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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°

Solution:
i) sin 60° cos 30° + sin 30° cos 60°

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

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°
Solution:
A) sin 60°

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

iii) sin 2A = 2 sin A is true when A ;
A) 0°
B) 30°
C) 45°
D) 60°
Solution:
A) 0°
LHS = sin 2A = sin 0° = 0
RHS = 2sin A = 2.sin0° = 0

iv) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) =
A) cos 60°
B) sin 60°
C) tan 60°
D) sin 30°
Solution:
C) tan 60°

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
Solution:
False
Take A = 45° and B = 45°
LHS: sin (45° + 45°) = sin 90° = 1
RHS: sin 45 + sin 45 = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}\)
LHS ≠ RHS.

ii) The value of sin θ increases as θ increases.
Solution:
True.

∴ The value of sin θ increases as θ increases.

iii) The value of cos θ increases as θ increases.
Solution:
False.
cos 0° = 1 cos 30° = \(\frac{\sqrt{3}}{2}\) =0.87
cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.7 cos 60° = \(\frac{1}{2}=\) = 0.5
cos 90° = 1
∴ The value of cos 0 increases as 0 increases – the statement is False.

iv) sin θ = cos θ for all values of θ.
Solution:
False.
sin 30° = \(\frac{1}{2}\) cos 30° = \(\frac{\sqrt{3}}{2}\)
sin 30° ≠ cos 30°
But sin 45° = cos 45° = \(\frac{1}{\sqrt{2}}\)

v) cot A is not defined for A = 0°.
Solution:
True.

∴ When A = 0°, cot A is 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 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:

In ⊥∆ABC, ∠B = 90°
As per Pythagoras theorem
AC² = AB² + BC²
= (24)² + (7) ²
= 576 + 49
AC² = 625
∴ AC = 25 cm

Question 2.
In the given figure, find tan P – cot R.
Solution:

In ⊥∆PQR, ∠Q = 90°
∴ PQ² + QR² = PR²
(12)² + QR² = (13)²
144 + QR² = 169
QR² = 169 – 144
QR² = 25
∴ QR = 5 cm.

Question 3.
If sin A = \(\frac{3}{4}\) calculate cos A and tan A.
Solution:

Question 4.
Given 15 cot A = 8, find sin A and sec A.
Solution:

Question 5.
Given sec θ = \(\frac{13}{12}\), calculate all other trigonometric ratios
Solution;

Question 6.
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
In ⊥∆ABC, ∠A and ∠B are acute angles. CD ⊥ AB is drawn.

CD is common.
S.S.S. Postulate :
∴ ∆ADC ~ ∆ADB
∴ ∠A = ∠B
∵ “Angles of similar triangles are equiangular.”

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:

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:

Question 9.
In ∆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:

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

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

Question 11.
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
Answer:
False. 60° = \(\sqrt{3}\) > 1

(ii) If sec A = \(\frac{12}{5}\) for some value of angle A.
Answer:
True. because sec A > 1.

(iii) cos A is the abbreviation used for the cosecant of angle A.
Answer:
False. Because cos A is simplified as cos.

(iv) cot A is the product of cot and A.
Answer:
False. cot is ∠A or meaningless. Here,
cot A = \(\frac{\text { Adjacent side }}{\text { Opposite side }}\)

(v) sin θ = \(\frac{4}{3}\) for some angle θ
Answer:
False. because sin θ ≯ 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.

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 = ?

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.


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.

∴ 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.


∴ 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

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.

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

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

∴ Inner volume of hemisphere dome, V

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

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.

KSEEB Solutions for Class 9 Maths

 

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.

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.

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

∴ A = 1386cm²
Radius of cylinder, r₁ = 1.5 cm
height, h = 7 cm
The curved surface area of cylinder support,
A = 2πrh

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,

KSEEB Solutions for Class 9 Maths

 

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.

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.

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 :

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:

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:

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° :

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:

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:

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.

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.

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.

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.

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?

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.

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

Question 5.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

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.

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.

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.

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 = \(\frac{\angle \mathrm{A}}{2}\)

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.

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.

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 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.

‘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°.

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.

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?

Solution:
Fig. (i): Regular Hexagonal with side 5 cm. is constructed. It has 6 equal sides.
Measure of each side is 5 cm.

Number of equilateral ∆ with 1 cm side

In Fig. (ii) : there are 12 equilateral triangles with side 5 cm

∴ Number of equilateral A with 1 cm side

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 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.

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.

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

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

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 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.

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.

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.

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.

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

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.

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.

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.

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.4, 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 10.3 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 10.3.

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3

Question 1.
Draw the graph of each of the following linear equations in two variables :
i) x + y = 4
ii) x – y = 2
iii) y = 3x
iv) 3 = 2x + y
Solution:
i) x + y = 4
This can be written in the form of y = mx + c, then
y = 4 – x.

x	0	1	4	-1
y	4	3	0	5

ii) x – y = 2
-y = 2 – x
y = -2 + x
y = x – 2

x	0	1	2	4
y	-2	-1	0	2

iii) y = 3x

x	0	1	2	-1
y	0	3	6	-3

iv) 3 = 2x + y
2x + y = 3
y = 3 – 2x

x	0	1	2	-1
y	3	1	-1	5

Question 2.
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why ?
Solution:
E.g. i) y = 7x
ii) x = \(\frac{y}{7}\)
iii) x + y = 16 .
We may write infinite equation, because Graph of a linear equation in two variables has many solutions.

Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
3y = ax + 7
If point (3, 4) then a = ?
3y = ax + 7
3(4) = a(3) + 7 ∵ x = 3, y = 4
12 = 3a + 7
3a + 7 = 12
3a = 12 – 7
3a = 5
∴ a = \(\frac{5}{3}\)

Question 4.
The taxi fare in a city is as follows:
For the first kilometre, the fare is Rs. 8 and for the subsequent distance, it is Rs. 5 per km. Taking the distance covered as x km and the total fare is Rs. y. write a linear equation for this information, and draw its graph.
Solution:
Let the distance travelled be x km.
Let the total distance be y km.

For the first kilometre, the fare is Rs. 8 + fare for the remaining distance, i.e. (x – 1) km.
Rs. 5 for 1 km.
if it travels (x – 1) km … ? Rs.
Rs. (x – 1)5
∴ Total cost = Rs. y
Fare for first km + fare for subsequent distance means (x-1) km = Rs. y
8 + (x – 1)5 = y
8 + 5x – 5 = y
8 – 5 + 5x = y
3 + 5x = y
∴ y = 3 + 5x linear equation.

x	0	1	-1	-2
y	3	8	-2	-7

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. 10.6 and Fig. 10. 7.

Solution:
Solution for graph of linear equation of Fig. 10.6 :

Sum of (x + y) = 0.
Equation for this is

(ii) x + y = 0
because x = -y we have OR -x = y
0 = 0
1 = -1
-1 = 1
Solutions for graph for linear equation in Fig. 10.7 :

x + y = 2 is common in all.
∴ y = -x + 2
∴ Ans: (iii) y = -x + 2 equation.

Question 6.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
(i) 2 units
(ii) 0 unit.
Solution:
Let the work done be, W
Distance travelled be D.
If constant force is K, then
Force is directly proportional to the distance travelled by the body.
∴ W ∝ D.
W = K × D
Constant force, K = 5 units.
W = 5 × D linear equation.
If D = 2, then W = 5D
W = 5 × 2
W = 10
If D = 0, then W = 5D
W = 5 × 0
W = 0

Question 7.
Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs. x and Rs. y.) Draw the graph of the same.
Solution:
Yamini and Fatima together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims.
In that contribution of Yamini is R. x Contribution of Fatima is Rs. y Together Rs. 100.
∴ x + y = 100
∴ y = 100 – x.

x	20	40	50	60
y	80	60	50	40

Question 8.
In countries like USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius :
F = \(\left(\frac{9}{5}\right)\) C + 32.
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.
(ii) If the temperature is 30° C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95° F, what is the temperature in Celsius?
(iv) If the temperature is 0° C, what is the temperature in Fahrenheit, and if the temperature is 0° F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution:
Let Temperature be Celsius C,
The temperature be Fahrenheit F.
F = \(\left(\frac{9}{5}\right)\) C + 32

C	10	30	0	35
F	50	86	32	95

In the x-axis Celsius,
In the y-axis Fahrenheit.
(i) Graph

(ii) If temperature is 30° C, then 86° F.
(iii) If the temperature is 95° F, then 35° C.
(iv) If temperature is 0° C, then 32° F.
If 0°F, then -32°C
(v) Temperature is not equal in Celsius and Fahrenheit.
because C ≠ F

We hope the KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.3, drop a comment below and we will get back to you at the earliest.