KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1

   

KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 1
Which method did you use for finding the mean, and why ?
Solution:
Average by Direct Method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 1.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 1.2

Question 2.
Consider the following distribution of daily wages of 30 workers of a factory.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 2
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Step deviation method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 tabe 2.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 2.2
∴ Average = 545.20.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 3
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 3.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 3.2

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 4
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 4.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 4.2
∴ Mean heartbeats per minute = 75.9

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 5
Find the mean number of mangeos kept in a packing box. Which method of finding the mean did you choose ?
Solution:
Step Deviation Method :
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 5.1
∴ Mean Number of mangoes in the boxes = 57.19

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 6
Solution:
By Step deviation method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 6.1
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 6.2
∴ Mean Daily expenditure for food = Rs. 211.

Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 7
Find the mean concentration of SO2 in the air.
Solution:
By step Deviation method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 7.1
∴ Mean Concentration of SO2 = 0.099 ppm.

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 8
Solution:
Mean Deviation method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 8.1
Here, a = 17
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 8.2
∴ Mean number of days a student was absent = 12.48 Days.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 9
Solution:
Step Deviation Method:
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 9.1
By Step Deviation Method
KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 Q 9.2
∴ Mean Literacy rate = 69.43%

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 13 Statistics Ex 13.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 13 Statistics Exercise 13.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4

   

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4

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

Question 1.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
Height of drinking glass, h = 14 cm.
Diameter of circular ends
D = 4 cm ∴ r = 2 cm.
d = 2 cm ∴ r = 1 cm.
Capacity of glass = ?
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 1
Volume of glass = Radius with 2 cm + Volume of frustum with radius 1 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 1.1
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 1.2

Question 2.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its ciruclar ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution:
Slant height of a i frustum of a cone, l = 4 cm.
Circumference of its circular ends = 18 cm. and 6 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 2
2πr =18 cm.
∴ R = \(\frac{9}{\pi}\) cm.
2πr = 6
∴ r = \(\frac{3}{\pi}\) cm.
∴ Lateral surface area of a frustum of a Cone
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 2.1

Question 3.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the figure).
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3
If its radius on the open side is 10 cm, radius at the upper base is 4 cm, and its slant height is 15 cm, find the area of material used for making it.
Solution:
Radius on the open side of the cap, R = 10 cm.
Radius of upper end, r = 4 cm.
Slant height l = 15 cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3.1
R = 10 cm, r = 4 cm, l = 15 cm
∴ Curved surface = π × l × (R + r)
= π × 15 × (10 + 4)
= \(\frac{22}{7}\) × 15 × 14
= 660 cm2
∴ Area of the cap at the end
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3.2
∴ Area of material used for making it.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 3.3

Question 4.
A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm2. (Take π = 3.14).
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 4
R = 20 cm
r = 8 cm
h = 20 cm
∴ Slant height,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 4.1
∴ Volume of metallic sheet,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 4.2
Quantity of milk in the container
= \(\frac{10449.82}{1000}\)
Cost of 1 litre of milk is Rs. 20,
Cost of 10.45 litres of milk ……. ??
∴ 20 × 10.45 = Rs. 209.
Cost of metal sheet = π(R + r) + πr2
= π {20 × (20 + 8) + (8)2}
= 3.14 × 624
= 1959.36 cm2.
∴ Cost of preparing metallic container: For 100 cm2 Rs. 8
∴ For 1959.36 cm2 ……?
= 8 × \(\frac{1159.36}{100}\)
= Rs. 156.75.

Question 5.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \(\frac{1}{16}\) cm., find the length of the wire.
Solution:
Height of a cone, h = 20 cm.
Vertical angle = 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5
Diameter of frustum wire = \(\frac{1}{16}\) cm.
i) In ⊥∆AQC, ∠Q = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.1
ii) In ⊥∆PRA, ∠R = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.2
Height of frustum, h = 10 cm.
∴ Volume of metallic frustum, ABCD,
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.3
Let the length of wire which has diameter \(\frac{1}{16}\) cm be ‘x’ cm
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 Q 5.4
∴ Length of wire = 7964.4 m

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3

   

KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3

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

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
Let the height of cylinder be ‘h’ cm.
Radius of cylinder, r = 6 cm.
i) Volume of metallic sphere, V = \(\frac{4}{3} \pi r^{3}\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 1
ii) Volume of Cylinder, V = \(\pi r^{2} h\)
\(\frac{22}{7}\) × 36 × h = 310.46
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 1.1
∴ Height of cylinder is 2.74 cm.

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Let the radius of the resulting sphere be ‘r’ cm.
(i) Volume of 3 spheres having radii 8 cm and 10 cm.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 2
(i) Volume of the resulting Sphere, V
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 2.1
∴ Volume of the resulting sphere = Volume of three spheres
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 2.2
∴ Radius of the resulting sphere is 12 cm.

Question 3.
A 20 m deep well with a diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution:
Radius of well, ‘r’ = 3.5 m.
Height of well, h = 20 m.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 3
(i) Volume of soil dug out of well, V = \(\pi r^{2} h\) (∵ cylinder sphere)
= \(\frac{22}{7} \times(3.5)^{2} \times 20\)
= \(=\frac{22}{7} \times \frac{35}{10} \times \frac{35}{10} \times 20\)
= 770m3
(ii) Length of platform, l = 22 m.
Breadth, b = 14 m.
Height, h =?
Volume of Platform, V= l × b × h (∵ Cuboid)
= 22 × 14 × h
∴ Volume of Platform = Volume of soil dug out of well.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 3.1
∴ Height of platform is 2.5 m

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution:
Let the height of the embankment be ‘h’ m.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 4
Dimeter of well, d = 3 m.
Radius of well, r =1.5 m.
Height of well, h = 14 m.
Volume of an embankment = Volume of soil dug from well.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 4.1
∴ Height of embankment = 1.125 m.

Question 5.
A container shaped like a right cylinder diameter 12height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones cones which can be filled with ice cream.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 5
Solution:
Diameter of right circular cylinder is 12 cm ➝ r = 6 cm.
Height: 15 cm. ➝ h = 15 cm.
Diameter of Cone, 6 cm. ➝ r = 3 cm.
Height 12 cm. ➝ h = 12 cm.
(i) Volume of the ice cream which has cone-shaped and hemisphere.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 5.1
(ii) Volume of cylindrical vessel, V = \(\pi r^{2} h\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 5.2
Let number of cones be ‘n’,
n × 54π = π × 36 × 15
54n = 36 × 15
\(n=\frac{36 \times 15}{54}=\frac{540}{54}\)
∴ n = 10
Ice cream can be filled in 10 cones.

Question 6.
How many silver coins, 1.75 cm. in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Solution:
Diameter of the silver coins is 1.75 cm.
Thickness = 2 mm.
Measurement of cuboid: 5.5 × 10 × 3.5 cm.
∴ Let the required silver coins be ‘n’,
Volume of ‘n’ coins = Volume of cuboid
\(\pi r^{2} h\) = l × b × h
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 6
∴ n = 400
∴ 400 silver coins are required.

Question 7.
A cylindrical bucket, 32 cm high and with a radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
Height of cylindrical bucket, h = 32 cm.
Radius, r = 18 cm.
Height of conical heap of sand = 24 cm.
Radius, r =?
Slant height, l =?
Volume of the conical heap of sand = Volume of Cylinder.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 7
Radius of Cone, r = 36 cm.
Slant height of cone, l = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}\)
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 7.1
∴ Radius of cone, r = 36 cm.
Slant height of cone, h = \(12 \sqrt{13}\) cm

Question 8.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution:
Breadth of a canal, b = 6 m.
Height, h = 1.5 m
Speed of water =10 km/hr.
Quantity of water = 8 cm.
Quantity of water flowing in 30 minutes?
Volume of water flowing in 60 minutes,
= (10 × 1000) × 6 × 1.5 cm3.
( ∵ Speed = 10 km/hr.; 10 × 1000 m/hr.)
Volume of water flowing in 30 minutes,
= \(10000 \times 9 \times \frac{30}{60} \mathrm{m}^{3}\)
∴ Let the area of land be ‘x’ sq.m.
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 8
x= 562500 sq.m.

Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Solution:
Internal diameter of a pipe = 20 cm.
Radius of pipe =10 cm. = \(\frac{1}{10} \mathrm{m}\)
Diameter of cylindrical tank = 10 m; r=5 m.
Depth of cylindrical tank = 2 m.
Speed of water is 3 km/hr.
Time required to fill the water tank =?
Speed of water = 3 km/hr.
= \(\frac{3000}{60}\) m/minute
= 50 m/min.
Let the time required to fill the tank be n’ minutes.
∴ Flowing water in ‘n’ minutes = Volume of the water tank
KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 Q 9
∴ n = 100 minutes
∴ Time required to fill the water tank = 100 minutes

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Ex 15.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 15 Surface Areas and Volumes Exercise 15.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1

   

KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. (see the fig. given)
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 1
Solution:
Length of rope is 20 m.
Height of the pole = ?, AB = ?
Let the height of the pole AB be ‘A’.
Length of rope AC = l = 20 m.
Angle of elevation, ∠BCA = 30°.
∴ Let ⊥∆AABC, ∠B = 90°.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 1.1
∴ AB = 10 m.
∴ Height of the pole 10 m.

Question 2.
A tree breaks due to storm and the broken part bends, so that the tope of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 2
Solution:
Let the Height of the broken part be ‘x’ m.
BC is the distance between the foot of the tree to the point where the top touches the ground.
In ⊥∆ABC, ∠B = 90°.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 2.1
∴ Height of broken part of the tree, AB = x
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 2.2
= 4.62 m
Let AC = AD = y m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 2.3
∴ y = 2x = 2 × 4.62
y = 9.24 m.
Full height of the tree = x + y
∴ Height of the tree = x + y
= 4.62 + 9.24
= 13.86 m.

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case ?
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 3
Solution:
i) AB = 1.5 m., AC = l1 m., ∠ACB = 30°
In ⊥∆ABC, ∠B = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 3.1
∴ l1 = 2 × 1.5 = 3 m
∴ Length of slide of younger children = 3 m
ii) PQ = 3 m, OP = 12 m, ∠POQ = 60°
In ⊥∆PQR, ∠Q = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 3.2
∴ Length of slide of elder children = \(2 \sqrt{3}\) m

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 4
Solution:
Let the height of a tower. PQ = h m.
Distance of QO from base of tower = 30 m.
Angle of elevation, ∠POQ = 30°
In ⊥∆APQO, ∠Q = 90°,
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 4.1
∴ Height of tower, PQ = h = \(10 \sqrt{3}\) m.

Question 5.
A kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 5
Solution:
P is the position of Kite. It is flying at a height of 60 m. above the ground.
The inclination of the string with the ground, ∠POQ = 60°
Let length of string be OP = L
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 5.1
∴ Length of slide of elder children \(40 \sqrt{3}\) m.

Question 6.
A 1.5 m. tall boy is standing at some distance from a 30 m. tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 6
Solution:
Height of the building, PQ = 30 m.
Height of the boy, OA = 1.5 m.
If the position of Boy at OA, angle of elevation, ∠POR = 30°, His second position, OO’, ∠POR = 60°.
RQ = OA = 1.5 m.
∴ PR = 30 – RQ = 30 – 1.5
PR = 28.5 m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 6.1
Distance he walked towards the building:
OO’ = OR – O’R
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 6.2
Distance he walked towards the building:
OO’ = \(19 \sqrt{3}\) m

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 7
Solution:
Let the height of building, PQ = 20 m.
PR= h m.
Bottom of transmission tower, P, vertex point ‘R’.
∠POQ = 45°, ∠ROQ = 60°.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 7.1
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 7.2
∴ Height of transmission tower is = \(20(\sqrt{3}-1)\) m

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 8
Solution:
Height of pedestal, PQ = h m.
PR = 1.6 m.
∠POQ = 45°
∠ROQ = 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 8.1
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 8.2
∴ Height of pedestal, h = \(0.8 \times(\sqrt{3}+1)\) m

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high. find the height of the building.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 9
Solution:
Height of the tower, PQ = 50 m.
Let Height of the building, AB = h m.
Angle of elevation to top of tower, = ∠PBQ = 60°
Angle of elevation to building. = ∠AQB = 30°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 9.1
h = 16.6 m

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 10
Solution:
Height of AB and PQ be h m.
Breadth of Road, BQ = 80 m.
‘O’ is the point on road.
∠POQ = 60°, ∠AOB = 30°
OQ = ‘x’ m. OB = ‘y’ m.
x + y = 80 m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 10.1
∴ Height of pole, AB = PQ = h = \(20 \sqrt{3}\) m.
Distance of the points c from the poles 20 m. and 60 m.

Question 11.
A TV tower stands vertically on a bank Of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the figure given). Find the height of the tower and the width of the canal.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 11
Solution:
Let the height of TV tower, AB = h m.
Breadth of a canal, BC = x m.
∠ACB = 60° ∠ADB = 30°,
DC = 20 m..
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 11.1
From eqtn. (ii).
\(\mathrm{h}=\mathrm{x} \sqrt{3}=10 \sqrt{3}\)
∴ Height of tower = \(10 \sqrt{3}\) m
Breadth of canal = 10 m.

Question 12.
From the top of a 7m. high building, the angle of elevation of the top of a cable tower is 60° and th angle of depression is its foot is 45°. Determine the height of the tower.
Solution:
Let the height of tower, PQ = h m.
The height of building, AB = 7 m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 12
Angle of elevation, ∠PAR = 60°
Angle of depression, ∠RAQ = 45°
∠AQB = 45° (∵ Alternate angle)
Let, BQ = AR = x m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 12.1
∴ Height of tower, PQ = \(7(\sqrt{3}-1)\) m.

Question 13.
As observed from the top of a 75 m. high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 13
Solution:
Height of lighthouse, PQ = 75 m.
A and B are positions of ship.
Px is Horizontal line.
∠APx = 30° and ∠BPx = 45°
∠PAQ = 30° and ∠PBQ = 45°
Let AB = x m, BQ = y m.
Then AQ = AB + BQ = (x + y) m
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 13.1
∴ Distance between the two ships, AQ = x + y
= \(75 \sqrt{3}\) m.

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see the figure given below). Find the distance travelled by the balloon during the interval.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 14
Solution:
Height of the girl, OO’ = 1.2 m
∠AOB = 60°
∠POQ = 30°
Let OB = x m, BQ = d m, O’Q’ = y m.
AB = PQ = Q’P – Q’Q
= 88.2 – O’O
= 88.2 – 1.2 = 87 m.
Let OQ = ‘y’.
Distance balloon travelled, d = BQ
= (y – x)
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 14.1

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 15
Solution:
Let the height of tower, PQ = h m.
Px = Horizontal line
Let the positions of 1st car and 2nd car be A and B.
∠APx = 30° ∴ ∠PAQ = 30°
∠BPx = 60° ∠PBQ = 60°
Let speed of car be x m/s.
∴ Distance, AB = 6x m.
Time required to travel from B to Q be ‘n’ seconds,
∴ Distance, BQ = nx m
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 15.1
\(n+6=n \sqrt{3} \times \sqrt{3}\)
3n = n+6
2n=6
∴ n = 3 seconds

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 16
Solution:
Let the height of tower, PQ = h m.
AQ = 4 m, BQ = 9 m.
∠PAQ = θ ∠PBQ = 90 – θ
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 16.1
From eqn. (i) and eqn. (ii),
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 16.2
∴ h2 = 36
∴ h = 6 m
∴ Height of tower, h = 6 m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

   

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.2

Question 1.
In Fig. 3.28. find the values of x and y and then shows that AB||CD.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 1
Answer:
i) ∠x = ?
ii) ∠y = ?
iii) Show tht AB||CD.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 2
(i) PQ Bisects AB at R.
∠PRA and ∠ARS are on straight line PQ.
∴ ∠PRA + ∠ARS = 180° Adjacent angles
50 + ∠ARS = 180°
∠ARS = 180 – 50
∴ x = ∠ARS = 130° (i)

(ii) Similarly, PQ straight line intersects straight line CD at S.
∴ ∠CSQ = ∠RSD . Vertically opposite angles
∴ y = ∠RSD = 130° (ii)
From (i) and (ii),
∠ARS = ∠RYD
x = y = 130°

(iii) Therse are mutual alternate angles
∴ x = y = 130°
∴ AB || CD.

Question 2.
In Fig. 3.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 3
Answer:
PQ straight line intersects AB||CD]|EF at points R, S, T.
If y : z = 3 : 7, then value of ‘x’ = ?
PQ straight line intersects CD||EF at S and T.
∴ ∠CSR = ∠DST = y (∵ vertically opposite angle)
But, ∠DST + ∠STF = 180°
(∵ Sum of interior angles in the same side)
∴ y + z = 180°
But, y : z = 3 : 7.
Sum of ratio 3 + 7 = 10
If ratio 10 means 180
If ratio 9 means ?
\(\frac{180 \times 3}{10}=54^{\circ}\)
∴ y = 54°
y + z = 180
54 + z = 180
∴ z = 180 – 54
∴ z = 126°
But, x + y = 180
(∵ Sum of interior angles on one side)
x + 54 = 180
∵ x = 180 – 54
∵ x = 126°.

Question 3.
In Fig. 3.30, If AB||CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 4
Answer:
AB||CD and EF⊥CD and ∠GED = 126°
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 5
∠AGE =?
∠GEF =?
∠FGE =?
∠GED = 126°.
∴ ∠GEF + ∠ZFED = 126°
∠GEF + 90° = 126° ∵ EF ⊥ CD.
∴ GEF = 126-90
∴  GEF = 36°.
∴ ∠GFE = 90° . EF ⊥ AB
∴ In ∆GEF
∠FGE + ∠GFE + ∠GEF = 180°
∠FGE + 90° + 36° = 180°
∠FGE + 126° = 180°
∴ ∠FGE = 180-126
∴ ∠FGE = 54°.
∠AGE + ∠FGE = 180° (Adjacent angles)
∠AGE + 54° = 180°
∴ ∠AGE =180-54
∠AGE = 126°
∴ ∠AGE = 126°
∠GEF = 36°
∠FGE = 54°.

Question 4.
In Fig. 3.31, If PQ||ST, ∠PQR =110°, and ∠RST = 130°, find ∠QRS.
(Hint: Draw a line parallel to ST through point R).
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 6
Answer:
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 7
Data: PQ||ST and ∠PQR= 110° and ∠RST= 130°.
To Prove: ∠QRS =?
Construction: Draw ST||UV through ‘R’
Proof: PQ||ST (Data)
ST || UV (Construction)
∴ PQ||ST||UV
PQ||UV.
∴ ∠PQR + ∠URQ = 180° (Sum of interior angles)
110 + ∠URQ = 180°
∠URQ = 180- 110
∴ ∠URQ = 70° (i)
Similarly, ST||UV.
∴ ∠RST + ∠SRV = 180°
130 + ∠SRV = 180
∴ ∠SRV = 180 – 130
∴ ∠SRV = 50° (ii)
But, URV is a straight line.
∴ ∠URV = 180°.
∠URQ + ∠QRS + ∠RSV = 180°
70 + ∠QRS + 50 = 180
∠QRS + 120 = 180°
∴ ∠QRS = 180 – 120
∴ ∠QRS = 60°.

Question 5.
In Fig. 3.32. if AB||CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 8
Answer:
If AB || CD and ∠APQ= 50° and ∠PRD = 127°, then ∠x = ? ∠y = ?
AB || CD PQ straight line intersects these at P and Q.
∴ ∠APQ = ∠PQR . alternate angles.
50 = ∠PQR = x
∴ x = 50°
Similarly, AB || CD PR straight line intersects these at P and R.
∴ ∠APR = ∠PRD
∠APQ + ∠QPR = ∠PRD
50 + ∠QPR = 127
∠QPR = 127 – 50
∴∠QPR = 77 °
∠QPR = y = 77 °
∴ x = 50 °,
y = 77 °.

Question 6.
In Fig. 3.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
Prove that AB||CD.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 9
Answer:
Data: PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B. the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD.
To Prove: AB || CD.
Proof: PQ is a mirror.
AB is an incident ray, BC is a reflected ray. ∠ABC is the angle of incidence.
∠BCD is an angle of reflection.
The incident ray formed in the PQ mirror is equal to the reflected ray formed in the RS mirror.
∴ ∠ABC = ∠BCD.
These are pair of alternate interior angles, then AB || CD.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4

   

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4.

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.4.

Question 1.
Determine which of the following polynomials has (x+1) a factor :
i) x3 + x2 + x + 1
ii) x4 + x3 + x2 + x + 1
iii) x4 + 3x3 + 3x2 + x + 1
iv) x3 – x2 – (2 + \(\sqrt{2}\) )x + \(\sqrt{2}\)
Answer:
i) x- 1 is a factor of p(x)
x + 1 = x – a
a = -1
For the value of p(a),
value of r(x) = 0.
∴ p(x) = x3 + x2 + x + 1
p(-1) = (-1)3 + (-1)2 + (-1) + 1
= -1 + 1 -1 + 1
p(-1) = 0
Here, p(a) = r(x) = 0
∴ x + 1 is a factor.

ii) If x + 1 = x – a, then a = -1
p(x) = x4 + x3 + x2 + x + 1
p(-1)= (-1)4 + (-1)3 +(-1)2 + (-1)+ 1
= 1 – 1 + 1 – 1 + 1
= 3 – 2 p(-1)= 1
Here, r(x) = p(a)= 1 Reminder is not zero.
∴ x+1 is not a factor.

iii) If x + 1 = x – a then
a = -1
p(x) = x4 + 3x3 + 3x2 + x + 1
p(-1) = (-1)4 + 3(-1 )3 +3(-1 )2 + (-1) + 1
= 1 + 3(-1) + 3(1) + 1(-1) + 1
= 1- 3 + 3 – 1 + 1
= 5 – 4
P(-1)= 1
Here, r(x) = p(a)=l Remainder is not zero
∴ x+1 is not a factor.

iv) If x + 1 = x – a then,
a = -1
p(x) = x3 – x2 – (2 + \(\sqrt{2}\))x+ \(\sqrt{2}\)
p(-1) = (-1)3 – (-1)2 -(2 + \(\sqrt{2}\))(-1) + \(\sqrt{2}\)
= -1 -(+1) – (2 – \(\sqrt{2}\))+ \(\sqrt{2}\)
= -1 – 1 + 2 + \(\sqrt{2}\) + \(\sqrt{2}\)
= -2 + 2 + \(2 \sqrt{2}\)
= = + \(2 \sqrt{2}\)
p(-1) = \(2 \sqrt{2}\)
Here, r(x) = p(a) = \(2 \sqrt{2}\) Value of remainder r(x) is not zero.
∴ x + 1 is not a factor.

Question 2.
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x+1
ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3
Answer:
i) p(x) = 2x3 + x2 – 2x – 1
g(x) = x + 1
If x + 1 = 0, then x = -1
p(x) = 2x3 + x2 – 2x – 1
p(-1)= 2(-1)3 + (-1)2 -2(- 1) – 1
= 2(-1) + (1) + 2 – 1
= -2 + 1 + 2 – 1
P(-1)= 0
Here, r(x) = p(a) = 0,
∴ g(x) is the a factor f(x)

ii) p(x) = x3 + 3x2 + 3x + 1
g(x) = x + 2
If x + 2 = 0, then
x = -2
∴ p(x) = x3 + 3x2 + 3x + 1
p(-2) = (-2)3 + 3(-2)2 + 3 (-2) + 1
= -8 + 3(4) + 3(-2) + 1
= -8 + 12 – 6 + 1
= 13 – 24
p(-2)= -11
Here we have r(x) = p(a) =-11.
Value of r(x) is not equal to zero.
∴ g(x) is not a factor of f(x).

iii) p(x) = x3 – 4x2 + x + 6
g(x) = x – 3
Let, x – 3 = 0, then
x = 3
p(x) = x3 – 4x2 + x + 6
p(3) = (3)2 – 4(3)2 + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6
= 36 – 36
p(3) = 0
Here, we have r(x) = p(a) = 0
∴ (x – 3) is the factor of p(x).

Question 3.
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases :
i) p(x) = x2 + x + k
ii) p(x) = 2x2 + kx + \(\sqrt{2}\)
iii) p(x) = kx2 – \(\sqrt{2} \mathrm{x}\) + 1
iv) p(x) = kx2 – 3x + k
Answer:
i) p(x) = x2 + x + k
g(x) = x – 1
k = ?
If x – 1 = 0, then
x = 1
p(x) = x2 + x + k
p(1) = (1)2 + 1 + k
p( 1) = 1 + 1 + k
p( 1) = 2 + k
If g(x) is a factor, then we have r(x) = 0
∴ p(1) = 0
2 + k= 0
∴ k = 0 – 2
k = -2

ii) p(x) = 2x2 + kx + \(\sqrt{2}\)
g(x) = x – 1 k = ?
If x – 1 = 0, then x = 1
p(x) = 2x2 + kx + \(\sqrt{2}\)
p(1) = 2(1)2 + k(1) + \(\sqrt{2}\)
= 2(1) + k(l) + \(\sqrt{2}\)
p(1) = 2 + k + \(\sqrt{2}\)
If (x – 1) is the factor of p(x), then we have p(1) = 0.
∴ 2 + k + \(\sqrt{2}\) = 0
k = -2 – \(\sqrt{2}\)

iii) p(x) = kx2 – \(\sqrt{2} x\) + 1
g(x) = x – 1 k = ?
If x – 1 = 0, then
x = 1
p(x) = kx2 – \(\sqrt{2} x\) + 1
p(1) = k(1)2– \(\sqrt{2}\)(1) + 1
= k( 1) – \(\sqrt{2}\) + 1
p(1) = k- \(\sqrt{2}\) + 1
If (x – 1) is the factor of p(x) then we have p(1) = 0.
∴ p(1) = k – \(\sqrt{2}\) + 1 = 0
∴ k= \(\sqrt{2}\) – 1

iv) p(x) = kx2 – 3x + k
g(x) = x – 1 k = ?
If x – 1 = 0, then x – 1
p(x) = kx2 – 3x + k
p(1) = k(1)2 – 3(1) + k
= k(1) – 3(1) + k
= k – 3 + k
p(1) = 2k – 3
If (x – 1) is the factor of p(x), then we have p(1) = 0.

Question 4.
Factorise :
i) 12x2 – 7x + 1
ii) 2x2 + 7x + 3
iii) 6x2 + 5x – 6
iv) 3x2 – x – 4
Answer:
i) 12x2 – 7x + 1
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 1
= 12x2 – 4x – 3x + 1
= 4x (3x – 1) – 1(3x – 1)
= (3x – 1) (4x – 1)

ii) 2x2 + 7x + 3
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 2
= 2x2 + 6x + x + 3
= 2x(x + 3) + 1 (x + 3)
= (x + 3) (2x + 1)

iii) 6x2 + 5x – 6
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 3
= 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3) (3x – 2)

iv) 3x2 – x – 4
KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 4
= 3x2 – 4x + 3x – 4
=x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)

Question 5.
Factorise :
i) x2 – 2x2 – x + 2
ii) x2 – 3x2 – 9x – 5
iii) x3 + 13x2 + 32x + 20
iv) 2y3 + y2 – 2y – 1
Answer:
i) x3 – 2x2 – x + 2
= x2(x – 2) – 1 (x – 2)
= (x – 2) (x2 – 1)
= (x – 2) (x + 1) (x- 1)

ii) x3 – 3x2 – 9x – 5
= x3 – 5x2 + 2x2 – 10x + x – 5
= x2(x – 5) + 2x(x – 5) + 1 (x – 5)
= (x – 5) (x2 + 2x + 1)
= (x – 5) {x2 + x + x + 1}
= (x – 5) (x(x + 1) + 1(x + 1)}
= (x- 5)(x + 1) (x + 1)

iii) x3 + 13x2 + 32x + 20
= x3 + 10x2 + 3x2 + 30x + 2x + 26
= x2(x + 10) + 3x(x + 10) + 2(x + 10)
= (x + 10) (x2 + 3x + 2)
= (x + 10) {x2 + 2x + x + 2)
= (x + 10) {x(x + 2) + 1 (x + 2))
= (x + 10) (x + 2) (x + 1)

iv) 2y3 + y2 – 2y – 1
= y2(2y + 1) – 1(2y + 1)
= (2y + 1) (y2– 1)
= (2y + 1) {(y)2 – (1)2}
= (2y+ 1) (y + 1) (y- 1)

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

   

KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.6.

Karnataka Board Class 9 Maths Chapter 1 Number Systems Ex 1.6

Question 1.
Find:
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 1
Answer:
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 2

Question 2.
Find:
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 3
Answer:
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 4

Question 3.
Simplify:
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 5
Answer:
KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 6

We hope the KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.6, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1

   

KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1.

Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Ex 3.1

Question 1.
In Fig 3.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 1
Answer:
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 2
∠AOC = ∠BOD = 40°
∠BOD = 40° (Date)
∠AOC = ∠BOD = 40° because vertifically opposite angles.
∴ ∠AOC + ∠BOE = 70°
40° + ∠BOE = 70°
∴ ∠BOE = 70° – 40° = 30°, and
∠AOD = 180° – ∠BOD = 180° – 40° = 140°
Reflex angle COE = ∠COA + ∠AOD + ∠BOD + ∠BOE
= 40° + 140° + 40° + 30°
∴ Reflex angle, ∠COE = 250°.

Question 2.
In Fig. 3.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 3
Answer:
∠XOP + ∠POY = 180°
∵ straight supplementary
∠XOP + 90° =180°
∴ ∠XOP = 180° – 90° = 90°
But, ∠XOP = a + b = 90°
a : b = 2 : 3
2 + 3 = 5 Ratio
Ratio 5 means 90°
\(=\frac{2 \times 90}{5}=2 \times 18=36^{\circ}\)
∴ If a = 36° then, ∠b = 54°.
∠XOM = ∠YON = b = 54° (∵ Vertically opposite angles)
∠XON + ∠YON = 180° (∵ Straight angle)
∴ c + 54° = 180°
c = 180 – 54
∴ c = 126°.

Question 3.
In Fig. 3.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 4
Answer:
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 5
Data: In this figure, ∠PQR = ∠PRQ.
To prove: ∠PQS = ∠PRT
Proof: In ∆PQR, ∠Q = ∠R.
∴ This is an isosceles triangle.
Let ∠Q = 70°, then ∠R = 70° .
∠PQS + ∠PQR = 180°
∠PQS + 70° = 180°
∴ ∠PQS =180 – 70
∠PQS = 110° … (i)
Similarly,
∠PRT + ∠PRQ = 180°
∠PRT + 70° = 180°
∴ ∠PRT = 180 – 70
∠PRT =100 … (ii)
from (i) and (ii)
∠PQS = ∠PRT =110°
∴ ∠PQS = ∠PRS proved.

Question 4.
If Fig. 3.16, if x + y = w + z, then prove that AOB is a line.
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 6
Answer:
Data: In this fiure, ∠BOC = x°
∠AOC = y°
∠BOD = w°
∠AOD = z° and
x + y = w + z.
To Prove: AOB is a straight line.
Proof: x + y + w + z = 360° (∵ one completre angle)
But, x + y = w + z.
∴ x + y = w + z= \(\frac{360}{2}\) = 180°
∴ x + y = 180°
∴ w + z = 180° proved.
But, ∠AOC and ∠BOC are adjacent angles,
∠AOC + ∠BOC = 180°
x = y = 180°
∴ ∠AOB = 180°.
∴ AOB is a straight line.

Question 5.
In Fig. 3.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS).
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 7
Answer:
Data: POQ is a straight line. Ray OR is perpendicular on straight line PQ. OS Ray is in between Rays OP and OR.
To Prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)
Proof : ∠QOR = ∠POR = 90° (Data)
∠POR = ∠POS + ∠ROS
∴ ∠ROS = ∠POR – ∠POS
∠ROS = 90°- ∠POS (i)
Now, QOS = ∠QOR + ∠ROS
∠QOS = 90° + ∠ROS
∴ ∠ROS = ∠QOS – 90° (ii)
By adding equation (i) and (ii)
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 8
∴ ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)

Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYO and reflex ∠QYP.
Answer:
KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 9

Data: ∠XYZ = 64° and XY are produced up to P. ∠ZYP is bisected.
To Prove: ∠XYQ = ? Reflex ∠QYP = ?
Proof: YQ bisects ∠ZYP
∴ Let ∠PYQ = ∠ZYQ = x°.
∠XYZ + ∠ZYQ + ∠QYP = 180° (∵ ∠XYP is straight angle)
64 + x + x = 180
∴ 64 + 2x = 180
∴ 2x = 180 – 64
2x = 116
∴ \(x=\frac{116}{2}=58^{\circ}\)
∴ ∠PYQ = ∠ZYQ = 58°

(i) ∴ ∠XYQ = ∠XYZ + ∠ZYQ
= 64 + 58
∠XYQ = 122°

(ii) Reflex ∠QYP = ∠PYX + ∠XYZ + ∠ZYQ
= 180 ° + 64° + 58°
= 180 + 122
∴ Reflex ∠QYP = 302°.

We hope the KSEEB Solutions for Class 9 Maths Chapter 3 Lines and Angles Ex 3.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 3 Lines and Angles Exercise 3.1, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2

   

KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.2.

Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2

Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Answer:
Euclid’s fifth postulate states that “If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely meet on that side on which the sum of angles is less than two right angles.”
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 1
For example, the line PQ falls on lines AB and CD such that the sum of the interior angles 1 and 2 is less than 180° on the left side of PQ. Therefore, the lines AB and CD will eventually intersect on the left side of PQ.
In the above figure, ∠APQ + ∠PQC < 180° then AB and CD meet at the side of A and C.

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines ? Explain.
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 2
Answer:
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 3
‘n’ is a straight line which intersects l and m straight lines.
∠1 + ∠4 < 180° and ∠2 + ∠3 > 180°.
If l and m, produced by the side of ∠1 and ∠4 they meet at p.
If ∠1 + ∠4 < 180° and ∠2 + ∠3 > 180°,
then l and m straight lines do not meet even produced, which means they are parallel lines.

We hope the KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1

   

KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1.

Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1

Question 1.
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
Answer:
False. A number of lines can pass through a single point.

(ii) There is an infinite number of lines which pass through two distinct points.
Answer:
False. Because there is a unique line that passes through two points.

(iii) A terminated line can be produced indefinitely on both sides.
Answer:
True. A terminated line can be produced indefinitely on both sides.

(iv) If two circles’ are equal, then their radii are equal.
Answer:
True. Because the circumference of equal circles is equal their distance is the same from centre. Hence radii of two circles are equal.

(v) In Fig 2.9, if AB = PQ and PQ = XY, then AB = XY.
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 1
Answer:
True. Because E.g., If AB = 4 cm, PQ = 4 cm., then, AB = PQ = 4 cm.
PQ = XY = 4 cm.
∴ AB = PQ = XY
∴ AB = XY.

Question 2.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?
(i) parallel lines
(ii) perpendicular lines
(iii) line segment
(iv) the radius of a circle
(v) square.
Answer:
(i) Parallel Lines :
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 2
If endpoints of two straight lines do not meet even if they are produced on both sides, they are called parallel lines.
Before the definition concept of point, a straight line is necessary.
Point: A point is that which has no part.
Straight-line: A straight line is a line which lies evenly with the points on itself.

(ii) Perpendicular Lines:
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 3
If Angle between \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) is 90°, then only lines are perpendicular mutually.
Before the definition concept of Ray and Angle is necessary.
Ray: Joining one endpoint and the non-end point is called Ray.
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 4
Angle:
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 5
Ray \(\overrightarrow{\mathrm{OA}}\) and Ray \(\overrightarrow{\mathrm{OB}}\) revolve and forms an angle.

(iii) Segment:
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 6
The segment is formed when two endpoints are joined.
Here we must know the point and straight line.
Point: It has only dimension, no length, breadth, and thickness.
A – Starting point
B – Endpoint
AB – Segment
Straight-line: When two endpoints of a segment is produced on both sides, we get a straight line.

iv) The radius of a Circle :
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 7
A circle is a set of points that moves from a fixed point.
That equal distance is the radius of the circle.
‘O’ – Fixed point
OA – radius of the circle.
Necessary terms – Point, Segment.

v) Square :
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 8
A Square is a figure in which all sides and angles are equal.
AB = BC = CD = DA = 4 cms.
∠A = ∠B = ∠C = ∠D = 90°.
Necessary terms – Quadrilateral, angle.
Quadrilateral In a plane, if 3 points in 4 points are non-collinear and meet in different points, such a figure is called a quadrilateral.
Angle: ‘O’ is common point between \(\overrightarrow{\mathrm{OA}}\) and \(\overrightarrow{\mathrm{OB}}\) Such figure is called an angle.

Question 3.
Consider two ‘postulates’ given below :
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Answer:
There are many undefined terms for students.
They are consistent because they deal with two different situations.
(i) When two points A and B are given, C is in between two.
(ii) When two points A and B are given take point C which passes through A and B.
These postulates do not follow from Euclid’s postulates. But they follow Axiom 2.1

Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that \(A C=\frac{1}{2} A B\). Explain by drawing the figure.
Answer:
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 9
Data: C lies between A and B such that AC = BC.
To Prove: AC = \(\frac{1}{2}\)AB
Proof: In this figure AC = BC.
Adding AC on both sides,
AC + AC = BC + AC (equals are added to equals)
2AC = BC + AC
2AC = AB (∵ BC + AC = AB)
∴ AC = \(\frac{1}{2}\)AB.

Question 5.
In Question 4, point C is called a mid¬point of line segment AB. Prove that every line segment has one and only one mid-point.
Answer:
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 10
Let C and D are two points lies on AB.
C is the mid-point of AB.
AC = BC
Adding AC on both sides,
AC + AC = BC + AC
2AC = AB → (i)
D is the mid-point of AB,
AD = DB
Adding AD on both sides,
AD + AD = DB + AD
2AD = AB → (ii)
From equations (i) and (ii),
2AC = 2 AD.
∴ AC = AD.
Then C and D are not different, they are the same.
∴ “Every segment has one and only one mid-point.”

Question 6.
In Fig. 2.10, if AC = BD, then prove that AB = CD.
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 11
Answer:
Data: In this figure AC = BD.
To Prove : AB = CD
Proof: AC = BD (Data) → (i)
But, AC = AB + BC
and BD = BC + CD.
Substituting the value of (i)
AC = BD
KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 12
∴ AB = CD.

Question 7.
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’ ? (Note that the question is not about the fifth postulate).
Answer:
Euclid’s 5th Axiom of Euclid states that “The whole is greater than the part.”
This is a universal truth. Because it is not applicable to Mathematics only. It is useful for all.
E.g. 1: ‘a’ is whole, ’b’ and ‘c’ are its parts.
a = b + c.
Now, a > b and a > c.
It means a is greater than b.
a is greater than c.
E.g 2: If the human body is full, fingers are its parts.
∴ “Human body is greater than-his fingers.”

We hope the KSEEB Solutions for Class 9 Maths Chapter 2 Introduction to Euclid Geometry Ex 2.1 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 2 Introduction to Euclid Geometry Exercise 2.1, drop a comment below and we will get back to you at the earliest.

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