ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths Solutions

ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths APC Understanding Solutions. Solutions of  Exercise-12. 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 Exe-12 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 Exe-12 Questions
Edition 2021-2022

Exe-12 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 Exe-12

ML Aggarwal Class 9 ICSE Maths Solutions

Page 253

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

Answer :

We have to use h2 = b2+a2 [Pythagoras theorem]

(i) Given sides are 3 cm, 8 cm and 6 cm

b2+a2 = 32+ 62 = 9 + 36 = 45

⇒ h2 = 82 = 64

Here, 45 ≠ 64

Therefore, the given triangle is not a right triangle.

(ii) Given sides are 13 cm, 12 cm and 5 cm

b2+a2 = 122+ 52 = 144+25 = 169

⇒ h2 = 132 = 169

Here, b2+a2 = h2

Therefore, the given triangle is a right triangle.

Length of the hypotenuse is 13 cm.

(iii) Given sides are 1.4 cm, 4.8 cm and 5 cm

b2+a2 = 1.42+ 4.82 = 1.96+23.04 = 25

⇒ h2 = 52 = 25

Here, b2+a2 = h2

Therefore, the given triangle is a right triangle.

Length of the hypotenuse is 5 cm.


Pythagoras Theorem Exe-12

ML Aggarwal Class 9 ICSE Maths Solutions

Page 254

Question 2. Foot of a 10 m long ladder leaning against a vertical well is 6 m away from the base of the wail. Find the height of the point on the wall where the top of the ladder reaches.

 

Answer :

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

Let PR be the ladder and QR be the vertical wall.

Length of the ladder PR = 10 m

PQ = 6 m

Let height of the wall, QR = h

PR2 = PQ2+QR2

⇒ 102 = 62+QR2

⇒ 100 = 36+QR2

⇒ QR2 = 100-36

⇒ QR2 = 64

Taking square root on both sides,

QR = 8

Therefore, the height of the wall where the top of the ladder reaches is 8 m.

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 taught?

Answer :

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?

Let AC be the wire and AB be the height of the pole.

AC = 24 cm

AB = 18 cm

AC2 = AB2+BC2

⇒ 242 = 182+BC2

⇒ 576 = 324+BC2

⇒ BC2 = 576-324

⇒ BC2 = 252

Taking square root on both sides,

BC = √252

= √(4×9×7)

= 2×3√7

= 6√7 cm

Therefore, the distance is 6√7 cm.

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.

 

Answer :

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.

Let AB and CD be the poles which are 12 m apart.

AB = 6 m

CD = 11 m

BD = 12 m

Draw AE BD

CE = 11-6 = 5 m

AE = 12 m

AC2 = AE2+CE2

⇒ AC2 = 122+52

⇒ AC2 = 144+25

⇒ AC2 = 169

Taking square root on both sides

AC = 13

Therefore, the distance between their tops is 13 m.

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.

Answer :

Hypotenuse, h = 20 cm

Ratio of other two sides, a:b = 4:3

Let altitude of the triangle be 4x and base be 3x.

h2 = b2+a2

⇒ 202 = (3x)2+(4x)2

⇒ 400 = 9x2+16x2

⇒ 25x2 = 400

⇒ x2 = 400/25

⇒ x2 = 16

Taking square root on both sides

x = 4

base, b = 3x = 3×4 = 12

altitude, a = 4x = 4×4 = 16

Therefore, the other sides are 12 cm and 16 cm

Question 6. If the sides of a triangle are in the ratio 3 : 4 : 5, prove that it is right-angled triangle.

Answer :

Pythagoras theorem class 9 chapter 11 ing 4

The sides are in the ratio 3:4:5.

Let ABC be the given triangle.

Let the sides be 3x, 4x and hypotenuse be 5x.

AC2 = BC2+AB2

BC2+AB2= (3x)2+(4x)2

= 9x2+16x2

= 25x2

AC2 = (5x)2 = 25x2

⇒ AC2 = BC2+AB2

Therefore, ABC is a right angled triangle.

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.

Answer :

Pythagoras theorem class 9 chapter 11 ing 5

AC = 2x km

CB = 2(x+7)km

AB = 26

Given AC CB.

AB2 = CB2+AC2

⇒ 262 = ( 2(x+7))2+(2x)2

⇒ 676 = 4(x2+14x+49) + 4x2

⇒ 4x2+56x+196+4x2 = 676

⇒ 8x2+56x+196 = 676

⇒ 8x2+56x +196-676 = 0

⇒ 8x2+56x -480 = 0

⇒ x2+7x -60 = 0

⇒ (x-5)(x+12) = 0

⇒ (x-5) = 0 or (x+12) = 0

⇒ x = 5 or x = -12

Length cannot be negative. So x = 5

BC = 2(x+7) = 2(5+7) = 2×12 = 24 km

AC = 2x = 2×5 = 10 km

Total distance = AC + BC = 10+24 = 34 km

Distance saved = 34-26 = 8 km

Therefore, the distance saved is 8 km.

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.

Answer :

Let the shortest side be x.

Then hypotenuse = 2x+6

Third side = 2x+6-2 = 2x+4

AB2 = CB2+AC2

⇒ (2x+6)2 = x2+(2x+4)2

⇒ 4x2+24x+36 = x2+4x2+16x+16

⇒ x2-8x-20 = 0

⇒ (x-10)(x+2) = 0

⇒ x-10 = 0 or x+2 = 0

⇒ x = 10 or x = -2

So shortest side is 10 m.

Hypotenuse = 2x+6

= 2×10 +6

= 20+6

= 26 m

Third side = 2x+4

= = 2×10 +4

= 20+4

= 24 m

Therefore, the shortest side, hypotenuse and third side of the triangle are 10 m, 26 m and 24 m respectively.

Question 9. ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Answer :

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

Let ABC be the isosceles right angled triangle .

C = 90˚

AC = BC 

AB2 = BC2+AC2

⇒ AB2 = AC2+AC[∵AC = BC]

⇒ AB2 = 2AC2

Hence proved.

Question 10. In a triangle ABC, AD is perpendicular to BC. Prove that AB² + CD² = AC² + BD².

Answer :

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

Given AD BC.

So ADB and ADC are right triangles.

In ADB,

AB2 = AD2+BD2

AD2 = AB2– BD2 …(i)

In ADC,

AC2 = AD2+CD2

AD2 = AC2– CD2 …(ii)

Comparing (i) and (ii)

AB2– BD= AC2– CD2

AB2+ CD2 = AC2+ BD2

Hence proved.

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).

Answer :

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).

PQ = a, PR = b, QD = c and DR = d.

PD QR.

So PDQ and PDR are right triangles.

In PDQ,

PQ2 = PD2+QD2

⇒ PD= PQ2– QD2

⇒ PD2 = a2– c2 …(i) [∵ PQ = a and QD = c]

In PDR,

PR2 = PD2+DR2

⇒ PD2 = PR2– DR2

⇒ PD2 = b2– d2 …(ii) [∵ PR = b and DR = d]

Comparing (i) and (ii)

a2– c2= b2– d2

⇒ a2– b2= c2– d2

⇒ (a + b)(a – b) = (c + d)(c – d)

Hence proved.

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

Answer :

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

Let AD be the altitude of ABC.

Given AB = AC = 12 cm

BC = 8 cm

The altitude to the base of an isosceles triangle bisects the base.

So BD = DC

BD = 8/2 = 4 cm

DC = 4 cm

ADC is a right triangle.

⇒ AD2 = AB2 -BD2

⇒ AD2 = 122-42

⇒ AD2 = 144-16

⇒ AD2 = 128

Taking square root on both sides,

AD = √128 = √(2×64) = 8√2 cm

Area of ABC = ½ ×base ×height

= ½ ×8×8√2

= 4×8√2

= 32√2 cm2

Therefore, the area of triangle is 32√2 cm2.

Question 13. Find the area and the perimeter of a square whose diagonal is 10 cm long.

Answer :

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

Given length of the diagonal of the square is 10 cm.

AC = 10

Let AB = BC = x [Sides of square are equal in measure]

B = 90°

ABC is a right triangle.

AC2 = AB2+BC2

⇒ 102 = x2+x2

⇒ 100 = 2x2

⇒ x2 = 50

⇒ x = √50 = √(25×2)

⇒ x = 5√2

area of square = x2

= (5√2)2 = 50 cm2

Perimeter = 4x

= 4×5√2

= 20√2 cm

Therefore, area and perimeter of the square are 50 cm2 and 20√2 cm.

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.

(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.

Answer :

(i) Given AD = 13 cm, DC = 12 m

BC = 3 cm

ABD = BCD = 90°

BCD is a right triangle.

⇒ BD2 = 32+122

⇒ BD2 = 9+144

⇒ BD2 = 153

ABD is a right triangle.

AD2 = AB2+BD2

⇒ 132 = AB2+153

⇒ 169 = AB2+153

⇒ AB2 = 169-153

⇒ AB2 = 16

Taking square root on both sides,

AB = 4 cm

Therefore, the length of AB is 4 cm.

(ii) Given AB = AD, A = 90° = C, BC = 8 cm and CD = 6 cm

BCD is a right triangle.

BD2 = BC2+DC2

⇒ BD2 = 82+62

⇒ BD2 = 64+36

⇒ BD2 = 100

Taking square root on both sides,

BD = 10 cm

ABD is a right triangle.

BD2 = AB2+AD2

⇒ 102 = 2AB2 [∵AB = AD]

⇒ 100 = 2AB2

⇒ AB2 = 100/2

⇒ AB2 = 50

Taking square root on both sides,

AB = √50

⇒ AB = √(2×25)

⇒ AB = 5√2 cm

Therefore, the length of AB is 5√2 cm.

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.

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.

Answer :

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

ABC is a right triangle.

AC2 = AB2+BC2

⇒ 132 = 122+BC2

⇒ BC2 = 132-122

⇒ BC2 = 169-144

⇒ BC2 = 25

Taking square root on both sides,

BC = 5 cm

CDE is a right triangle.

CE2 = CD2+DE2 

⇒ 102 = CD2+62

⇒ 100 = CD2+36

⇒ CD2 = 100-36

⇒ CD2 = 64

Taking square root on both sides,

CD = 8 cm

BD = BC +CD

⇒ BD = 5+8

⇒ BD = 13 cm

Hence the length of BD is 13 cm.

(b) Given PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm

PSQ is a right triangle.

PQ2 = PS2+QS2

⇒ 102 = PS2+62

⇒ 100 = PS2+36

⇒ PS2 = 100-36

⇒ PS2 = 64

Taking square root on both sides,

PS = 8 cm

PSR is a right triangle.

RS = RQ+QS

⇒ RS = 9+6

⇒ RS = 15 cm

⇒ PR2 = PS2+RS2

⇒ PR2 = 82+152

⇒ PR2 = 64+225

⇒ PR2 = 289

Taking square root on both sides,

PR = 17 cm

Hence the length of PR is 17 cm.

(c) D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm

ADC is a right triangle.

AC2 = AD2+CD2

⇒ 62 = AD2+CD2 …(i)

ABD is a right triangle.

AB2 = AD2+BD2

⇒ 162 = AD2+(BC+CD)2

⇒ 162 = AD2+(12+CD)2

⇒ 256 = AD2+144+24CD+CD2

⇒ 256-144 = AD2+CD2+24CD

⇒ AD2+CD2 = 112-24CD

⇒ 62 = 112-24CD [from (i)]

⇒ 36 = 112-24CD

⇒ 24CD = 112-36

24CD = 76

⇒ CD = 76/24 = 19/6

CD = 3(1/6)

Hence, the length of CD is 3(1/6) cm


Pythagoras Theorem Exe-12

ML Aggarwal Class 9 ICSE Maths Solutions

Page 255

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.

(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.

Answer :

(a) Given BC = 5 cm,

B =90°, AB = 5AE,

CD = 2AE and AC = ED

ABC is a right triangle.

AC2 = AB2+BC

BED is a right triangle.

ED2 = BE2+BD2

⇒ AC2 = BE2+BD…(ii)

Comparing (i) and (ii)

AB2+BC2 = BE2+BD2

⇒ (5AE)2+52 = (4AE)2+(BC+CD)2 [∵BE = AB-AE = 5AE-AE = 4AE]

⇒ (5AE)2+25 = (4AE)2+(5+2AE)2 …(iii)[∵BC = 5, CD = 2AE]

Let AE = x.

So (iii) becomes,

(5x)2+25 = (4x)2+(5+2x)2

⇒ 25x2+25 = 16x2+25+20x+4x2

⇒ 25x2 = 20x2+20x

⇒ 5x2 = 20x

⇒ x = 20/5 = 4

AE = 4 cm

CD = 2AE = 2×4 = 8 cm

AB = 5AE

AB = 5×4 = 20 cm

ABC is a right triangle.

AC2 = AB2+BC2

⇒ AC2 = 202+52

⇒ AC2 = 400+25

⇒ AC2 = 425

Taking square root on both sides,

AC = √425 = √(25×17)

⇒ AC = 5√17 cm

Hence EA = 4 cm, CD = 8 cm, AB = 20 cm and AC = 5√17 cm.

(b) Given D is the midpoint of BC.

DC = ½ BC

ABC is a right triangle.

AB2 = AC2+BC2 (i)

ADC is a right triangle.

AD2 = AC2+DC2 …(ii)

⇒ AC2 = AD2-DC2

⇒ AC2 = AD2– (½ BC)2 [∵DC = ½ BC]

⇒ AC2 = AD2– ¼ BC2

⇒ 4AC2 = 4AD2– BC2

⇒ AC2+3AC2 = 4AD2– BC2

⇒ AC2+BC2 = 4AD2-3AC2

⇒ AB2 = 4AD2-3AC2 [from (i)]

Hence proved.

Question 17. In ∆ ABC, AB = AC = x, BC = 10 cm and the area of ∆ ABC is 60 cm². Find x.

Answer :

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

AB = AC = x

So ABC is an isosceles triangle.

AD ⟂ BC

BD = DC = 10/2 = 5 cm

Given area = 60 cm2

½ ×base ×height = ½ ×10×AD = 60

⇒ AD = 60×2/10

⇒ AD = 60/5

⇒ AD = 12cm

ADC is a right triangle.

AC2 = AD2+DC2

⇒ x2 = 122+52

⇒ x2 = 144+25

⇒ x2 = 169

Taking square root on both sides

x = 13 cm

Therefore, the value of x is 13 cm.

Question 18. In a rhombus, If diagonals are 30 cm and 40 cm, find its perimeter.

Answer :

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

Let ABCD be the rhombus.

Given AC = 30cm

BD = 40 cm

Diagonals of a rhombus are perpendicular bisectors of each other.

OB = ½ BD = ½ ×40 = 20 cm

OC = ½ AC = ½ ×30 = 15 cm

OCB is a right triangle.

BC2 = OC2+OB2

⇒ BC= 152+202

⇒ BC= 225+400

⇒ BC= 625

Taking square root on both sides

BC = 25 cm

So, side of a rhombus, a = 25 cm.

Perimeter = 4a = 4×25 = 100 cm

Hence the perimeter of the rhombus is 100 cm.

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.

(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. 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.

Answer :

(a) AB || DC, BC = AD = 13 cm. (Given )

AB = 22 cm and DC = 12cm

Here DC = 12

MN = 12 cm

AM = BN

AB = AM+MN+BN

⇒ 22 = AM+12+AM

⇒ 2AM = 22-12 = 10

⇒ AM = 10/2

⇒ AM = 5 cm

AMD is a right triangle.

⇒ 132 = 52+DM2

⇒ DM2 = 132-52

⇒ DM2 = 169-25

⇒ DM2 = 144

Taking square root on both sides,

DM = 12 cm

Hence the height of the trapezium is 12 cm.

(b) Given AB || DC, A = 90°, DC = 7 cm,

AB = 17 cm and AC = 25 cm

ADC is a right triangle.

AC2 = AD2+DC

⇒ 252 = AD2+72

⇒ AD2 = 252-72

⇒ AD2 = 625-49

⇒ AD2 = 576

Taking square root on both sides

AD = 24 cm

CM = 24 cm

DC = 7 cm

AM = 7 cm

BM = AB-AM

⇒ BM = 17-7 = 10 cm

BMC is a right triangle.

BC2 = BM2+CM2

⇒ BC2 = 102+242

⇒ BC2 = 100+576

⇒ BC2 = 676

Taking square root on both sides

BC = 26 cm

Hence length of BC is 26 cm.

(c) (i) Proof:

Given ABCD is a square of side 7 cm.

So AB = BC = CD = AD = 7 cm

Also given AE = FC = CG = HA = 3 cm

BE = AB-AE = 7-3 = 4 cm

BF = BC-FC = 7-3 = 4 cm

GD = CD-CG = 7-3 = 4 cm

DH = AD-HA = 7-3 = 4 cm

A = 90˚

AHE is a right triangle.

HE2 = AE2+AH2

⇒ HE2 = 32+32

⇒ HE2 = 9+9 = 18

⇒ HE = √(9×2) = 3√2 cm

Similarly GF = 3√2 cm

EBF is a right triangle.

⇒ EF2 = 42+42

⇒ EF2 = 16+16 = 32

Taking square root on both sides

EF = √(16×2) = 4√2 cm

HG = 4√2 cm

Now join EG

In EFG

EG2 = EF2+GF2

⇒ EG2 = (4√2)2+(3√2)2

⇒ EG2 = 32+18 = 50

⇒ EG = √50 = 5√2 cm …(i)

Join HF.

Also HF2 = EH2+HG2

= (3√2)2+(4√2)2

= 18+32 = 50

HF = √50 = 5√2 cm …(ii)

From (i) and (ii)

EG= HF

Diagonals of the quadrilateral are congruent. So EFGH is a rectangle.

Hence proved.

(ii) Area of rectangle EFGH = length × breadth

= HE ×EF

= 3√2×4√2

= 24 cm2

Perimeter of rectangle EFGH = 2(length + breadth)

= 2×(4√2+3√2)

= 2×7√2

= 14√2 cm

Therefore, area of the rectangle is 24 cm2 and perimeter is 14√2 cm.

Question 20. AD is perpendicular to the side BC of an equilateral Δ ABC. Prove that 4AD² = 3AB².

Answer :

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

Given AD BC

D = 90˚

Proof:

Since ABC is an equilateral triangle,

AB = AC = BC

ABD is a right triangle.

AB2 = AD2+BD2

BD = ½ BC

AB2 = AD2+( ½ BC)2

⇒ AB2 = AD2+( ½ AB)2 [∵BC = AB]

⇒ AB2 = AD2+ ¼ AB2

⇒ AB2 = (4AD2+ AB2)/4

⇒ 4AB2 = 4AD2+ AB2

⇒ 4AD= 4AB2– AB2

⇒ 4AD= 3AB2

Hence proved.


Pythagoras Theorem Exe-12

ML Aggarwal Class 9 ICSE Maths Solutions

Page 256

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.

Prove that :

(i)4AD2 = 4AC2+BC2

(ii)4BE2 = 4BC2+AC2

(iii)4(AD2+BE2) = 5AB2

 

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.

Answer :

Proof:

(i) C = 90°

So ACD is a right triangle.

Multiply both sides by 4, we get

4AD2 = 4AC2+4CD2

⇒ 4AD2 = 4AC2+4BD2 (∵D is the midpoint of BC, CD = BD = ½ BC)

⇒ 4AD2 = 4AC2+(2BD)2

⇒ 4AD2 = 4AC2+BC2…(i) [∵BC = 2BD]

Hence proved.

(ii) BCE is a right triangle.

BE2 = BC2+CE2

Multiply both sides by 4

4BE2 = 4BC2+4CE2

⇒ 4BE2 = 4BC2+(2CE)2

⇒ 4BE2 = 4BC2+AC2 …(ii) (∵E is the midpoint of AC, AE = CE = ½ AC)

Hence proved.

(iii) Adding (i) and (ii)

4AD2+4BE2 = 4AC2+BC2+4BC2+AC2

⇒ 4AD2+4BE2 = 5AC2+5BC2

⇒ 4(AD2+BE2 ) = 5(AC2+BC2)

⇒ 4(AD2+BE2 ) = 5(AB2) (∵ABC is a right triangle, AB2 = AC2+BC2)

Hence proved.

Question 22. If AD, BE and CF are medians of ΕABC, prove that 3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

 

Answer :

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

Draw APBC

Proof:

APB is a right triangle.

⇒ AB2 = AP2+(BD-PD)2

⇒ AB2 = AP2+BD2+PD2-2BD×PD

⇒ AB2 = (AP2+PD2)+BD2-2BD×PD

⇒ AB= AD2+ (½ BC)2-2×( ½ BC)×PD [∵AP2+PD2 = AD2 and BD = ½ BC]

⇒ AB= AD2+ ¼ BC2– BC×PD …(i)

APC is a right triangle.

⇒ AC2 = AP2+(PD2+DC2)

⇒ AC2 = AP2+PD2+DC2+ 2×PD×DC

⇒ AC2 = (AP2+PD2)+ (½ BC)2+2×PD×( ½ BC) [DC = ½ BC]

⇒ AC2 = (AD)2+ ¼ BC2+PD× BC …(ii) [In APD, AP2+PD2 = AD2]

Adding (i) and (ii), we get

AB2+AC2 = 2AD2+ ½ BC2 …(iii)

Draw perpendicular from B and C to AC and AB respectively.

Similarly we get,

BC2+CA2 = 2CF2+ ½ AB2 …(iv)

AB2+BC2 = 2BE2+ ½ AC2 …(v)

Adding (iii), (iv) and (v)

2(AB2+BC2+CA2) = 2(AD2+BE2+CF2)+ ½ (BC2+AB2+AC2)

⇒ 2(AB2+BC2+CA2) = 2(AB2+BC2+CA2) – ½ (AB2+BC2+CA2)

⇒ 2(AD2+BE2+CF2) = (3/2)× (AB2+BC2+CA2)

⇒ 4(AD2+BE2+CF2) = 3(AB2+BC2+CA2)

Hence proved.

Question 23. 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.

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.

Answer :

Given diagonals of quadrilateral ABCD, AC and BD intersect at O at right angles.

Proof:

AOB is a right triangle.

COD is a right triangle.

CD2 = OC2+OD2 …(ii)

Adding (i) and (ii)

AB2+ CD2 = OB2+OA2+ OC2+OD2

⇒ AB2+ CD2 = (OA2+OD2)+ (OC2+OB2) …(iii)

AOD is a right triangle.

AD2 = OA2+OD2 …(iv)

BOC is a right triangle.

BC2 = OC2+OB2 …(v)

Substitute (iv) and (v) in (iii), we get

AB2+ CD2 = AD2+BC2

Hence proved.

Question 24. In a quadrilateral, ABCD∠B = 90° = ∠D. Prove that 2 AC² – BC2 = AB² + AD² + DC².

Answer 

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

Given B = D = 90˚

So ABC and ADC are right triangles.

In ABC,

AC2 = AB2+BC2 …(i)

In ADC,

AC2 = AD2+DC2 …(ii)

Adding (i) and (ii)

2AC2 = AB2+BC2+ AD2+DC2

⇒ 2AC2 -BC2 = AB2+AD2+DC2

Hence proved.

Question 25. In a ∆ ABC, ∠ A = 90°, CA = AB and D is a point on AB produced. Prove that :

DC² – BD² = 2AB. AD.

Answer :

In a ∆ ABC, ∠ A = 90°, CA = AB and D is a point on AB produced. Prove that : DC² – BD² = 2AB. AD.

Given A = 90°

CA = AB

Proof:

In ACD,

⇒ DC2 = CA2+(AB+BD)2

⇒ DC2 = CA2+AB2+BD2+2AB×BD

⇒ DC2 -BD= CA2+AB2+2AB×BD

⇒ DC2 -BD2 = AB2+AB2+2AB×BD

⇒ DC2 -BD2 = 2AB2+2AB×BD

⇒ DC2 -BD2 = 2AB(AB+BD)

⇒ DC2 -BD2 = 2AB×AD [A-B-D]

Hence proved.

Question 26.  In an isosceles triangle ABC, AB = AC and D is a point on BC produced. Prove that AD² = AC² + BD.CD.

Answer :

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

Given ABC is an isosceles triangle.

AB = AC

Draw AP BC

Proof:

APD is a right triangle.

⇒ AD2 = AP2+(PC+CD)2 [PD = PC+CD]

⇒ AD2 = AP2+PC2+CD2 +2PC×CD …(i)

APC is a right triangle.

AC2 = AP2+PC2 …(ii)

Substitute (ii) in (i)

AD2 = AC2 +CD2+2PC×CD …(iii)

PC = ½ BC [The altitude to the base of an isosceles triangle bisects the base]

AD2 = AC2 +CD2+2× ½ BC ×CD

⇒ AD2 = AC2 +CD2+BC×CD

⇒ AD2 = AC2 +CD(CD+BC)

⇒ AD2 = AC2 +CD×BD [CD+BC = BD]

⇒ AD2 = AC2 +BD×CD

Hence proved.

—  : End of ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths Solutions :–

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

Thanks

Please Share with Your Friends

2 thoughts on “ML Aggarwal Pythagoras Theorem Exe-12 Class 9 ICSE Maths Solutions”

  1. can you pls try to give this chapters answers fast i really need help for some problems in this chapters
    by the way thnx for existing 🙂 what would i do without u guys’s help

    Reply

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.