Class 9

Karnataka Board Class 9 Maths Chapter 13 Surface Area and Volumes Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine :
(i) The area of the sheet required for making the box.
(ii) The cost of the sheet for it, if a sheet measuring 1 m² costs Rs. 20.
Solution:
(i) Length of plastic box, l = 1.5 m
breadth b = 1.25 m
height, h = 65 cm = 0.65 m.
= (1.5 × 1.25) + (2 × 1.25 × 0.65) + 2(1.5)(0.65)
= 1.875 + 1.625 + 1.950
A= 5.45 m².
(ii) Cost of 1 sq. m sheet is Rs. 20
cost of 5.45 sq.m, the sheet is = 5.45 × 20 = Rs. 109.

Question 2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 7.50 per m².
Solution:
Length of room, l = 5 m
breadth, b = 4m
height, h = 3 m
White washing the room and the ceiling except floor to be made.
∴ Area, A = lb + 2bh + 21h
= lb + 2h(b + l)
= (5 × 4) + 2 × 3(4 + 5)
= 20 + 6(9)
= 20 + 54
= 74 m².
Cost of white washing for 1 sq.m. is Rs. 7.50
∴ Cost of white washing 74 sq. m. is ?
= 74 × 7.50
= Rs. 555.

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per m² is Rs. 15000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Solution:
Perimeter of a rectangle room = 250 m
= 2(l + b)
Cost of Painting = Rs. 15000
For Rs. 10, 1 sq.m.
For Rs.15000 … ? \(=\frac{15000}{10}\)
= 1500m²
Area of four walls = Lateral surface area
Lateral surface area of rectangle = 1500 m²
2h(l + b) = 1500
h × 2(l + b) = 1500
h × 250 = 1500

∴ h = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution:
Length of Brick l = 22.5 cm
breadth, b = 10 cm
height, h = 7.5 cm.
Area to be painted, A = 9.375 m²
= 93750 cm²
Total surface area of brick, A
A= 2(lb + bh + lh)
= 2[22 × 10 + 10 × 7.5 + 22.5 × 7.5]
= 2[468.75]
= 937.5 cm².
Number of bricks to be painted, For 937.5 cm², 1 brick is required For 93750 cm² …??…

= 100
∴ 100 bricks have to be painted.

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
Each edge of cubical box, a = 10 cm.
(i) Curved Surface Area = 4a² = 4(10)²
= 4 × 100
= 400 cm².
Length of cuboid box, l = 12.5 cm.
b = 10 cm
h = 8 cm.
Curved Surface area of Cuboid, A = 2h (l+ b)
A = 2 × 8(12.5 + 10)
= 16 × 22.5
= 360 cm².
∴ 400 – 360 = 40 cm².
L.S.A. of cuboid box is 40 cm² greater than rectangular cuboid.

(ii) Total surface area of cube (T.S.A.) = 6a²
= 6(10)²
= 6 × 100
= 600 cm².
T.S.A. of cuboid = 2(Ib + lh + bh)
= 2[ 12.5 × 10 + 12.5 × 8 + 10 × 8]
= 2 (125 + 100 + 80)
= 2 × 305
= 610 cm².
Here, cuboid box’s T.S.A. is more than T.S.A. of cubical box, i.e,
610 – 600 = 10 cm² more.

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Solution:
(i) Length of Herbarium, l = 30 cm,
breadth, b = 25 cm.
height, h = 25 cm.
∴ TSA of Herbarium = 2 (lb + bh + Hi)
= 2 [30 × 25 + 25 × 25 + 30 × 25]
= 2 [750 + 625 + 750]
= 2 (2125)
= 4250 cm².

(ii) Tape required for all 12 edges means, it has 4 length, 4 breadth and 4 heights.
∴ Length of tape required = 4 (l + b + h)
= 4 (30 + 25 + 25)
= 4 × 80
= 320 cm.

Question 7.
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Measurement of big box, l = 25 cm.
b = 20 cm.
h = 5 cm.
∴ TSA of eacxh big box = 2(lb + bh + lh)
= 2 [25 × 20 + 20 × 5 + 25 × 5]
= 2 [500 + 100 + 120]
= 2 (725)
T.S.A. =1450 cm².
For all the overlaps 5% of the total surface area is required extra.

= 72.5 cm².
∴ Total surface area of big box, TSA = 1450 + 72.5 = 1522.5 cm².
Area required for 1 box is 1522.5 cm2 Area required for 25 boxes … ?…
= 1522.5 × 250
= 380625 cm².
Dimension of small box, l = 15 cm.
b = 12 cm.
h = 5 cm.
∴ TSA of each small box = 2(lb + bh + lh)
= 2 [15 × 12 + 12 × 5 + 15 × 5]
= 2 [180 + 60 + 75]
= 2 × 315
= 630 cm².
More area to overlap cardboard is 5% more.
For 100 cm², 5 cm²
For 630 cm², …?…

= 31.5 cm².
∴ T.S.A.= 630 + 31.5= 661.5 cm².
TSA required for 1 box is 661.5 cm²
TSA required for 250 boxes …?…
= 661.5 × 250
= 165375 cm².
∴ To prepare 500 boxes, TSA = 380625 + 165375
= 546000 cm².
Cost of 1000 sq.cm, of cardboard is Rs.4 Cost of 546000 sq.cm of cardboard is ?

= Rs. 2184

Question 8.
Parveen wanted to make a temporary shelter for her car, by making a box-like structure with a tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how many tarpaulins would be required to make the shelter of height 2.5 m, with base dimensions 4m × 3m?
Solution:
Measurement of the roof, l = 4 m.
b = 3 m.
h = 2.5 m.
∴ Area of Tarpaulin = 2 (lb + bh) + lh
= 2 [4 × 2.5 + 3 × 2.5] +4 × 3
= 2 [10 + 7.5] + 12
= 2 × 17.5 + 12
= 35 + 12
= 47 m².

KSEEB Solutions for Class 9 Maths

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 Area and Volumes Ex 13.7

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

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm.
(ii) radius 3.5 cm, slant height 12 cm.
Solution:
(i) Volume of Cone, V = \(\frac{1}{3} \pi r^{2} h\)

∴ V = 264 cm³.
(ii) r = 3.5 cm, h = 12 cm.
Volume of Cone, V = \(\frac{1}{3} \pi r^{2} h\)

= 11 × 14
∴ V = 154 cm³.

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm.
(ii) height 12 cm, slant height 13 cm.
Solution:
(i) radius, r = 7 cm, slant height, l = 25 cm
Volume of conical vessel, V = \(\frac{1}{3} \pi r^{2} h\)

∴ h = 24 cm.
V = \(\frac{1}{3} \pi r^{2} h\)

= 22 × 7 × 8
∴ V = 1232 cm³
∴ Volume of conical vessel
= \(\frac{1232}{1000}\) = 1.232 litres
(ii) height, h =12 cm, slant height, l = 13cm. r = ?, V = ? .
h² = l² – r²
∴ r² = l² – h²

∴ r = 5 cm.
Volume of vessel, V = \(\frac{1}{3} \pi r^{2} h\)

∴ V = 314.28 cm³.
In litres, = \(\frac{314.28}{1000}\) = 0.31428 litres.

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)
Solution:
Height of the cone, h = 15 cm.
Volume, V = 1570 cm³, radius, r = ?
Volume of a cone. V = \(\frac{1}{3} \pi r^{2} h\)

r² = 100
∴ r = 10 cm.

Question 4.
If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Solution:
Volume of a right circular cone,V= 48π cm³,
height, h = 9 cm.
diameter, d = ?
Volume of cone, V = \(\frac{1}{3} \pi r^{2} h\)

∴ r² = 16
∴ r = 4 cm
∴ diameter, d = 2r
= 2 × 4
∴ d = 8 cm.

Question 5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Solution:
d = 3.5m., h = 12 m., V = ?

Volume of Conical pit, V

= 22 × 0.5 × 3.5
∴ V = 38.5 m³
V = 38.5 kilolitres

Question 6.
The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
Solution:
Volume of Cone, V = 9856 cm³
diameter of the base, d = 28 cm.
(i) h = ?,
(ii) I = ?,
(iii) C.S.A. = ?
(i) V = 9856 cm³.
d = 28 cm. ∴ r = \(\frac{28}{2}\) = 14 cm.
Volume of cone, V= \(\frac{1}{3} \pi r^{2} h\)

h = \(\frac{2688}{56}\)
h = 48 cm.

(ii) l² = h² + r²

l = 50 cm.

(iii) Curved surface area of the cone, A
A = πrl

= 22 × 100
A= 2200 cm².

Question 7.
A right triangle ABC with sides 5 cm, 12 cm, and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Solution:
Cone is formed ∆ABC is revolved about the side 12 cm.

r = 5 cm,
h = 12 cm,
V = ?
Volume of Cone, V

∴ V = 314.29 cm³.

Question 8.
If the triangle ABC in Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.
Solution:
Cone is formed when ∆ABC is revolved about 5 cm.
r = 12 cm,
h = 5 cm.

Volume of Cone, V
v = \(\frac{1}{3} \pi r^{2} h\)

V = 754.28 cm³.
Volume of Cone in problem 7 = 314.29 cm³
Volume of Cone in problem 8 = 754.28 cm³
∴ ratio of both Volumes = \(\frac{314.29}{754.28}\)
= \(\frac{5}{2}\) = 5 : 12

Question 9.
A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find ifs volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Solution:
Diameter of heap of wheat in the form of Cone,
diameter, d = 10.5 m = \(\frac{21}{2}\)

height, h = 3 m. V = ?
Volume of Canvas Cover, V = \(\frac{1}{3} \pi r^{2} h\)

= 86.625 m³.
Area of Canvas required to cover a heap of wheat, C.S.A.
C.S.A.= πrl

∴ Area of Canvas, A = πrl

∴ A = 99.83 m².

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.

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.

KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3.

Karnataka Board Class 9 Maths Chapter 5 Triangles Ex 5.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that

(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Solution:
Data : ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
To Prove:
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC
(v) AD is the angular bisector of ∠A.
Proof:
(i) In ∆ABD and ∆ACD,
AB = AC (data)
BD = DC (data)
AD is common.
S.S.S. Congruence rule.
∴ ∆ABD ≅ ∆ACD

(ii) In ∆ABP and ∆ACP,
AB = AC (data)
∠ABP = ∠ACP (Opposite angles)
∠BAP = ∠CAP (∵ ∆ABD ≅ ∆ACD proved)
Now ASA postulate.
∆ABP ≅ ∆ACP.

(iii) ∆BAD ≅ ∆CAD proved.
AP bisects ∠A.
In ∆BDP and ∆CDP,
BD = DC (data)
BP = PC (proved)
DP is common.
∴ ∆BDP ≅ ∆CDP (SSS postulate)
∴ ∠BDP = ∠CDP
∴ DP bisects ∠D.
∴ AP bisects ∠D.

(iv) Now, ∠APB + ∠APC = 180° (Linear pair)
∠APB + ∠APB = 180°
2 ∠APB = 180
∴ ∠APB = \(\frac{180}{2}\)
∴∠APB = 90°
∠APB = ∠APC = 90°
BP = PC (proved)
∴ AP is the perpendicular bisector BC.

(v) AP is the angular bisector of ∠A.
Angular bisector of ∠A is aD, because AD, AP is in one line.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that A

(i) AD bisects BC
(ii) AD bisects ∠A.
Solution:
Data: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To Prove:
(i) AD bisects BC.
(ii) AD bisects ∠A.
Proof: i) In ∆ABD and ∆ACD,
∠ADB = ∠ADC (∵ AD ⊥ BC)
AB = AC (data)
AD is common.
∴ ∆ABD ≅ ∆ACD
∴ BD = DC
∴ AD bisects BC.

(ii) ∠BAD = ∠CAD (∵ ∆ADB ≅ ∆ADC)
∴ AD bisects ∠A.

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR. Show that :

(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Solution:
Data: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR.
To Prove:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR.
Proof: (i) In ∆ABC,
AM is the median drawn to BC.
∴ BM = \(\frac{1}{2} \) BC
Similarly, in ∆PQR,
QN = \(\frac{1}{2}\) QR
But, BC = QR
\(\frac{1}{2} \) BC = \(\frac{1}{2}\) QR
∴ BM = QN
In ∆ABM and ∆PQN,
AB = PQ (data)
BM = QN (data)
AM = PN (proved)
∴ ∆ABM ≅ ∆PQN (SSS postulate)

(ii) In ∆ABC and ∆PQR,
AB = PQ (data)
∠ABC = ∠PQR (proved)
BC = QR (data)
∴ ∆ABC ≅ ∆PQR (SSS postulate)

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:
Data: BE and CF are two equal altitudes of a triangle ABC.
To Prove: ABC is an isosceles triangle.
Proof : BE = CF (data)
In ∆BCF and ∆CBE,
∠BFC = ∠CEB = 90° (data)
BC is a common hypotenuse.
As per Right angle, hypotenuse, side postulate,
∴ ∆BCF ≅ ∆CBE
∴ ∠CBF = ∠BCE
∴ ∠CBA = ∠BCA
∴ AB = AC
∴ ∆ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:
Data: ABC is an isosceles triangle with AB = AC.
To Prove : ∠B = ∠C
Construction: Draw AP ⊥ BC.
Proof: In ∆ABC, AP ⊥ BC and AB = BC.
∴ In ∆ABP and ∆ACP
∠APB = ∠APC = 90° ( ∵ AP ⊥ BC)
Hypotenuse AB = Hypotenuse AC
AP is common.
As per RHS Postulate,
∆ABP ≅ ∆ACP
∴ ∠ABP = ∠ACP
∴ ∠ABC = ∠ACB
∴∠B = ∠C.

We hope the KSEEB Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 helps you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 5 Triangles Exercise 5.3, drop a comment below and we will get back to you at the earliest.