Interactive ICSE Class 9 Maths Solutions S Chand Chapter 10 Pythagoras Theorem Ex 10(A) engage students in active learning and exploration.

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

Question 1.

In a right angled triangle ABC, c is the length of the hypotenuse, and a and 6 are other two sides.

(a) If a = 6 and b = 8, then find c.

(b) If c = 25 and a = 24, then find b.

(c) If c = 13 and 6 = 5, then find a.

(d) If a = 10 and c = 21, then find b.

(e) If a = 9 and 6 = 9, then find c.

Solution:

In right angled ∆ABC

(a) a = 6, 6 = 8, then

c² = a² + b² (by pythagoras theorem)

∴ c = \(\sqrt{a^2+b^2}\)

= \(\sqrt{(6)^2+(8)^2}=\sqrt{36+64}\)

= \(\sqrt{100}\) = 100

∴ c = 10

(b) c = 25, a = 24

∵ c² = a² + b²

⇒ (25)² = (24)² + b²

⇒ 625 = 526 + b²

⇒ b² = 625 – 576 = 49 = (7)²

∴ b = 7

(c) c = 13, b = 5

∵ c² = a² + b²

⇒ (13)² = a² + (5)²

⇒ 169 = a² + 25

⇒ a² = 169 – 25

⇒ a² = 144 = (12)²

∴ a = 12

(d) a = 10, c = 21

∵ c² = a² + b²

∴ (21)² = (10)² + b²

⇒ 441 = 100 + b²

⇒ b² = 441 – 100

⇒ b² = 314 ⇒ b = \(\sqrt{341}\)

(e) a = 9, b = 9

∵ c² = a² + b²

⇒ (9)² + (9)² = 81 + 81 = 162

= 2 x 81 = (\(\sqrt{2}\) x 9)²

= (9\(\sqrt{2}\))²

c = 9\(\sqrt{2}\)

Question 2.

A rectangular field is 30 m by 40 m. What distance is saved by walking diagonally across the field.

Solution:

In the rectangular field, length (l) = 40 m

Breadth (b) = 30 m

BD is its diagonal

∴ In right ∆ABD,

BD² = AB² + AD²

(Pythagoras Theorem)

= (40)² + (30)² = 1600 + 900

= 2500 = (50)²

∴ BD = 50 m

Actual saving = (AB + AD) – BD

= (40 + 30 – 50) = 20 m

Question 3.

A man travels 7 km due north, then goes 3 km due east and then 3 km due south. How far is he from his starting point?

Solution:

A man travels 7 km due north and then 3 km east then 3 km due south

OA = 7 km, AB = 3 km, BC = 3 km

Join OC and draw CD || AB

Then OD = AO – AD

= 7 – 3 = 4 km

Now in right ∆OCD,

OC² = OD² + CD² (Pythagoras Theorem)

= (4)² + (3)²

= 16 + 9 = 25 = (5)²

∴ OC = 5

∴ He is 5 km from the starting point.

Question 4.

The diagonals of a rhombus are 12 cm and 9 cm long. Calculate the length of one side of the rhombus.

Solution:

In a rhombus ∆BCD, its diagonals AC and BD bisect each other at right angles at O and AC = 12 cm, BD = 9 cm

∴ AO = OC = \(\frac { 12 }{ 2 }\) = 6 cm

and BO = OD = \(\frac { 9 }{ 2 }\) cm = 4.5 cm

Now in right ∆AOB,

AB² = AO² + BO² (Pythagoras Theorem)

= (6)² + (4.5)² = 36 + 20.25

= 56.25 = (7.5)²

∴ AB = 7.5

Hence length of each side = 7.5 cm

Question 5.

(a) In a right-angled triangle ABC it is given that the hypotenuse, AC = 2.5 cm, and the side AB = 1.5 cm. Calculate the side BC.

(b) AD is drawn perpendicular to BC, base of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.

Solution:

(a) In right angled ∆ABC, ∠B = 90°

∴ AC² = AB² + BC² (Pythagoras Theorem)

⇒ (2.5)² = (1.5)² + BC²

⇒ 6.25 = 2.25 + BC²

⇒ BC² = 6.25 – 2.25 = 4.00 = (2)²

∴ BC = 2 cm

(b) In an equilateral ∆ABC

AD ⊥ BC

∴ D is mid-point of BC

BC = 10 cm

∴ BD = \(\frac { 10 }{ 2 }\) = 5 cm

and AB = AC = BC = 10 cm

Now in right ∆ABD

AB² = AD² + BD² (Pythagoras Theorem)

(10)² = AD² + (5)²

⇒ AD² = 10² – 5²

⇒ AD² = 100 – 25 = 75 = 5 x 25

AD = \(\sqrt{5 \times 25}=5 \sqrt{3}\) cm = 8.7 cm

Question 6.

ABC is right-angled triangle. Angle ABC = 90°, AC = 25 cm and AB = 24 cm. Calculate the area of a ∆ABC.

Solution:

In right ∆ABC, ∠ABC = 90°

AC = 25 cm and AB = 24 cm

∴ AC² = AB² + BC² (Pythagoras Theorem)

⇒ (25)² = (24)² + BC²

⇒ 625 = 576 + BC²

⇒ BC² = 625 – 576 = 49 = (7)²

∴ BC = 7 cm

Now area of ∆ABC = \(\frac { 1 }{ 2 }\) BC x AB (∆ = \(\frac { 1 }{ 2 }\) Base x Altitude)

= \(\frac { 1 }{ 2 }\) x 7 x 24 = 84 cm²

Question 7.

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 ground.

Solution:

Let AC be ladder and its foot C is 5 away from the wall AB

AC = 13 m and BC = 5 m

∴ AC² = AB² + BC²

⇒ (13)² = (AB)² + (5)²

⇒ 169 = AB² + 25

⇒ AB² = 169 – 25 = 144

⇒ AB² = (12)² ⇒ AB = 12

∴ Height of the wall = 12 m

Question 8.

Use the given information to make a neat diagram of the figure having given properties and write down the name of the figure.

In triangle ABC, AB² = BC² + Ac² and AC = 2 BC.

Solution:

We are given, in the triangle ABC,

AB² = BC² + AC²

and AC = 2BC

∵ AB² = BC² + AC²

∴ ∆ABC is a right angled at C

(i) Draw a line segment BC of some length

(ii) At C, draw a ray CX making 90° at BC and cut off CA = 2BC

Join AB,

Then ∆ABC is the required triangle

Question 9.

In the figure, ABCD represents a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm, and ∠ABD = ∠BCD = 90°. Calculate the length of AB.

Solution:

In the figure ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90°

In right ∆BCD

BD² = CD² + BC² (Pythagoras Theorem)

= (12)² + (3)² = 144 + 9 = 153

Now in right ∆ABD,

AD² = AB² – BD²

(13)² = AB² – 153 ⇒ 169 = AB² + 153

⇒ AB² = 169 – 153 = 16 = (4)²

∴ AB = 4 cm

Question 10.

Which of the triangles whose sides are given below are right angled?

(i) 4 cm, 5 cm, 6 cm

(ii) 1.2 cm, 3.7 cm, 3.5 cm

(iii) 4 cm, 9.6 cm, 10.4 cm

(iv) 2.2 cm, 3.3 cm, 4.4 cm

Solution:

We know that in a right angled triangle,

(Hypotenuse)² = Sunt of squares of other two sides and hypotenuse is the longest side i.e. If square of longest side = Sum of the square of other two sides

Then the triangle is a right angled

(Converse of Pythagoras Theorem)

(a) 4 cm, 5 cm, 6 cm

Here longest side = 6 cm

Now (6)² = 36

and sum of squres of other two sides

= (4)² + (5)² = 16 + 25 = 41

∵ (36) ≠ 41

∴ It is not a right angled triangle

(b) 1.2 cm, 3.7 cm, 3.5 cm Here longest side = 3.7

∴ (3.7)² = 13.69

and (1.2)² + (3.5)² = 1.44 + 12.25 = 13.69

∵ 13.69 = 13.69

∴ It is a right angled triangle

(c) 4 cm, 9.6 cm, 10.4 cm

Here longest side = 10.4 cm

∴ (10.4)² = 108.16

and (4)² + (9.6)² = 16.00 + 92.16 = 108.16

∵ 108.16 = 108.16

∴ It is a right angled triangle

(d) 2.2 cm, 3.3 cm, 4.4 cm

Here longest side = 4.4 cm

∴ (4.4)² = 19.36

and (2.2)² + (3.3)² = 4.84 + 10.89 = 15.73

∵ 19.36 ≠ 15.73

∴ It is not a right angled tirangle

Question 11.

In the figure, find the distance of D from A, unit of length is 1 cm, (AB = 2. BC = 4, CD = 2).

Solution:

In the figure, AB = 2 units, BC = 4 units, CD = 2 units

and ∠ABC = ∠BCD = 90°

Join AD which intersects BC at E

In ∆ABE and ∆CDE

AB = CD (each equal to 2 units)

∠AEB = ∠DEC (vertically opposite angle)

and ∠ABE = ∠DCE (each = 90°)

∴ ∆ABE ≅ ∆CDE (ASA axiom)

∴ AE = ED (c.p.c.t.)

and BE = EC = \(\frac { 4 }{ 2 }\) = 2 units

Now in right ∆ABE

AE² = AB² + BE²

= (2)² + (2)² (Pythagoras Theorem)

= 4 + 4 = 8 = (2\(\sqrt{2}\))²

∴ AE = 2\(\sqrt{2}\) units

AD = 2AE = 2 x 2\(\sqrt{2}\) = 4\(\sqrt{2}\) units

= 4 (1.414) = 5.656 = 5.66 cm

Question 12.

The sides of a right-angled triangle containing the right angle are 5x and (3x – 1) cm. If the area of the triangle be 60 cm², calculate the lengths of the sides of the triangle.

Solution:

In right angled ∆ABC,

AB = 5x cm and BC = (3x – 1) cm

Area of ∆ABC = 60 cm²

Either x – 3 = 0, then x = 3

or 3x + 8 = 0 then 3x = – 8 ⇒ x = \(\frac { -8 }{ 3 }\)

Which is not possible being negative

∴ x = 3

∴ AB = 5x = 5 x 3 = 15

BC = 3x – 1 = 3 x 3 – 1 = 9 – 1 = 8

Now in right ∆ABC

AC² = AB² + BC² (Pythagoras Theorem)

= (15)² + (8)² = 225 + 64

= 289 = (17)²

∴ AC= 17

Hence sides are 15 cm, 8 cm and 17 cm

Question 13.

If the figure, it is given that AB = BC = 25 m. If AE = 7 m, and CD = 24 m, find the length of DE and also show that triangle ABE and triangle BDC are congruent.

Solution:

In the given figure,

AB = BC = 25 m.

If AE = 7 m and CD = 24 m

In right ∆BCD

BC² = CD² + BD² (Pythagoras Theorem)

⇒ (25)² = (24)² + DB² ⇒ DB² = 576 + DB²

⇒ DB² = 625 – 576 = 49 = (7)²

∴ DB = 7m

Similarly in right ∆ABE

AB² = AE² – BE²

⇒ (25)² = (7)² + BE² ⇒ 625 = 49 + BE²

⇒ BE² = 625 – 49 = 576 = (24)²

∴ BE = 24 m

∴ DE = DB + BE = 7 + 24 = 31 m

Now in ∆ABE and ∆ABCD

AB = BC (each = 25 m)

AE = DB (each = 7 m)

BE = CD (each = 24 m)

∆ABE ≅ ∆BCD (SSS axiom)

Hence proved.

Question 14.

A ladder rests against a vertical wall at a height of 12 m from the ground with its foot at a distance of 9 m from the wall on the ground. If the foot of the ladder is shifted 3 m away from the wall, how much lower will the ladder slide down?

Solution:

Let AB be the ladder and AC be the wall, then

AC = 12 m and BC = 9 m

∴ AB² = Ac² + BC² (Pythagoras Theorem)

= (12)² + (9)²

= 144 + 81 = 225 = (15)² = 15 m

∴ Length of ladder = 15 m

Again DE is the ladder = AB = 15 m

and on shifting its foot 3 m away from the foot of the wall,

The distance EC = 9 + 3 = 12 m

Now in right ∆DEC

DE² = EC² + DC²

(15)² = (12)² + DC²

⇒ 225 = 144 + DC²

⇒ DC² = 225 – 144

⇒ DC² = 81 = (9)²

∴ DC = 9 m

∴ AD = AC – DC = 12 – 9 = 3 m 3 m

∴ 3 m is lower will be ladder

Question 15.

AD is an altitude of a AABC and AD is 12 cm, BD = 9 cm and DC = 16 cm long respectively. Prove that the angle BAC is a right angle.

Solution:

Given : In ∆ABC, AD ⊥ BC

AD = 12 cm, BD = 9 cm and DC = 16 cm

To prove : ∠BAC = 1 rt angle

Proof: In right ∆ABD,

AB² = AD² + BD² (Pythagoras Theorem)

= (12)² + (9)² = 144 + 81 = 225 = (15)²

∴ AB = 15 cm

Similarly in right ∆ACD,

AC² = AD² + DC²

= (12)² + (16)² = 144 + 256 = 400 = (20)²

∴ AC = 20 cm

and BC = BD + DC = 9 + 16 = 25 cm

Now (BC)² = (25)² = 625

and AB² + Ac² = (15)² + (20)²

= 225 + 400 = 625

∴ AB² + AC² = BC²

∴ ∆ABC is a right angle, right angled at

∴ ∠BAC = 90°

Hence proved.

Question 16.

The shortest distance AP from a point A to a straight line QR is 12 cm, and Q, R are 15 cm and 20 cm distance from A on opposite sides of AP. Prove that QAR is a right angle.

Solution:

Given : In ∆AQR, AP ⊥ QR

AP = 12 cm, AQ = 15 cm and AR = 20 cm

To prove : ∠QAR is a right angle

Proof: ∵ AP ⊥ QR

∴ In right ∆APQ

AQ² = AP² + QP²

(15)² = (12)² + QP² (Pythagoras Theorem)

∴ 225 = 144 + QP² ⇒ QP² = 225 – 144 = 81

⇒ QP² = (9)² ⇒ QP = 9 cm

Similarly in right ∆APR

AR² = AP² + PR²

⇒ (20)² = (12)² + PR² ⇒ 400 = 144 + PR²

⇒ PR² = 400 – 144 = 256 = (16)²

∴ PR = 16 cm

Now QR = QP + PR = 9 + 16 = 25 cm

In ∆QAR,

Now (QR)² = (25)² = 625

and AQ² + AR² = (15)² + (20)² = 225 + 400 = 625

∵ QR² = AQ² + AR²

∴ ∆AQR is a right angled triangle

and ∠QAR = 1 right angle

Hence proved.

Question 17.

In the figure, PT is an altitude of the triangle PQR in which PQ = 25 cm, PR = 17 cm, PT = 15 cm, and QR = x cm. Calculate x.

Solution:

In the figure, PQR is a triangle in which PT ⊥ BR

PQ = 25 cm, PR = 17 cm, PR = 15 cm, QR = x cm

In right ∆PQR

PQ² = PT² + QT² (Pythagoras Theorem)

⇒ (25)² = (15)² + QT²

⇒ 625 = 225 + QT²

⇒ QT² = 625 – 225 = 400 = (20)²

∴ QT = 20 cm

Similarly in right ∆PTR,

PR² = PT² + TR²

(17)² = (15)² + (TR)²

⇒ TR² = 289 – 225 + TR²

⇒ TR² = 289 – 225 = 64 = (8)²

∴ TR = 8 cm

Now QR = x = QT + TR

= 20 + 8 = 28 cm

Question 18.

In the figure, AB = 8 cm, BC = 6 cm, AC = 3 cm, and the angle ADC = 90°. Calculate CD.

Solution:

In the figure, AB = 8 cm, BC = 6 cm, AC = 3 cm and ∠ABC = 90°

Let CD = x

∴ BD = (6 + x)

Now in right ∆ABD

AB² = BD² + AD² (Pythagoras Theorem)

⇒ (8)² = (6 + x)² + AD² ⇒ AD² = (8)² – (6 + x)²

AD² = 64 – (6 + x)² … (i)

Similarly in right ∆ACD,

AC² = CD² + AD²

⇒ (3)² = x² + AD² ⇒ 9 = x² + AD²

⇒ AD² = 9 – x² … (ii)

From (i) and (ii)

64 – (6 + x)² = 9 – x²

64 – (36 + 12x + x²) = 9 – x²

⇒ 64 – 36 – 12x – x² = 9 – x²

28 – 12x – x² + x² = 9

– 12x = 9 – 28 = – 19

⇒ x = \(\frac { -19 }{ -12 }\) = \(\frac { 19 }{ 12 }\)

∴ CD = \(\frac { 19 }{ 12 }\) = 1\(\frac { 7 }{ 12 }\)

Question 19.

In the figure, the angle BAC is a right angle and AD is perpendicular to BC; AB = 4 cm, AC = 3 cm and BD = x. Calculate x.

Solution:

In right ∆ABC, ∠A = 90°

AD ⊥ BC

AB = 4, AC = 3, BD = x

In right ∆ABC

BC² = AB² + AC² (Pythagoras Theorem)

= (4)²+ (3)²= 16 + 9 = 25

= (5)²

∴ BC = 5

BD = x

∴ DC = (5 – x)

Now in right ∆ABD,

AD² = AB²- BD²

= (4)² – x² = 16 – x² … (i)

and in right ∆ACD

AD² = Ac² – (DC)²

= (3)² – (5 – x)²

= 9 – (5 – x)² … (ii)

From (i) and (ii),

16 – x² = 9 – (5 – x)²

⇒ 16 – x² = 9 – (25 – 10x + x²)

⇒ 16 – x² = 9 – 25 + 10x – x²

⇒ 16 – x² – 9 + 25 – 1

⇒ – 10x + x² = 0 ⇒ 10x + 32 = 0

⇒ x = \(\frac { 32 }{ 10 }\) = 32

Question 20.

In the figure, the angle BAD and ADC are right angles and AX is parallel to BC. If AB = BC = 5 cm, and DC = 8 cm. Calculate the area of ABCX.

Solution:

In the figure, AB = BC = 5 cm

∴ AX = BC = 5 cm

CD = 8 cm

AX || BC and ∠ADC = ∠DAB = 90°

∵ AX || BC

∴ AB = XC = 5 cm

∴ DX = 8 – 5 = 3 cm

Now in right ∆ADX

Ax² = AD² + Dx² (Pythagoras Theorem)

⇒ (5)² = AD² + (3)² ⇒ 25 = AD² + 9

⇒ AD² = 25 – 9 = 16 = (4)²

∴ AD = 4

Now area of ABCX which is a rhombus

= Base x Height

= 5 x 4 = 20 cm²