Class 10

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Ex 9.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Exercise 9.1

Question 1.
Fill in the blanks :
(i) A tangent to a circle intersects it in __________ points
Solution:
one

(ii) A line intersecting a circle in two points is called a _________
Solution:
secant

(iii) A circle can have _________ parallel tangents at the most,
Solution:
two

(iv) The common point of a tangent to a circle and the circle is called _______
Solution:
point of contact

(v) We can draw __________ tangents to a given circle
Solution:
infinite

Question 2.
A tangent PQ of a point P’ of a circle of radius 5 cm meets a line through the centre ‘O’ at a point Q so that OQ = 12 cm. Find the length of PQ.
Solution:
PQ is a tangent to a circle whose centre is ‘O’.
P is the point of contact. PO is the radius of the circle.
Hence ∠OPQ = 90°
Given that OP = 5 cm and OQ = 12 cm

By pythagorus theorem.
(OP)² + (PQ)² = (OQ)²
(PQ)² = (OQ)² – (OP)²
(PQ)² = 12² – 5² = 119
PQ = \(\sqrt{119}\)
Hence, length of PQ = \(\sqrt{119}\) cm

Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Solution:
The adjacent figure is a circle with centre ‘O’
AB is the given line CD is a tangent to the circle of P parallel to AB.

EF is a secant of the circle, drawn parallel to AB

Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm. (A.P. Mar. ’16, 15)
Solution:

In ∆OTR ∠OTP = 90°
OT = 9 cm
OP = 15 cm
By pythagorus theorem
(OP)² = (OT)² + (PT)²
⇒ 15² = 9² + (PT)²
⇒ PT² = 15² – 92 = 225 – 81
(PT)² = 144 ⇒ PT = \(\sqrt{144}\) = 12
Length of the required tangent = PT = 12 cm

Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel. (A.P. June ’15)
Solution:
PQ is the diameter of a circle with centre ‘O’. APB and CQD are tangents to the circle at P and Q respectively.
We have to prove that AB || CD
AB is a tangent to the circle ‘P’ is the point of contact and
PO is the radius drawn through P
∴ ∠APO = 90° ……………… (1)
CD is a tangent to the circle. Q is the point of contact and QO is the radius draw through Q.
∴ ∠DQO = 90°
From (1) and (2), we have
∠APO = ∠DQO
AB, CD are two lines. PQ is a transversal ∠APO and ∠DQO are a pair alternate angles. Since the alternate angles are equal AB || CD

Telangana SCERT 10th Class Physics Study Material Telangana 1st Lesson Reflection of Light at Curved Surfaces Textbook Questions and Answers.

TS 10th Class Physical Science 1st Lesson Questions and Answers Reflection of Light at Curved Surfaces

Improve Your Learning
I. Reflections on concepts

Question 1.
Where will the image be formed when we place an object, on the principal axis of a concave mirror at a point between focus and centre of curvature?
Answer:
1. When we place an object on the principal axis of concave mirror at a point between focus and centre of curvature, the image is formed beyond centre of curvature.

2. The image so formed is real, inverted and magnified.

Question 2.
State the differences between convex and concave mirrors.
Answer:

Question 3.
Distinguish between real and virtual images.
Answer:

Question 4.
How do you get a virtual image using a concave mirror?
Answer:
A. When an object is kept between pole and focus of a concave mirror virtual image is formed behind the mirror.

Question 5.
What do you know about the terms given below related to spherical mirrors?
(a) pole
(b) centre of curvature
(c) focus
(d) Radius of curvature
(e) Focal length
(f) principal axis
(g) object distance
(h) image distance
(i) Magnification.
Answer:
(a) Pole: The point on the principal axis of spherical mirror with respect to which all the measurements are made. Usually, it is the mid point of the curvature of mirror.

(b) Centre of curvature: The centre of the sphere of which the curved surface of the mirror is a part.

(c) Focus: The light rays coming from a source parallel to the principal axis converge at a point after reflection. This point is called focus or focal point.

(d) Radius of curvature: The radius of the sphere of which the curved surface is a part is called radius of curvature.

(e) Focal length: The distance between the pole of the mirror and focus is called focal length of mirror.

(f) Principal axis: The straight line passing through the centre of curvature and pole of curved mirror is called principal axis.

(g) Object distance: The distance between the pole of the mirror and object position of object is known as object distance.

(h) Image distance: The distance between the pole of the mirror and position of image is called image distance.

(i) Magnification: Magnification of a spherical mirror is the ratio between size (height) of image to the size (height) of object. Also, m= v/u.

Question 6.
What do you infer from the experiment which you did to measure the object distance and image distance?
Answer:
Inference from the experiment which I did with concave mirror :

Inference:

1. As the object moves away from the mirror the image approaches the mirror.
2. As the object moves away from mirror the size of the image becomes smaller.

Question 7.
Write the rules for sign convention.
Answer:
Rules for sign convension:

1. AN distances should be measured from the pole.
2. The distances measured in the direction of incident light, are taken positive and the opposite direction of incident light are taken negative.
3. Height of object (H₀) and height of image (H₁) are positive if measured upward
   from the axis and negative if measured downward.
4. For a concave mirror ‘f’ and ‘R’ are negative and for a convex mirror these are positive.

II. Application of concepts

Question 1.
Find the distance of the image, when an object is placed on the principal axis, at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.
Answer:
Object distance u = -10 cm
Radius of curvature (r) = -8cm
Image distance v=?

The image distance (v) = 6.7 cm.
i.e., Real image is formed at same side of the mirror.

Question 2.
The magnification product by a plane mirror is +1. What does it mean?
Answer:

1. Magnification = \(\frac{\text { height of the image }}{\text { height of the object }}=\frac{\text { distance of the image }}{\text { distance of the object }}\)
2. The magnification produced by a plane mirror is +1 means then the size of the image is equal to the size of the object.
3. + sign indicates that the image is erect. Magnification ‘+1’ indicates the image is erect and size of the image is equal to size of the object.

Question 3.
If the spherical mirrors were not known to human beings, guess the consequences.
Answer:
If spherical mirrors are not known to human beings

1. Many optical instruments would not have been invented.
2. We cannot increase the size of images of the objects.
3. The problem of lateral inversion of images will not be solved.
4. Now a days spherical mirrors are used as shaving mirrors, head mirrors for ENT specialists, in headlights of motor vehicles, in solar furnaces and as rearview mirror. If spherical mirrors are not known all these are not possible.

Question 4.
Draw suitable rays by which we can guess the position of the imge formed by a concave mirror.
Answer:
The following rays are used to guess the position of the image formed by a concave mirror.
(i) A ray parallel to the principal axis passes through principal focus (F) after reflection from a concave mirror.

(ii) A ray passing through ‘F’ becomes parallel to principal axis after reflection from a concave mirror.

(iii) A ray passing through ‘C’ is reflected back along the same path after reflection from a concave mirror.

(iv) A ray incident obliquely to the principal axis towards the pole P, on the concave mirror is reflected obliquely, following the laws of reflection.

Question 5.
Show the formation of image with a ray diagram, when an object is placed on the principal axis of a concave mirror away from the centre of curvature.
Answer:
When an object is placed on the principal axis of a concave mirror and beyond (away from) its centre of curvature ‘C’, the image is formed between the focus (F) and the centre of curvature (C). The ray which is parallel to principle axis will pass through focus after reflection and the ray which passes through focus will travels parallel to principal axis after reflection. These two rays will converge between F and C of the mirror.

The image is real, inverted and diminished in size.

Question 6.
Why do we prefer a convex mirror as a rear-view mirror in the vehicles?
Answer:
We use convex mirror as a rear-view mirror in the vehicles because

1. Convex mirror always forms virtual, erect, and diminished images irrespective of distance of the object.
2. A convex mirror enables a driver to view large area of the traffic behind him.
3. Convex mirror forms very small image than the object. Due to this reason convex mirrors are used as rear-view mirrors in vehicles.

III. Higher Order Thinking Questions

Question 1.
A convex mirror with a radius of curvature of 3 m is used as a rearview mirror for a vehicle. If a bus is located at 5m from this mirror, find the position, nature, and size of the image.
Answer:
According to the sign convention :
Radius of curvature = R = + 3m
Object distance = u = -5 m (negative sign)
Image distance = v =?
Focal length, f = \(\frac{R}{2}=\frac{3}{2} m\) = 1.5 m
Formula : \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

The image is formed at a distance of 1.15 m at the back of the mirror.
Magnification, m = \(\frac{h_i}{h_o}=-\frac{v}{u}=\frac{-1.15 m}{-5 m}=\frac{1.15}{5}\) = 0.23
The image is virtual, erect, and diminished to 0.23 times of the size of the object.

Question 2.
To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.
Answer:
To form the image on the object itself, the object should be kept at center of curvature of a concave mirror.

1. An object AB has been placed at the centre of curvature ‘C’ on the concave mirror.
2. A ray of light AD which is parallel to principle axis passes through the focus ‘F after reflection as DA’.
   
3. A ray of light passing through the focus of the concave mirror becomes parallel to the principle axis after reflection.
4. Here AE ray passing through focus and reflected as EA’.
5. The reflected rays DA’ and EA’ meet at A’ point. So the real image formed at point A’ of the object.
   We get couple image AB perpendicular to the principal axis.
6. Thus A’B’ is the real inverted image of the object AB.

IV. Multiple choices questions

Question 1.
If an object is placed at ‘C’ on the principal axis in front of a concave mirror, the position of the image is ………………. . [ ]
(a) at infinity
(b) between F and C
(c) at C
(d) beyond C
Answer:
(c) at C

Question 2.
We get a diminished image with a concave mirror when the object is placed ………………………. . [ ]
(a) at F
(b) between the pole and F
(c) at C
(d) beyond C
Answer:
(d) beyond C

Question 3.
We get a virtual image in a concave mirror when the object is placed …………………….. . [ ]
(a) at F
(b) between the pole and F
(c) at C
(d) beyond C
Answer:
(b) between the pole and F

Question 4.
Which of the following represents Magnification? [ ]
(i) \(\frac{v}{u}\)
(ii) \(\frac{-v}{u}\)
(iii) \(\frac{h_i}{h_0}\)
(iv) \(\frac{h_0}{h_i}\)
(a) i, ii
(b) ii, iii
(c) iii, iv
(d) iv, i
Answer:
(d) iv, i

Question 5.
ray which seems to be traveling through the focus of a convex mirror, path of the reflected ray of an incident ……………… . [ ]
(a) parallel to the axis
(b) along the same path in opposite direction
(c) through F
(d) through C
Answer:
(a) parallel to the axis

Question 6.
Size of image formed by a convex mirror is always …………………… .[ ]
(a) enlarged
(b) diminished
(c) equal to the size of object
(d) depends on position of object
Answer:
(b) diminished

Question 7.
An object is placed at a certain distance on the principal axis of a concave mirror. The image is formed at a distance of 30 cm from the mirror. Find the object distance if radius of curvature R = 15cm. [ ]
(a) 15 cm
(b) 10 cm
(c) 30 cm
(d) 7.5 cm
Answer:
(c) 30 cm

Question 8.
All the distances related to spherical mirror will be measured from …………………… .[ ]
(a) object to image
(b) focus of the mirror
(c) pole of the mirror
(d) image to object
Answer:
(c) pole of the mirror

Question 9.
The minimum distance from real object to a real image in a concave mirror is ………………….. . [ ]
(a) 2f
(b) f
(c) 0
(d) f/2
Answer:
(c) 0

Suggested Experiments

Question 1.
Conduct an experiment to find the focal length of concave mirror.
(or)
How can you find out the focal length of concave mirror experimentally when there is no sunlight?
Answer:
Aim: To find the focal length of a concave mirror,
Materials required : (i) A concave mirror (ii) V-shape stand (iii) A candle (iv) A meter scale.

Procedure :

1. Place the concave mirror on the V-shape stand.
2. Keep a burning candle in front of the concave mirror.
3. Place a thick white paper behind the candle. This acts as a screen.
4. Adjust distances between candle and mirror, screen and mirror by moving them either forward or backward till a clear well-defined image appears on the screen.
5. Measure the distance between the mirror and candle (object distance u) and the distance between mirror and screen (image distance v).
6. Using the mirror formula, \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v} \) or f = \(\frac{u v}{u+v} \)
   This gives the focal length of the concave mirror.

Question 2.
Find the nature and position of images when an object is placed at different places on the principal axis of a concave mirror.
Answer:
Aim: Observing the types of images and measuring the object distance and image distance from the concave mirror.
Material required: A candle, paper, concave mirror (known focal length), V- stand, measuring tape or meter scale.

Procedure:

1. Place the concave mirror on V-stand, a candle and meter scale as shown in figure.
   
2. Keep the candle at different distances from the mirror (10 cm to 80 cm) along the axis and by moving the paper screen find the position where you get the sharp image on paper.
3. Note down your observations in the following table.
4. Since we know the focal point and centre of curvature, we can classify our above observations as shown in the following table.

Suggested Project Works

Question 1.
Collect information about the history of spherical mirrors in human civilization, write a report on it.
Answer:

1. The first mirrors used by people were most likely pools of water or still water. The earliest manufactured mirrors were pieces of polished stones.
2. Parabolic mirrors were described and studied in classical antiquity by the mathematician Archimedes in his work on burning mirrors.
3. Ptolemy conducted a number of experiments with curved polished iron mirrors. He also discussed plane, convex, concave, and spherical mirrors in his optics.
4. In China, people began making mirrors with the use of silver mercury amalgams as early as 500 AD.
5. In 16th century, Venice, a big city popular for its glass-making expertise, became a centre of mirror production using this new technique.
6. The invention of the silvered-glass mirror is credited to German Chemist Justus Von Liebig in 1835.

Question 2.
Think about the objects which act as concave or convex mirrors in your surroundings. Make a table of these objects and display in your classroom.
Answer:

Question 3.
Collect photographs from your daily life where you use convex and concave mirrors and display in your classroom.
Answer:

TS 10th Class Physical Science Reflection of Light at Curved Surfaces Intext Questions

Page 1

Question 1.
Is the image formed by a bulged surface same as the image formed by a plane mirror?
Answer:
No, the image formed by a bulged surface is virtual, created, and diminished image.

Question 2.
Is the mirror used in automobiles a plane mirror? Why it is showing small images?
Answer:
No, the mirror used in automobile is convex mirror. At it bulging outwards it forms small images.

Question 3.
Why does our image appear thin or bulged out in some mirrors?
Answer:
The image in a mirror appears thin or bulged out because the thickness of the mirror may vary or the reflecting surface may not be flat.

Question 4.
Can be see inverted image in any mirror?
Answer:
Can we see inverted image in any mirror? Yes, we can see inverted image in concave mirror for a distant objects.

Question 5.
Can we focus the sunlight at a point using a mirror instead of a magnifying glass?
Answer:
Yes. By using a black paper with a tiny hole at its centre.

Question 6.
Are the angle of reflection and angle of incidence also equal for reflection by curved surfaces?
Answer:
No.

Page 4

Question 7.
Does this help you to verify the conclusions you arrived at with your drawing?
Answer:
Yes.

Question 8.
What happens if you hold the paper at a distance shorter than the focal length from the mirror and move it away?
Answer:
We find there is no point at which the reflected rays converge at a point. But as we move the paper away from the focal point, we find images formed at different distances from the mirror.

Question 9.
Does the image of the sun become smaller or bigger?
Answer:
We notice that the image of the sun keeps on becoming smaller. Beyond the focal point, it will become bigger.

Page 5

Question 10.
Do we get an image with a concave mirror at the focus every time?
Answer:
We get the images not only at the focus every time, we get different ¡mages by keeping the object at different points depending on focal length of mirror.

Page 6

Question 11.
It is inverted or erect, enlarged or diminished?
Answer:

Question 12.
What do you infer from the table?
Answer:
From the table 2. I infer that images can be formed at other positions different from focal point.

Page 7

Question 13.
Why only at point A?
Answer:
If we hold the screen at any point before or beyond point A (for example at point B), we see that the rays will meet the screen at different points due to these rays. If we draw more rays emanating from the same tip we will see that
at point A they will meet but at point B they do not. So the image of the tip of the flame will be sharp at point A.

Page 9

Question 14.
Where is the base of the candle expected to be in the image when the candle is placed on the axis of the mirror?
Answer:
The base of the candle is going to be on the principle axis ¡n the image when the object is placed on the axis of the mirror.

Question 15.
During the experiment, did you get any positions where you could not get an image on the screen?
Answer:
Yes. When the object is placed at a distance less than the focal length of the mirror we do not get an ¡mage on the screen.

Page 11

Question 16.
Have you observed the rearview mirrors of a car?
Answer:
Yes. I have observed rear view mirror of a car.

Question 17.
What type of surface do they have?
Answer:
A concave mirror will be like the rubber sole bent inwards and the reflecting surface will be curved inwards.

Question 18.
Can we draw ray diagrams for convex surface?
Answer:
We can draw ray diagrams with convex surface by making use of “easy” rays that we have identified, with small modifications.

Think and discuss 

Question 1.
See the figure – 5 In text. A set of parallel rays are falling on a convex mirror. What conclusions can you draw from this?
Answer:
On seeing figure 5, the following conclusions can be drawn;

1. The parallel beam of rays meet at infinity.
2. So the Image Is not visible. :
3. This parallel beam of rays forms a virtual image which is smaller than the size of object.

Question 2.
Will you get a point Image If you place a paper at the focal point?
Answer:
No. These parallel beam of rays do not meet at a visible point and we do not get a point image.

Question 3.
Do you get an image when object is placed at F? Draw a ray diagram. Do the experiment.
Answer:
We did not get the image when the object is placed at Focus ‘F’ of concave mirror.
Experiment:
Aim: Observing the Image formed by the object which is placed at ‘F’ of concave mirror.
Materal required: A candle, paper, a concave mirror (known focal length), V-stand, measuring
tape or meter scale.

Procedure:
1. Place the concave mirror on V-stand, a candle and meter scale as shown in figure.

2. Keep the candle at a distance equal to the focal length of the mirror. (which Is known)
3. Now move the paper screen away from the mirror along the axis to observe the image.
4. You will notice that the image cannot be seen because it Is formed at infinity.

TS 10th Class Physical Science Reflection of Light at Curved Surfaces Activities

Activity 1

Question 1.
Explain an activity to find the normal to a curved surface.
Answer:

1. Take a small piece of thin foam or rubber.
2. Put some pins in a straight line on the foam.
3. All these pins are perpendicular to the foam.
4. If the foam is considered as a mirror, each pin would represent the normal at that point.
5. Any ray incident at the point where the pin makes contact with the surface will reflect at the same angle the incident ray made with the pin-normal.
6. Now bend the foam piece inwards.
7. The pins still represent the normal at various poInts.
8. You will observe that all the pins tend to coverage at a point.
9. This will be appear like a concave mirror.
10. Now bend the foam piece outwards.
11. The pins seem to move away from each other that means they diverge.
12. The pins still represent the normal at various points.
13. This will be appear like a convex mirror.
    

Activity 2

Question 2.
How do you Identify the focal point and foal length of a concave mirror?
Answer:
Hold a concave mirror perpendicular to me direction of sunlight

• Take a small paper and slowly move it in front of the mirror.
• Find the point where you get smallest and brigh – test spot, which is the image of the sun.
• The rays coming from the sun parallel to the concave mirror are converging at a point. This point is called focus or focal point (F) of the concave mirror.
• Measure the distance of this spot from the pole (P) of the mirror.
• ThIs distance Is the focal length (f) of the mirror

Lab Activity

Question 1.
Describe an experiment to observe types of mages formed by e concave mirror and measure the object end image distances.
(OR)
Write the experimental method in measuring the distances of object and image using concave mirror. And write the table For observations.
Answer:
Aim: Observing the types of images and reasoning the object distance and image distance from the concave mirror.
Material required: A candle, paper, concave minor (known focal length), V- stand, moesunng tape or meter scale.
Procedure
1. Place the concave mirror on V-stand, a candle and meter scale as shown in figure.

2. Keep the candle at different distances from the mIrror (10 cm to 80 cm) along the axis and by moving the paper screen fd the position where you get the sharp mage on paper.
3. Note down your observations in the following table.

4. Since we know the focal point and centre of curvature, we can re-classify our above observations as shown in the following table.

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?

i) The taxi fare after each km when the fare is ₹ 20 for the first km and rises by ₹ 8 for each additional km.
Solution:
Fare for the first km = ₹ 20 = a
Fare for each km after the first = ₹ 8 = d
∴ The fares would be 20, 28, 36, 44, …….
The above list forms an A.P.
Since each term in the list, starting from the second can be obtained by adding ‘8’ to its preceding term.

ii) The amount of air present in a cylinder when a vacuum pump removes 1/4^th of the air remaining in the cylinder at a time.
Solution:
Let the amount of air initially present in the cylinder be 1024 lit.
First it removes \(\frac{1}{4}\)^th of the volume
i.e., \(\frac{1}{4}\) × 1024 = 256
Remaining air present in the cylinder = 768
At second time it removes \(\frac{1}{4}\)^th of 768
i.e., \(\frac{1}{4}\)^th × 768 = 192
Remaining air in the cylinder = 768 – 192 = 576
Again at third time it removes \(\frac{1}{4}\)^th of 576
i.e., \(\frac{1}{4}\) × 576 = 144
Remaining air in the cylinder
= 576 – 144 = 432
i.e., the volume of the air present in the cylinder after 1^st, 2^nd, 3^rd …. times is 1024, 768, 576, 432,…….
Here, a₂ – a₁ = 768 – 1024 = -256
a₃ – a₂ = 576 – 768 = – 192
a₄ – a₃ = 432 – 576 = – 144
Thus the difference between any two successive terms is not equal to a fixed number.
∴ The given situation doesn’t show an A.R

iii) The cost of digging a well, after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
Solution:
Cost for digging the first metre = ₹ 150
Cost for digging subsequent metres = ₹ 50 each i.e.,

The list is 150, 200, 250, 300, 350, …..
Here d = a₂ – a₁ = a₃ – a₂
= a₄ – a₃ = ……… = 50
∴ The given situation represents an A.P

iv) The amount of money in the account every year, when ₹ 10,000 is deposited at compound interest at 8% per annum.
Solution:
Amount deposited initially = P = ₹ 10,000
Rate of interest = R = 8% p.a [at C.I.]
∴ A = P(1 + \(\frac{\mathrm{R}}{100}\))ⁿ

The terms 10800, 11664, 12597.12, ….
a₂ – a₂ = 800
a₃ – a₂ = 864

a₄ – a₃ = 933.12
…………………….

Here a = 10,000
But a₂ – a₁ ≠ a₃
– a₂ ≠ a₄ – a₃

∴ The given situation doesn’t represent an A.P.

Question 2.
Write first four of the AP, when first term a and the common difference d are given as follows :
i) a = 10, d = 10
ii) a = -2, d = 0
iii) a = 4, d = -3
iv) a = -1, d = 1/2
v) a = -1.25, d = -0.25
Solution:

Question 3.
For the following A.Ps, write the first term and the common difference :

i) 3, 1, -1, – 3, …..
ii) -5,-1, 3, 7, …….
iii) \(\frac{1}{3}\), \(\frac{5}{3}\), \(\frac{9}{3}\), \(\frac{13}{3}\),………
iv) 0.6, 1.7, 2.8, 3.9,…….. (A.P. Mar ’15)
Solution:

Question 4.
Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.

i) 2, 4, 8, 16, ……..
ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\),………..
iii) – 1.2, -3.2, -5.2, -7.2, …….
iv) -10, -6, -2, 2,……….
v) 3, 3 + \(\sqrt{2}\), 3 + 2\(\sqrt{2}\), 3 + 3\(\sqrt{2}\),……….
vi) 0.2, 0.22, 0.222, 0.2222,……….
vii) 0, -4, -8, -12,……….
viii) –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\), –\(\frac{1}{2}\),…………
ix) 1, 3, 9, 27,……………….
x) a, 2a, 3a, 4a,………
xi) a, a², a³, a⁴,………..
xii) \(\sqrt{2}\), \(\sqrt{8}\), \(\sqrt{18}\), \(\sqrt{32}\),…………
xiii) \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{9}\), \(\sqrt{12}\),…………
Solution:

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration InText Questions

Try This

Question 1.
Consider the following situations. In each find out whether you need volume or area and why ? (Page No.245)

1. Quantity of water inside a bottle.
2. Canvas needed for making a tent.
3. Number of bags inside the lorry.
4. Gas filled in a cylinder.
5. Number of match sticks that can be put in match box.

Solution:

1. Volume : 3-d shape
2. Area : L.S.A. / T.S.A
3. Volume : 3-d shape
4. Volume : 3-d shape
5. Volume : 3-d shape

Question 2.
State 5 more such examples and ask your friends to choose what they need ? (AS₃) (Page No. 245)
Solution:

1. To paint a pillar in the shape of a cylinder.
2. To white wash the walls of a house.
3. To find the quantity of rich in a heap of rice.
4. An object is the form of a cone having a herispherical shape on its top. Quantity of ice-cream to be filled in it.
5. Later flowing through a cylindricalipipe.

Question 3.
Break the pictures in the previous figure into solids of known shapes. (AS₅) (Page No. 246)
Solution:

Question 4.
Think of 5 more objects around you that can be seen as a combination of shapes. Name the shapes that combined to make them. (AS₃) (Page No. 246)
Solution:
Student’s Activity.

Try This

Question 1.
Use known solid shapes and make as many objects (by combining more than two) as possible that you come across in
your daily life.
(Hint : Use clay, or balls, pip€s. paper cones, boxes like cube, cuboid etc.) (AS4, AS5) (Page No. 252)
Solution:
Student’s Activity.

Think – Discuss

Question 1.
A sphere is inscribed in a cylinder. Is the surface of the sphere equal to the curved surface of the cylinder ? If yes, explain how. (AS2, AS3) (Page No. 252)

Solution:
Yes, the surface area of the sphere is equal to the curved surface area of the cylinder.
Let the radius of this cylinder be ‘r’ and its height ‘h’
Then its curved surface area = 2πrh
= 2πr (r + r)
∴ height = diameter of the sphere
= diameter of the cylinder
= 2r
= 2πr(2r)
= 4πr²
And surface area of the sphere = 4πr²
∴ C.S.A of cylinder = Surface area of sphere.

Try This

Question 1.
If the diameter of the cross-section of a wire is decreased by 5% by what percentage should the length be increased so that the volume remains the same ? (AS4) (Page No. 257)
Solution:
Radius of wire r = r and length = h₁
diameter of cross – section of wire d₁ = 2r
decreased diameter 5%
= 2r × \(\frac{5}{100}\) = \(\frac{\mathrm{r}}{10}\)
∴ decreased radius r₂ = 2r – \(\frac{\mathrm{r}}{20}\)
= \(\frac{19\mathrm{r}}{10}\) × \(\frac{1}{2}\) = \(\frac{19 \mathrm{r}}{20}\)
volume of the wire V₁ = πr₁²h₁, length = h₂ (after increased) volume of the wire V₁ (after increased) = πr₂²h₂ volumes are equal. So v₁ = v₂
πr₁²h₁ = πr₂²h₂
πr²h₁ = π \(\left(\frac{19 r}{20}\right)^2\) × h₂
h₁ = \(\frac{361}{400}\) × h₂
h₁ = \(\frac{361}{400}\) h₂
h₂ = \(\frac{400 \mathrm{~h}_1}{361}\)
increased length = h₂ – h₁
= \(\frac{400 \mathrm{~h}_1}{361}\) – h₁
= \(\frac{400 \mathrm{~h}_1-361 \mathrm{~h}_1}{361}\) = \(\frac{39 \mathrm{~h}_1}{361}\)
increased percentage
= 100 × \(\frac{39 \mathrm{~h}_1}{361}\) × \(\frac{\mathrm{h}_1}{\mathrm{~h}_1}\)
= \(\frac{3900}{361}\) = 10.8%

Question 2.
Surface areas of a sphere and cube are equal then find the ratio of their volumes. (AS4)(Page No. 257)
Solution:
radius of sphere = r, and side of cube = a
surface area of sphere = 4πr²
surface area of cube = 6a²
surface area of sphere = surface area of cube (given)
4πr² = 6a² ⇒ 4 × \(\frac{22}{7}\) × π² = 6 × a²

Think – Discuss

Question 1.
Which barrel shown in the below figure can hold more water ? Discuss with your friends. (AS₂, AS₃) (Page No. 262)

Solution:
r₁ = \(\frac{1}{2}\) = 0.5 cm; h₁ = 4 cm
Volume of the 1^st barrel = πr²h
= \(\frac{22}{7}\) × 0.5 × 0.5 × 4 = 3.142 cm³

r₂ = \(\frac{4}{2}\) = 2 cm ; h = 1 cm
Volume of the 2^nd barrel
V = πr²h = \(\frac{22}{7}\) × 2 × 2 × 1
= 12.57 cm³
Hence, the volume of the 2^nd barrel is more than the first barrel.

Do This

Question 1.
A copper rod of diameter 1cm and length 8 cm is drawn into a wire of length 18m of uniform thickness. Find the thickness of wire. (AS₄) (Page No. 263)
Solution:
Volume of the copper rod(Cylinder) = πr²h
= \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 8
= \(\frac{44}{7}\) cm²
If ‘r’ is the radius of the wire, then its volume = πr²h
∴ The volume of rod is equal to the volume of the wire. We have
⇒ \(\frac{22}{7}\) × r² × 18 m = \(\frac{44}{7}\) cm³
⇒ r² = \(\frac{44}{7}\) × \(\frac{7}{22}\) × \(\frac{1}{1800}\)
[∴ 18m = 18 × 100 cm]
⇒ r² = \(\frac{1}{900}\)
⇒ r = \(\frac{1}{30}\) cm = \(0.0 \overline{3}\) cm
∴ Thickness = d = 2 × 0.03 = 0.06 cm

Question 2.
Pravali house has a water tank in the shape of a cylinder on the roof. This is filled by pumping water from a sumpfan underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44m × 9.5 cm. The water tank has radius 60 cm. and height 95 cm. Find the height of the water left in the sump after the water tank has been completely filled with water from the sump which had been full of water. Compare the ca¬pacity of the tank with that of the sump. (AS₄) (Page No. 263)
Solution:
Volume of the water in the sump = [v = lbh]
= 1.57 × 1.44 × 0.95
(∵ 9.5 cm = \(\frac{9.5}{100}\) m = 0.95 m).
= 2.14776 m³ = 2147160 cm³
Volume of the tank on the roof = πr²h
= 3.14 × 60 × 60 × 95 = 1073880 cm³
∴ Volume of the water left in the sump after filling the tank
= 2147760 – 1073880
= 1073880 cm³
Let the height of the water in the tank be h.
∴ 157 × 144 × h = 1073880
h = \(\frac{1073880}{157 \times 144}\) = 47.5 cm
∴ Ratio of the volume of the sump and tank
= 21447760 : 1073880 = 2 : 1
∴ Sump can hold two times the water that can be hold in the tank.

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.2

Question 1.
Find the roots of the following quadratic equations by factorisation.

i) x² – 3x – 10 = 0
Solution:
Given : x² – 3x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x(x – 5) + 2 (x – 5) = 0
⇒ (x – 5) (x + 2) = 0
⇒ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = -2
⇒ x = 5 or – 2
are the roots of the given Q.E.

ii) 2x² + x – 6 = 0 (A.P.Mar. 16)
Solution:
Given : 2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or 2x = 3
⇒ x = -2 or \(\frac{3}{2}\)
are the roots of the given Q.E.

iii) \(\sqrt{2}\)x² + 7x + 5\(\sqrt{2}\) = 0
Solution:
Given : \(\sqrt{2}\)x² + 7x + 5\(\sqrt{2}\) = 0
⇒ \(\sqrt{2}\)x² + 5x + 2x + 5\(\sqrt{2}\) = 0
⇒ x(\(\sqrt{2}\)x + 5) + \(\sqrt{2}\)(\(\sqrt{2}\)x + 5) = 0
⇒ (\(\sqrt{2}\)x + 5)(x + \(\sqrt{2}\)) = 0
⇒ \(\sqrt{2}\)x + 5 = 0 or x + \(\sqrt{2}\) = 0
⇒ \(\sqrt{2}\)x = – 5 or x = –\(\sqrt{2}\)
⇒ x = \(\frac{-5}{\sqrt{2}}\) or –\(\sqrt{2}\) are the roots of the given Q.E.

iv) 2x² – x + \(\frac{1}{8}\) = 0
Solution:
Given : 2x² – x + \(\frac{1}{8}\) = 0
⇒ \(\frac{16 x^2-8 x+1}{8}\) = 0
⇒ 16x² – 8x + 1 = 0
⇒ 16x² – 4x – 4x + 1 = 0
⇒ 4x(4x – 1) – 1(4x – 1) = 0
⇒ (4x – 1) (4x- 1) = 0
⇒ 4x – 1 = 0
⇒ 4x = 1
⇒ x = \(\frac{1}{4}\), \(\frac{1}{4}\)
are the roots of given Q.E.

v) 100x² – 20x + 1 = 0
Solution:
Given : 100x² – 20x + 1 = 0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x (10x – 1) – 1(10x – 1) = 0
⇒ (10x – 1) (10x – 1) = 0
⇒ 10x – 1 = 0
⇒ 10x = 1
x = \(\frac{1}{10}\), \(\frac{1}{10}\)
are the roots of the given Q.E.

vi) x(x + 4) = 12
Solution:
Given : x(x + 4) = 12
⇒ x² + 4x = 12
⇒ x² + 4x – 12 = 0
⇒ x² + 6x – 2x – 12 = 0
⇒ x(x + 6) – 2(x + 6) = 0
⇒ (x + 6) (x – 2) = 0
⇒ x + 6 = 0 or x – 2 = 0
⇒ x = -6 or x = 2
⇒ x = -6 or 2
are the roots of the given Q.E.

vii) 3x² – 5x + 2 = 0
Solution:
Given : 3x² – 5x + 2 = 0
⇒ 3x² – 3x – 2x + 2 = 0
⇒ 3x(x – 1) – 2(x – 1) = 0
⇒ (x – 1) (3x – 2) = 0
⇒ x – 1 = 0 or 3x – 2 = 0
⇒ x = 1 or x = \(\frac{2}{3}\)
⇒ x = 1 or \(\frac{2}{3}\)
are the roots of the given Q.E.

viii) x – \(\frac{3}{x}\) = 2
Solution:
Given : x – \(\frac{3}{x}\) = 2
⇒ \(\frac{x^2-3}{x}\) = 2
⇒ x² – 3 = 2x
⇒ x² – 2x – 3 = 0
⇒ x² – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x – 3) (x + 1) = 0
⇒ x = 3 or x = -1
⇒ x = 3 or – 1
are the roots of the given Q.E.

ix) 3(x – 4)² – 5(x – 4) = 12
Solution:
Take (x -4) = a, then the given Q.E.
⇒ reduces to 3a² – 5a = 12
⇒ 3a² – 5a – 12 = 0
⇒ 3a² – 9a + 4a – 12 = 0
⇒ 3a(a – 3) + 4(a – 3) = 0
⇒ (a – 3) (3a + 4) = 0 = 0
⇒ a – 3 = 0 or 3a + 4 = 0
⇒ a = 3 or a = \(\frac{-4}{3}\)
but a = x – 4
∴ x – 4 = 3 (or) x – 4 = \(\frac{-4}{3}\)
⇒ x = 7 or x = 4 – \(\frac{4}{3}\)
⇒ x = 7 or \(\frac{8}{3}\)
are the roots of the given Q.E.

Question 2.
Find two numbers whose sum is 27 and product is 182. (A.P.Mar. 15)
Solution:
Let a number be x.
Then the other number = 27 – x
Product of the numbers = x(27 – x)
= 27x – x²
By problem 27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14) = 0
⇒ (x – 13) (x – 14) = 0
⇒ x – 13 = 0 or x – 14 = 0
⇒ x = 13 or 14.
The numbers are 13; 27 – 13 = 14 or 14 and 27 – 14 = 13.

Question 3.
Find two consecutive positive integers, sum of whose squares is 613.
Solution:
Let a positive integer be x.
Then the second integer = x + 1
Sum of the squares of the above integers = x² + (x + 1)²
= x² + x² + 2x + 1
= 2x² + 2x + 1
By problem 2x² + 2x + 1 = 613
⇒ 2x² + 2x – 612 = 0
⇒ x² + x – 306 = 0
⇒ x² + 18x – 17x – 306 = 0
⇒ x(x + 18) – 17(x + 18) = 0
⇒ (x – 17) (x + 18) = 0
⇒ x – 17 = 0 (or) x + 18 = 0
⇒ x = 17 (or) -18
Then the numbers are x = (17; 17 + 1) or x = (-18; -18 + 1)
i.e., 17, 18 or -17, -18.

Question 4.
The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Solution:
Let the base of the right triangle = x cm
Then its altitude = x – 7 cm
By Pythagoras Theorem

(base)² + (height)² = (hypotenuse)²
⇒ x² + (x – 7)² = 13²
⇒ x² + x² – 14x + 49 = 169
⇒ 2x² – 14x + 49 – 169 = 0
⇒ 2x² – 14x – 120 = 0
⇒ x² – 7x – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
⇒ x = 12 (or) x = – 5
But x can’t be negative.
∴ x = 12
x – 7 = 12 – 7 = 5
The two sides are 12 cm and 5 cm.

Question 5.
A cottage industry produces a certain number of pottery articles in a day. It was observed in a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. if the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced be x.
Then the cost of each article = 2x + 3
Total cost of the articles produced = x[2x + 3] = 2x² + 3x
By problem 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x(2x + 15) – 6(2x + 15) = 0
⇒ (2x + 15) (x – 6) = 0
⇒ 2x + 15 = 0 (or) x – 6 = 0
⇒ x = \(\frac{-15}{2}\) or x = 6
But x can’t be negative, x = 6
∴ x = 6
2x + 3 = 2 × 6 + 3 = 15
∴ Number of articles produced = 6
Cost of each article = ₹ 15.

Question 6.
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Solution:
Let the length of the rectangle = x
Given perimeter = 2(l + b) = 28
⇒ l + b = \(\frac{28}{2}\) = 14
∴ Breadth of the rectangle = 14 – x
Area = Length × Breadth = x(14 – x)
= 14x – x²
By problem, 14x – x² = 40
⇒ x² – 14x + 40 = 0
⇒ x² – 10x – 4x + 40 = 0
⇒ x(x – 10) – 4(x – 10) = 0
⇒ (x – 10) (x – 4) = 0
⇒ x – 10 = 0 (or) x – 4 = 0
⇒ x = 10 (or) 4
∴ Length = 10 m or 4 m
Then breadth 14 – 10 = 4 m (or) = 14 – 4 = 10 m

Question 7.
The base of a triangel is 4 cm longer than its altitude. If the area of the triangle is 48 sq.cm, then find its base and altitude.
Solution:
Let the altitude of the triangle h = x cm
Then its base ‘b’ = x + 4
Area = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)(x + 4)(x) = \(\frac{x^2+4 x}{2}\)
By problem \(\frac{x^2+4 x}{2}\) = 48
⇒ x² + 4x = 2 × 48
⇒ x² + 4x – 96 = 0
⇒ x² + 12x – 8x – 96 = 0
⇒ x(x + 12) – 8(x + 12) = 0
⇒ (x + 12) (x – 8) = 0
⇒ x + 12 = 0 (or) x – 8 = 0
⇒ x = -12 (or) x = 8
But x can’t be negative.
∴ x = 8 and x + 4 = 8 + 4 = 12
Hence altitude = 8 cm and base = 12 cm.

Question 8.
Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train.
Solution:
Let the speed of the slower train = x kmph
Then speed of the faster train = x + 5 kmph
Distance = Speed × Time
Distance travelled by the first train = 2(x + 5) = 2x + 10
Distance travelled by the second train = 2.x = 2x
By Pythagoras Theorem
(hypotenuse)² = (side)² + (side)²
⇒ (2x)² + (2x + 10)² = 50²
⇒ 4x² + (4x² + 40x + 100) = 2500
⇒ 4x² + 4x² + 40x + 100 = 2500
⇒ 8x² + 40x – 2400 = 0
⇒ x² + 5x – 300 = 0
⇒ x² + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
⇒ (x + 20) (x – 15) = 0
∴ x – 15 = 0 (or) x + 20 = 0
⇒ x = 15 (or) x – 20
⇒ x = 15 (or) – 20
But x can’t be negative.
∴ Speed of the slower train x = 15 kmph.
Speed of the faster train x + 5 = 15 + 5 = 20 kmph.

Question 9.
In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was ₹ 1600, how many boys are there in the class ?
Solution:
Let the number of boys in the class = x
Then number of girls in the class = 60 – x [∵ total students = 60]
Money contributed by the boys = x(60 – x) = 60x – x² [∵ given]
Money contributed by the girls = (60 – x) x = 60x – x²
Money contributed by the class = 120x – 2x²
By problem 120x – 2x² = 1600
⇒ x² – 60x + 800 = 0
⇒ x² – 40x – 20x + 800 = 0
⇒ x(x – 40) – 20(x – 40) = 0
⇒ (x – 40) (x – 20) = 0
⇒ x = 40 (or) 20
∴ Boys = 40 or 20
Girls = 20 or 40.

Question 10.
A motor boat heads upstream a distance of 24 km on a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed ?
Solution:
Let the speed of the boat in still water be x kmph.
Speed of the current = 3 kmph
Then speed of the boat in upstream = (x – 3)kmph
Speed of the boat in down stream = (x + 3) kmph
By problem total time taken = 6hrs
∴ \(\frac{24}{x-3}\) + \(\frac{24}{x+3}\) = 6
⇒ 24[\(\frac{1}{x-3}\) + \(\frac{1}{x+3}\)] = 6
⇒ 24\(\left[\frac{x+3+x-3}{(x+3)(x-3)}\right]\) = 6
⇒ 24(2x) = 6(x² – 9)
⇒ 8x = x² – 9
⇒ x² – 8x – 9 = 0
⇒ x² – 9x + x – 9 = 0
⇒ x(x – 9) + 1(x – 9) = 0
⇒ (x – 9) (x + 1) = 0
⇒ x – 9 = 0 or x + 1 = 0
x can’t be negative.
∴ x = 9
i.e., speed of the boal in still water = 9 kmph.

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.1

Question 1.
A Joker’s cap is in the form of right circular cone whose base radius is 7cm and height is 24 cm. Find the area of the sheet required to make 10 such caps. (AS4)
Solution:
Radius of the cap (r) = 7 cm
Height of the cap(h) = 24 cm
Slant height of the cap (l) = \(\sqrt{\mathrm{r}^2+\mathrm{h}^2}\)
= \(\sqrt{7^2+24^2}\)
= \(\sqrt{49+576}\)
= \(\sqrt{625}\) = 25
∴ l = 25 cm

Lateral surface area of the cap (cone) = πrl
L.S.A = \(\frac{22}{7}\) × 7 × 25 = 550 cm²
∴ Area of the sheet required for 10 caps
= 10 × 550 = 5500 cm².

Question 2.
A sports Company was ordered to prepare 100 paper cylinders for packing shuttle cocks. The required dimensions of the cylinder are 35 cm length/height and its radius is 7cm. Find the required area of thick paper sheet needed to make 100 cylinders. (AS4)
Solution:
Radius of the cylinder r = 7 cm
Height of the cylinder h = 35 cm
L.S.A of the cylinder with lids at both ends = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 35 = 1,540 cm²
∴ Area of thick paper required for 100 cylinders
= 100 × 1,540
= 1,54,000 cm²
= \(\frac{1,54,000}{100 \times 100}\) m² = 15.40 m².

Question 3.
Find the volume of right circular cone with radius 6 cm and height 7 cm. (AS1) (Mar. ’16 (A.P.))
Solution:
Radius of the cone (r) = 6 cm
Height of the cone (h) = 7 cm

Volume of the cone = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7 = 264 cm³.

Question 4.
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their bases be the same, find the ratio of the height of the cylinder and slant height of the cone. (AS4)
Solution:
Lateral surface area of a cylinder = 2πrh
Curved surface area of the cone = πrl
Given that 2πrh = πrl
⇒ 2h = l
h = l/2
∴ The ratio of the height of the cylinder and slant height of the cone = h : l
⇒ h : 2h [∵ l = 2h]
⇒ 1 : 2.

Question 5.
A self help group wants to manufacture joker’s caps of 3 cm radius and 4 cm height. If the available paper sheet is 1000 cm² then how many caps can be manufactured from that paper sheet ?
Solution:
Radius of the cap (conical cap)(r) = 3 cm
Height of the cap(h) = 4 cm

Slant height l = \(\sqrt{\mathrm{r}^2+\mathrm{h}^2}\)
(by pythagoras theorem)
= \(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\) = 5 cm
C.S.A of the cap = πrl = \(\frac{22}{7}\) × 3 × 5
= 47.14 cm²
Number of caps that can be made out of
1000 cm² = \(\frac{1000}{47.14}\) ≈ 21.21
∴ Number of caps = 21

Question 6.
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1 (AS4) (June ’15 (A.P.))
Solution:
Given that the bases of a cylinder and a cone are of equal radii.
The height are also equal (Given).
∴ Volume of the cylinder = πr²h
Volume of the cone = \(\frac{1}{3}\) πr²h

∴ The ratio of their volumes
= πr²h : \(\frac{1}{3}\) πr²h
= 1 : \(\frac{1}{3}\)
= 3 : 1
(cancelling the common factor πr²h)

Question 7.
The shape of solid iron rod is a cylindrical. Its height is 11cm. and base diameter is 7 cm. Then find the total volume of 50 such rods. (AS4)
Solution:
Diameter of the cylinder (d) = 7 cm
Radius of the base (r) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 11 cm
Volume of the cylinder V = πr²h

= \(\frac{22}{7}\) × 3.5 × 3.5 × 11
= 423.5 cm³
∴ The total volume of 50 rods
= 50 × 423.5 cm³
= 21,175 cm³.

Question 8.
A heap of rice is in the form of a cone of diameter 12m and height 8m. Find its volume ? How much canvas cloth is required to cover the heap ? (AS4)
Solution:
Diameter of the heap (conical)

(d) = 12 m
∴ Radius = \(\frac{\mathrm{d}}{2}\) = \(\frac{12}{2}\) = 6 m
Height of the cone (h) = 8 m 1 2
Volume of the cone V = \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 8
= 301.71 m³.

Question 9.
The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height ? (AS1) (Mar ’15 (A.P.))
Solution:
Diameter of the base of the cone (d) = 70 cm
∴ Radius of the base(r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{70}{2}\) = 35 cm
Curved surface area of the cone = πrl

By problem = πrl = 4070
= \(\frac{22}{7}\) × 35 × l = 4070
∴ l = 4070 × \(\frac{7}{22}\) × \(\frac{1}{35}\) = 37 cm
Hence, the slant height of the cone = 37cm.

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.1

Question 1.
Check whether the following are quadratic equations or not.
i) (x + 1)² = 2(x – 3)
Solution:
Given: (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2(x – 3)
⇒ x² + 2x + 1 – 2x + 6 = 0
⇒ x² + 7 = 0 is a Q.E.

ii) x² – 2x = -2(3 – x)
Solution:
Given : x² – 2x = -2(3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x² – 4x + 6 = 0 is a Q.E.

iii) (x – 2) (x + 1) = (x – 1) (x + 3)
Solution:
Given : (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x(x + 1) -2(x + 1)
= x(x + 3) – 1(x + 3)
Note : Compare the coefficients of x² on both sides. If they are equal it is not a Q.E.
⇒ x² + x – 2x – 2 = x² + 3x – x – 3
⇒ x² – x – 2 = x² + 2x – 3
⇒ 3x – 1 = 0 is not a Q.E.

iv) (x – 3) (2x + 1) = x(x + 5)
Solution:
Given : (x – 3) (2x + 1) = x(x + 5)
⇒ x(2x + 1) – 3(2x + 1) = x. x + 5.x
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ x² – 10x – 3 = 0 is a Q.E.
(or)
Comparing the coefficients of x² on both sides, x . 2x and x. x ⇒ 2x² and x² 2x² ≠ x²
Hence it’s a Q.E.

v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Solution:
Given : (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x(x – 3) – 1(x – 3) = x(x – 1)+ 5(x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Hence it’s a Q.E.
(or)
Co.eff. of x² on L.H.S. = 2 × 1 = 2
Co.eff. of x² on R.H.S = 1 × 1 = 1
LHS ≠ RHS
Hence it is a Q.E.

vi) x² + 3x + 1 = (x – 2)²
Solution:
Given : x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0 is not a Q.E.

vii) (x + 2)³ = 2x (x² – 1)
Solution:
Given : (x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
⇒ x³ + 6x² + 14x + 8 = 0 is not a Q.E. [∵ degree = 3]

viii) x³ – 4x² – x + 1 = (x – 2)³
Solution:
Given : x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = (x – 2)³
= x³ – 6x² + 12x – 8
⇒ 6x² – 12x + 8 – 4x² – x + 1 = 0
⇒ 2x² – 13x + 9 = 0 is a Q.E.

Question 2.
Represent the following situations in the form-of quadratic equations :
i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth of the rectangular plot be x m.
Then its length (by problem) = 2x + 1.
Area = l . b = (2x + 1). x = 2x² + x but area = 528 m² (∵ given)
∴ 2x² + x = 528
⇒ 2x² + x – 528 = 0 where x is the breadth of the rectangle.

ii) The product of the consecutive positive integers is 306. We need to find the integers.
Solution:
Let the consecutive integers be x and x + 1
Their product = x(x + 1) = x² + x
By problem x² + x = 306
⇒ x² + x – 306 = 0
Where x is the smaller integer.

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Solution:
Let the present age of Rohan be x years. Then age of Rohan’s mother = x + 26
After 3 years :
Age of Rohan would be = x + 3
Rohan’s mother’s age would be = (x + 26) + 3 = x + 29
By problem (x + 3) (x + 29) = 360
⇒ x(x + 29) + 3(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Where x is Rohan’s present age.

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the speed of the train be x kmph. Then time taken to travel a distance

It the speed is 8km/hr less, then time needed to cover the same distance
would be \(\frac{480}{x-8}\)
By problem = \(\frac{480}{x-8}\) – \(\frac{480}{x}\) = 3

⇒ x² – 8x = 1280
⇒ x² – 8x – 1280 = 0
Where x is the speed of the train.

Students can practice 10th Class Maths Solutions Telangana Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Exercise 4.3

Question 1.
Solve each of the following pairs of equations by reducing them to a pair of linear equations. (T.S. Mar. 15) (A.P. Mar. 15)

i) \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) – \(\frac{3}{y-2}\) = 1
Solution:
Given, \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2 and
\(\frac{6}{x-1}\) – \(\frac{3}{y-2}\) = 1
Put \(\frac{1}{x-1}\) = a and \(\frac{1}{y-2}\) = b,
then the given equations reduce to
5a + b = 2 —– (1)
6a – 3b = 1 —– (2)

Substituting b = \(\frac{1}{3}\) in equation(1), we get
5a + \(\frac{1}{3}\) = 2 ⇒ 5a = 2 – \(\frac{1}{3}\)
⇒ 5a = \(\frac{3 \times 2-1}{3}\)
⇒ 5a = \(\frac{6-1}{3}\)
∴ a = \(\frac{5}{3}\) × \(\frac{1}{5}\) = \(\frac{1}{3}\)
But a = \(\frac{1}{x-1}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{x-1}\)
⇒ (x – 1) . 1 = 3 x 1
x – 1 = 3 ⇒ x = 3 + 1 = 4
b = \(\frac{1}{y-2}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{y-2}\)
⇒ y – 2 = 3 ⇒ y = 3 + 2 = 5
∴ Solution (x, y) = (4, 5)

ii) \(\frac{x+y}{x y}\) = 2, \(\frac{x-y}{x y}\) = 6
Solution:

Then the given equations reduces to

Substituting b = 4 in equation (1), we get
a + 4 = 2 ⇒ a = 2 – 4 = -2
But a = \(\frac{1}{x}\) = -2 ⇒ x = \(\frac{-1}{2}\)
b = \(\frac{1}{y}\) = 4 ⇒ y = \(\frac{1}{4}\)
∴ The solution (x, y) = \(\left[\frac{-1}{2}, \frac{1}{4}\right]\)

iii) \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2; \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Solution:
Given, \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2 and \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Take \(\frac{1}{\sqrt{x}}\) = a and \(\frac{1}{\sqrt{y}}\) = a
Then the given equations reduce to
2a + 3b = 2 —- (1)
4a – 9b = – 1 —– (2)

Substituting, b = \(\frac{1}{3}\) in equation (1), we get

iv) 6x + 3y = 6xy
2x + 4y = 5xy
Solution:
Given, 6x + 3y = 6xy

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b
3a + 6b = 6 —– (1)
4a + 2b = 5 —- (2)

v) \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1, \(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10 where x ≠ 0, y ≠ 0.
Solution:
Given, \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1 and \(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then the given equations reduce to
5a – 2b = -1 —— (1)
15a + 7b = 10 —- (2)

Substituting b = 1 in equation (1), we get
5a – 2(1) = -1 ⇒ 5a = -1 + 2
5a = 1 ⇒ a = \(\frac{1}{5}\)
But a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1
Solving the above equations

Substituting x = 3 in x + y = 5, we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ The solution (x, y) = (3, 2)

vi) \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13; \(\frac{5}{x}\) – \(\frac{4}{y}\) = -2 where x ≠ 0, y ≠ 0. (A.P. Jun. 15)
Solution:
Given, \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 and \(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
Put \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b
Then the given equations reduce to
2a + 3b = 13 —– (1)
5a – 4b = -2 —– (2)
Solving (1) & (2), we get

Substituting b = 3 in equation (1), we get
2a + 3(3) = 13 ⇒ 2a = 13 – 9
⇒ a = \(\frac{4}{2}\) = 2
But, a = \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)
b = \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)
∴ The solution (x, y) = \(\left(\frac{1}{2}, \frac{1}{3}\right)\)

vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4; \(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Solution:
Given, \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4; \(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Taking \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b then the given equations reduce to
10a + 2b = 4 —– (1)
15a – 5b = -2 —- (2)

Substituting b = 1 in equation (1), we get 10a + 2(1) = 4
⇒ 10a = 4 – 2 ⇒ a = \(\frac{2}{10}\) = \(\frac{1}{5}\)

Substituting x = 3 in x + y = 5, we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ The solution (x, y) = (3, 2)

viii) \(\frac{1}{3 x+y}\) + \(\frac{1}{3 x-y}\) = \(\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}\) – \(\frac{1}{2(3 x-y)}\) = \(\frac{-1}{8}\)
Solution:
\(\frac{1}{3 x+y}\) + \(\frac{1}{3 x-y}\) = \(\frac{3}{4}\) and \(\frac{1}{2(3 x+y)}\) – \(\frac{1}{2(3 x-y)}\) = \(\frac{-1}{8}\)
Take \(\frac{1}{3 x+y}\) = a; \(\frac{1}{3 x-y}\) = b,
then the given equations reduce to

Sustituting a = \(\frac{1}{4}\) in equation (1), we get

Substituting x = 1 in 3x + y = 4
⇒ 3(1) + y = 4 ⇒ y = 4 – 3 = 1
∴ The solution (x, y) = (1, 1)

Question 2.
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and speed of the boat in still water.
Solution:
Let, the speed of the boat in still water = x kmph
and the speed of the stream = y kmph.
then speed in downstream = x + y Speed in upstream = x – y

then the above equations reduce to
30a + 44b = 10 —– (1)
40a + 55b = 13 —- (2)

Substituting b = \(\frac{1}{11}\) in equation (1), we get
30a + 44(\(\frac{1}{11}\)) = 10
⇒ 30a = 10 – 4 ⇒ a = \(\frac{6}{30}\) = \(\frac{1}{5}\)
but a = \(\frac{1}{x-y}\) = \(\frac{1}{5}\) ⇒ x – y = 5 —- (3)
b = \(\frac{1}{x+y}\) = \(\frac{1}{11}\) ⇒ x + y = 11 —- (4)
Adding (3) and (4)

Substituting x = 8 in x – y = 5, we get
8 – y = 5 ⇒ y = 8 – 5 = 3
∴ The solution (x, y) = (8, 3)
Speed of the boat in still water = 8 kmph
Speed of the stream = 3kmph

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours If he travels 120 km by train and rest by car. He takes 20 minutes more If he travels 200 km by train and rest by car. Find the speed of the train and the car.
Solution:
Let, the speed of the train be x kmph, and the speed of the car be y kmph
By problem,

then the equations reduce to
15a + 60b = 1 —- (1)
8a + 16b = \(\frac{1}{3}\) ⇒ 24a + 48b = 1 —- (2)

Substituting a = \(\frac{1}{36}\) in Equation (1)
we get,

∴ Time taken by 1 woman 18 days
Time taken by 1 man = 36 days

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.4

Question 1.
Find the nature of the roots of the following quadratic equations. If real roots exist, find them. (A.P June is)

i) 2x² – 3x + 5 = 0
Solution:
Given : 2x² – 3x + 5 = 0
a = 2; b = -3; c = 5
Discriminant = b² – 4ac
b² – 4ac = (-3)² – 4(2)(5)
= 9 – 40
= -31 < 0
Roots are imaginary,

ii) 3x² – 4 \(\sqrt{3}\)x + 4 = 0
Solution:
Given : 3x² – 4\(\sqrt{3}\)x + 4 = 0
a = 3; b = -4\(\sqrt{3}\) ; c = 4
b² – 4ac = (-\(\sqrt{3}\))² – 4(3)(4)
= 48 – 48 = 0
∴ Roots are real and equal and they are
\(\frac{-b}{2 a}\), \(\frac{-b}{2 a}\)
= \(\frac{-(-4 \sqrt{3})}{2 \times 3}\) = \(\frac{4 \sqrt{3}}{6}\) = \(\frac{2}{\sqrt{3}}\), \(\frac{2}{\sqrt{3}}\)

iii) 2x² – 6x + 30
Solution:
Given : 2x² – 6x + 3 = 0
a = 2; b = -6; c = 3
b² – 4ac = (-6)² – 4(2)(3)
= 36 – 24 = 12 > 0
The roots are real and distinct.
They are

Question 2.
Find the values of k for each of the following quadratic equations so that they have two equal roots.

i) 2x² + kx + 3 = 0
Solution:
Given : 2x² + kx + 3 = 0 has equal roots
∴ b² – 4ac = 0
Here a = 2; b = k; c = 3
b² – 4ac = (k)² – 4(2) (3) = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = \(\sqrt{24}\)
= ±2 \(\sqrt{6}\)

ii) kx(x – 2) + 6 = 0
Solution:
Given : kx(x – 2) + 6 = 0
kx² – 2kx + 6 = 0
As this Q.E. has equal roots.
b² – 4ac = 0
Here a = k; b = -2k; c = 6
∴ b² – 4ac = (-2k)² – 4(k) (6) = 0
⇒ 4k² – 24 k = 0
⇒ 4k(k – 6) = 0
⇒ k = 0 (or) k – 6 = 0
⇒ k = 0 (or) 6
But k = 0 is trivial
∴ k = 6

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m² ? If so, find its length and breadth. (A.P. June 15)
Solution:
Let the breadth = x m
Then length = 2x m
Area = length × breadth
= x. (2x) = 2x² m²
By problem 2x² = 800
⇒ x² = 400
and x = \(\sqrt{400}\) = ± 20
∴ Breadth x = 20 m and
Length 2x = 2 × 20 = 40 m

Question 4.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the situation possible ? If so, determine their present ages.
Solution:
Let the age one of the two friends be x years.
Then the age of the other = 20 – x
Then, 4 years ago their ages would be (x – 4) and (20 – x – 4) = 16 – x
∴ Problem (x – 4) (16 – x) = 48
⇒ x(16 – x) – 4(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
Here a = 1; b = – 20; c = 112
b² – 4ac = (-20)² – 4(1) (112)
= 400 – 448 = -48 < 0
Thus the roots are not real.
∴ The situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80m and area 400m² ? If so, find its length and breadth.
Solution:
Given : Perimeter of a rectangle 2(1l + b) = 80
⇒ l + b = \(\frac{80}{2}\) = 40 —– (1)
Area of the rectangle, l × b = 400
It possible, let us suppose that length of the rectangle = x m say
Then its breadth by equation (1) = 40 – x
By problem area = x. (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
Here a = 1; b = – 40; c = 400
b² – 4ac = (-40)² – 4(1)(400)
= 1600 – 1600
= 0
∴ The roots are real and equal.
They are \(\frac{-b}{2 a}\), \(\frac{-b}{2 a}\)
i.e., \(\frac{-(-40)}{2 \times 1}\) = \(\frac{40}{2}\) = 20
∴ The dimensions are 20m, 20m.
(∴ The park is in square shape)

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.3

Question 1.
An iron pillar consists of a cylindrical portion of 2.8 m height and 20 cm in di-ameter and a cone of 42 cm height sur-mounting it. Find the weight of the pillar if 1cm3 of iron weighs 7.5g. (AS4)
Solution:
Volume of the iron pillar = Volume of the cylinder + volume of the cone
Cylinder : Radius = \(\frac{\text { diameter }}{2}\)
= \(\frac{20}{2}\) = 10 cm
Height = 2.8 m
= 280 cm
Volume = πr²h
= \(\frac{22}{7}\) × 10 × 10 × 280
= 88,000 cm³

Cone : Radius ‘r’ = \(\frac{\text { diameter }}{2}\)
= \(\frac{20}{2}\) = 10 cm
Height ‘h’ = 42 cm
Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 10 × 10 × 42
= 4,400 cm³.
Total volume = 88,000 + 4,400
= 92,400 cm³
∴ Total weight of the pillar at a weight of 7.5g per 1 cm³ = 92,400 × 7.5
= 6,93,000 gms
= \(\frac{6,93,000}{1,000}\) kg = 693 kg

Question 2.
A toy is made in the form of hemisphere surmounted by a right cone whose cicular, base is joined with the plane surface of hemisphere. The radius of the base of the cone is 7cm and its volume is \(\frac{3}{2}\) of hemisphere. Calculate the height of the cone and surface area of the toy correct to 2 places of decimal (take π = 3\(\frac{1}{7}\)) (AS4)
Solution:
Radius of the base of the cone(r) = 7 cm
Radius of hemisphere = 7cm
Volume of hemisphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7cm³
Volume of the cone
= \(\frac{3}{2}\) × volume of the hemisphere (given)
= \(\frac{3}{2}\) × \(\frac{2}{3}\) × \(\frac{22}{7}\) × 7 × 7 × 7
= 1078

Volume of cone = \(\frac{1}{3}\) πr²h = 1,078
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 7 × 7 × h = 1078
∴ h = 1,078 × \(\frac{3}{1}\) × \(\frac{7}{22}\) × \(\frac{1}{7}\) × \(\frac{1}{7}\) = 21cm
l² = h² + r² = 21² + 7²
= 441 + 49 = 490
l = \(\sqrt{490}\) = \(\sqrt{49 \times 10}\)
= 7\(\sqrt{10}\) = 7 × 3.16 = 22.12 cm
Surface area of the toy = curved surface area of the cone + curved surface area of the hemisphere = πrl + 2 πr²
= \(\frac{22}{7}\) × 7 × 7 × \(\sqrt{10}\) + 2 × \(\frac{22}{7}\) × 7 × 7
= \(\frac{22}{7}\) × 7 [7\(\sqrt{10}\) + 14]
= 22 × (22.12 + 14)
= 22 × 36.12 = 794.64 cm²

Question 3.
Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7cm. (AS4)
Solution:
The base of the largest right circular cone will be the circle inscribed in a face of the cube and its height will be equal to an edge of the cube.
∴ r = Radius of the base of the cone
= \(\frac{7}{2}\) cm (∴ edge = 7cm)
h = height of the cone = 7cm
Hence, volume of the cone
= \(\frac{1}{3}\) πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 7
= \(\frac{539}{6}\) cm³ = 89.83 cm³

Question 4.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemisphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in tub (Take π = \(\frac{22}{7}\)) (AS4)
Solution:
Radius of the cylinder(r) = 5 cm
Height of the cylinder (h) = 9.8 cm
Volume of the cylinder = πr²h

= \(\frac{22}{7}\) × 5 × 5 × 9.8
= 770 cm³.
Radius of hemi-sphere(r) = 3.5 cm
Volume of the hemi-sphere = \(\frac{2}{3}\) πr³
= \(\frac{2}{3}\) × \(\frac{22}{7}\) × 35 × 35 × 35
= \(\frac{539}{6}\) cm³

Radius of the base of the cone(r) = 3.5 cm
Height of the cone = 5 cm
∴ Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3.5 × 3.5 × 5
= \(\frac{385}{6}\) cm³
Total volume of the solid = volume of the hemisphere + volume of the cone
= \(\frac{539}{6}\) + \(\frac{385}{6}\) = \(\frac{924}{6}\) = 154 cm²
The volume of water left in the tub = volume of the cylindrical tube – volume of the solid immersed
= 770 – 154 = 616 cm³

Question 5.
In the adjacent figure, the height of a solid cylinder is 10 cm and diameter is 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in figure. Find the volume of the remaining solid. (AS4)

Solution:
Radius of the solid cylinder (r) = \(\frac{7}{2}\) cm
(∴ diameter = 7cm)
Height of the solid cylinder(h) = 10 cm
Volume of the solid cylinder = πr²h
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\) × 10 = 385 cm³
Radius of concial hole(r) = 3 cm
Height(depth) of the conical hole(h) = 4 cm
Volume of the conical part = \(\frac{1}{3}\) pr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 4
Total volume of two conical holes
= 2 × \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3 × 3 × 4
= \(\frac{528}{7}\) = 75.43 cm³
Hence, volume of the remaining solid
= [Volume of the Cylindrical part] – [Total volume of two conical holes]
= 385 – 75.43
= 309.57cm³

Question 6.
Spherical Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7cm, which contains some water. Find the number of marbles that should be dropped into the beaker, so that water level rises by 5.6 cm. (AS4) Solution:
Rise in the water level is seem in cylinderial shape of Radius = Beaker radius

= \(\frac{\text { diameter }}{2}\) = \(\frac{7}{2}\) = 3.5 cm
Height ‘h’ of the rise = 5.6 cm
∴ Volume of the ‘water rise’ = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 5.6
= \(\frac{22 \times 12.25 \times 5.6}{7}\) = 215.6
Volume of each marble dropped
= \(\frac{4 \pi r^3}{3}\)
Where radius r = \(\frac{\mathrm{d}}{2}\) = \(\frac{1.4}{2}\) = 0.7 cm
∴ V = \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.7 × 0.7 × 0.7
≈ 1.4373 cm³
∴ Volume of the ‘rise’ = Total volume of the marbles.
Let the number of marbles be ‘n’ then
n × volume of each marble = Volume of the rise
n × 1.4373 = 215.6
n = \(\frac{215.6}{1.4373}\) = 150
∴ Number of marbles = 150

Question 7.
A pen stand is made of wood in the shape of cuboid with three conical depression to hold the pens. The dimensions of the cuboid are 15 cm by 10cm by 3.5cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand. (AS4)

Solution:
The dimensions of the cuboid are
15 cm × 10 cm × 3.5 cm
∴ Volume of the ‘cuboid’ = \(\frac{15 \times 10 \times 7}{2}\)
= 525 cm³
Radius of the conical depression (r) = 0.5 cm
Depth of conical hole (h) = 1.4 cm
= \(\frac{14}{10}\) cm
Volume of one conical hole
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) ×\(\frac{14}{10}\)
Volume of three conical holes
= 3 × \(\frac{1}{3}\) × \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{14}{10}\)
= \(\frac{11}{10}\) = 1.1 m³
∴ The volume of wood in the entire stand
= (Volume of the cuboid) – (volume of 3 conical holes)
= 525 – 1.1 = 523.9cm³.

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square. (A.P.Mar. 16, 15)

i) 2x² + x – 4 = 0
Solution:
Given : 2x² + x – 4 = 0
⇒ 2x² + x = 4
⇒ (\(\sqrt{2}\))² + x = 4
⇒ (\(\sqrt{2}\)x)² + 2.\(\sqrt{2}\). x. \(\frac{1}{2 \sqrt{2}}\) = 4
Now LHS is in the form a² + 2ab Where b = \(\frac{1}{2 \sqrt{2}}\)
Adding b² = \(\left(\frac{1}{2 \sqrt{2}}\right)^2\) on both sides we get

ii) 4x² + 4\(\sqrt{3}\)x + 3= 0
Solution:
Given : 4x² + 4\(\sqrt{3}\)x + 3 = 0
⇒ 4x² + 4\(\sqrt{3}\)x = – 3
⇒ (2x)² + 2(2x) \(\sqrt{3}\) = -3
LHS is of the form a² + 2ab where b = \(\sqrt{3}\)
∴ Adding b² = (\(\sqrt{3}\))² = 3 on both sides, we get
⇒ (2x)² + 2(2x)(\(\sqrt{3}\)) + (\(\sqrt{3}\))²
= -3 + (\(\sqrt{3}\))²
⇒ (2x + (\(\sqrt{3}\)))² = -3 + 3 = 0
⇒ (2x + \(\sqrt{3}\))² = 0
⇒ 2x + \(\sqrt{3}\) – 0
⇒ 2x + \(\sqrt{3}\) = 0
⇒ 2x = –\(\sqrt{3}\)
⇒ x = \(\frac{-\sqrt{3}}{2}\)
∴ The roots are \(\frac{-\sqrt{3}}{2}\), \(\frac{-\sqrt{3}}{2}\)

iii) 5x² – 7x – 6 = 0
Solution:
The given Q.E. is 5x² – 7x – 6 = 0
⇒ 5x² – 7x = 6
(\(\sqrt{5}\))² – 2\(\sqrt{5}\). x.\(\left(\frac{7}{2 \sqrt{5}}\right)\) = 6
The LHS is of the form a² – 2ab
where b = \(\frac{7}{2 \sqrt{5}}\)
Now adding b² = \(\left(\frac{7}{2 \sqrt{5}}\right)^2\) on both sides we get

Note : If we take the Q.E. as 5x² – 7x + 6 = 0, then we get the T.B. answer.

iv) x² + 5 = -6x
Solution:
The given Q.E. is x² + 5 = -6x
⇒ x² + 6x = -5
⇒ (x)² + 2. (x). 3 = -5
Now L.H.S. is of the form a² + 2ab where b = 3.
Adding b² = 3² on both sides we get
x² + 2(x)(3) + 3² = -5 + 3²
(x + 3)² = -5 + 9 = 4
∴ x + 3 = \(\sqrt{4}\) = ±2
⇒ x = +2 – 3 or -2 -3
= -1 or -5 are the roots of the given Q.E.

Question 2.
Find the roots of the quadratic equations given in Q.l above by applying the quadratic formula,
i) 2x² + x – 4 = 0
Solution:
Comparing this Q.E.with ax² + bx + c = 0
a = 2, b = 1, c = -4

ii) 4x² + 4\(\sqrt{3}\)x + 3 = 0
Solution:
Given : 4x² + 4 \(\sqrt{3}\) x + 3 = 0
Here a = 4, b = \(\sqrt{3}\); c = 3

iii) 5x² – 7x – 6 = 0
Solution:
Given : 5x² – 7x – 6 = 0
Here a = 5, b = -7 and c = -6

iv) x² + 5 = -6x
Solution:
Given : x² + 5 = -6x
⇒ x² + 6x + 5 = 0
Here a = 1; b = 6; c = 5
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Question 3.
Find the roots of the following equations :

i) x – \(\frac{1}{x}\) = 3, x ≠ 0
Solution:
Given x – \(\frac{1}{x}\) = 3
⇒ \(\frac{x^2-1}{x}\) = 3
⇒ x² – 1 = 3x
⇒ x² – 3x – 1 = 0
Here a = 1; b = -3; c = 1

ii) \(\frac{1}{x+4}\) – \(\frac{1}{x-7}\) = \(\frac{11}{30}\), x ≠ -4, 7
Solution:

⇒ x² – 3x – 28 = -30
⇒ x² – 3x – 28 + 30 = 0
⇒ x² – 3x + 2 = 0
⇒ x² – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x – 2 = 0 (or) x – 1 = O
⇒ x = 2 or x = 1
∴ x = 2 or 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\), Find his present age.
Solution:
Let the present age of Rehman be x years.
3 years age Rehman’s age = x – 3 and its reciprocal is \(\frac{1}{x+5}\)
Rehman’s age 5 years from now = x + 5 and its reciprocal is \(\frac{1}{x+5}\)
The sum of the reciprocals

⇒ x² + 2x – 15 = 3(2x + 2)
⇒ x² + 2x – 15 = 6x + 6
⇒ x² + 2x – 15 – 6x – 6 = 0
⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
⇒ x = -7 = 0 or x + 3 = 0
⇒ x = 7 or x = -3
But x can’t be negative
∴ x = 7
ie., present age of Rehman = 7 years.

Question 5.
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.
Solution:
Sum of the marks in Mathematics and English = 30
Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x
If she got 2 more marks in Mathematics, then her marks would be x + 2.
It she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x
By problem (x + 2) (27 – x ) = 210
⇒ x(27 – x) + 2(27 – x) = 210
⇒ 27x – x² + 54 – 2x = 210
⇒ -x² + 25x + 54 = 210
⇒ x² – 25x – 54 + 210 = 0
⇒ x² – 25x + 156 = 0
⇒ x² – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x – 12) = 0
⇒ x – 12 = 0 or x – 13 = 0
⇒ x = 12 or x = 13
If x = 12, then marks in Mathematics =12
English = 30 – 12 = 18
If x = 13, then marks in Mathematics = 13
English = 30 – 13 = 17

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangular field = x m.
Then its longer side = x + 30 m

The diagonal of a rectangle is also the hypotenuse of the lower triangle.
Here the diagonal = x + 60
∴ By Pythagoras Theorem
(side)² + (side)² = (hypotenuse)²
⇒ (x + 30)² + x² =(x + 60)²
⇒ x² + 60x + 900 + x² = x² + 120 x + 3600
⇒ x² – 60x – 2700 = 0
⇒ x² – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ x – 90 = 0(or) x + 30 = 0
⇒ x = 90 (or) x = -30
But ‘x’ can’t be negative.
∴ x = 90 m
i.e., the shorter side x = 90 m
Longer side x + 30 = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the large number be x.
8 times larger number = square of the small number = 8x
Square of the large number = x²
By problem, x² – 8x = 180
⇒ x² – 8x – 180 = 0
⇒ x² – 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10) (x – 18) = 0
⇒ x + 10 = 0(or) x – 18 = 0
⇒ x = – 10(or) x = 18
If x = 18, then larger number = 18;
(small number)² = 8 × (18) = 144
∴ Small number = \(\sqrt{144}\) = 12
The numbers are 18, 12
Note : Discard x = – 1o.

Question 8.
A train travels 360 km at a uniform speed. It the speed had been 5 km/br more, It would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
The distnce travelled = 360 km.
Let the speed of the train = x kmph.
Time taken to complete a journey =

By problem \(\frac{360}{x}\) – \(\frac{360}{x+5}\) = 1
⇒ 360\(\left(\frac{1}{x}-\frac{1}{x+5}\right)\) = 1
⇒ 360\(\left(\frac{x+5-x}{x(x+5)}\right)\)
⇒ \(\frac{5}{x^2+5 x}\) = \(\frac{1}{360}\)
⇒ x² + 5x = 1800
⇒ x² + 5x – 1800 = 0
⇒ x² + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
⇒ x + 45 = 0 or x – 40 = 0
⇒ x = – 45 or x = 40
But x can’t be negative
∴ The speed of the train = 40 kmph.

Question 9.
Two water taps together can fill a tank in 9\(\frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the time taken by tap with smaller diameter alone to fill the tank = x hrs.
Then time taken by the tap with larger diameter = (x – 10) hrs.
By problem both taps worked for
9\(\frac{3}{8}\) = \(\frac{75}{8}\) hours


⇒ 75(x – 5) = 4(x² – 10x)
⇒ 4x² – 40x = 75x – 375
⇒ 4x² – 115x + 375 = 0
⇒ 4x² – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
⇒ x – 25 = 0 (or) 4x – 15 = 0
⇒ x = 25 (or) x = \(\frac{15}{4}\)
x can’t be \(\frac{15}{4}\)
∴ x = 25
i.e., Time taken by the first tap = 25hrs
Time taken by the second tap
= 25 – 10 = 15 hrs.

Question 10.
An express train takes 1 hour less then a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the speed of the passenger train = x kmph
Then speed of the express train = x + 11 kmph.
Distance travelled = 132 km

⇒ x² + 11x = 132 × 11
⇒ x² + 11x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33(x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ x + 44 = 0 (or) x – 33 = 0
⇒ x = -44 (or) x = 33
But x cant be negative.
∴ Speed of the passenger train
= x = 33 kmph.
Speed of the express train
= x + 11 = 44 kmph.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of the perimeters is 24m, find the sides of the two square.
Solution:
Let the side of first square = x m say
Then perimeter of the first square
= 4x [∵ P = 4. side]
By Problem, perimeter of the second square = 4x + 24 (or) 4x – 24
∴ Side of the second square
⇒ \(\frac{4 x+24}{4}\) = \(\frac{4(x+6)}{4}\) = x + 6 (or)
\(\frac{4 x-24}{4}\) = x – 6
Now sum of the areas of the two squares is given as 468 m²
x² + (x + 6)² = 468
x² + x² + 12x + 36 = 468
2x² + 12x + 36 – 468 = 0
2x² + 12x – 432 = 0
x² + 6x – 216 = 0
x² + 18x – 12x – 216 = 0
x(x + 18) – 12(x + 18) = 0
(x + 18) (x – 12) = 0
⇒ x + 18 = 0 (or) x – 12 = 0
⇒ x = – 18 (or) 12
But x can’t be negative
∴ x = 12
i.e., side of the first square = 12
∴ Perimeter = 4 × 12 = 48
∴ Perimeter of the second square = 48 + 24 = 72
∴ Side of the second square = \(\frac{72}{4}\) = 18 m.
(or)
x² + (x – 6)² = 468
x² + x² – 12x + 36 = 468
2x² – 12x – 432 = 0
x² – 6x – 216 = 0
x²– 18x + 12x – 216 = 0
x(x – 18) + 12(x – 18) = 0
(x – 18) (x + 12) = 0
⇒ x – 18 = 0 (or) x + 12 = 0
⇒ x = 18 (or) -12
But x can’t be negative
∴ x = 18
i.e., side of the first square = 18 m
Perimeter = 4 × 18 = 72
Perimeter of the second square
= 72 – 24 = 48
∴ Side of the second square = \(\frac{48}{4}\) = 12 m
i. e., In any way, the sides of the squares are 12m, 18m.

Question 12.
S = ut – \(\frac{1}{2}\)gt² is a formula which gives the distance S in meters travelled by a ball from the thrower’s hands if it is thrown upwards with an initial velocity of u m/s after a time of t seconds, g is the acceleration due to gravity and is 9.8 m/s²
(i) If a ball is thrown upwards at 14 m/s, how high has it gone after 1 second ?
(ii) How long does it take for the ball to reach a height of 5 meters ?
(iii) Why are there two possible times to reach a height of 5 meters ?
Solution:
S = ut – \(\frac{1}{2}\)gt²
i) u = 14 m/s; t = 1 sec; g = 9.8 m/s²
S = 14 × 1 – \(\frac{1}{2}\) × 9.8 × (1)²
= 14 – \(\frac{1}{2}\) × 9.8 = 14 – 4.9 = 9.1m.

ii) S = 5 m.
5 = 14t – 4.9 t²
4.9 t² – 14t + 5 = 0
\(\frac{49}{10}\)t² – 14t + 5 = 0
49t² – 140t + 50 = 0
a = 49, b = -140, c = 50

iii) When the body goes upwards t value is 2.4 sec.
When the body comes downwards t value is 0.4 sec.
∴ So there are two possible times to reach a height of 5 meters.

Question 13.
If a polygon of ‘n’ sides has \(\frac{1}{2}\)n(n – 3) diagonals. How many sides will a polygon having 65 diagonals ? Is there a polygon with 50 diagonals ?
Solution:
Given : Number of diagonals of a polygon with n-sides = \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\)
No. of diagonals of a given polygon = 65
i.e., \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 65
where n is number of sides of the polygon
⇒ n² – 3n = 2 × 65
⇒ n² – 3n – 130 = 0
⇒ n² – 13n + 10n – 130 = 0
⇒ n(n – 13) + 10(n – 13) = 0
⇒ (n – 13) (n + 10) = 0
⇒ (n – 13) = 0 (or) n + 10 = 0
⇒ n = 13 (or) n = -10
But n can’t be negative.
∴ n = 13 (i.e.) number of sides = 13.
Also to check 50 as the number of diagonals of a polygon
∴ \(\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) = 50
⇒ n² – 3n = 100
⇒ n² – 3n – 100 = 0
There is no real value of n for which the above equation is satisfied.
∴ There can’t be a polygon with 50 diagonals.