Theorem for ICSE Maths Class-9.  APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-12 Pythagoras Theorem for ICSE Board Class-9Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Pythagoras Theorem Chapter Test Class 9 ICSE Maths Solutions

Board ICSE
Publications Avichal Publishig Company (APC)
Subject Maths
Class 9th
Chapter-12 Pythagoras Theorem
Writer ML Aggarwal
Book Name Understanding
Topics Solution of Chapter Test
Edition 2021-2022

Chapter Test Solutions of ML Aggarwal for ICSE Class-9 Ch-12, Pythagoras Theorem

Note:- Before viewing Solutions of Chapter -12 Pythagoras Theorem Class-9 of ML AggarwaSolutions .  Read the Chapter Carefully. Then solve all example given in Exercise-12, MCQs, Chapter Test.


Pythagoras Theorem Chapter Test

ML Aggarwal Class 9 ICSE Maths Solutions

Page 258

Question 1.

(a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC.

(b) In figure (ii) given below, ∠BAC = 90°, ∠ADC = 90°, AD = 6 cm, CD = 8 cm and BC = 26 cm. Find :
(i) AC (ii) AB (iii) area of the shaded region.

(c) In figure (iii) given below, triangle ABC is right angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are mid-points of the sides AB and AC respectively, calculate
(i) the length of BC (ii) the area of ∆ ADE.

1. a) In fig. (i) given below, AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm. Find the length of BC

Answer :

(a) Given AD ⊥ BC, AB = 25 cm, AC = 17 cm and AD = 15 cm

ADC is a right triangle.

⇒ 172 = 152+DC2

⇒ 289 = 225+DC2

⇒ DC2 = 289-225

⇒ DC2 = 64

Taking square root on both sides,

DC = 8 cm

ADB is a right triangle.

⇒ 252 = 152+BD2

⇒ 625 = 225+ BD2

⇒ BD2 = 625-225 = 400

BD = 20 cm

⇒ BC = BD+DC

= 20+8

= 28 cm

Hence the length of BC is 28 cm.

(b) Given BAC = 90°, ADC = 90°

AD = 6 cm, CD = 8 cm and BC = 26 cm.

(i) ADC is a right triangle.

⇒ AC2 = 62+82

⇒ AC2 = 36+64

⇒ AC2 = 100

Taking square root on both sides,

AC = 10 cm

Hence length of AC is 10 cm.

(ii) ABC is a right triangle.

⇒ 262 = 102+AB2

⇒ AB2 = 262-102

⇒ AB2 = 676-100

⇒ AB2 = 576

Taking square root on both sides,

AB = 24 cm

Hence length of AB is 24 cm.

(iii) Area of ABC = ½ ×AB×AC

= ½ ×24×10

= 120 cm2

Area of ADC = ½ ×AD×DC

= ½ ×6×8

= 24 cm2

Area of shaded region = area of ABC- area of ADC

= 120-24

= 96 cm2

Hence, the area of shaded region is 96 cm2.

(c) Given B = 90°

AB = 9 cm, AC = 15 cm .

D, E are mid-points of the sides AB and AC respectively.

(i) ABC is a right triangle.

⇒ 152 = 92+BC2

⇒ 225 = 81+BC2

⇒ BC2 = 225-81

⇒ BC2 = 144

Taking square root on both sides,

BC = 12 cm

Hence, the length of BC is 12 cm.

(ii) AD = ½ AB [D is the midpoint of AB]

AD = ½ ×9 = 9/2

⇒ AE = ½ AC [E is the midpoint of AC]

⇒ AE = ½ ×15 = 15/2

ADE is a right triangle.

⇒ (15/2)2 = (9/2)2+DE2

⇒ DE2 = (15/2)2 – (9/2)2

⇒ DE2 = 225/4 -81/4

⇒ DE2 = 144/4

Taking square root on both sides,

DE = 12/2 = 6 cm.

Area of ADE = ½ ×DE×AD

= ½ ×6×9/2

= 13.5 cm2

Therefore, the area of the ADE is 13.5 cm2.

Question 2. If in ∆ ABC, AB > AC and DI BC, prove that AB² – AC² = BD² – CD².

Answer :

2. If in ∆ ABC, AB AC and AD BC, prove that AB

Given AD BC, AB>AC

So ADB and ADC are right triangles.

Proof:

In ADB,

⇒ AD2 = AB2-BD2 …(i)

In ADC,

AC2 = AD2+CD2

⇒ AD2 = AC2-CD2 …(ii)

Equating (i) and (ii)

AB2-BD2 = AC2-CD2

AB2-AC2 = BD2– CD2

Hence proved.

Question 3.  In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2:1. Prove that

(i) 9AQ² = 9AC² + 4BC²
(ii) 9BP² = 9BC² + 4AC²
(iii) 9(AQ² + BP²) = 13AB².

Answer :

In a right angled triangle ABC, right angled at C, P and Q are the points on the sides CA and CB respectively which divide these sides in the ratio 2:1. Prove that

Join AQ and BP.

Given C = 90°

Proof:

(i) In ACQ,

Multiplying both sides by 9

9AQ2 = 9AC2+9CQ2

⇒ 9AQ= 9AC2+(3CQ)2 …(i)

Given BQ: CQ = 1:2

⇒ CQ/BC = CQ/(BQ+CQ)

⇒ CQ/BC = 2/3

⇒ 3CQ = 2BC …(ii)

Substitute (ii) in (i)

9AQ= 9AC2+(2BC)2

⇒ 9AQ= 9AC2+4BC2 …(iii)

Hence, proved.

(ii) In BPC,

Multiplying both sides by 9, we get

9BP2 = 9BC2+9CP2

⇒ 9BP2 = 9BC2+(3CP)2 …(iv)

Given AP: PC = 1:2

CP/AC = CP/AP+PC

⇒ CP/AC = 2/3

⇒ 3CP = 2AC …(v)

Substitute (v) in (iv)

9BP2 = 9BC2+(2AC)2

⇒ 9BP2 = 9BC2+4AC2 …(vi)

Hence proved.

(iii) Adding (iii) and (vi), we get

9AQ2+9BP2 = 9AC2+4BC2+9BC2+4AC2

⇒ 9(AQ2+BP)2 = 13AC2+13BC2

⇒ 9(AQ2+BP)2 = 13(AC2+BC2…(vii)

In ABC,

AB2 = AC2+BC2 …(viii)

Substitute (viii) in (viii), we get

9(AQ2+BP)2 = 13AB2

Hence proved.

Question 4. In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² – 3PR² + 5PS².

 

Answer :

In the given figure, ∆PQR is right angled at Q and points S and T trisect side QR. Prove that 8PT² – 3PR² + 5PS².

Q = 90°

S and T are points on RQ such that these points trisect it.

So RT = TS = SQ

To prove : 8PT² = 3PR² + 5PS².

Proof:

Let RT = TS = SQ = x

In PRQ,

⇒ PR2 = (3x)2+PQ2

⇒ PR2 = 9x2+PQ2

Multiply above equation by 3

3PR2 = 27x2+3PQ2 …(i)

Similarly in PTS,

⇒ PT2 = (2x)2+PQ2

⇒ PT2 = 4x2+PQ2

Multiply above equation by 8

8PT2 = 32x2+8PQ2 …(ii)

Similarly in PSQ,

⇒ PS2 = x2+PQ2

Multiply above equation by 5

5PS2 = 5x2+5PQ2 …(iii)

Add (i) and (iii), we get

3PR2 +5PS2 = 27x2+3PQ2+5x2+5PQ2

⇒ 3PR2 +5PS2 = 32x2+8PQ2

⇒ 3PR2 +5PS2 = 8PT2 [From (ii)]

8PT2 = 3PR2 +5PS2

Hence proved.

Question 5. In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.

Answer :

In a quadrilateral ABCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.

B = 90˚ in quadrilateral ABCD

AD² = AB² + BC² + CD²

To prove: ACD = 90°

Proof:

In ABC,

AC2 = AB2+BC2 …(i) 

Given,

AD² = AB² + BC² + CD²

⇒ AD² = AC2+CD2 [from (i)]

In ACD, ACD = 90° [Converse of Pythagoras theorem]

Hence proved.

Question 6.  In the adjoining figure, find the length of AD in terms of b and c.

In the given figure, find the length of AD in terms of b and c.

Answer :

A = 90°

AB = c

AC = b

ADB = 90°

In ABC,

⇒ BC2 = b2+c2

⇒ BC = √( b2+c2…(i)

Area of ABC = ½ ×AB×AC

= ½ ×bc …(ii)

Also, Area of ABC = ½ ×BC×AD

= ½ ×√( b2+c2) ×AD …(iii)

Equating (ii) and (iii)

½ ×bc = ½ ×√( b2+c2) ×AD

⇒ AD = bc /(√( b2+c2)

Therefore,

AD is bc /(√( b2+c2).

Question 7. ABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of ∆FBE is 108 cm², find the length of AC.

Answer :

ABCD is a square, F is mid-point of AB and BE is one-third of BC. If area of ∆FBE is 108 cm², find the length of AC.

Let x be each side of the square ABCD.

FB = ½ AB [∵ F is the midpoint of AB]

FB = ½ x …(i)

BE = (1/3) BC

⇒ BE = (1/3) x …(ii)

AC = √2 ×side

⇒ AC = √2x

Area of FBE = ½ FB×BE

108 = ½ × ½ x ×(1/3)x [given area of FBE = 108 cm2]

⇒ 108 = (1/12)x2

⇒ x2 = 108×12

⇒ x2 = 1296

Taking square root on both sides.

x = 36

AC = √2×36 = 36√2

Therefore,

length of AC is 36√2 cm.

Question 8. In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC x CD, Prove that BD = BC.

Answer :

In a triangle ABC, AB = AC and D is a point on side AC such that BC² = AC x CD, Prove that BD = BC.

In ABC, AB = AC

D is a point on side AC such that BC² = AC×CD

To prove :

BD = BC

Construction: Draw BEAC

Proof:

In BCE ,

⇒ BC2 = BE2+(AC-AE)2

⇒ BC2 = BE2+AC2+AE2– 2 AC×AE

⇒ BC2 = BE2+AE2+AC2– 2 AC×AE …(i)

In ABC,

AB2 = BE2+AE2 …(ii)

Substitute (ii) in (i)

BC2 = AB2+AC2– 2 AC×AE

⇒ BC2 = AC2+AC2– 2 AC×AE [∵AB = AC]

⇒ BC2 = 2AC2-2 AC×AE

⇒ BC2 = 2AC(AC-AE)

⇒ BC2 = 2AC×EC

BC² = AC × CD

⇒ 2AC×EC = AC × CD

⇒ 2EC = CD …(ii)

E is the midpoint of CD.

EC = DE …(iii)

In BED and BEC,

EC = DE [From (iii)]

BE = BE [common side]

BED = BEC [By SAS congruency rule]

BD = BD [c.p.c.t]

Hence proved.

—  : End of ML Aggarwal Pythagoras Theorem Chapter Test Class 9 ICSE Maths Solutions :–

Return to :-  ML Aggarawal Maths Solutions for ICSE  Class-9

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