Class 10

Haryana State Board HBSE 10th Class Maths Notes Chapter 9 त्रिकोणमिति का अनुप्रयोग Notes.

Haryana Board 10th Class Maths Notes Chapter 9 त्रिकोणमिति का अनुप्रयोग

→ उन्नयन कोण- यदि वस्तु हमारी आँख के स्तर से ऊपर हो अर्थात् आँख की क्षैतिज रेखा से ऊपर हो तो हमें वस्तु को देखने के लिए अपनी आँखों को ऊपर की ओर उठाना पड़ेगा। इस प्रक्रिया में हमारी दृष्टि रेखा क्षतिज रेखा से एक कोण से ऊपर की ओर घूम जाती है। इस कोण को, वस्तु का उन्नयन कोण (Angle of elevation) कहते हैं।

→ अवनमन कोण-यदि वस्तु हमारी आँख के स्तर से नीचे हो अर्थात् वस्तु हमारी आँख की क्षैतिज रेखा से नीचे हो तो हमें वस्तु को देखने के लिए अपनी आँख को नीचे की ओर झुकाना पड़ेगा। इस प्रक्रिया में हमारी दृष्टि रेखा क्षैतिज रेखा से एक कोण पर नीचे की ओर घूम जाती है। इस कोण को वस्तु का अवनमन कोण (Angle of depression) कहा जाता है।

→ सर्वसमिका-एक या अधिक चरों वाले उस समीकरण को सर्वसमिका कहा जाता है जोकि संबंधित सभी मानों के लिए संतुष्ट हो जाता है अर्थात् चरों के सभी मानों के लिए समीकरण का बायाँ पक्ष, दाएं पक्ष के समान होता है।

प्रमुख त्रिकोणमितीय कोणों के मान निम्नलिखित सारणी में दिए गए हैं-

Haryana State Board HBSE 10th Class Maths Notes Chapter 14 Statistics Notes.

Haryana Board 10th Class Maths Notes Chapter 14 Statistics

Introduction
In the earlier classes, we have already learnt about the representation of given data into ungrouped as well as grouped frequency distributions and its representation through various graphs such as bar graphs, histograms, frequency polygons etc and you have also studied about measures of central tendency such as mean, median and mode of an ungrouped data. In this chapter, we shall learn how to calculate mean, median and mode for the grouped data. We shall also discuss the concept of cumulative frequency distribution and learn how to draw cumulative frequency curves, called Ogives.

1. Statistics: The branch of mathematics in which we study to extract meaningful information from the collected data. It is the area of study dealing with the presentation, analysis and interpretation of the data. It seems to have been derived from the Latin word ‘status’ or German word ‘statistik’ or Italian word ‘statista’.
2. Data: The facts or figures, which are numerical or otherwise, collected with a definite purpose are called data. Data is the plural form of the Latin word datum.
3. Observation: Every factor figure of the data is called an observation.
4. Frequency: The number of times a particular observation occurs is called the frequency of the observation.
5. Grouped frequency distribution: If the data are very large and the range is large, we put the data in groups of suitable size and mention the frequency of each group. Such a distribution is called grouped frequency distribution.

6. An inclusive frequency distribution: The upper limit of one class does not coincide with the lower limit of the next class. Such as, 1 – 10, 11 – 20, ……… is known as an inclusive frequency distribution.
7. An exclusive frequency distribution: The upper limit of one class coincides with the lower limit of the next class such as 1 – 10, 10 – 20, ……… is known as an exclusive frequency distribution.
8. Measures of Central Tendency: The numerical expressions which represent the characteristic of a group are called Measures of Central Tendency or Average. Mean, Median and Mode are three measures of central tendency (averages).
9. Class interval: Each group into which the raw data is condensed is called a class interval. Each class is bounded by two figures, which are called the class limits. The figures on the left side of the classes are called lower limits while figures on the right are known as upper limits.
10. Class size: The difference between the true upper limit and true lower limit of a class is called its class size.
11. Class Mark: The class mark of the class interval is the value midway between its true lower limit and true upper limit.
Class mark of a class = \(\frac{\text { True upper limit + True lower limit }}{2}\)
12. Cumulative Frequency: The cumulative frequency of a class interval is the sum of frequencies of all classes up to that class (including the frequency of that particular class).
13. Mean of grouped data: We know that if x₁, x₂, x₃, ………., x_n be n observations with respective frequencies f₁, f₂, f₃, …….., f_n, their mean is given by
\(\bar{x}=\frac{f_1 x_1+f_2 x_2+f_3 x_3+\ldots \ldots+f_n x_n}{f_1+f_2+f_3+\ldots \ldots+f_n}\)
We can write this in short form
\(\bar{x}=\frac{\sum_{i=1}^n f_i x_i}{\sum_{i=1}^n f_i}\)
It is more briefly written as \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}\), It is understood that varies from 1 to n. The Greek letter ‘Σ’ (capital sigma) is particularly used for writing summations.
With this assumption we can have the following three methods to calculate the mean of grouped data.

(a) Direct Method
1. For each class, find the class mark x_i, as
\(x_i=\frac{\text { lower limit }+\text { upper limit }}{2}\)
2. Find the product of each x_i with the corresponding fi, and find the algebraic sum of these products, i.e. Σf_ix_i.
3. Find the sum of all the frequencies i.e., Σf_i
4. Calculate the value of \(\bar{x}\), using the formula.
\(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

(b) Assumed Mean Method (Shortcut Method)
In case the values of the variable are very large in magnitude ie., the values of xi are very large in magnitude then computation of the mean \(\bar{x}\) becomes rather tedious and lengthy. To make calculation easier we use assumed mean method to find \(\bar{x}\).

Here, we choose an arbitrary constant a, also called assumed mean and subtract it from each of the value x_i. The reduced value d_i = x_i – a is called the deviation of x from a.

While using this method, we go through the following steps:
1. For each class interval find class mark x_i, as x_i = \(\frac{1}{2}\)(lower limit + upper limit)
2. Assume a suitable value of x_i in the middle of x_i‘ s as the assumed mean.
3. Find out the deviations of the mid value of each from the assumed mean (d_i = x_i – a).
4. Calculate the product of deviation (d_i) with corresponding frequency (f_i) for each class.
5. Find the algebraic sum of these products ie., Σf_id_i.
6. Find the sum of all the frequencies ie., Σf_i
7. Calculate the value of \(\bar{x}\), using the formula
\(\bar{x}\) = \(a+\frac{\sum f_i d_i}{\sum f_i}\)

(c) Step Deviation Method
The shortcut method discussed is further simplified or calculations are reduced to a great extent by adopting step deviation method. Scaling down the deviation (from the assumed mean) by a step (further dividing by a common factor), will reduce the calculation to a minimum.

Here we choose an arbitrary constant a (also called assumed mean) and subtract it from each of the value x_i. The reduced value (x_i – a) is called the deviation of x_i from ‘a’. These deviations are then divided by constant h, where h is the suitable divisor of all the d_i‘s.

In this method; we go through the following steps:
1. For each class interval, calculate the class mark x_i by using the formula,
x_i = \(\frac{1}{2}\)(lower limit + upper limit)
2. Choose the assumed mean ‘a’ in the middle of x_i.
3. Calculate the values of d_i (d_i = x_i – a)
4. Calculate the values of u_i {u_i = \(\frac{x_i-a}{h}\), where h is the class width}
5. Find the product of each u_i with the corresponding f_i.
6. Find the algebraic sum of these products i.e. Σf_iu_i
7. Find the sum of all the frequencies i.e., Σf_i.
8. Calculate the mean \(\bar{x}\) by using formula.
\(\bar{x}\) = \(a+\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) h\)

Mode of Grouped Data
Recall from class IX that the mode of statistical data is the value among the observations which occurs most frequently. In other words, mode of a statistical data is the value of the observations which has maximum frequency.

In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode of grouped data is a value inside the modal class, and is given by the formula.
Mode = \(l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)
Where,
l = Lower limit of the modal class
f₁ = Frequency of the modal class
f₀ = Frequency of the class preceding the modal class
f₂ = Frequency of the class succeeding the modal class
h = Size of the class interval (assuming all class sizes to be equal)

Remark: In some cases, It is possible that more than one value may have the same maximum frequency. In such a case the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to unimodal data only i.e. the data having a single mode.

Median of Grouped Data
Median of a distribution is the value of the middle-most observation which divides it exactly in two equal parts when the data are arranged in ascending (or descending order).

(a) Median of an ungrouped data:
Arrange the data in ascending or descending order. Let the total number of observations be n
(i) If n is odd, the median is the value of the \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation.
(ii) If n is even, the median is mean of the \(\left(\frac{n}{2}\right)^{\text {th }}\) and \(\left(\frac{n}{2}+1\right)^{\text {th }}\) observations.

(b) Median of discrete series:
Arrange the terms in ascending or descending order. Then prepare a cumulative frequency distribution table. Let the total frequency be n
(i) If n is odd, then median = size of the \(\left(\frac{n+1}{2}\right)^{\text {th }}\) term.
(ii) If n is even then
median = \(\frac{1}{2}\)[size of the \(\left(\frac{n}{2}\right)^{\text {th }}\) term + size of the \(\left(\frac{n}{2}+1\right)^{\text {th }}\) term]

(c) Cumulative frequency distribution:
The frequencies are expressed as cumulative total against the class intervals in cumulative frequency distribution. It is of two types:
For example
(i) Less than type :

(ii) More than type :

(d) Median of grouped or continuous frequency distribution:
In order to calculate the median of the grouped data or continuous frequency distribution, we go through the following ahead steps:
(1) Prepare a cumulative frequency distribution and obtain n = Σf_i
(2) Find \(\frac{n}{2}\)
(3) Locate the class whose cumulative frequency is greater than (and nearest to) \(\frac{n}{2}\). This class is the median class.
(4) Calculate the median using the formula given by:
Median = \(l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h\)
Where l = lower limit of median class
n = number of observations
c_f = Cumulative frequency of class preceding the median class
f = frequency of median class
h = class size (assuming class size to be equal)

(e) Empirical relation between mean, median and mode.
3 Median = Mode + 2 Mean

Graphical Representation of Cumulative Frequency Distribution
In class IX, we have learnt about representing statistical data by using bar graphs, histograms and frequency polygons. In this section, we shall learn about to represent cumulative frequency distribution through cumulative frequency curves or ogives.

As we already know that cumulative frequency distribution are of two types, namely, less than type and more than type, accordingly there are two types of cumulatives frequency curves (or ogives).
(a) Less than ogive:
To draw a less than ogive, we go through the following steps:

• Prepare a less than cumulative frequency distribution from the given ordinary frequency distribution.
• Mark the upper class limits along the x-axis choosing a suitable scale.
• Mark the cumulative frequencies along the y-axis choosing a suitable scale.
• On joining these points successively by a free hand smooth curve, we get a cumulative frequency curve or an ogive (of less than type).

(b) More than ogive:
To draw a more than ogive, we go through the following steps:

• Prepare a more than frequency distribution from the given ordinary frequency distribution.
• Mark the lower class limits along the x-axis choosing a suitable scale.
• Mark the cumulative frequencies along the y-axis choosing a suitable scale.
• On joining these points successively by free hand smooth curve, we get a cumulative frequency curve or an ogive of more than type.

Remark 1: Draw any one of the two types of ogives for the given distribution. Take a point P(0, \(\frac{n}{2}\)) on the y-axis and draw PM || x-axis cutting the above curve at a point M. Again draw MN perpendicular to x-axis, cutting the x-axis at point N. Then median = x co-ordinate of point N.

Remark 2: Draw both types of ogives i.e., less than type and more than type for the given distribution on the same graph paper. Mark A as the point of intersection of these two ogives. Draw AP perpendicular to x-axis, cutting x-axis at P. Then median = x co-ordinate of point P.

Remark 3: For drawing ogive, it should be ensured that the class intervals are continuous.

Haryana State Board HBSE 10th Class Maths Notes Chapter 9 Some Applications of Trigonometry Notes.

Haryana Board 10th Class Maths Notes Chapter 9 Some Applications of Trigonometry

Introduction
We have studied about trigonometric ratios in earlier classes. In this chapter, we shall use the trigonometric ratios, to solve the problems regarding the heights and distances of various objects. Two common terms used in this chapter are angle of elevation and angle of depression.
1. Elevation: The height to which something is raised above a point of reference.
2. Depression: The depth to which something is lowered below a point of reference.
3. Line of sight: The line drawn from the eye of an observer to the point in the object viewed by the observer.
4. Complementary angles: Two angles having a sum of 90°, are called complementary angles.

5. Speed = Distance/Time
6. Alternate interior angle: The pair of angles on opposite sides of the transversal but inside the two lines are called alternate interior angles.
If the two lines are parallel, then the alternate interior angles formed are equal.

Heights and Distances
Angle of Elevation: If a person is looking up at an object, the acute angle measured from the horizontal level to the line of sight when the object being viewed is called the angle of elevation. Here

O is the point of observation, P is the position of the object, OP is the line of sight. OA is horizontal level and α is the angle of elevation.

Angle of Depression: The angle between the line of sight and horizontal level through the eye of the observer, when the object being viewed is below the horizontal level is called angle of depression. Here position of observer is at O. OM is the horizontal level through O. OP is the line of sight. β is the angle of depression when the object at P is observed from O.

Remarks:
(a) Numerically the angle of elevation is equal to the angle of depression.
(b) The angle of elevation and angle of depression both are measured with the horizontal.
(c) The angle of elevation or depression increases as the observer moves towards the object.

Haryana State Board HBSE 10th Class Maths Notes Chapter 8 त्रिकोणमिति का परिचय Notes.

Haryana Board 10th Class Maths Notes Chapter 8 त्रिकोणमिति का परिचय

→

→ विभिन्न त्रिकोणमितीय अनुपातों में संबंध-
cosec θ = \(\frac{1}{\sin \theta}\);
sec θ = \(\frac{1}{\cos \theta}\);
cot θ = \(\frac{1}{\tan \theta}\);
tan θ = \(\frac{\sin \theta}{\cos \theta}\);
cot θ = \(\frac{\cos \theta}{\sin \theta}\).

→ मूलभूत त्रिकोणमितीय सर्वसमिकाएँ-
मुख्य सूत्र
(i) sin² θ + cos² θ = 1
(ii) sec² θ = 1 + tan² θ
(iii) cosec² θ = 1 + cot² θ
मुख्य सूत्र से प्राप्त अन्य सूत्र
(i) sin² θ = 1 – cos² θ
cos² θ = 1 – sin² θ
(ii) sec² θ – tan² θ = 1
sec² θ – 1 = tan² θ
(iii) cosec² θ – cot² θ = 1
cosec² θ – 1 = cot² θ

→ पूरक कोणों के त्रिकोणमितीय अनुपात- समकोण ΔABC में यदि 0° ≤ θ ≤ 90°
तो,
(i) sin (90° – θ) = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = cos θ,
(ii) cos (90° – θ) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = sin θ,
(iii) tan (90° – θ) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) = cotθ,
(iv) cot (90° – θ) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = tan θ,
(v) sec (90° – θ) = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = cosec θ.
(vi) cosec (90° – θ) = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) = sec θ.

→ sin A या cos A का मान कभी भी 1 से अधिक नहीं होता, जबकि sec A या cosec A का मान सदैव 1 से अधिक या 1 के बराबर होता है।

→ विशिष्ट कोणों के त्रिकोणमितीय अनुपात-

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 7 Coordinate Geometry Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

Short/Long Answer Type Questions

Question 1.
The distance between the point A(5, -3) and B(13, m) is 10 units. Calculate the value of m.
Solution :
Here, x₁ = 5, y₁ = – 3, x₂ = 13, y₂ = m
Distance (AB) = 10
⇒ \(\sqrt{(13-5)^2+(m+3)^2}\) = 10
⇒ \(\sqrt{8^2+m^2+9+6 m}\) = 10
⇒ \(\sqrt{64+m^2+9+6 m}\) = 10
⇒ \(\sqrt{73+m^2+6 m}\) = 10
⇒ [Squaring both sides]
⇒ m² + 6m + 73 – 100 = 0
⇒ m² + 6m – 27 = 0
⇒ m² + (9m – 3m) – 27 = 0
⇒ m² + 9m – 3m – 27 = 0
⇒ m(m + 9) – 3(m + 9) = 0
⇒ (m + 9) (m – 3) = 0
⇒ m + 9 = 0 or m – 3 = 0
⇒ m = – 9 or m = 3

Question 2.
If the distances of P(x, y) from A(5, 1) and B(-1, 5) are equal, then prove that 3x
Solution :
Since, of from A(5, 1) and B(-1, 5) are equal

So, PA = PB
⇒ PA² = PB²
⇒ (5 – x)² + (1 – y)² = (-1 – x)² + (5 – y)²
⇒ 25 + x² – 10x + 1 + y² – 2y = 1 + x² + 2x + 25 + y² – 10y
⇒ 26 – 10x – 2y = 26 + 2x – 10y
⇒ – 10x – 2x = – 10y + 2y
⇒ – 12x = – 8y
⇒ 3x = 2y Hence proved.

Question 3.
It k(5, 4) is the mid point of line segment PQ and coordinates of Q are (2, 3) then find the coordinates of point P.
Solution :
Let coordinate of point P be (x, y)

Since k(5, 4) is mid point of P(x, y) and Q(2, 3)
∴ 5 = \(\frac{x+2}{2}\) and 4 = \(\frac{y+3}{2}\)
⇒ 10 = x + 2 and 8 = y + 3
⇒ x = 10 – 2 and y = 8 – 3
⇒ x = 8 and y = 5
∴ coordinates of point P are (8, 5)

Question 4.
If the midpoint of two points A(2, 5) and B(-5, y) is (-\(\frac {7}{2}\), 3). then find the distance between points A and B.
Solution :
Let point C(-\(\frac {7}{2}\), 3) is the midpoint of line AB.

∴ 3 = \(\frac{5+y}{2}\)
⇒ 6 = 5 + y
⇒ y = 6 – 5 = 1
Coordinates of point is (-5, 1)
Length of AB = \(\sqrt{(-5+2)^2+(1-5)^2}\)
= \(\sqrt{(-3)^2+(-4)^2}\)
= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5
Hence, length of AB = 5 units.

Question 5.
In a parallelogram ABCD A(3, 1), B(5, 1), C(a, b) and D(4, 3) are the vertices. Find vertex C(a, b).
Solution :
We know that diagonals of a parallelogram bisects each other so, O is the mid point of AC as well as that of BD

∴ Coordinates of mid point of AC = coordinates of mid point of BD
⇒ (\(\left(\frac{3+a}{2}, \frac{1+b}{2}\right)\)) = (\(\left(\frac{5+4}{2}, \frac{1+3}{2}\right)\))
⇒ (\(\left(\frac{3+a}{2}, \frac{1+b}{2}\right)\)) = (\(\frac {9}{2}\), \(\frac {4}{2}\))
⇒ \(\frac{3+a}{2}\) = \(\frac {9}{2}\) and \(\frac{1+b}{2}\) = \(\frac {4}{2}\)
⇒ 3 + a = 9 and l + b = 4
⇒ a = 9 – 3 and b = 4 – 1
⇒ a = 6 and b = 3
Hence, coordinates of vertex C is (6, 3)

Question 6.
If the coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect each other at (2, – 5), find the coordinates of the other two vertices.
Solution :
Diagonals of a parallelogram bisect each other.
Let coordinates of vertices C be (x₁, y₁) and d be (x₂, y₂) O is the mid point of AC as well as that of BD.

∴ coordinates of mid point of AC = (2, – 5)
⇒ (\(\frac{3+x_1}{2}, \frac{2+y_1}{2}\)) = (2, – 5)
⇒ \(\frac{3+x_1}{2}\) = 2, \(\frac{2+y_1}{2}\) = – 5
⇒ 3 + x₁ = 4, 2 + y₁ = – 10
⇒ x₁ = 1, y₁ = – 12
And coordinates of mid point of BD = (2, – 5)
⇒ (\(\frac{1+x_2}{2}, \frac{0+y_2}{2}\)) = (2, – 5)
⇒ \(\frac{1+x_2}{2}\) = 2, \(\frac{0+y_2}{2}\) = – 5
⇒ 1 + x₂ = 4, y₂ = – 10
⇒ x₂ = 3, y₂ = – 10
Coordinates of vertice Care (1, -12) and vertice D are (3, – 10).

Question 7.
If the midpoint of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y) and x + y – 10 = 0. Find the value of k.
Solution :
Since, mid point’s coordinates are C(x, y)
∴ x = \(\frac{3+k}{2}\) and y = \(\frac{4+6}{2}\) = 5
Since (x, y) lies the equation x + y – 10, so

substituting the value, A(3, 4) of x, y in this equ., we get
\(\frac{3+k}{2}\) + 5 – 10 = 0
⇒ \(\frac{3+k}{2}\) – 5=0
⇒ \(\frac{3+k-10}{2}\) = 0
⇒ k – 7 = 0
⇒ k = 7
Hence, value of k is 7.

Question 8.
The coordinate of the vertices of ΔABC are A (7, 2), B (9, 10) and C(1, 4). If E and F are the mid points of AB and AC respectvely, prove that EF = \(\frac {1}{2}\)BC
Solution :
Since, E and f are the mid points of AB and AC respectvely coordinates of points are
\(\frac{9+7}{2}, \frac{10+2}{2}\) = (8, 6)
Coordinates of points Fare
\(\frac{1+7}{2}, \frac{4+2}{2}\) = (4, 3)

Question 9.
If the point C(-1, 2) divides in ternally the time segments joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
Solution :
Hence, x₁ = 2, y₁ = 5, x₂ = x, y₂ = y, x = – 1, y = 2, m₁ = 3, m₂ = – 4

⇒ 3x + 8 = – 7 and 3y + 20 = 14
⇒ 3x = – 7 – 8 and 3y = 14 – 20
⇒ 3x = -15 and 3y = -6
⇒ x = –\(\frac {15}{3}\) and y = – \(\frac {6}{3}\)
⇒ x = – 5 and y = – 2
Hence, coordinate of point Bare (-5, -2)

Question 10.
In what ratio does the point P (\(\frac {24}{11}\), y) divide the line segment joining the points P(2, -2) and Q(3, 7) ? Also find the value y.
Solution :
Let the required ratio be k : 1
Here, x₁ = 2, y = – 2, x₂ = 3, y₂ = 7, x = \(\frac {24}{11}\), y = y
m₁ = k, m₂ = 1
By section formula, we have

Question 11.
Find the value of P, if the points A(2, 3) B(4, P) and C(6, – 3) are collinear.
Solution :
The given points are collinear So, the area of ΔABC will be zero.
Here, x₁ = 2, y₁ = 3, x₂ = 4, y₂ = P, x₃ = 6, y₃ = – 3
Area of ΔABC = 0
⇒ \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac {1}{2}\)[2 (P + 3) + 4 (-3 – 3) + 6 (3 – P)]= 0
⇒ \(\frac {1}{2}\)[2P + 6 – 24 + 18 – 6P] = 0
⇒ \(\frac {1}{2}\)[-4P + 0] = 0
⇒ – 4P = 0
⇒ P = 0

Question 12.
Find the area of triangle ABC with A(1, – 4) and the mid points of sides through A being (2, – 1) and (0, – 1).
Solution :
Let the coordinates of vertices B be (x₂, y₂) and C be (x₃, y₃)

Let E (2, – 1) and f(0, – 1) are mid points of AB and AC respectively,
∴ 2 = \(\frac{1+x_2}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 4 = 1 + x₂ and – 2 = – 4 + y₂
⇒ x₂ = 3 and y₂ = 2
∴ (x₂, y₂) = (3, 2)
Again 0 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 0 = 1 + x₃ and – 2 = – 4 + y₃
⇒ x₃ = – 1 and y₃ = 2
∴ (x₃, y₃) = (-1, 2)
Here, x₁ = 1, y₁ = – 4, x₂ = 3, y₂ = 2, x₃ = -1, y₃ = 2
Area of ABC
= \(\frac {1}{2}\)[x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac {1}{2}\)[1(2 – 2) + 3 (2 + 4) + (-1)(-4 – 2)]
= \(\frac {1}{2}\)[1 × 0 + 3 × 6 + (-1) × (- 6)]
= \(\frac {1}{2}\)[0 + 18 + 6]
= \(\frac {1}{2}\) × 24 = 12
Hence, area of ABC = 12sq. units.

Question 13.
Prove that the points (2, -2), (2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the area of this trian.
Solution :
Let the points A (2, -2), B (-2, 1) and C (5, 2) are the vertices of right angled triangle.
AB² = (-2 – 2)² + (1 + 2)²
= 16 + 9 = 25
BC² = (5 + 2)² + (2 – 1)² = 49 + 1 = 50
And AC² = (5 – 2)² + (2 + 2)²
= 9 + 16 = 25
AB² + AC² = 25 + 25 = 50 = BC²

Since, BC² = AB² + AC², So by converse of phythagores theorem ∠A = 90° there fore point A (2, -2), B (-2, 1) and C (5, 2) are the vertices of a right angled triangle
Here, x₁ = 2, y₁ = -2, x₂ = -2, y₂ = – 1, x₃ = 5, y₃ = 2
Area of ΔABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[2(1 – 2) + (-2)(2 + 2) + 5(-2 – 1)]
= \(\frac {1}{2}\)[2 × (-1) + (-2) × 4 + 5 × (-3)]
= \(\frac {1}{2}\)[- 2 – 8 – 15]
= \(\frac {1}{2}\)[-25]
= \(\frac {-25}{2}\)
Hence, area of Δ = \(\frac {25}{2}\)sq. units.

Question 14.
Find the area of that triangle whose vertices are (-3, -2), (5, -2) and (5, 4). Also prove that it is a right-angled triangle
Solution :
Let the points of vertices be A(-3, – 2), B (5, -2) and C (5, 4)
Here, x₁ = – 3, y₁ = – 2, x₂ = 5, y₂ = – 2, x₃ = 5 and y₃ = 4
Area of D ABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[(-3)(-2 – 4) + 5 (4 + 2) + 5(-2 + 2)]
= \(\frac {1}{2}\)[(-3) × (-6) + 5 × 6 + 5 × 0)]
= \(\frac {1}{2}\)[18 + 30 + 0]
= \(\frac {1}{2}\) × 48 = 24 sq. units
AB² = (5 + 3)² + (-2 + 2)² = 8² + 0² = 64
BC² = (5 – 5)² + (4 + 2)² = 0² + 6² = 36
AC² = (5 + 3)² + (4 +2)² = 8² + 6² = 64 + 36 = 100
AB² + BC² = 64 + 36 = 100 = AC²

By converse of phythagoras theorem ∠B = 90° Hence, ΔABC is a right angled triangle.

Question 15.
In a ΔABC, A is (1, – 4). If E (0, -1) and Δ(2,-1) are the mid points of AB and AC. Calculate the area of ΔABC.
Solution :
Let the coordinates of vertices B is (x₂, y₂) and C is (x₃, y₃) since, E is the mid point of AB

∴ 0 = \(\frac{1+x_2}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 0 = \(\frac{1+x_2}{2}\) and – 2 = – 4 + y₂
⇒ x₂ = -1 and y₂ = – 2 + 4 = 2
(x₂, y₂) = (-1, 2)
Since, D is the midpoint of AC
∴ 2 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 0 = 1 + x₂ and – 2 = – 4 + y₂
⇒ x₂ = -1 and y₂ = – 2 + 4 = 2
∴ (x₃, y₃) = (-1, 2)
Since, D is the midpoint of AC.
∴ 2 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_3}{2}\)
4 = 1 + x₃ and -2 = – 4 + y₃
x₃ = 3 and y₃ = 4 – 2 = 2
∴ (x₃, y₃) = (3, 2)
Here, x₁ = 1, y₁ = – 4, x₂ = – 1, y₂ = 2, x₃ = 3, y₃ = 2
Area of ΔABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[1(2 – 2) + (-1) (2 + 4) + 3(-4 – 2)]
= \(\frac {1}{2}\)[1 × 0 + (-1) × 6 + 3 × (-6)]
= \(\frac {1}{2}\)[0 – 6 – 18]
= \(\frac {1}{2}\) × (-24) = – 12
Hence, area of ΔABC = 12 sq. units.

Question 16.
Find the area of the triangle formed by joining the mid points of the sides of a triangle, whose coordinates of vertices are (0, -1), (2, 1) and (0, 3).
Solution :
Let vertices of triangle be A (0, -1), B (2, 1) and (0, 3) in which of sides AB, BC and AC respectvely.

∴ Coordinates of point
(D) = \(\frac{1+0}{2}, \frac{1-1}{2}\) = (1, 0)
Coordinates of point
E = \(\frac{2+0}{2}, \frac{1+3}{2}\) = (1, 2)
And Coordinates of point
F = \(\frac{0+0}{2}, \frac{3-1}{2}\) = (0, 1)
Here, x₁ = – 1, y₁ = 0, x₂ = 1, y₂ = 2, x₃ = 0, y₃ = 1
Area of ΔDEF
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[1 (2 – 1) + 1(1 – 0) + 0 (0 – 2)]
= \(\frac {1}{2}\)[1 × 1 + 1 × 1 + 0]
= \(\frac {1}{2}\) × (1 + 1)
= \(\frac {1}{2}\) × 2 = 1 sq. unit.
Hence, area of ΔDEF = 1sq. unit.

Fill in the Blanks

Question 1.
The point of intersection of two axes is called ……….
Solution :
origin

Question 2.
The distance of a point from the y axis is known its ………….
Solution :
abscissa

Question 3.
The ……… of a line segment divides the line segment in the ratio 1 : 1
Solution :
midpoint

Question 4.
If the area of a triangle is ……… sequare unit, then its vertices will be collinear.
Solution :
zero

Question 5.
The distance of a point from the x axis is called its …….
Solution :
ordinate

Question 6.
The distance of a point P(x, y) from the ………. is \(\sqrt{x^2+y^2}\)
Solution :
origin.

Multiple Choice Questions

Question 1.
The coordinates of the point which is reflection of point (-3, 5) in x axis are :
(a) (3, 5)
(b) (3, -5)
(c) (-3, -5)
(d) (-3, 5)

Solution :
(d) (-3, 5)

In graph paper we observe reflection of (-3, 5) in axis is (-3, -5).

Question 2.
The point P on x axis equidistant from the A(-1, 0) and B(5, 0) is:
(a) (2, 0)
(b) (0, 2)
(c) (3, 0)
(d) (2, 2)
Solution :
(a) (2, 0)

Since point P on x axis. So the coordinates of P(x, 0)

∵ Point P is equidistant from point A and point B.
∴ PA = PB
⇒ PA² = PB²
⇒ (-1 – x)² + (0 – 0)² = (5 – x)² + (0 – 0)²
⇒ 1 + x² + 2x + 0 = 25 + x² – 10x + 0
⇒ x² + 2x – x² + 10x = 25 – 1
⇒ 12x = 24
⇒ x = \(\frac {24}{12}\) = 2
Coordinates of point P(2, 0).

Question 3.
The distance between the points (acosθ + b sinθ, 0) and (0, asinθ – bcosθ), is :
(a) a² + b²
(b) a² – b²
(c) \(\sqrt{a^2+b^2}\)
(d) \(\sqrt{a^2-b^2}\)
Solution :
(c) \(\sqrt{a^2+b^2}\)

Let the given points A (a cosθ + b sinθ, 0) and (o, a sinθ – b cosθ)
The distance AB

Question 4.
If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is:
(a) 4
(b) 3
(c) 2
(d) 1
Solution :
(d) 1

The point P(6, 2) divides the line segment joining A(6, 2) and B(4, y) in the ratio 3 : 1

Here x₁ = 6, y₁ = 5, x₂ = 4, y₂ = y, m₁ = 3, m₂ = 1, x = 6, y = 2
By section formula

Question 5.
If the point P(k, 0) divides the line segment joining the points A(2, – 2) and B(-7, 4) in the ratio 1 : 2 then the value of k is :
(a) 1
(b) 2
(c) – 2
(d) – 1
Solution :
(d) – 1

Here, x₁ = 2, y₁ = – 2, x₂ = – 7, y₂ = 4, x = k, y = 0, m₁ = 1, m₂ = 2.
By section formula,

Question 6.
The value of P, for which the points A(3, 1), B(5, P) and C (7, – 5) are collinear is :
(a) – 2
(b) 2
(c) – 1
(d) 1.
Solution :
(a) – 2

the given points are collinear. So, area of ΔABC is zero.
Here, x₁ = 3, y₁ = – 1, x₂ = 5, y₂ = P, x₃ = 7, y₃ = – 5
Area of ΔABC = 0
⇒ \(\frac {1}{2}\)[x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac {1}{2}\)[3(P + 5) + 5(-5 – 1) + 7(1 – P)] = 0
⇒ \(\frac {1}{2}\)[3P + 15 + 5(-6) + 7 – 7P] = 0
⇒ \(\frac {1}{2}\)[3P + 15 – 30 + 7 – 7P] = 0
⇒ \(\frac {1}{2}\)[-4P – 8] = 0
⇒ -4P = 8
⇒ P = \(\frac {8}{- 4}\)
⇒ P = – 2

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 1 Real Numbers Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 1 Real Numbers

Short/Long Answer Type Questions

Question 1.
Show that square of any positive integer cannot be of the form (54 + 2) or (5q + 3) for any integer q.
OR
Prove that one of three consecutive positive integer is divisible by 3.
Solution :
Let be any positive integer and applying Euclid’s division Lemma it is of the form 5. For 5P + 1 or 5P + 2 or 5P + 3 or 5P + 4.
So, we have the following cases
Case I : When x = 5P
⇒ x² = 25P² = 5 (5P²)
⇒ x² = 5q [Where q = 5P²]

Case II : When
x = 5P + 1
x² = (5P + 1)²
= 25p² + 10P + 1
= 5(5P² + 2P) + 1
= 59 +1
[Where q = 5P² + 2P]

Case III : When x = 5P + 2
x² = (5P + 2)²
= 25P² + 20P + 4
= 5(5P² + 4P) + 4
= 59 + 4
(Where q = 5P² + 4P)

Case IV : When x = 5P + 3
⇒ x² = (5P + 3)²
= 25P² + 30P + 4
= 25P² + 30P + 5 + 4
= 5 (5P² + 6P + 1) + 4
= 5 + 4
(Where q = 5P² + 6P + 1)

Case V : When x = 5P + 4
⇒ x² = (5P + 4)²
= 25P² + 40P + 16
= 25P² + 40P + 15 + 1
= 5(5P² + 8P + 3) + 1
= 5q + 1
(Where q = 5P² + 8P + 3)
So, square of any positive integer cannot be of the form (5q + 2) or (5q + 3).
Or
Solution :
Let x be any positive integer. By Euclid’s division lemma x = 3q + r, where 0 < r ≤ 3
[∴ r = 0, 1, 2]
Putting r = 0, we get
x = 3q + 0 = 3q which is divisible by 3.

Putting r = 1, we get
x = 3q + 1 which is not divisible by 3.

Putting r = 2, we get
x = 3q + 2, which is not divisible by 3.
So, one of every three consecutive positive integers is divisible by 3.

Question 2.
If n is an odd integer, then show that n² – 1 is divisible by 8.
Solution :
We know that any odd positive integer x can be written in form 4q + 1 or 4q + 3
So, according to the question
Case I : When x = 4q + 1
Then, x² – 1 = (4q + 1)² – 1
= 16q² + 8q + 1 – 1
= 89 (2q + 2) …(1)
Which is divisible by 8.

Case II : When x = 4q + 3
Then, x² – 1 = (4q + 3)² – 1
= 16q² + 24q + 9 – 1
= 16q² + 24q + 8
= 8(2q² + 3 + 1) ………(2)
Which is divisible by 8. Therefore, from equations (1) and (2), it is clear that, if x is an odd positive integer. x² – 1 is divisible by 8.

Question 3.
What is the HCF of the smallest prime number and the smallest composite number,
Solution :
Smallest prime number = 2
Smallest composite number = 4
∴ 2 = 2
and 4 = 2 × 2 = 2²
HCF of (2, 4) 2.

Question 4.
Write the smallest number which is divisible by both 306 and 657.
Solution :
The required smallest number is the LCM of 306 and 657.
We have 306 = 2 × 3 × 3 × 17
= 2 × 3² × 17
And 657 = 3 × 3 × 73
= 3² × 73
LCM (306, 657) = 2 × 3² × 17 × 73
= 22938

Hence, the required smallest number = 22338

Question 5.
The length, breadth and height of a room are 8m 50cm, 6m 26cm and 4m 75cm respectively. Find the length of longest rod that can measure the dimensions of the room exactly.
Solution :
Dimensions of a room are:
Length = 8m 50cm = 850 cm.
Breadth = 6m 25cm = 625 cm
And Height = 4m 75cm = 475 cm.
The required length of longest rod is the HCF of 850 cm, 625 cm and 475 cm.
850 = 2 × 5 × 5 × 17
= 2 × 5² × 17
625 = 5 × 5 × 5 × 5 = 5⁴ And
475 = 5 × 5 × 19 = 5² × 19

HCF of (850, 625, 475) = 5² = 25 cm
Hence, required length of longest rod = 25 cm.

Question 6.
State the fundamental theorem of Arithmetic. Find the LCM of numbers 2520 and 10530 by prime factorization method.
Solution :
Fundamental Theorem of Arithmetic : Every composite number can be expressed (or factorized) as a product of primes, and this factorisation is unque, apart from the order in which the prime factors occur. Then factorisation of x can be written as x = P₁ × P₂ × P₃, × ………. P_x, Where P₁, P₂, ……..P_x are primes and written in ascending order.
So, we have
2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7
= 2³ × 3² × 5 × 7
And 10530 = 2 × 3 × 3 × 3 × 3 × 5 × 13
= 2 × 3⁴ × 5 × 13
LCM (2520, 10530) = 2³ × 3⁴ × 5 × 7 × 13
= 294840

Question 7.
Show that 3\(\sqrt{7}\) is an irrational number.
Solution :
Let us assume that 3\(\sqrt{7}\) is a rational. It can be expressed as form of \(\frac {a}{b}\),where a and b are coprime positive integers and b ≠ 0.
∴ 3\(\sqrt{7}\) = \(\frac {a}{b}\)[HCF of a and b is 1 and b ≠ 0]
⇒ \(\sqrt{7}\) = \(\frac {a}{3b}\) ……………(1)
⇒ \(\frac {a}{3b}\) = rational
[∵ a and b are positive integers]
So, from equ. (1) \(\sqrt{7}\) is rational number. But this contradicts the fact \(\sqrt{7}\) is irrational number. So, our asumption that 3\(\sqrt{7}\) is an irrational number, is wrong.
Hence, that \(\sqrt{6}\) is an irrational numbers.

Question 8.
Prove that \(\sqrt{6}\) is an irrational number.
Solution :
Let us assume that \(\sqrt{6}\) is rational. It can be express in the form of \(\frac {a}{3b}\), where a and b are coprime positive integers and b ≠ 0.
∴ \(\sqrt{6}\) = \(\frac {a}{b}\) (Where a and b are coprime ∴ HCF of a and b = 1 ……..(i)
⇒ 6 = \(\frac{a^2}{b^2}\) (Squaring both sides)
⇒ 6b² = a² ……………(i)
Therefore 6 divides a². It follows that 6 divides a.
[By theorem 1.3]
Let a = 6c and put this value in equ. (i) we get
6b² = (6c)²
⇒ 6b² = 36c²
⇒ \(\frac{36 c^2}{6}\) = b²
⇒ 6c² = b² ………..(ii)

It means b² is divisible by 6. It follows that b, is divisible by 6.
(By theorem 1.3)
From equations (i) and (ii) we say that 6 is a common factor of both a and b. But this contradicts the fact that a and b are coprime, so we have no common factor. So, our assumption that \(\sqrt{6}\) is a rational number is wrong. Therefore, \(\sqrt{6}\) is an irrational number. Proved.

Question 9.
Given that \(\sqrt{2}\) is an irrational number, then prove that (5 + 3\(\sqrt{2}\)) is an irrational number.
Solution :
Let us assume that 5 + 3\(\sqrt{2}\) is a rational number. It can be express in the form of \(\frac {a}{b}\), where a and b are coprime positive integers and b ≠ 0.
∴ 5 + 3\(\sqrt{2}\) = \(\frac {a}{b}\)
[Where HCF of a and b = 1]
⇒ 5 – \(\frac {a}{b}\) = 3\(\sqrt{2}\)
⇒ \(\frac{5 b-a}{b}\) = 3\(\sqrt{2}\)
⇒ \(\frac{5 b-a}{3}\) = \(\sqrt{2}\)
∵ a and b are positive integers
⇒ \(\frac{5 b-a}{3}\) is rational
Therefore, \(\sqrt{2}\) is rational But given that \(\sqrt{2}\) is irrational. So, our assumption that 5 + 3\(\sqrt{2}\) is rational is wrong.
Hence, 5 + 3\(\sqrt{2}\) is an irrational number.
Proved.

Question 10.
What type of decimal expansion does a rational number has? How can you distinguish it from decimal expansion of irrational numbers?
Solution :
A rational number may has its decimal expansion either terminating decimal expansion or a non-terminating repeating. But an irrational number has its decimal expansion non-repeating and non-terminating.

Question 11.
Write whether rational number \(\frac {7}{75}\) will have terminating decimal expansion or a non-terminating decimal.
Solution :
= \(\frac{7}{3 \times 5^2}\)
Since, denominator of given rational number is not form 2^m × 5ⁿ.
Hence, it is non-terminating decimal expansion

Question 12.
After how many decimal places will the decimal expansion of \(\frac{23}{2^4 \times 5^3}\) terminate?
Solution :
We have
\(\frac{23}{2^4 \times 5^3}=\frac{23 \times 5}{2^4 \times 5^3 \times 5}=\frac{23 \times 5}{2^4 \times 5^4}\)
= \(\frac{115}{(10)^4}=\frac{115}{10000}\) = 0.0115
Hence, \(\frac{23}{2^4 \times 5^3}\) will terminate after 4 decimal places.

Fill in the Blanks

Question 1.
The sum or difference of a rational and an irrational number is.
Solution :
irrational

Question 2.
Every composite number can be factorized as the product of ………….
Solution :
primes

Question 3.
The product and quotient of a …………..rational and irrational number is irrational.
Solution :
non-zero

Question 4.
………………is the least prime and…………is the least composite number.
Solution :
2, 4

Question 5.
The HCF of two co-prime numbers is always …………..
Solution :
1

Question 6.
If a and b are co-primes, then a² and b² are …………….
Solution :
co-prime.

Multiple Choice Questions

Question 1.
Euclid’s division Lemma states that for two positive integers a and b, there exists unque integers q and r satisfying a = bq + r, and :
(a) o < r < b (b) 0 > r ≤ b
(c) 0 ≤ r < b
(d) 0 ≤ r ≤ b
Solution :
(c) 0 ≤ r < b

Question 2.
The total number of factors of prime numbers is :
(a) 1
(b) 0
(c) 2
(d) 3
Solution :
(c) 2
∵ We know that prime numbers h only two factors 1 and number itself.

Question 3.
Sum of the exponents of prime factors in the prime factorization of 196 is :
(a) 3
(b) 4
(c) 5
(d) 2
Solution :
(b) 4
∵ We have 196 = 2 × 2 × 7 × 7
= 2² × 7²
It’s sum of exponents
= 2 + 2 = 4

Question 4.
The HCF and LCM of 12, 21, 15 respectvely are :
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Solution :
(c) 3, 420
We have 12 = 2 × 2 × 3 = 2² × 3
21 = 3 × 7 × 3 × 7 and
15 = 3 × 5 = 3 × 5
LCM (12, 21, 15) = 2² × 3 × 5 × 7 = 420
And HCF (12, 21, 15) = 3
So, HCF and LCM of (12, 21, 15) = 3, 420

Question 5.
Which of the following rational number.
(a) \(\frac {1}{2}\)
(b) \(\frac {1}{2}\)
(c) \(\frac{343}{2^3 \times 5^2 \times 7^3}\)
(d) \(\frac{31}{2^4 \times 3^5}\)
Solution :
(c) \(\frac{343}{2^3 \times 5^2 \times 7^3}\)

Question 6.
The decimal representation of \(\frac{11}{2^3 \times 5}\) will :
(a) terminate after 1 decimal place
(b) terminate after 2 decimal place
(c) terminate after 3 decimal place
(d) not terminate
Solution :
(c) terminate after 3 decimal place
∵ Since, \(\frac{11}{2^3 \times 5}=\frac{11}{8 \times 5}=\frac{11}{40}\) = 0.275
So, \(\frac{11}{2^3 \times 5}\) will terminate after 3 decimal places.

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 10 Circles Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 10 Circles

Short / Long Answer Type Questions

Question 1.
Two concetric circles of raddi a and b (a > b) are given. Find the length of chord of the larger circle which touches the smaller circle.
Solution :
Let two concetric circles of centre O with raddi a and b (a > b)
AO = OB = a and OM = b
OM ⊥ AB (By theorem 10.1)

In right triangle AMO, we have
AO² = AM² + OM²
(By Pythagoras theorem)
⇒ a² = AM² + b²
⇒ AM² = a² – b²
⇒ AM = \(\sqrt{a^2-b^2}\) units
we know that perpendicular drawn from centre to chord bisect the chord
∴ AM = MB
AB = 2 × AM
= 2 × \(\sqrt{a^2-b^2}\) units
Hence, length of chord of larger circle
= 2\(\sqrt{a^2-b^2}\) units

Question 2.
How many tangents can be drawn on the circle of radius 5 cm from a point lying outside the circle at distance 9 cm from the centre.
Solution :
Draw a circle of radius 5 cm with centre O. Let P be the point outside 9 cm from the centre O.

From external point P, we can draw two tangents PA and PB only.

Question 3.
In the given figure, two circles touch each other at the point C. Prove that common tangent to the circles at C, bisects the common tangents at P and Q.

Solution :
PT and CT are tangents at circle with centre A from external point T
∴ PT = CT …… (1) [by theorem 10.2]
Similarly
QT = CT ……… (2)
From equ. (1) and (2)
PT = QT
Now, PQ = PT + QT
⇒ PQ = 2 PT
⇒ PT = QT = \(\frac {1}{2}\)PQ
Hence, common tangent CT bisects the common tangents at P and Q. Hence proved

Question 4.
In the given figure ΔABC is circumscribing a circle, the length of BC is ….. cm.

Solution :
BQ = BP = 3 cm
[by theorem 10.2]
AR = AP = 4 cm
[by theorem 10.2]
Now,
CR = AC – AR
= 11 – 4 = 7
CR = QC
⇒ QC = 7 cm
BC = BQ + QC
= 3 + 7 = 10 cm

Question 5.
In the given figure, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°, then find ∠OQR.

Solution :
Join QR
PQ = PR by theorem 10.2]
⇒ ∠PQR = ∠PRQ [Angles Opp. to equal sides are equal] … (1)
In ΔPQR, we have
∠PQR + ∠PRO + ∠QPR = 180°
⇒ ∠PQR + ∠PRO + 50° = 180° [using equ. (1)]
⇒ 2∠PQR = 180° – 50°
⇒ ∠PQR = \(\frac {130°}{2}\)
⇒ ∠PQR = 65°
Now, OQ ⊥ PQ
[By theorem 10.1]
⇒ ∠PRO = 90°
⇒ ∠OQR = ∠PQO – ∠PQR
= 90° – 65° = 25°
Hence, ∠OQR = 25°

Question 6.
In the given figure PQ is a chord of length 6 cm of the circle of radius 6 cm. TP and TQ are tangents to the circle at points P and Q respectively. Find the ∠PTQ.

Solution :
Here PQ = 6 cm
PO = OQ = 6 cm
[equal raddi]
∴ PO = OQ = PQ
⇒ POQ is a equilateral triangle
∠POQ = 60°
Now, OP ⊥ PT and OQ ⊥ TQ
[By theorem 10.1]
In quadrilateral POQT, we have
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°
⇒ 60° + 90° + ∠PTQ + 90° = 360°
⇒ 240° + ∠PTQ = 360°
⇒ ∠PTQ = 360° – 240°
⇒ ∠PTQ = 120°

Question 7.
In the figure, AB and CD are common tangents to two circles of unequal raddi. Prove that AB = CD.

Solution :
Produced AB and CD to meet at P
Now, PA = PC
(By theorem 10.2) …. (1)

PB = PD
(By theorem 10.2) … (2)
Subtracting equ. (2) from equ. (1), we get
PA – PB = PC – PD
⇒ AB = CD Hence proved

Question 8.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q.

If PA = 12 cm, QC = DQ = 3 cm, then find PC + PD.
Solution :
AC = CQ
(By theorem 10.2) ……. (1)
BD = DQ
(By theorem 10.2) …….(2)
But CQ = DQ …….(3)
By equ. (1). (2) and (3), we get
AC = BD ……….(4)
Now,
AP = PB
(by theorem 10.2)
⇒ CP + AC = PD + BD
⇒ CP + AC = PD + AC
[using equ. (4)]
⇒ CP = PD
CP = AP – AC = 12 – CQ [∴ CQ = AC]
⇒ CP = 12 – 3 = 9 cm
[CQ = DQ = 3 cm]
∴ DP = CP = 9 cm
PC + PD = 9 + 9 = 18 cm

Question 9.
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ΔABC is 54 cm², then find the lengths of sides AB and AC.

Solution :
Join OA, OB and OC
Draw OE ⊥ AC and OF ⊥ AB
Area of ΔABC = 54 cm² (given)
⇒ ar (ΔBOC) + ar (ΔAOC) + ar (ΔAOB) = 54
⇒ \(\frac {1}{2}\)BC × OD + \(\frac {1}{2}\)AC × OE + \(\frac {1}{2}\)AB × OF = 54
⇒ \(\frac {1}{2}\)[BC × 3 + AC × 3 + AB × 3] = 54
⇒ \(\frac {1}{2}\)[BC + AC + AB] = 54
⇒ AB + BC + AC = \(\frac{54 \times 2}{3}\)
⇒ AB + BC + AC = 36 cm ……….(1)
AF = AE [by theorem 10.2]
Let AF = AE = x cm
BD = BF and CD = CE [by theorem 10.2]
⇒ BF = 6 cm and CE = 9 cm

Now AB + BC + AC = 36
⇒ AF + BF + BD + CD + CE + AE = 36
⇒ x + 6 + 6 + 6 + 9 + 9 + 1 = 36
⇒ 2x + 30 = 36
⇒ 2x = 6
⇒ x = 3
∴ AB = 6 + 3 = 9 cm, AC = 9 + 3 = 12 cm.

Question 10.
Prove that the parallelogram circular scribing a circle is rhombus.
Solution :
Let ABCD be a parallelogram
∴ AB = CD and AD = BC
we know that length of tangents drawn to a circle from an exterior point are equal in length.
AP = AS …(1)
PB = BQ …(2)
CR = CQ …(3)
DR = DS ……(4)

Adding equations (1), (2), (3) and (4), we get
AP + PB + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
⇒ AB+ AB = BC + BC
⇒ 2 AB = 2 BC
⇒ AB = BC
since, in a parallelogram ABCD adjacent sides AB and BC are equal.
So, ABCD is a rhombus.

Fill in the Blanks

Question 1.
Tangent is perpendicular to the …….. through the point of contact.
Solution :
radius

Question 2.
Only …….. tangents can be drawn to a circle from an external point.
Solution :
two

Question 3.
Lengths of tangents from an external point to a circle are ……..
Solution :
equal

Question 4.
The line containing the radius through the point of contact is also sometimes called the …….. to the circle at the point.
Solution :
normal

Question 5.
Circles having the same …….. are called concentric circles.
Solution :
centre

Question 6.
The word ……. to circle has been deribed from the latin word “……..”
Solution :
tangent

Question 7.
The point at which the tangent line meets the circles is called the ……..
Solution :
point of contact

Multiple Choice Questions

Choose the correct answer each of the following :

Question 1.
In the adjoining figure, if PA and PB are tangents to the circle with centre such that ∠APB = 50°, then ∠OAB is equal to :

(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution :
From figure
∠1 + ∠P = 180°

[Opposite angles of cyclic ∠1]
⇒ ∠1 + 50° = 180°
⇒ ∠1 = 180° – 50° = 130°
In ΔOAB,
OA = OB (radius)
⇒ ∠2 = ∠3 (Isosceles Δprop)
∠2 + ∠3 + ∠AOB = 180°
(Angle sum property of Δ)
⇒ ∠2 + ∠2 + 130° = 180°
⇒ 2∠2 = 180° – 130° = 50°
⇒ ∠2 = ∠OAB = \(\frac {50°}{2}\) = 25°
Hence correct choice is (a).

Question 2.
In the adjoining figure, if AP = 4 cm, CR = 5 cm and BQ = 6 cm the perimeter of ΔABC is:

(a) 15 cm
(b) 30 cm
(c) 25 cm
(d) 20 cm
Solution :
From Fig. BQ = BP = 6 cm
CQ = CR = 5 cm
PA = AR = 4 cm
The perimeter of ΔABC = BQ + CQ + CR + AR + AP + BP
= 60 + 5 + 5 + 4 + 4 + 6
= 30 cm
Hence correct choice is (b).

Question 3.
In the adjoining figure, if ∠AOB = 125°, then ∠COD is equal to :
(a) 62.5°
(b) 45°
(c) 35°
(d) 55°

Solution :
From fig. ∠AOB + ∠COD = 180°
125° + ∠COD = 180°
⇒ ∠COD = 180° – 125°
= 55°
Hence correct choice is (d).

Question 4.
In the adjoining figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to :
(a) 4 cm
(b) 2 cm
(c) 2\(\sqrt{3}\) cm
(d) 4\(\sqrt{3}\) cm

Solution :
∠OAT = 90° (Angle between radius and tangent)
Cos 30° = \(\frac {AT}{OT}\)

\(\frac{\sqrt{3}}{2}\) = \(\frac {AT}{OT}\)
⇒ AT = 2\(\sqrt{3}\)cm
Hence correct choice is (c).

Question 5.
In the given figure QR is a common tangent to the given circle, which touch externally at P. If QP = 3.8 cm, the length of QR is :
(a) 38 cm
(b) 7.6 cm
(c) 5 cm
(d) 1.9 cm

Solution :
From Fig QP = PT = PR = 3.8 cm
So, QR = QP + PR
= 3.8 + 3.8
= 7.6 cm
Hence correct choice is (b).

Question 6.
In the given figure, if ∠APO = 40°. Then ∠AOB is :
(a) 100°
(b) 80°
(c) 50°
(d) 40°

Solution :
Given ∠APO = 40°
and ∠OAP = 90°
(radius is ⊥ to tangent)
∴ In ΔPAQ
∠AOP = 180 – (40 + 90)
= 50°
∴ ∠AOB = 2 × ∠AOP
= 2 × 50°
= 100°
So correct choice is (a)

Question 7.
In the adjoining figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(c) 35°
(d) 45°

Solution :
∠DQR = 90°
(Angle between radius and tangent)
∠BQD = ∠DQR – ∠BQR
= 90° – 70°
= 20°
Similarly ∠AQD = 20
∠AQB = 20° + 20°
So correct choice is (b).

Question 8.
If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm then length of each tangent is equal to :
(a) \(\frac {3}{2}\)\(\sqrt{3}\) cm
(b) 6 cm
(c) 3 cm
(d) 3\(\sqrt{3}\) cm
Solution :
∠OBA = 90°
(Angle between radius and tangent)
In right ΔOBA
Tan 30° = \(\frac {OB}{AB}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac {3}{AB}\)

AB = 3\(\sqrt{3}\)
∴ AC = AB = 3\(\sqrt{3}\) (tangents from same external point)
So correct choice is (d).

Haryana State Board HBSE 10th Class Maths Notes Chapter 8 Introduction to Trigonometry Notes.

Haryana Board 10th Class Maths Notes Chapter 8 Introduction to Trigonometry

Introduction
In our daily life we measure the distances or heights by using some mathematical techniques which come under a branch of mathematics called trigonometry The word trigonometry is derived from the Greek words tri means three), gon (means sides), metron (means measure). Thus, trigonometry is the study of relationship between the sides and angles of a triangle Early even today trigonometrical technique used to find out the distances of the stars and planets from the Earth by astronomers. It is also used in Engineering and Physical science. In the present chapter, we shall study some ratios of the sides of a right triangle with respect to its acute angles, called trignometric ratios of the angle or briefly T-ratios. We shall also calculate the values of the trigonometric ratios for some specific angles, define them for angles 0° and 90° and establish some identities involving these ratios called trigonometric identities.

1. Trigonometry: Trigonometry is the study of relationship between the sides and angles of a triangle.
2. Trigonometric Ratios: The trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
3. Trigonometrie Identity: An equation involving trigonometrie ration of an angle is cnlled a trigonometric identity, if it is true for all values of the angles involved.
4. sin θ: sin θ is the abbreviation used for sine of angle θ.
5. cos θ: cos θ is the abbreviation used for cosine of angle θ.
6. tan θ: tan θ is the abbreviation used for tangent of angle θ.
7. cosec θ: cosec θ is the abbreviation used for cosecant of angle θ.
8. sec θ: sec θ is the abbreviation used for secant of angle θ.
9. cot θ: cot θ is the abbreviation used for cotangent of angle θ.
10. Complementary angles: Two angles having sum 90° are called complementary angles.

Trigonometric Ratios
We shall define the trigonometric ratios for an acute angle of a right angled triangle. Let us take a right triangle ABC right angled at B.
Here, ∠CAB and ∠ACB both are acute angles.
The side BC which is opposite to the ∠A we call it the opposite side to ∠A and side AB which is adjacent to ∠A is called adjacent side to ∠A and AC is the hypotenuse of right triangle.

There are six trigonometric ratios of ∠A in right triangle ABC which are defined as follows:


The ratios defined above are generally written in short form as sin A, cos A, tan A, cosec A, sec A and cot A respectively and ratios cosec A, sec A, and cot A are respectively the reciprocals of the ratios sin A, cos A and tan A.
Note: Position of the sides changes when we consider ∠C in place of ∠A as follows:

1. sine C = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)
2. consine C = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
3. tangent C = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
4. cosecant C = \(\frac{\mathrm{AC}}{\mathrm{AB}}\)
5. secant C = \(\frac{\mathrm{AC}}{\mathrm{BC}}\)
6. cotangent C = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
So, the trigonometric ratios of an acute angle in a right triangle express the relationship between the angles and lengths of its sides.

Remark:
(1) sin A is not the product of ‘sin’ and A. The symbol ‘sin’ separated from the angle A has no meaning. Similarly, cos A is not the Product of cos’and A. Similar interpretations follows for other trigonometric ratios also.
(2) Since the hypotenuse is the longest side in a right triangle, the value of sin A or cos A is always less than 1 (or in particular, equal to 1).
(3) The values of cosec A and sec A cannot be smaller than 1.

(a) Relationship between the trignometric ratios of an angle:
Consider a right triangle ABC in which ∠B = 90° and ∠A = θ
Then, tan θ = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)

Trigonometric Ratios of Some Specific Angles
In this section, we will find the values of the trigonometric ratio of 30°, 45°, 60°, 90° and 0°.
(a) Trigonometric Ratios of 30° and 60°

Consider an equilateral ΔABC with each side of length 2a. We know that each angle of an equilateral triangle is 60°. Therefore, ∠A = ∠B = ∠C = 60°. Draw AD ⊥ BC.
We know that in equilateral Δ, the perpendicular from vertex on the base, bisects the bruse and also the vertical angle.
∴ D is the midpoint of BC
⇒ BD = DC = a.
Also, AD is the bisector of ∠A,
∴ ∠BAD = ∠CAD = \(\frac{60^{\circ}}{2}\) = 30°
Now, in right ΔADB,
AB² = BD² + AD²
[By Pythagoras theorem]
(2a)² = a² + AD²
4a² = a² + AD²
AD² = 4a² – a² = 3a²
AD = \(\sqrt{3}\)a
Trigonometric Ratios of 30°
In right ΔADB, we have
Base (AD) = \(\sqrt{3}\)a, Perpendicular (BD) = a, Hypotenuse (AB) = 2a and ∠BAD = 30°.

Trigonometric Ratios of 60°
In right ΔABD, we have
Base (BD) = a, Perpendicular (AD) = \(\sqrt{3}\)a, Hypotenuse (AB) = 2a and ∠ABD = 60°.

(b) Trigonometric Ratios of 45°
Consider a right triangle ABC, right angled at B such that ∠A = 45°.
∵ ∠A + ∠B + ∠C = 180°
[sum of ∠s of a triangle]
⇒ 45° + 90° + ∠C = 180°
⇒ ∠C + 135° = 180°
⇒ ∠C = 180° – 135°
= 45°
∴ ∠A = ∠C
⇒ AB = BC

Let AB = BC = a. Then,
AC² = AB² + BC²
[By Pythagoras theorem]
AC² = a² + a²
AC² = 2a²
AC = \(\sqrt{2}\)a
Now, in right ΔABC, we have
∠A = 45°, Base (AB) = a, Perpendicular (BC) = a, Hypotenuse (AC) = \(\sqrt{2}\)a.

(c) Trigonometric Ratios of 0° and 90°
Let ∠BAC = θ be an acute angle and let P be a point on its side AB. Draw PM ⊥ AC.
It is evidence from ΔPMA, that as becomes smaller and smaller, line segment PM also becomes smaller and smaller, and finally when θ becomes 0°, the point P will coincide with M. Consequently, we have,
PM = 0, AP = AM


(Not defined)
From ΔPMA, it is evident that as θ increases, the line segment AM becomes smaller and smaller and finally when θ becomes 90°, the point M will coincide A. Consequently, we have
AM = 0, AP = PM

Table of Values of Trigonometric ratios of 0°, 30°, 45°, 60° and 90°

Trigonometric Ratios of Complementary Angles
Complementary Angles: Two angles are said to be complementary, if their sum is 90°.
Thus, θ and (90° – θ) are complementary angles.
Trigonometric Ratios of Complementary Angles:

Consider a ΔOMP such that ∠M = 90°, ∠POM = 0.
Then, ∠OPM = (90° – θ) is its o complementary angle.
For the reference angle θ, we have

For the reference angle (90° – θ), we have
Base = PM, Perpendicular = OM and Hypotenuse = OP.

From (1) and (2), we get
sin (90° – θ) = cos θ
cos (90° – θ) = sin θ
tan (90° – θ) = cot θ
cosec (90° – θ) = sec θ
sec (90° – θ) = cosec θ
cot (90° – θ) = tan θ

Trigonometric Identities
An equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angles involved.
Fundamental Trigonometric Identities
1. sin² θ + cos² θ = 1
2. sec² θ = 1 + tan² θ
3. cosec² θ = 1 + cot² θ
In right ΔABC, right angled at B, we have
BC² + AB² = AC² ……(1) [By Pythagoras theorem]

1. Dividing each term of (1) by AC², we get

These fundamental identities can be expressed in the following forms:
(i) sin² θ = 1 – cos² θ
or sin θ = \(\sqrt{\left(1-\cos ^2 \theta\right)}\)
(ii) cos² θ = 1 – sin² θ
or cos θ = \(\sqrt{\left(1-\sin ^2 \theta\right)}\)
(iii) sec² θ – tan² θ = 1
(iv) tan² θ = sec² θ – 1
or tan θ = \(\sqrt{\sec ^2 \theta-1}\)
(v) cosec² θ – cot² θ = 1
(vi) cot² θ = cosec² θ – 1
or cot θ =

Haryana State Board HBSE 10th Class Maths Notes Chapter 7 निर्देशांक ज्यामिति Notes.

Haryana Board 10th Class Maths Notes Chapter 7 निर्देशांक ज्यामिति

→ निर्देशांक अक्ष-X’OX तथा Y’OY को निर्देशांक अक्ष (Coordinate Axes) कहते हैं। X’OX को x-अक्ष, Y’OY को y-अक्ष तथा O को मूल बिंदु (Origin point) कहते हैं। चूँकि X’OX तथा Y’OY परस्पर लंब हैं, अतः X’OX तथा Y’OY को कभी-कभी समकोणिक अक्ष (Perpendicular Axes) भी कहते हैं।

→ चतुर्थांश-लंबवत् रेखाएं X’OX तथा Y’OY तल को चार भागों में विभक्त करती हैं, जिन्हें चतुर्थांश (Quadrant) कहते हैं। XOY, YOX’, X’OY’ और Y’OX को क्रमशः प्रथम, द्वितीय, तृतीय और चतुर्थ चतुर्थांश कहते हैं। हम OX, OY दिशाओं को प्रायः धनात्मक और OX’, OY’ को ऋणात्मक लेते हैं। चतुर्थांशों में स्थित बिंदुओं के निर्देशांक इस प्रकार हैं-

→ दो बिंदुओं के बीच की दूरी-दो बिंदुओं P (x₁, y₁) तथा Q(x₂, y₂) के बीच की दूरी के लिए सूत्र निम्नलिखित है-
PQ = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

→ बिंदु P(x, y) की मूल बिंदु O(0, 0) से दूरी निम्न होती है-
OP = \(\sqrt{x^2+y^2}\)

→ विभाजन सूत्र- बिंदु P, जो बिंदुओं A (x₁, y₁) और B(x₂, y₂) को मिलाने वाले रेखाखंड को m₁ : m₂ के अनुपात में आंतरिक रूप में विभाजित करता है, के निर्देशांक-
\(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)

→ बिंदुओं P (x₁, y₁) और Q (x₂, y₂) को जोड़ने वाले रेखाखंड PQ के मध्य-बिंदु के निर्देशांक \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\) होते हैं।

→ बिंदुओं (x₁, y₁), (x₂, y₂) और (x₃, y₃) से बनने वाले त्रिभुज का क्षेत्रफल व्यंजक \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₁(y₃ – y₁) + x₃(y₁ – y₂)] का संख्यात्मक मान होता है।

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 4 Quadratic Equations Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 4 Quadratic Equations

Short/Long Answer Type Questions

Question 1.
If x = 3 is one root of quadratic equation x² – 2kx – 6 = 0, then find the value of k.
Solution :
Since, x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0. So, we put x = 3 in the given equation, we get
(3)² – 2k × 3 – 6 = 0
⇒ 9 – 6k – 6 = 0
⇒ – 6k + 3 = 0
⇒ k = \(\frac {3}{6}\)
⇒ k = \(\frac {1}{2}\)

Question 2.
If one root of the quadratic equation 6x² – x – k = 0 is \(\frac {2}{3}\) then find the value of k.
Solution :
Since, \(\frac {2}{3}\) is one root of the quadratic equation 6x² – x – k = 0.
So, we put x = \(\frac {2}{3}\) in the given equation, we get
6 × (\(\frac {2}{3}\))² – \(\frac {2}{3}\) – k = 0
⇒ 6 × \(\frac{4}{9}-\frac{2}{3}\) – k = 0
⇒ \(\frac{8}{3}-\frac{2}{3}\) – k = 0
⇒ \(\frac {6}{3}\) – k = 0
⇒ 2 – k = 0
⇒ k = 2

Question 3.
Find the possible root of \(\sqrt{3 x^2+6}\) = 9.
Solution :
The given equation is
\(\sqrt{3 x^2+6}\) = 9
Squaring on both sides, we get
3x² + 6 = 81
⇒ 3x² = 81 – 6
⇒ 3x² = 75
⇒ x² = \(\frac {75}{3}\)
⇒ x² = 25
Taking square root both sides, we get
\(\sqrt{x^2}\) = \(\sqrt{25}\)
x = ± 5

Question 4.
Solve : \(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}=\frac{1}{x}\) where a + b ≠ 0.
Solution :
The given equation is :

⇒ – (a + b) × ab = (a + b)x (a + b + x)
⇒ (a + b)x (a + b + x) + ab (a + b) = 0
⇒ (a + b) [x(a + b + x) + ab] = 0
⇒ x (a + b + x) + ab = 0
⇒ ax + bx + x² + ab = 0
⇒ x² + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
⇒ (x + a) = 0 and (x + b) = 0
⇒ x = – a and x = – b

Question 5.
A train travels 300 km at a uniform speed. If the speed had been 10 km/hr more, it would have 1 hr less for same journey. Find the speed of train.
Solution :
Let the speed of train be x km/hr
Distance = 300 km (given)
Time taken by train = \(\frac {300}{x}\)hrs.
[∵ Time = Distance / Speed]
If speed had been 10 km/hr more then new speed = (x + 10) km/hr.
Time taken = \(\frac{360}{x+10}\) hrs.
According to question

Question 6.
The diagonal of a rectangular field is 25 metres more than the shorter side. If longer side is 23 metres more than the shorter side, find the sides of the field.
Solution :
Let the shorter side be x m. Then diagonal = (x + 25)m and longer side = (x + 23)m.
We known that each angle of a rectangle is 90°. In a right ΔABC, we have (x + 25)² = (x + 23)² + x².

⇒ x² + 50x + 625 = x² + 46x + 529 + x²
⇒ 2x² + 46x + 529 – x² – 50x – 625 = 0
⇒ x² – 4x – 96 = 0
⇒ x² – (12 – 8)x – 96 = 0
⇒ x² – 12x + 8x – 96 = 0
⇒ (x² – 12x) + (8x – 96) = 0
⇒ x(x – 12) + 8 (x – 12) = 0
⇒ (x – 12) (x + 8) = 0
⇒ x – 12 = 0 and x + 8 = 0
⇒ x = 12 and x = – 8 (Reject)
Hence, longer side 12 + 23 i.e., 35 m and shorter side = 12 m.

Question 7.
In a flight of 600 km, an discraft was slowed down due to bad weather. The average speed of the strip was reduced by 200 km/hr and time of flight increased by 30 minutes. Find the duration of flight.
Solution :
Distance = 600 km.
Let the speed of discraft be x km/hr. Time taken by discraft (T₁) = \(\frac {600}{x}\) hrs.
It speed of is craft was slow down by 200 km/ hr then new speed = (x – 200) km/hr.
Time taken by discraft (T₂) = \(\frac{600}{x-200}\) hrs
According to question,

⇒ 2,40,000 = x² – 200x
⇒ x² – 200x – 2,40,000 = 0
⇒ x² – (600 – 400)x – 2,40,000 = 0
⇒ x² – 600x + 400x – 2,40,000 = 0
⇒ x(x – 600) + 400(x – 600) = 0
⇒ (x – 600) (x + 400) = 0
⇒ x – 600 = 0 and x + 400 = 0
⇒ x = 600 and x = – 400 (Reject)
So, speed of discraft = 600 km/hr and duration of flight = \(\frac{600}{600-200}\) hrs
= \(\frac{600}{400}=\frac{3}{2}\) = 1\(\frac {1}{2}\)hrs.

Question 8.
The speed of a boat in still water is 18 km/hr. It takes \(\frac {1}{2}\) an hours extra in going 12 km upstream instread of going the same distance down stream. Find the speed of the stream.
Solution :
Let the speed of stream be x km/hr. Speed of boat in still water = 18 km/hr (given)
Speed of boat in down stream = (18 + x) km/hr
Time taken in going down stream (T₁) = \(\frac{12}{(18+x)}\) hrs
Speed of boat in upstream (18 – x) km/hr.
Time taken in going upstream (T₂) = \(\frac{12}{(18-x)}\) hrs
According to question,

⇒ 48x = – x² + 324
⇒ x² + 48x – 324 = 0
⇒ x² + (54 – 6)x – 324 = 0
⇒ x² + 54x – 6x – 324 = 0
⇒ x(x + 54) – 6(x + 54) = 0
⇒ (x + 54) (x – 6) = 0
⇒ x + 54 = 0 and x – 6 = 0
⇒ x = – 54 (Reject) and x = 6
Hence, speed of stream = 6 km/hr.

Question 9.
I can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately ?
Solution :
Let the time taken by larger pipe be x hours and time taken by smaller pipe be y hours
The portion of tank filled by larger pipe in 1 hour = \(\frac {1}{x}\)part
The portion of tank filled by smaller pipe in 1 hour = \(\frac {1}{y}\)part
Two pipe can fill the tank in 12 hours.
∴ The portion of tank filled by both pipe in 1 hr = \(\frac {1}{12}\)part
According to condition 1st
\(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ……………(1)
If larger pipe used for 4 hours, then the portion of tank filled by larger pipe in 4 hours = \(\frac {4}{x}\)part
And smaller pipe used for 9 hours, then. Portion of tank filled by smaller pipe in 9 hours = \(\frac {9}{y}\)part
According to condition 2nd
\(\frac{4}{x}+\frac{9}{y}=\frac{1}{2}\) ………..(2)
Let \(\frac{1}{x}\) be a and \(\frac{1}{y}\) be b, from equations (1) and (2) we get
a + b = \(\frac {1}{12}\) ………(3)
4a + 9b = \(\frac {1}{2}\) ………….(4)
By cross-multiplication, we get

Putting the values of a and b, we get
\(\frac{1}{x}=\frac{1}{20}\) and \(\frac{1}{y}=\frac{1}{30}\)
⇒ x = 20 and y = 30
Hence, time taken by larger pipe is 20 hours and smaller pipe is 30 hours.

Question 10.
Find the value of k for which are roots of the equation 3x² – 10x + k = 0 are reciprocal of each other.
Solution :
The given equation is :
3x² – 10x + k = 0
Let one root of the equation be a since roots are reciprocal each other so, other root is \(\frac {1}{α}\)
Product of two roots = \(\frac {c}{a}\)
⇒ α × \(\frac {1}{α}\) = \(\frac {k}{3}\)
⇒ 1 = \(\frac {k}{3}\)
⇒ k = 3

Question 11.
For what value of k, the given quadratic equation kx² – 6x – 1 = 0 has no real roots ?
Solution :
The given equation is :
kx² – 6x – 1 = 0
The condition for no real roots is :
D < 0
⇒ b² – 4ac < 0
⇒ (-6)² – 4 × k × -1 < 0
⇒ 36 + 4k < 0
⇒ 4k < – 36
⇒ k < –\(\frac {36}{4}\)
⇒ k < – 9
Hence, k should be less than – 9 i.e., (-10, – 11, …..).

Question 12.
For what values of k, the roots of the equation kx² + 4x + k = 0 are real ?
Solution :
The given equation is :
x² + 4x + k = 0
∵ The given equation has real roots
∴ D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ (4)² – 4 × 1 × k ≥ 0
⇒ 16 – 4k ≥ 0
⇒ – 4k ≥ – 16
⇒ k ≤ \(\frac {-16}{-4}\)
⇒ k ≤ + 4
Therefore, k should be ≤ 4 i.e., (4, 3, 2, 1, ….)

Question 13.
Find the k so that the quadratie equation (k + 1)x² – 2 (k + 1)x + 1 = 0 has equal roots.
Solution :
The given equation is :
(k + 1)x² – 2(k + 1) x + 1 = 0
Since, equation has equal roots
∴ D = 0
⇒ b² – 4ac = 0
⇒ [-2(k + 1)]² – 4 × (k + 1) × 1 = 0
⇒ 4(k² + 2k + 1) – 4k – 4 = 0
⇒ 4k² + 8k + 4 – 4k – 4 = 0
⇒ 4k² + 4k = 0
⇒ 4k(k + 1) = 0 and k + 1 = 9
⇒ k = – 1
k = – 1 (Reject since if we put the k = – 1 in the equation (-1 + 1)x will be zero
k = 0

Question 14.
If – 3 is a root of the quadratic equation 2x² + Px – 15 = 0 while the quadratic equation x² – 4Px + k = 0 has equal root. Find the value of k.
Solution :
Since, – 3 is a root of the equation
2x² + Px – 15 = 0.
So, we put x = – 3 in this equation, we get
2(-3)² + P(-3) – 15 = 0
⇒ 18 – 3P – 15 = 0
⇒ 3 – 3P = 0
⇒ P = \(\frac {-3}{-3}\) = 1
Putting the value of P in the equation, we get
x² – 4 × 1 + k = 0
⇒ x² – 4x + k = 0
The equation x² – 4x + k = 0 has equal roots.
So, D = 0
⇒ b² – 4ac = 0
⇒ (-4)² – 4 × 1 × k = 0
⇒ 16 – 4k = 0
⇒ k = \(\frac {-16}{-4}\)
⇒ k = 4

Question 15.
If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c)(x – a) = 0 are equal, then show that a = b = c.
Solution :
The given equation is :
(x – a) (x – b) + (x – b) (x – c) + (x – c)(x – a) = 0
x² – ax – bx + ab + x² – bx – xc + bc + x² – cx – ax + ac = 0
⇒ 3x² – 2ax – 2x – 2x + ab + bc + ac = 0
⇒ 3x² – 2 (a + b + c) x + (ab + bc + ac) = 0
Since, roots are equal, then
D = 0
⇒ b² – 4ac = 0
⇒ [- 2 (a + b + c)]² – 4 × 3 × (ab + bc + ac) = 0
⇒ 4(a² + b² + c² + 2ab + 2bc + 2ca) – 12 (ab + bc + ac) = 0
⇒ 4[a² + b² + c² + 2ab + 2bc + 2ca – 3ab – 3bc – 3ac] = 0
⇒ a² + b² + c² – ab – bc – ac = 0
⇒ \(\frac {1}{2}\)[2a² + 2b² + 2c² – 2ab – 2bc – 2ac] = 0
⇒ \(\frac {1}{2}\)[(a² + b² – 2ab) + (b² + c² – 2bc) + (a² + c² – 2ac)] = 0
⇒ (a – b)² + (b – c)² + (a – c)² = 0
⇒ (a – b)² = 0, (b – c)² = 0, (a – c)² = 0
⇒ a = b, b = c, a = c
∴ a = b = c
Hence Proved.

Question 16.
If the root of the quadratic equation (c² – ab)x² – 2 (a² – bc)x + b² – ac = 0 in x are equal, then show that either a = 0 or a³ + b³ + c³ = 3abc.
Solution :
The given equation is:
(c² – ab)x² – 2(a² – bc)x + (b² – ac) = 0
Since, roots are equal, then
D = 0
⇒ b² – 4ac = 0
⇒ [- 2(a² – bc)]² – 4 × (c² – ab) (b² – ac) = 0
⇒ 4(a⁴ + b²c² – 2abc) – 4(b²c² – ac³ – ab³ + a²bc) = 0
⇒ 4[a⁴ + b²c² – 2a²bc – b²c² + ac³ + ab³ – a²bc) = 0
⇒ [a⁴ – 3a²bc + ac³ + ab³] = 0
⇒ a[a³ + b³ + c³ – 3abc] = 0
⇒ a = 0 or a³ + b³ + c³ – 3abc = 0
⇒ a = 0 or a³ + b³ + c³ = 3abc
Hence proved.

Question 17.
A pole has to erected at a point on the boundary of a circular park of diameter 17 metres in such a way that the differences of its distances from two dimetrically opposite fixed gates A and B on the boundary is 7 metres. It is possible to do so? If yes, at what distances from the two gates should the pole be erected ?
Solution :
Let us first draw the diagram. Let P be location of the pole. Let the distance of pole from the gate B be x m ie., BP = x m. Now distances of the pole from the two gates = AP – BP (or BP – AP) = 7 m.
Therefore AP = (x + 7) m.
Now,
AB = 17 m.

∠APB = 90° (angle in a pamicircle is 90°) In a right triangle APB, we have
AB² = AP² + BP²
[By Pythagoaes theorem]
⇒ 17² = (x + 7)² + x²
⇒ 289 = x² + 49 + 14x + x²
⇒ 289 = 2x² + 14x + 49
⇒ 2x² + 14x + 49 – 289 = 0
⇒ x² + 7x – 120 = 0
⇒ x² + (15 – 8) x – 120 = 0
⇒ x² + 15x – 8x – 120 = 0
⇒ x(x + 15) – 8(x + 15) = 0
⇒ (x – 8) (x + 15) = 0
⇒ x – 8 = 0 or x + 15 = 0
⇒ x = 8 or x = – 15 (Reject)
Hence, the pole has to be erected on the boundary of the park at a distance of 8 m from gate Band 8 + 7 i.e., 15 m from gate A.

Fill in the Blanks

Question 1.
A quardratic equation ax² + bx + c = 0 has two………..real roots, if b² – 4ac > 0.
Solution :
Distinct

Question 2.
A quadratic equation ax² + bx + c = 0 has no ………… roots, if b² – 4ac < 0.
Solution :
Real

Question 3.
If we can factorise ax² + bx + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation ax² + bx + c = 0 can be found by equating each ………. to zero.
Solution :
Factor

Question 4.
A quadratic equation can also be solved by the method of completing the …………..
Solution :
Square

Question 5.
ar² + bx + c = 0, a ≠ 0 is called the standard form in a quadratic ………..
Solution :
Equation

Question 6.
A quadratic equation has atmost ……….. roots.
Solution :
Two.

Multiple Choice Questions

Question 1.
The root of the quadratic equation x² – 0.04 = 0 are:
(a) ± 0.2
(b) ± 0.02
(c) 0.4
(d) 2
Solution :
(a) ± 0.2

x² – 0.04 = 0
⇒ x² – (0.2)² = 0
⇒ (x + 0.2) (x – 0.2) = 0
⇒ x + 0.2 = 0
and x – 0.2 = 0
⇒ x = – 0.2
and x = 0.2
⇒ x = ± 0.2

Question 2.
If x² + 2kx + 4 = 0 has a root x = 2, then value of k is:
(a) – 1
(b) – 2
(c) 2
(d) – 4.
Solution :
(b) – 2

x² + 2kx + 4 = 0
one root x = 2 then
(2)² + 2k(2) + 4 = 0
4k + 8 = 0
⇒ k = – 2
So correct choice is (b).

Question 3.
(x² + 1)² – x² = 0 has
[NCERT Exemplar Problems]
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real roots.
Solution :
(c) no real roots

(x² + 1)² – x² = 0 get x² = y, then
(y + 1)² – y = 0
y² + 1 + 2y – y = 0
y² + y + 1 = 0
∵ b² < 4ac
So, no real root.
Hence correct choice is (c).

Question 4.
Which of the following equations has two distinct real roots ?
[NCERT Exemplar Problems]
(a) 2x² – 3\(\sqrt{2}\)x + \(\frac {9}{4}\)
(b) x² + x – 5 = 0
(c) x² + 3x + 2\(\sqrt{2}\) = 0
(d) 5x² – 3x + 1 = 0.
Solution :
(b) x² + x – 5 = 0

(a) 2x² – 3\(\sqrt{2}\)x + \(\frac {9}{4}\) = 0
D = (-3\(\sqrt{2}\))² – 4 × 2 × \(\frac {9}{4}\) = 0

(b) x² + x – 5 = 0
D = (1)² – (4) (1) (-5) = 21
D > 1
Hence correct choice is (b).

Question 5.
The roots of the equation 3x² – 4x + 3 = 0 are :
(a) real and unequal
(b) real and equal
(c) imaginary
(d) none of these.
Solution :
(c) imaginary

3x² – 4x + 3 = 0
a = 3, b = – 4, c = 3
b² = (-4)² = 16
4ac = 4 × 3 × 3 = 36
∵ b² < 4ac
So roots are imaginary Hence correct choice (c).

Haryana State Board HBSE 10th Class Maths Notes Chapter 15 प्रायिकता Notes.

Haryana Board 10th Class Maths Notes Chapter 15 प्रायिकता

→ घटना E की सैद्धांतिक (या परंपरागत) प्रायिकता P(E) को निम्नलिखित रूप में परिभाषित किया जाता है-

जहाँ हम कल्पना करते हैं कि प्रयोग के सभी परिणाम समप्रायिक हैं।

→ एक निश्चित या निर्धारित घटना की प्रायिकता 1 होती है।

→ एक असंभव घटना की प्रायिकता 0 होती है।

→ घटना E की प्रायिकता एक ऐसी संख्या P(E) है कि 0 ≤ P(E) ≤ 1

→ वह घटना जिसका केवल एक ही परिणाम हो एक प्रारंभिक घटना कहलाती है। किसी प्रयोग की सभी प्रारंभिक घटनाओं की प्रायिकता का योग 1 होता है।

→ किसी भी घटना E के लिए P(E) + P(\(\bar{E}\)) = 1 होता है, जहाँ E घटना ‘E नहीं’ को व्यक्त करता है। E और \(\bar{E}\) पूरक घटनाएँ कहलाती हैं।