NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 13 Surface Areas and Volumes Class 10 NCERT Solutions Ex 13.1.

Students find NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes quite helpful for their CBSE Board Preparation. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam. All questions and answers from the NCERT Book of class 10 Math Chapter 13 are provided here for you for free.

• Surface Areas and Volumes Class 10 Ex 13.2
• Surface Areas and Volumes Class 10 Ex 13.3
• Surface Areas and Volumes Class 10 Ex 13.4
• Surface Areas and Volumes Class 10 Ex 13.5

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

NCERT Solutions for Class 10 Maths

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Question 6.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)

Solution:

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution:

Question 9.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1, drop a comment below and we will get back to you at the earliest.

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 11 Constructions Class 10 NCERT Solutions Ex 11.1.

In this NCERT For Class 10 Maths Chapter 11 Constructions students will learn the following topics given below. Here Students can find all NCERT Questions for class 10 Constructions, Additional Questions of Constructions and Solved Previous Year Questions. Our free NCERT Textbook Solutions for Chapter 11 – Constructions will strengthen your fundamentals in this chapter and can help you to score more marks in the examination. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Page No: 219

Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Solution:
The steps of construction are as follows:
1. Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with side AB.
2. Locate 13 (= 5 + 8) points A₁, A₂, A₃, A_4….. A₁₃ on AX such that AA₁ = A₁A₂ = A₂A₃ … = A₁₂A₁₃
3. Join BA₁₃.
4. Through the point A5 draw a line parallel to BA₁₃ (by making an angle equal to ∠AA₁₃B) at A₅ intersecting AB at point C.
Now C is the point dividing line segment AB of 7.6 cm in the required ratio of 5: 8.
We can measure the lengths of AC and CB. The length of AC and CB comes to 2.9 cm and 4.7 cm respectively.

Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Solution:
The steps of construction are as follows:
1. Draw a line segment AB = 4cm. Taking point A as centre draw an arc of 5 cm. radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other point C. Now AC = 5 cm. and BC = 6 cm and ∆ABC is the required triangle.
2. Draw a ray AX making acute angle with line AB on opposite side of vertex C.
3. Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁= A₁A₂ = A₂A₃.
4. Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B’.
5. Draw a line through B’ parallel to the line BC to intersect AC at C’. ∆AB’C’ is the required triangle.

• Constructions Class 10 Ex 11.2

Question 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.

Solution:

Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1.5 times the corresponding sides of the isosceles triangle.

Solution:
Let ∆ABC be an isosceles triangle having CA and CB of equal lengths, base AB is 8 cm and AD is the attitude of length 4 cm.
Now, the steps of construction are as follows:
1. Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of line segment while taking point A and B as its centre. Let these arcs intersect each other at 0 and O’. Join 00′. Let 00’ intersect AB at D.
2. Take D as centre and draw an arc of 4 cm radius which cuts the extended line segment 00’ at point C. Now an isosceles ∆ABC is formed, having CD (attitude) as 4 cm and AB (base) as 8 cm.
3. Draw a ray AX making an acute angle with line segment AB on opposite side of vertex C.
4. Locate 3 points (as 3 is greater between 3 and 2) on AX such that AA₁ = A₁A₂ = A₂A₃.
5. Join BA₂ and draw a line through A₃ parallel to BA₂ to intersect extended line segment AB at point B’.
6. Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’ ∆AB’C’ is the required triangle.

Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Solution:
The steps of construction are as follows:
1. Draw a line segment BC of length 6 cm. Draw an arc of any radius while taking B as centre. Let it intersect line BC at point O. Now taking O as centre draw another arc to cut the previous arc at point O’. Joint BO’ which is the ray making 60⁰ with line BC.
2. Now draw an arc of 5 cm. radius, while taking, B as centre, intersecting extended line segment BO’ at point A. Join AC. ∆ABC is having AB = 5 cm. BC = 6 cm and ∠ABC = 60⁰.
3. Draw a ray BX making an acute angle with BC on opposite side of vertex A.
4. Locate 4 points (as 4 is greater in 3 and 4). B₁, B₂, B₃, B₄ on line segment BX.
5. Join B₄C and draw a line through B₃, parallel to B₄C intersecting BC at C’.
6. Draw a line through C’ parallel to AC intersecting AB at A’. ∆A’BC’ is the required triangle.

Question 6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.
Solution:
∠B = 45°, ∠A = 105°
It is known that the sum of all interior angles in a triangle is 180°.
∠B + ∠C = 180°.
105° + 45° + ∠C = 180°
∠C = 180° — 150° = 30°
Now, the steps of construction are as follows:
1. Draw a line segment BC = 7 cm. Draw an arc of any radius while taking B as centre. Let it intersects BC at P. Draw an arc from P, of same radius as before, to intersect this arc at Q. From Q, again draw an arc, of same radius as before, to cut the arc at R. Now from points Q and R draw arcs of same radius as before, to intersect each other at S. Join BS.
Let BS intersect the arc at T. from T and P draw arcs of same radius as before to intersect each other at U. Join BU which is making 45° with BC.
2. Draw an arc of any radius taking C as its centre. Let ¡t intersects BC at O. Taking O as centre, draw an arc of same radius intersecting the previous arc at O’. Now taking O and O’ as centre, draw arcs of same radius as before, to intersect each other at Y. Join CY which is making 30° to BC.
3. Extend line segment CY and BU. Let they intersect each other at A.
∆ABC is the triangle having ∠A = 105°, ∠B = 45° and BC = 7 cm.
4. Draw a ray BX making an acute angle with BC on opposite side of vertex A.
5. Locate 4 points (as 4 is greater in 4 and 3) B₁, B₂, B₃, B₄ on BX.
6. Join B₃C. Draw a line through B₄ parallel to B₃C intersecting extended BC at C’.
7. Through C’ draw a line parallel to AC intersecting extended line segment at C’. ∆’BC’ is required triangle.

Question 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Solution:
The steps of construction are as follows:
1. Draw a line segment AB = 4 cm draw a ray SA making 90° with it.
2. Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ∆ABC is required triangle.
3. Draw a ray AX making an acute angle with AB, opposite to vertex C.
4. Locate 5 points (as 5 is greater in 5 and 3) A₁, A₂, A₃, A₄, A₅ on line segment AX.
5. Join A₃B, Draw a line through A₅ parallel to A₃B intersecting extended line segment AB at B’.
6. Through B’ draw a line parallel to BC intersecting extended line segment AC at C’. ∆ABC’ is required triangle.

We hope the NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1, drop a comment below and we will get back to you at the earliest.

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— ObulReddy cbse (@ObulReddyCBSE) August 3, 2018

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 14 Statistics Class 10 NCERT Solutions Ex 14.1.

Students find NCERT Solutions for Class 10 Maths Chapter 14 Statistics quite helpful for their CBSE Board Preparation. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam. All questions and answers from the NCERT Book of class 10 Math Chapter 14 are provided here for you for free.

• Statistics Class 10 Ex 14.2
• Statistics Class 10 Ex 14.3
• Statistics Class 10 Ex 14.4

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

NCERT Solutions for Class 10 Maths

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.

Which method did you use for finding the mean, and why?
Solution:
Let us find class marks (x_i) for each interval by using the relation.

Now we may compute x_i and f_ix_i as following

Number of plants	Number of houses (fi)	xi	fixi
0 – 2	1	1	1— 1 = 1
2 – 4	2	3	2 — 3 = 6
4 – 6	1	5	1 — 5 = 5
6 – 8	5	7	5 — 7 = 35
8 – 10	6	9	6 — 9 = 54
10 – 12	2	11	2 —11 = 22
12 – 14	3	13	3 — 13 = 39
Total	20		162

From the table we may observe that

So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (x_i) and f_i are small.

Question 2.
Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages (in Rs)	100 – 120	120 – 140	140 -160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

Solution:
Let us find class mark for each interval by using the relation.

Class size (h) of this data = 20
Now taking 150 as assured mean (a) we may calculate di, u_i and f_iu_i as following.

Daily wages (in Rs)	Number of workers (fi)	xi	di = xi – 150		fiui
100 -120	12	110	– 40	-2	– 24
120 – 140	14	130	– 20	-1	– 14
140 – 160	8	150	0	0	0
160 -180	6	170	20	1	6
180 – 200	10	190	40	2	20
Total	50				-12

From the table we may observe that

So mean daily wages of the workers of the factory is Rs.145.20

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.

Daily pocket allowance (in Rs)	11 – 13	13 – 15	15 -17	17 – 19	19 – 21	21 – 23	23 – 25
Number of workers	7	6	9	13	f	5	4

Solution:
We may find class mark (x_i) for each interval by using the relation.

Given that mean pocket allowance \(\overline { x } \) = Rs.18
Now taking 18 as assured mean (a) we may calculate d_i and f_id_i as following.

Daily pocket allowance (in Rs.)	Number of children fi	Class mark xi	di = xi – 18	fidi
11 – 13	7	12	– 6	– 42
13 – 15	6	14	– 4	– 24
15 – 17	9	16	– 2	– 18
17 – 19	13	18	0	0
19 – 21	f	20	2	2 f
21 – 23	5	22	4	20
23 – 25	4	24	6	24
Total				2f – 40

From the table we may obtain

Hence the missing frequency f is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	65 – 68	68 – 71	71-74	74 – 77	77 – 80	80 – 83	83 – 86
Number of women	2	4	3	8	7	4	2

Solution:
We may find class mark of each interval (x_i) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following.

Number of heart beats per minute	Number of women fi	xi	di = xi -75.5		fiui
65 – 68	2	66.5	– 9	– 3	– 6
68 – 71	4	69.5	– 6	– 2	– 8
71 – 74	3	72.5	– 3	– 1	– 3
74 – 77	8	75.5	0	0	0
77 – 80	7	78.5	3	1	7
80 – 83	4	81.5	6	2	8
83 – 86	2	84.5	9	3	6
Total	30				4

Now we may observe from table that

So mean heart beats per minute for these women are 75.9 beats per minute.

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.

Number of mangoes	50 – 52	53 – 55	56 – 58	59 – 61	62 – 64
Number of boxes	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

Number of mangoes	Number of boxes fi
50 – 52	15
53 – 55	110
56 – 58	135
59 – 61	115
62 – 64	25

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add 1/2 to upper class limit and subtract 1/2 from lower class limit of each interval.
And class mark (x_i) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following –

Class interval	fi	xi	di = xi – 57		fiui
49.5 – 52.5	15	51	-6	-2	-30
52.5 – 55.5	110	54	-3	-1	-110
55.5 – 58.5	135	57	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	400				25

Now we may observe that

Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of f_i, d_i are big and also there is a common multiple between all d_i.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)	100 – 150	150 – 200	200 – 250	250 – 300	300 – 350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate cla

mark (x_i) for each interval by using the relation

Class size = 50

Now taking 225 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following

Daily expenditure (in Rs)	fi	xi	di = xi – 225		fiui
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
Total	25				-7

Now we may observe that –

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:

concentration of SO2 (in pmm)	Frequency
0.00 – 0.04	4
0.04 – 0.08	9
0.08 – 0.12	9
0.12 – 0.16	2
0.16 – 0.20	4
0.20 – 0.24	2

Find the mean concentration of SO₂ in the air.

Solution:

Concentration of SO2 (in ppm)	Frequency	Class mark xi	di = xi – 0.14		fiui
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	30				-31

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.

Number of days	0 – 6	6 – 10	10 – 14	14 – 20	20 – 28	28 – 38	38 –  40
Number of students	11	10	7	4	4	3	1

Solution:
We may find class mark of each interval by using the relation

Now taking 16 as assumed mean (a) we may calculate d_i and f_id_i as following

Number of days	Number of students fi	xi	di= xi – 16	fidi
0 – 6	11	3	-13	-143
6 -10	10	8	-8	-80
10 – 14	7	12	-4	-28
14 – 20	4	16	0	0
20 – 28	4	24	8	32
28 – 38	3	33	17	51
38 – 40	1	39	23	23
Total	40			-145

Now we may observe that

So, mean number of days is 12.38 days, for which a student was absent.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45 – 55	55 – 65	65 – 75	75 – 85	85 – 95
Number of cities	3	10	11	8	3

Solution:
We may find class marks by using the relation

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate d_i, u_i, and f_iu_i as following

Literacy rate  (in %)	Number of cities  fi	xi	di= xi – 70	ui	fiui
45 – 55	3	50	-20	-2	-6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	35				-2

Now we may observe that

So, mean literacy rate is 69.43%.

 

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions For Class 10 Maths – AplusTopperhttps://t.co/q7hSKh3xmf#NCERTSolutionsForClass10Maths #NCERTSolutions #AplusTopper

— ObulReddy cbse (@ObulReddyCBSE) August 3, 2018

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 9 Some Applications of Trigonometry Class 10 NCERT Solutions Ex 9.1.

Trigonometry can be used in many ways in the things around us like we can use it for calculating the height and distance of some objects without calculating them actually. We are provides free comprehensive chapter wise class 10 Mathematics notes with proper images & diagram. Here you will find all the answers to the NCERT textbook questions of Chapter 9 Some Applications of Trigonometry.

NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1

NCERT Solutions for Class 10 Maths

Page No: 203

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°.

Solution:
In the Figure, AB is the pole.

Question 2. A tree breaks due to storm and the broken part bends so that the top 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.

Solution:

Let AC be the original tree and be the broken part which makes an angle of 30^o with the ground.

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?

Solution:
In the two figures, AC and PR are the slides for younger and elder children respectively

Page No: 204

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.

Solution:

Question 5.
A kite is flying at a height of 60 m 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.

Solution:

Let A be the position of the kite and the string is tied to point C on ground.

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.

Solution:

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.

Solution:

Let BC be the building, AB be the transmission tower, and D be the point on ground from where elevation angles are to be measured.

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.

Solution:

Let AB be the statue, BC be the pedestal and D be the point on ground from where elevation angles are to be measured.

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.

Solution:

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.

Solution:

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 m 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 Fig. 9.12). Find the height of the tower and the width of the canal.

Solution:

Question 12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution:

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.

Solution:

Let AB be the lighthouse and the two ships be at point C and D respectively.

Page No: 205

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 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°. Find the distance travelled by the balloon during the interval.

Solution:

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.

Solution:

Let AB be the tower. C is the original position of the car which changes to D after six seconds.

Question 16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution:

Let AQ be the tower and R, S respectively be the points which are 4m, 9m away from base of tower.

As the height can not be negative, the height of the tower is 6 m.

NCERT Solutions For Class 10 Maths – AplusTopperhttps://t.co/q7hSKh3xmf#NCERTSolutionsForClass10Maths #NCERTSolutions #AplusTopper

— ObulReddy cbse (@ObulReddyCBSE) August 3, 2018

We hope the NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 7 Coordinate Geometry Class 10 NCERT Solutions Ex 7.1.

Here you get the CBSE Class 10 Mathematics Chapter 7 Coordinate Geometry Ex 7.1: We are provides free comprehensive chapter wise class 10 Mathematics notes with proper images & diagram. Here you will find all the answers to the NCERT textbook questions of Chapter 7 Coordinate Geometry Ex 7.1

• Coordinate Geometry Class 10 Ex 7.2
• Coordinate Geometry Class 10 Ex 7.3
• Coordinate Geometry Class 10 Ex 7.4

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

NCERT Solutions for Class 10 Maths

Chapter 7 Coordinate Geometry

Page No: 161

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a, b), (− a, − b)

Solution:

Concept Insights: In the ordered pair (a, b) order is important coordinate a represent x coordinate and b represent y coordinate

Question 2.
Find the distance between the points (0, 0) and (36, 15).

Solution:
Distance between points (0,0) and (36,15)
Now Distance between two points (x₁,y₁) and (x₂,y₂) is given by

Question 3.
Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

Solution:
Three points are collinear if they lie on a line i.e one point lies in between any other two points.

Here sum of the distances of any two points is not equal to third point
Therefore points (1, 5), (2, 3) and (-2, -11) are not collinear.

Question 4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Solution:
Three non collinear points will represent the vertices of an isosceles triangle
if its two sides are of equal length.

Here AB = BC
As two sides are equal in length therefore ABC is an isosceles triangle.

Question 5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees.

Solution:
From the figure coordinates of points A,B,C and D are
A = (3,4), B = (6,7), C = (9,4), D = (6,1)

Here, all sides of this square are of same length and also diagonals are of same length. So, ABCD is a square and hence Champa was correct.

Concept Insight: For the Vertices of square all sides & both the diagonals are equal.

Question 6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution:
(i). Let A= (—1,—2),B= (1,O),C= (—1,2),D(—3,O)

(ii)Let A=(—3,5),B=(3,1),C=(O,3),D=(—1,—4)

Here, all sides of this quadrilateral are of same length and also diagonals are of same length. So, given points are vertices of a square.

(iii)    Let A = (4, 5), B = (7, 6), C = (4, 3), D = (1, 2)

Here opposite sides of this quadrilateral are of same length but diagonals are of different length. So, given points are vertices of a parallelogram.

Concept Insight: Recall the properties of various quadrilaterals.

Question 7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).

Solution:
We have to find point on x axis. So, its y coordinate will be O.
Let point on x-axis be (x, O)

Question 8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.

Solution:
Given that distance between (2,-3) and (10,y) is 10

Concept Insight: Any point on y axis will have x coordinate.

Question 9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Solution:

Question 10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Solution:

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