KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4

Question 1.
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution:
cosec2 A – cot2 A = 1
cosec2 A = 1 + cot2 A
cosec2 A = cot2 A + 1
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 1

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 2.
Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 2

Question 3.
Evaluate :
i) \(\frac{\sin ^{2} 63^{\circ}+\sin 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}\)
ii) sin 25° cos 65° + cos 25° sin 65°
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 4

ii) sin 25° cos 65° + cos 25° sin 65°.
= sin25 cos65° + cos25 sin65
= sin25 cos (90 – 25) + cos25 . sin (90 – 25)
= sin25 . sin25 + cos25 cos25
= sin225 + cos225 = 1

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 4.
Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(A) 1
(B) 9
(C) 8
(D) 0

(ii) (1 +tanθ + secθ) (1 + cotθ – cosec0) =
(A) 0
(B) 1
(C) 2
(D) – 1

(iii) (sec A + tan A) (1 – sinA) =
(A) sec A
(B) sin A
(C) cosec A
(D) cos A

(iv) \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\)
(A) sec2 A
(B) -1
(C) cot2 A
(D) tan2 A
Solution:
(i) (B): Since, 9 sec2 A – 9 tan2 A
= 9 (sec2 A – tan2 A) = 9 (1) = 9 [∵ sec2 A – tan2 A = 1]

(ii) (C): Here,
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 Q4
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.4 4

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 5.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 9
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 10

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 11
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 12

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 13
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 14
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 15

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 16
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 17

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 18
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 19
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 20
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 21
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 22
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 23
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 24
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 25
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 26
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 27
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 28
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 29
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 30
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 31
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 32

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.4, drop a comment below and we will get back to you at the earliest.

 

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2

Question 1.
Evaluate the following :
i) sin 60° cos 30° + sin 30° cos 60°
ii) 2 tan2 45° + cos2 30° – sin2 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 1
Solution:
i) sin 60° cos 30° + sin 30° cos 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 2

ii) 2 tan2 45° + cos2 30° – sin260°
= 2(tan 45°)2 + (cos 30°)2 – (sin 60°)2
= 2 (1)2 + \(\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= 2 × 1
= 2
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 3
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 4
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 5

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 2.
Choose the correct option and justify your choice:
(i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}=\)
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°

(ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=\)
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0

(iii) sin 2A = 2sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}=\)
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 Q2
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.2 7

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 3.
If tan (A + B) =\(\sqrt{3}\) and tan (A – B) = \(\frac{1}{\sqrt{3}}\) 0° < A + B ≤ 90°; A > B. find A and B.
Solution:
tan (A + B) = \(\sqrt{3}\)
tan (A + B) = tan 60°
A + B = 60°
tan (A – B) = \(\frac{1}{\sqrt{3}}\) = tan 30°
tan(A – B) = tan 30°
A – B = 30° → (2)
Adding (1) and (2)
A + B + A – B = 60 + 30
2A = 90
A = \(\frac{90}{2}\) = 45°
A = 45°
Put A = 45° in eqn (1)
A + B = 60
B = 60 – A= 60 – 45°
B = 15°.

Question 4.
State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin6 increases as θ increases.
(iii) The value of cosθ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Solution:
(i) False:
Let us take A = 30° and B = 60°, then
L.H.S = sin (30° + 60°) = sin 90° = 1
R.H.S.= sin 30° + sin 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 Q4
∴ L.H.S. ≠ R.H.S.
(ii) True:
Since, the values of sin θ increases from 0 to 1 as θ increases from 0° to 90°.
(iii) False:
Since, the value of cos θ decreases from 1 to 0 as θ increases from 0° to 90°.
(iv) False:
Let us take θ = 30°
sin 30° = \(\frac{1}{2}\) and cos 30° = \(\frac{\sqrt{3}}{2}\)
⇒ sin 30° ≠ cos 30°
(v) True:
We have, cot 0° = not defined

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1

Question 1.
In ∆ABC, right-angled at B, AB = 24 cm., BC = 7 cm. Determine:
i) sin A, cos A
ii) sin C, cos C
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 1
In ⊥∆ABC, ∠B = 90°
As per Pythagoras theorem
AC2 = AB2 + BC2
= (24)2 + (7) 2
= 576 + 49
AC2 = 625
∴ AC = 25 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 2

Question 2.
In the figure, find tan P – cotR?
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 3
Solution:
In right angle ∆PQR
Using the Pythagoras theorem, we get
QR2 = PR2 – PQ2
⇒ QR2 = 132 – 122 = (13 – 12)(13 + 12) = 1 × 25 = 25
∴ QR = \(\sqrt{25}\) = 5 cm
Now, tanP = \(\frac{Q R}{P Q}=\frac{5}{12}\) , cotR = \(\frac{Q R}{P Q}=\frac{5}{12}\)
∴ tanP – cotR = \(\frac{5}{12}-\frac{5}{12}\) = 0.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 5
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 6

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:
In the right angle triangle ABC, we have 15 cot A = 8
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 5

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios
Solution;
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 8

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 7
From equations (i) and (ii) we get:
\(\frac{\mathrm{CD}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{BE}}=\frac{\mathrm{AD}}{\mathrm{BF}}\)
⇒ ∆CDA ~ ∆EFB [By SSS similarity]
⇒ ∠A = ∠B Hence Proved

 

Question 7.
If cot θ = \(\frac{7}{8}\), evaluate :
(i) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
(ii) cot2θ
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 10
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 11

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 8.
If 3 cot A = 4, check whether \(\frac{1-\tan ^{2} A}{1+\tan ^{2} A}\) = cos2 A – sin2 A or not
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 12
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 13

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 9.
In triangle ABC, right angled at B, if tan A = \(\frac{1}{\sqrt{3}}\) find the value of:
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C
Solution:
In right ∆ABC, ∠B = 90°
For ∠A, we have
Base = AB, Perpendicular = BC,
Hypotenuse = AC
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 Q9
MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.1 13

Question 10.
In ∆PQR, right-angled at Q, PR + QR = 25 cm. and PQ = 5 cm. Determine the values of sin P, cos P and tan P
Solution:
PQ = 5 cm
PR + QR = 25 cm
∴ PR = 25 – QR
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 16
PR2 = PQ2 + QR2
QR2 = PR2 – PQ2
= (25 – QR)2 – (5)2
QR2 = 625 – 50QR + QR2 – 25
50QR = 600
∴ QR = 12 cm.
∴ PR = 25 – QR = 25 – 12 = 13 cm.
∴ QR = 12 cm
∴ PR = 25 – QR = 25 – 12 = 13 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 17

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) secA = \(\frac{12}{5}\) for some value of angle A.
(iii) cosA is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sinθ = \(\frac{4}{3}\) for some angle θ.
Solution:
(i) False
∵ A tangent of an angle is the ratio of sides other than hypotenuse, which may be equal or unequal to each other.
(ii) True
∵ cos A is always less than 1.
∴ \(\frac{1}{\cos A}\) i.e., sec A will always be greater than 1.
(iii) False
∵ ‘cosine A’ is abbreviated as ‘cosA’
(iv) False
∵ ‘cot A’ is a single and meaningful term whereas ‘cot’ alone has no meaning.
(v) False
∵ \(\frac{4}{3}\) is greater than 1 and sinθ cannot be greater than 1

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8

(Assume π = \(\frac{22}{7}\), unless stated otherwise)

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
Solution:
(i) r = 7 cm, V = ?
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 1

Question 2.
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm.
(ii) 0.21 m.
Solution:
(i) Diameter of solid spherical ball, diameter, d = 28 cm,
∴ radius, r = \(\frac{28}{2}\) = 14 cm.
Volume of solid spherical ball, V = \(\frac{4}{3}\) πr3.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 2
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 3
Amount of water displaced by the ball = 0.00485 m3.

Question 3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3 ?
Solution:
Diameter of metallic ball, d = 4.2 cm. density = 8.9 gm/cm3
mass, m = ? d = 4.2 cm.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 4
∴ V = 38.808 cm3.
∴ Mass = Density × Volume
= 8.9 × 38.808
= 345.40 gm.

Question 4.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Solution:
Let the diameter of earth be ‘d’ unit.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 5
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 6
∴ Volume of Moon = \(\frac{1}{64}\) × Volume of Earth

Question 5.
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Solution:
Diameter of a hemispherical bowl, d = 10.5 cm
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 7

Question 6.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Solution:
In a hemispherical tank,
Inner radius, r1 = 1 m
Outer radius, r2 = 1 + 0.01 = 1.01 m (∵ 1 cm = 0.01)
∴ Volume of hemispherical tank, V
= Volume of outer diameter – Volume of inner diameter.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 8
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 9

Question 7.
Find the volume of sphere whose surface area is 154 cm2.
Solution:
Surface area of sphere, 4πr2 =154 cm2.
Volume of Sphere, V = ?
A = 4πr2 = 154
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 10

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is Rs. 2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Solution:
(i) Cost of white-washing dome is Rs. 498.96
Cost of white-washing is Rs. 2 per sq. metre.
∴ Surface Area = \(\frac{498.96}{2}\) = 249.48 m2.

(ii) Surface Area of hemisphere, V = 2πr2
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 11
∴ Inner volume of hemisphere dome, V
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 12

Question 9.
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S1. Find the
(i) radius r of the new sphere,
(ii) ratio of S and S1
Solution:
(i) Volume of 1 sphere, V = \(\frac{4}{3}\)πr3
Volume of 27 solid sphere
= 27 × \(\frac{4}{3}\)πr3
Let r1is the radius of the new sphere.
Volume of new sphere = Volume of 27 solid sphere
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 13

Question 10.
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?
Solution:
Diameter of capsule of medicine, d
d = 3.5 mm = \(\frac{7}{2}\) mm.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.8 14

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Karnataka Board Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows:
Height = 110 cm, Depth = 25 cm, Breadth = 85 cm.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 1
The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.
Solution:
Outer length, l = 25 cm.
outer breadth, b = 85 cm.
outer height, h = 110 cm.
Outer surface area = lh + 2lb + 2bh = lh + 2(lb + bh)
= 85 × 110 + 2(85 × 25 + 25 × 110)
= 9350 + 9750
= 19100 cm2.
Area of front face,
= [85 × 110 – 75 × 100 + 2(75 × 5)]
= (9350 – 7500 + 2(375)]
= 9350 – 7500 + 750
= 11000 – 7500
= 3500 cm2.
Area to be polished,
= 19100 + 3500
= 22600 cm2.
Cost of polishing 1 cm3 is Rs. 0.20.
Cost of polishing 22600 cm3 … ? …
= 22600 × 0.20 = Rs. 4520.
Length of horizontal shelf, l = 75 cm.
breadth, b = 20 cm.
height, h = 30 cm.
Area of horizontal shelf
= 2(l + h)b + lh
= [2(75 + 30) × 20 + 75 × 30]
= (4200 + 2250] cm2.
= 6450 cm2.
∴ Area of painting 3 horizontal rows
= 3 × 6450
= 19350 cm2.
Cost of painting for 1 cm3 is Rs. 0.10.
∴ Cost of painting 19350 cm3 … ?
= 19350 × 0.10 = Rs. 1935.
∴ Total cost of polish and painting
= Rs. 4520 + Rs. 1935
= Rs. 6455.

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig.
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 2
Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
Solution:
Diameter of a wooden frame, d = 21 cm.
Radius r = \(\frac{21}{2}\)
Outer surface area of wooden spheres,
A = 4πr2
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 3
∴ A = 1386cm2
Radius of cylinder, r1 = 1.5 cm
height, h = 7 cm
The curved surface area of cylinder support,
A = 2πrh
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 4
Area of silver painted
= [8 × (1386 – 7.07)] cm2
= 8 × 1378.93
= 11031.44 cm2
Cost of silver paint = 11031.44 × 0.25
= Rs. 2757.86
Area of black paint = (8 × 66) cm2
= 528 cm2
Cost of black paint = 528 × 0.05
= Rs. 26.40
Total cost of silver and black paint.
= Rs. 2757.76 + Rs. 26.40
= Rs. 2784.26.

Question 3.
The diameter of a sphere is decreased by 25%. By what percent does its curved surface area decrease?
Solution:
Let the diametrer of sphere be ‘d’
Radius of sphere, r1 = \(\frac{\mathrm{d}}{2}\)
Radius of outer sphere, r2 = \(\frac{\mathrm{d}}{2}\left(1-\frac{25}{100}\right)\)
∴ r2 = \(\frac{3}{8}\)d
Outer Area of Sphere, S1 = 4πr12
= 4π\(\left(\frac{\mathrm{d}}{2}\right)^{2}\)
S1 = πd2
The diameter of sphere is decreased by 25%. Then its outer surface area,
KSSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9 5

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1

KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1.

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Data: Constructing an angle of 90° at the initial point of a given ray.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 1
Steps of Construction:

  1. Draw AB straight line.
  2. Taking A as centre and any radius draw an arc and intersect AB at C.
  3. Taking C as centre with the same radius draw an arc which intersects at D.
  4. With the same radius taking D as centre, it intersects at E.
  5. With centres E and D with the same radius draw two arcs which meet both at F.
  6. If AF is joined, AB straight line is the line in point A with 90°, i.e., AF.
    In the above construction ∠DAB = 60° and ∠DAE = 60°.
    Angular bisector of ∠DAE is AF.
    ∴ ∠DAF = ∠EAF = 30°
    ∴∠BAF = ∠DAB + ∠DAE = 60°+ 30°
    ∴∠BAF = 90°.

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:
Data: To construct an angle of 45° at the initial point of a given ray.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 2
Steps of Construction :

  1. Draw a straight line with any measurement.
  2. Taking A as centre with any radius draw an arc, which meets AB at C.
  3. With C centre, with the same radius draw an arc which meets at D.
  4. With D centre, with the same radius, draw an arc which meets at D.
  5. With centres E and D, with the previous radius draw two arcs which meet at F.
  6. Now AF is joined. Now AP1AF at A. Hence ∠BAF = 90°.
  7. Now, AG which is the angular bisector of ∠BAF is drawn.
  8. ∠BAG = 45° is constructed.
    ∠FAB = 90°
    AG is the angular bisector of ∠FAB
    ∴ ∠FAG = ∠GAB = 45°
    ∴ ∠GAB = 45°.

Question 3.
Construct the angles of the following measurements :
(i) 30°
(ii) \(22 \frac{1}{2}^{\circ}\)
(iii) 15°.
Solution:
(i) To construct 30° angle :
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 3
Steps of Construction :

  1. Draw PQ straight line.
  2. With P as centre with any radius draw \(\frac{1}{2}\) arc- intersects PQ at A.
  3. With A as centre with the same radius draw an arc which intersects at B. Join PB and produced.
  4. With A and B centres, with radius more than half of AB draw two arcs which intersect at C. Join PC.
  5. ∠BPC = ∠CPQ = 30°.

(ii) To draw an angle of \(22 \frac{1}{2}^{\circ}\)
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 4
Steps of Construction:

  1. Draw a straight line AB.
  2. With centre ‘A’, taking a convenient radius to draw an arc which intersect AB at P.
  3. With P as centre with the same radius draw an arc at Q, with Q as centre with the same radius draw an arc which intersects at R.
  4. With R and Q centres with the same radius draw two arcs which intersect at S. Join AS, ∠BAS = 90°.
  5. Now construct AT which is the angular bisector of ∠BAS, and joined, ∠TAB= 45°.
  6. Now AU which is the angular bisector of ∠TAB, AU is joined. Now, ∠UAB = \(22 \frac{1}{2}^{\circ}\)

(iii) To construct an angle of 15° :
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 5
Steps of Construction :

  1. Draw a straight line AB.
  2. With A as centre with a convenient radius draw an arc which intersect AB at C.
  3. With C as centre, with the same radius, draw an arc which intersects at D.
    Now. ∠DAB = 60°
  4. Construct AE, the bisector of ∠DAB. Join then, ∠EAB = 30°.
  5. Construct AF, the bisector of ∠EAB. Join then, ∠FAB = 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor :
(i) 75°
(ii) 105°
(iii) 135°.
Solution:
(i) To construct angle 75° :
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 6
Steps of Construction :

  1. Draw AB with any measurement.
  2. With A as centre, with convenient scale draw an arc AB which intersects at C.
  3. With C as centre with the same radius draw an arc which intersects at D. ∠DAB = 60°.
  4. With D as centre with the same radius draw an arc which intersects at E.
  5. With Centres E and D, by taking more than half of ED draw two arcs which meet at E. AF is joined.
    Now, ∠FAB = 90°.
  6. Draw AG bisector of ∠EAD, AG joined. ∠GAD = 15°.
  7. ∠GAD + ∠DAB = 15 + 60 = 75°
    ∴ ∠GAB = 75°.

(ii) To construct angle 105° :
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 7
Steps of Construction :

  1. Draw PQ with any measurement.
  2. With P as centre and with any radius, draw an arc which intersects PQ at A.
  3. With A as centre, with the same radius draw an arc which intersect at B.
  4. With B as centre with same radius draw another arc, it intersects at C.
  5. With centres C and B if two arcs are drawn these two meet at D. Join PD. ∠DPQ = 90°. Straight-line PD intersects the arc CB at E.
  6. Now taking radius half of CE, if the line is drawn from C to E it intersects at F. FP is joined. Now ∠FPD = 15°.
  7. ∠FPD + ∠DPQ = 15 + 90 = 105°.
    ∴ ∠FPQ = 105° is constructed.

(iii) To construct an angle of 135°
Solution:
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 8
Steps of Construction :

  1. Draw a straight line AB.
  2. With A as centre, a semicircle is drawn which meets AB at C and it intersects the produced BA line at G.
  3. With C as centre with the same radius and centre CD, draw a radius of arc DE.
  4. With centres E and D by taking a radius more than half draw two arcs which meet at F. Join AF. ∠FAB = 90°. AF intersects ED at H.
  5. With G and H centres by taking radius more than \(\frac{1}{2}\) of the radius GH draw arcs which meet at I. AI is joined. Now, ∠IAF = 45°.
  6. ∠IAF + ∠FAB = 45° + 90° = 135°
    ∴ ∠IAB = 135° is constructed.

Question 5.
Construct an equilateral triangle, given its side, and justify the construction.
Solution:
Data: Construct an equilateral triangle, given its side.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 9
Steps of Construction :

  1. Construct AB = 6 cm.
  2. With A as centre and convenient scale draw an arc, it meets AB at P.
  3. With P as centre, with same radius draw an arc, it intersects the first arc at Q, AQ is joined and produced. ∠QAB = 60°.
  4. With A as centre and taking 6 cm radius draw an arc. It intersects at C which is the produced line of AQ.
  5. Next, BC is joined, ABC is an equilateral triangle.

In ∆ABC, ∠A = ∠B = ∠C = 60°
AB = BC = CA = 6 cm.
In ∆ABC, AB = AC = 6 cm. ∠A = 60°
∠B = 60°
∠A + ∠B + ∠C = 180°
60 + ∠B + ∠B =180 (∵ ∠B = ∠C)
60 + 2∠B = 180
2∠B = 180 – 60
2∠B = 120°
∴ ∠B = \(\frac{120}{2}\)
∴ ∠B = 60°
∴∠B = ∠C = 60°
∴∠A = ∠B = ∠C = 60°
AC = BC (Opposite sides of equal angles)
But, AB = AC (known)
∴ AB = BC = CA = 6 cm.

We hope the KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 6 Constructions Exercise 6.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6

Karnataka Board Class 9 Maths Chapter 12 Circles Ex 12.6

Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 1
Solution:
Data: Two circles having centres A and B, intersect at C and D.
To Prove: ∠ACB = ∠ADB.
Construction: Join A and B.
Proof: In ∆ABC and ∆ABD,
AC = AD (∵ radii of same circle are equal)
BC = DD
AB is common.
∴ ∆ABC ≅ ∆ABD (SSS Postulate.)
∴ ∠ACB = ∠ADB.

Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm., find the radius of the circle.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 2
Solution:
Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre.
Distance between AB and CD is 6 cm.
To Prove: Radius of the circle, OP =?
Construction: Join OP and OQ, OB, and OD.
Proof: Chord AB || Chord CD.
AB = 5 cm, and CD =11 cm.
OP ⊥ AB
∴ BP = AP = \(\frac{5}{2}\) = 2.5 cm.
OQ⊥CD
∴ CQ = QD = \(\frac{11}{2}\) = 5.5 cm.
PQ = 6 cm. (Data)
Let OQ = 2 cm then, OP = (6 – x) cm.
In ∆BPO, ∠P = 90°
As per Pythagoras theorem,
OB2 = BP2 + PO2
= (2.5)2 + (6 – x)2
= 6.25 + 36 – 12x + x2
OB2 = x2– 12x + 42.25 …………….. (i)
In ∆OQD, ∠Q= 90°
∴ OD2 = OQ2 + QD2
= (x)2 + (5.5)2
OD2 = x2 + 30.25 ……………….. (ii)
OB = OD (∵ radii of same circle)
From (i) and (ii).
x2 – 12x + 42.25 = x2 + 30.25
-12x = 30.25 – 42.25
-12x = -12
12x = 12
∴ x = \(\frac{12}{12}\)
∴ x = 1 cm.
From (ii),
OD2 = x2 + 30.25
= (1)2 + 30.25
= 1 + 30.25
∴ OD2 = 31.25
OD = \(\sqrt{31.25}\)
∴ OD = 5.59 cm.
∴ Radius of circle OP = OD = 5.59 cm.

Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm, from the centre, what is the distance of the other chord from the center?
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 3
Solution:
Data: Chords of a circle are 6 cm. and 8 cm. are parallel. The smaller chord is at distance of 4 cm. from the centre.
To Prove: Distance between bigger chord and centre =?
Construction: Join OA and OC.
Proof: AB || CD, AB = 6 cm, CD = 8 cm.
OP⊥CD, OQ⊥CD.
In ∆OPA, ∠P = 90°
∴ OA2 = OP2 + PA2 (According to Pythagoras theorem)
= (4)2 + (3)2 = 16 + 9
OA2 = 25
∴ OA = 5 cm.
OA = OC = 5 cm. (radii of the same circle.)
Now, in ∆OQC,
OC2 = OQ2 + QC2
(5)2 = x2 + (4)2
25 = x2 + 16
x2 = 25 – 16 = 9
∴ x = \(\sqrt{9}\) ∴ x = 3 cm.
∴ Bigger chord is at a distance of 3 cm from the centre.

Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 4
Solution:
Data: The vertex of angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle.
To Prove: ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. OR
∠ABC= \(\frac{1}{2}\) [∠DOE – ∠AOC].
Construction: OA, OC, OE, OD are joined.
Proof: In ∆AOD and ∆COE,
OA = OC, OD = OE radii of same circle.
AD = CE (Data)
∴ ∆AOD ≅ ∆COE (SSS Postulate)
∠OAD = ∠OCE ……….. (i)
∴∠ODA = ∠OEC …………. (ii)
OA = OD
∴ ∠OAD = ∠ODA …………. (iii)
From (i) and (ii),
∠OAD = ∠OCE = ∠ODA = ∠OEC = x°.
In ∆ODE, OD = OE
∠ODE = ∠OED = y°.
ADEC is a cyclic quadrilateral.
∴ ∠CAD + ∠DEC = 180°
x + a + x + y = 180
2x + a + y = 180
y = 180 – 2x – a ……….. (iv)
But, ∠DOE = 180 – 2y
∠AOC = 180 – 2a
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 5

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 6
Solution:
Data: ABCD is a rhombus. Circle is drawn taking side CD diameter. Let the diagonals AC and BD intersect at ‘O’.
To Prove: Circle passes through the point ‘O’ of the intersection of its diagonals.
Proof: ∠DOC = 90° (Angle in the semicircle) and diagonals of rhombus bisect at right angles at ‘O’.
∴ ∠DOC = ∠COB = ∠BOA = ∠AOD = 90°
∴ Circle passes the point of intersection of its diagonal through ‘O’.

Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 7
Solution:
Data: ABCD is a parallelogram. The circle through A, B, and C intersect CD at E. AE is joined.
To Prove: AE = AD
Proof: ∠AEC + ∠AED = 180° …………. (i) (linear pair)
ABCE is a cyclic quadrilateral.
∴ ∠ABC + ∠AEC = 180° ………….(ii) (opposite angles)
Comparing (i) and (ii),
∠AEC = ∠AED = ∠ABC + ∠AEC
∠AED = ∠ABC ………….. (iii)
But, ∠ABC = ∠ADE (Opposite angles of quadrilateral)
Substituting in equation (iii),
∠AED = ∠ADE
∴ AE = AD.

Question 7.
AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
Solution:
Data : AC and BD are chords of a circle bisect each other at ‘O’.
To Prove:
i) AC and BD are diameters.
ii) ABCD is a rectangle.
Construction: AB, BC, CD and DA are joined.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 8
Proof: In ∆AOB and ∆COD,
AO = OC (Data)
BO = OD (Data)
∠AOB = ∠COD (vertically opposite angles)
∴ ∆AOB ≅ ∆COD (SAS postulate)
∴ ∠OAB = ∠OCD
These are pair of alternate angles.
∴ AB || CD and AB = CD.
∴ ABCD is a parallelogram.
∴ ∠BAD = ∠BCD (Opposite angles of parallelogram)
But. ∠BAD + ∠BCD = 180 (∵ Angles of cyclic quadrilateral)
∠BAD + ∠BAD = 180
2(∠BAD) = 180
∴ ∠BAD = \(\frac{180}{2}\)
∴ ∠BAD = 90°.
If angles of a quadrilateral are right angles it is rectangle. ABCD is a recrtangle.
∠BAD = 90°
∠BAD is separated from chord BD.
∴ This is the angl in semicircle.
∴ Chord BD is a diameter.
Similarly, ∠ADC = 90°
∴ Chord AC is a diameter.

Question 8.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – \(\frac{1}{2}\) A, 90° – \(\frac{1}{2}\) B and 90° – \(\frac{1}{2}\) C.
Solution:
Data: AD, BE and CF are angular bisectors of angles A, B and C of ∆ABC intersects its circumference at D, E and F respectively.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 9
To prove: Angles of ∆DEF are 90° – \(\frac{1}{2}\) A, 90° – \(\frac{1}{2}\) B and 90° – \(\frac{1}{2}\) C.
Proof: AD, BE and CF are angular bisectors of angles A, B and C of ∆ABC.
∴ ∠BAD = ∠CAD = \(\frac{\angle \mathrm{A}}{2}\)
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 10
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 11

Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lies on the two circles.
Prove that BP = BQ.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 12
Solution:
Data : Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.
To Prove: BP = BQ
Construction: Join AB.
Proof: Two congruent triangles with centres O and O’ intersects at A and B. Through A segment PAQ is drawn so that P, Q lie on the two circles.
Similarly, ∠AQB= 70° in circle subtended by chord AB. Because Angles subtended by circumference by same chord.
∴ ∠APB = ∠AQB = 70°.
Now, in ∆PBQ, ∠QPB = ∠PQB.
∴ Sides opposite to each other are equal.
∴ BP = BQ.

Question 10.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
KSSEB Solutions for Class 9 Maths Chapter 12 Circles Ex 12.6 13
Solution:
Data: In ∆ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect each other. O is the centre of the circle.
To Prove: Angle bisector of ∠A and perpendicular bisector of BC intersect at D.
Construction: Join OB, OC.
Proof: Angle subtended at the Centre
= 2 × angles subtended in the circumference.
∠BOC = 2 × ∠BAC
In ∆BOE and ∆COE,
∠OEB = ∠OEC = 90° (∵ OE⊥BC)
∴ BO = OC (radii)
OE is common.
∴ ∆BOE ≅ ∆COE (RHS postulate)
But, ∠BOE + ∠COE = ∠BOC
∠BOE + ∠BOE = ∠BOC
2∠BOE = ∠BOC
2∠BOE = 2∠BAC
∴ ∠BOE = ∠BAC
But, ∠BOE = ∠COE = ∠BAC
∠BAD = \(\frac{1}{2}\) ∠BAC
∠BAD = \(\frac{1}{2}\) ∠BOE
∠BAD = \(\frac{1}{2}\) ∠BOD
∴ ∠BOD = 2∠BAD
∴ The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference.
∴ Angle bisector of ∠A and perpendicular bisector of BC intersect at D.

KSEEB Solutions for Class 9 Maths

 

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.5.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
ABC is a triangle. To locate a point in the interior of ∆ABC which is equidistant from all the vertices of ∆ABC we have to find out circumcentre means the point where three perpendicular bisectors meet.
E.g. Three sides of ∆ABC are,
AB = 5 cm, BC = 4 cm, and AC = 6 cm.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 1
‘S’ is the circumcentre of ∆ABC

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Incentre is equidistant from three sides of a ∆. This is at the point where angular bisectors meet. This is called T.
E.g. In ∆ABC, AB = 6 cm, ∠B = 100°, ∠A = 50°.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 2
Point T is equidistant from three sides.

Question 3.
In a huge park, people are concentrated at three points
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an icecream parlour be set up so that the maximum number of persons can approach it?
(Hint: The parlour should be equidistant from A, B, and C).
Solution:
A: In a park where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 3
In ∆ABC, to locate a point equidistant from three sides, we have to find out perpendicular bisectors, which means where all perpendicular bisectors meet. This is called S’.
AB = BC = CA = 5 cm.
∴ The ice cream shop is at ‘S’.

Question 4.
Complete the hexagonal and star-shaped Rangolies by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 4
Solution:
Fig. (i): Regular Hexagonal with side 5 cm. is constructed. It has 6 equal sides.
Measure of each side is 5 cm.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 5
Number of equilateral ∆ with 1 cm side
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 6
In Fig. (ii) : there are 12 equilateral triangles with side 5 cm
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 7
∴ Number of equilateral A with 1 cm side
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 8

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.5, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.4.

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4

Question 1.
Give the geometric representations of y = 3 as an equation.
(i) in one variable
(ii) in two variables.
Solution:
x + 3 = 0
i) In one variable, y = 3.
KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 1

ii) In two variables, y = 3 straightline passes through (0, 3) parallel to x-axis. For any value of x, value of y is 3.
KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 2

Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation.
(i) in one variable
(ii) in two variables.
Solution:
2x + 9 = 0, this equation in
i) One variable 2x + 9 =0
2x = -9
x = \(-\frac{9}{2}\) = -4.5
KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 3

ii) 2x + 9 = 0 in two variables, 2x + 9 = 0. This equation passes through (-4.5, 0) and this is parallel ot y-axis. Coordinate of x = – 4.5
KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 4

We hope the KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 1
Solution:
In ∆ABC, ∠ABC = 90°
∠A + ∠C = 90°
∠ABC > ∠BAC and ∠ABC < ∠BCA
∴ D is the largest side of ∠ABC.
∴ AC is opposite side of larger angle.
∴ AC hypoternuse is largest side of ∆ABC.

Question 2.
In sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 2
Solution:
Data: AB and AC are the sides of ∆ABC and AB and AC are produced to P and Q respectively.
∠PBC < ∠QCB.
To Prove: AC > AB
Proof: ∠PBC < ∠QCB
Now, ∠PBC + ∠ABC = 180°
∠ABC = 180 – ∠PBC ………. (i)
Similarly, ∠QCB + ∠ACB = 180°
∠ACB = 180 – ∠QCB …………. (ii)
But, ∠PBC < ∠QCB (Data)
∴ ∠ABC > ∠ACB Comparing (i) and (ii), AC > AB
(∵ Angle opposite to larger side is larger)

Question 3.
In fig ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 3
Solution:
Data : ∠B < ∠A and ∠C < ∠D.
To Prove: AD < BC
Proof: ∠B < ∠A
∴ OA < OB …………. (i)
Similarly, ∠C < ∠D
∴ OD < OC …………. (ii)
Adding (i) and (ii), we have
OA + OD < OB + OC
∴ AD < BC.

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 4
Solution:
Data : AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD.
To Prove :
(i) ∠A > ∠C
(ii) ∠B > ∠D
Construction: Join AC and BD
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 5
Proof: In ∆ABC, AB < BC (data)
∴ ∠3 < ∠4 ………. (i)
In ∆ADC, AD < CD (data)
∴ ∠2 <∠1 …………. (ii)
Adding (i) and (ii),
∠3 + ∠2 < ∠4 + ∠1
∠C < ∠A
∠A > ∠C
By adding B,
In ∆ABD, AB < AD
∠5 < ∠8
In ∆BCD, BC < CD
∠6 < ∠7
∴ ∠5 + ∠6 < ∠8 + ∠7
∠D < ∠B
OR ∠B > ∠D.

Question 5.
In Fig. PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 6
Solution:
Data: PR > PQ and PS bisect ∠QPR.
To Prove: ∠PSR > ∠PSQ
Proof: In ∆PQR, PR > PQ
∴ ∠PQR > ∠PRQ ………. (i)
PS bisects ∠QPR.
∴ ∠QPS = ∠RPS ………… (ii)
By adding (i) and (ii),
∠PQR + ∠QPS > ∠PRQ + ∠RPS ………. (iii)
But, in ∆PQS, ∠PSR is an exterior angle.
∴ ∠PSR = ∠PQR + ∠QPS ………… (iv)
Similarly, in ∆PRS, ∠PSQ is the exterior angle.
∴ ∠PSQ = ∠PRS + ∠RPS ……… (v)
In (iii), subtracting (iv) and (v),
∠PSR > ∠PSQ.

Question 6.
Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.
KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 7
Solution:
P is a point on straight line l.
PR is the segment for l drawn such that PQ ⊥ l.
Now, in ∆PQR,
∠PQR = 90° (Construction)
∴ ∠QPR + ∠QRP = 90°
∴ ∠QRP < ∠PQR
∴ PQ < PR
Any line segment is drawn from P to l they form a small angle.
∴ PQ ⊥ l is smaller.

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