KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

Students can Download Class 10 Maths Chapter 9 Polynomials Ex 9.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2; \(\frac{1}{2}\), 1, – 2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Answer:
Let P(x) = 2x3 + x2 – 5x + 2 on comparing with general polynomial P(x) = ax3 + bx2 + cx + d we get
a = 2, b = 1, c = – 5 and d = 2
Given zeroes \(\frac{1}{2}\), 1, – 2
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 1
P(1) = 2 (1)3 + (1)2 – 5(1) + 2
= 2 × 1 + 1 – 5 + 2
= 2 + 1 – 5 + 2 = 5 – 5 = 0
P(1) = 0
P(- 2) = 2(- 2)3 + (-2)2 – 5(- 2) + 2
= 2 × – 8 + 4 + 10 + 2
= – 16 + 16
P(- 2) = 0
Hence 1/2, 1 & -2 are the zeroes of the given cubic polynomial
Again consider α = 1/2, β = 1 & γ = – 2
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 2

ii) Let P(x) = x3 – 4x2 + 5x – 2 on comparing with general polynomial
P(x) = ax3 + bx2 + cx + d
we get a = 1, b = – 4, c = 5 & d = 2.
Given zeroes 2, 1, 1
P(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 4 × 4 + 10 – 2
= 8 – 16 + 10 – 2
P(2) = 18 – 18 = 0
P(2) = 0
P(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
P(1) = 6 – 6 = 0
P(1) = 0
Hence, 2, 1 and 1 are the zeroes of the given cubic polynomial.
Again, consider α = 2, β = 1 & γ = 1
α + β + γ = 2 + 1 + 1 = 4
and α + β + γ = \(\frac{-b}{a}=\frac{-(-4)}{1}\) = 4
αβ + βγ + γα = 2(1) + (1) (1) + (1) × (2)
= 2 + 1 + 2 = 5
αβ + βγ + γα = \(\frac{c}{a}=\frac{5}{l}\) = 5
αβγ = 2 × 1 × 1 = 2
αβγ = \(\frac{-d}{a}=\frac{-(-2)}{1}=\frac{2}{1}\) = 2

KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
Answer:
Let the cubic polynomial be P(x) = ax3 + bx2 + cx + d
Then sum of zeroes = \(\frac{-b}{a}\) = 2
Sum of the product of zeroes taken two at a time = \(\frac{c}{a}\) = – 7
Product of zeroes = \(\frac{-d}{a}\) = – 14
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 3
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 4

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Answer:
The given Polynomial is x3 – 3x2 + x + 1
comparing with Ax3 + Bx2 + Cx + D,
A = 1, B = – 3, C = 1 & D = 1
Let α = a – b, β = a & γ = a + b
α + β + γ = \(\frac{B}{A}=\frac{-(-3)}{1}\) = 3
a – b + a + a + b = 3
3a = 3
a = \(\frac{3}{3}\) = 1
a = 1
αβγ = \(\frac{-D}{A}\) = – 1
(a – b) a (a + b) = – 1 Put a = 1
(1 – b) 1 (1 + b) = – 1
1 – b2 = – 1
1 + 1 = b2
b2 = 2
b = ±\(\sqrt{2}\)
∴ a = 1 & b = ±\(\sqrt{2}\)

KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ±\(\sqrt{3}\). find other zeroes.
Answer:
Two zeroes of the given polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± \(\sqrt{3}\).
∴ x = 2 ± \(\sqrt{3}\).
Therefore [(x – (2 + \(\sqrt{3}\))] [x – (2 – \(\sqrt{3}\))]
= (x – 2 – \(\sqrt{3}\)) (x – 2 + \(\sqrt{3}\))
= [(x – 2) – \(\sqrt{3}\)][(x – 2) + \(\sqrt{3}\))]
= (x – 2)2 – (\(\sqrt{3}\))2
= x2 – 4x + 4 – 3 = x2 – 4x + 1
∴ It is factor of Polynomial apply division algorithm.
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 5
Other two zeroes are given by the Quotient
x2 – 2x – 35 = 0
x2 – 7x + 5x – 35 = 0
x(x – 7) + 5(x – 7) = 0
(x – 7)(x + 5) = 0
x – 7 = 0 (or) x + 5 = 0
x = 7 (or) x = – 5
∴ x = 7, – 5

Question 5.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Answer:
Let us apply the division algorithm to the given Polynomial
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 6
But remainder is given to be x + a comparing coefficients of x in the remainder obtained and remainder given:
(2K – 9)x – K(8 – K)+ 10 = x + a
∴ 2K – 9 = 1
∴ 2K = 1 + 9
K = \(\frac{10}{2}\)
K = 5 and – K(8 – K) + 10 = a
Put K = 5
– 5(8 – 5) + 10 = a
– 5(3) + 10 = a
– 15 + 10 = a
a = – 5

KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 1.
Which term of the A.P: 121, 117, 113,……, is its first negative term?
[Hint: Find n for an < 0]
Answer:
The given AP is 121, 117, 113,…….
a = 121, d = 117 – 121 = – 4
Let the nth term of the AP be the first negative term.
∴ an < 0
a + ( n – 1 )d < 0
121 + (n – 1) (- 4) < 0
121 – 4n + 4 < 0
125 < 4n
4n > 125
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 1
∴ n = 32.
Hence, 32nd term of the given AP is the first negative term.

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Q2
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 3
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 4

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 3.
A ladder has rungs 25 cm apart. (see Fig.1.7 ). The rungs decrease uniformly ¡n length form 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 \(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs \(\frac{250}{25}\) + 1].

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 5
Answer:
Distance between top and bottom rung
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 6
Distance between every two rungs
= 25 cm [given]
Number of rungs (n) = + 1 for top most rung]
= 10 + 1 = 11
∴ n = 11
Because the rungs decrease uniformly in length from 45 cm at the bottom to 25cm at the top
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 7

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: Sx-1 = S49 – Sx]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Q4
MP Board Class 10th Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.4 10
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Q4 1

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m (see Fig 5.8). Calculate the total volume of concrete required to build the terrace.
Hint: Volume of concrete required to build the first step = \(\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{m}^{3}\)
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 20
Answer:
Volume of concrete required to build the first step length × breadth × Height.
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 21
∴ Total volume of concrete required to build the terrace of 15 steps
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 22
KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

Students can Download Class 10 Maths Chapter 14 Probability Ex 14.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Maths Chapter 14 Probability Ex 14.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Answer:
Number of all possible out comes
= 5 × 5 = 25
n(S) = 25
(i) Let E1 be a event favourable to visit same day for shop [Tue, Tue; Wed, Wed; Th, Thu; Fri, Fri; Sat, Sat]
n(E1) = 5
Probability of event both visit shop same day
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 1

(ii) Let E2 be a event favourable to visit shop both on consecutive days
E2 = {T, W; W, T; W, Th; Th, W; Th, F; F, Th; F, S; S, F} n(E2) = 8
Probability of a event favourable to visit both consecutive days
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 2

(iii) Probability of both will visit shop different days = 1 – Probability both visit shop same day
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 3

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

Question 2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score in the two throws:
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 Q2
What ist the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?
Solution:
The complete table as follows
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 Q2 1
∴ Number of all possible outcomes = 36
(i) For total score being even:
Favourable outcomes = 18
[∵ The even outcomes are: 2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8,12]
∴ The required probability = \(\frac{18}{36}=\frac{1}{2}\)

(ii) For total score being 6 :
In list of scores, we have four 6’s.
∴ Favourable outcomes = 4
∴ Required probability = \(\frac{4}{36}=\frac{1}{9}\)

(iii) For toal score being at least 6:
The favourable scores are : 7, 8, 8, 6, 6, 9, 6, 6, 9, 7, 8, 8, 9, 9 and 12
∴ Number of favourable outcomes = 15
∴ Required probability = \(\frac{15}{36}=\frac{5}{12}\)

 

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Answer:
Let the number of blue balls = x.
∴ total number of balls = 5 + x.
P (drawing red ball) = \(\frac{5}{x+5}\)
P (drawing blue ball) = \(\frac{x}{x+5}\)
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 9
x = 2 × 5
x = 10
Hence the number of blue balls in a bag is 10.

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

Question 4.
A box contains 12 balls out of whichxare black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution:
∵ The total number of balls in the box = 12
∴ Number of possible outcomes = 12
Case – I: For drawing a black ball
Number of favourable outcomes = x
∴ Probability of getting a black ball = \(\frac{x}{12}\)

Case – II: When 6 more black balls are added
Then, the total number of balls = 12 + 6 = 18
⇒ Number of possible outcomes = 18
Now, the number of black balls = (x + 6)
∴ Number of favourable outcomes = (x + 6)
∴ Required probability = \(\frac{x+6}{18}\)
According to the given condition,
\(\frac{x+6}{18}=2\left(\frac{x}{12}\right)\)
⇒ 12 (x + 6) = 36x ⇒ 12x + 72 = 36x
⇒ 36x – 12x = 72 ⇒ 24x = 72
⇒ x = \(\frac{72}{24}\) = 3
Thus, the required value of x is 3.

Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\) Find the number of blue balls in the jar.
Answer:
Total number of marbles in the jar = 24
Let the number of blue marbles is x and the number of green marbles is 24 – x
Probability of marble drawn is green
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 13
24 – x = 16
x = 24 – 16
x = 8
∴ Number of blue marble is 8.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 14 Probability Ex 14.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 14 Probability Exercise 14.2, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is tw ice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Answer:
Let the ages of Ani and Biju be ‘x’ years and ‘y’ years respectively.
x – y = 3 (x > y) (or) y – x = 3 (y > x) → (1)
Age of Ani’s father Dharm = 2x years
Age of Biju’s sister = \(\frac{y}{2}\) years.
According to the Question
2x – \(\frac{y}{2}\) = 30
4x – y = 60 → (2)
case (i) when x – y = 3
x – y= 3 → (1)
4x – y = 60 → (2)
Subtract equation (1) and (2)

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 1
x = 19 years
Put x = 19 in equation (1)
x – y = 3
19 – y = 3
– y = 3 – 19
y = 16
Ani’s age = x = 19 years.
Biju’s age = y = 16 years.
case (ii) y – x = 3
– x + y = 3 → (1)
4x – y = 60 → (2)
Adding equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 2
x = 21
Put x = 21 in equation (1)
– 21 + y = 3
y = 3 + 21
y = 24
Ani’s age is 21 years and Biju age is 24 years.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]
Solution:
Let the capital of 1st friend = ₹ x,
and the capital of 2nd friend = ₹ y
According to the condition,
x + 100 = 2(y -100)
⇒ x + 100 – 2y + 200 = 0
⇒ x – 2y + 300 = 0 … (1)
Also, 6(x – 10) = y + 10 ⇒ 6x – y – 70 = 0 …. (2)
From (1), x = -300 + 2y (3)
Substituting the value of x in equation (2), we get
6[-300 + 2y] – y – 70 = 0
⇒ -1870 + 11y = 0
\(\Rightarrow y=\frac{1870}{11}=170\)
Now, Substituting the value of y in equation (3), we get, x = – 300 + 2y
= – 300 + 2(170) = – 300 + 340 = 40
Thus, 1st friend has ₹ 40 and the 2nd friend has ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Answer:
Let the actual speed of the train be x km/hr and the actual time taken be ‘y’ hours.
Distance = speed x time = x × y
= xy km
xy = (x + 10) (y – 2)
xy = xy – 2x + 10y – 20
2x- 10y = – 20
xy = (x – 10) (y + 3)
xy = xy + 3x – 10y – 30
3x – 10y = 30 → (2)
Subtract equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 4
Put x = 50 in equation (1)
2(50) – 10y = – 20
– 10y = – 20 – 100
– 10y = – 120
y = \(\frac{-120}{-10}\)
y = 12
distance covered by train = xy
= 50 × 12 = 600 km.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Answer:
Let the number of students in the class be x and the number of rows be y. Then the number of students in each row \(\frac{x}{y}\) .
If 3 students are extra in row, then there would be 1 row less i,e when each row
hasKSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 5
Students then the number of rows is (y – 1)
∴ Total number of students
= number of rows x number of students in each row
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 6

and, if 3 students are less in a row, then there would be 2 rows more i.e, when each row has \(\left(\frac{x}{y}-3\right)\) students, then the number of rows is (y + 2).
∴ Total number of students = Number of rows x number of students in each row.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 7
Then equation (1) and (2) can be written as
a – 3y = – 3 → (3)
2a – 3y = 6 → (4)
Subtract equation (3) and (4)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 8
a = 9
Put a = 9 in equation (3)
9 – 3y = – 3
– 3y = – 3 – 9
– 3 y = – 12
x = 9 × 4
x = 36
Hence number of students in class is 36 and number of rows is 4.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Sum of angles of a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° … (1)
∵ ∠C = 3∠B = 2(∠A + ∠B) … (2)
From (1) and (2), we have ∠A + ∠B + 2 (∠A + ∠B) = 180°
⇒ ∠A + ∠B + 2∠A + 2∠B = 180°
⇒ ∠A + ∠B = 60° …. (3)
Also, ∠A + ∠B + 3∠B = 180°
⇒ ∠A + 4∠B = 180° ….(4)
Subtracting (3) from (4), we get
∠A + 4∠B – ∠A – ∠B = 180°- 60°
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 8
Substituting ∠B = 40° in (4) we get,
∠A + 4(40°) = 180°
⇒ ∠A = 180° – 160° = 20°
∴ ∠C = 3∠B = 3 × 40° = 120°
Thus, ∠A = 20°, ∠B = 40° and ∠C = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
Answer:
5x – y = 5 → (1)
3x – y = 3 → (2)
From equation (1)
y = 5x – 5
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 11
From equation (2)
y = 3x – 3
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 12
In the graph, we observe that the coordinates of the vertices’s of the triangle ABC formed by the two lines represented, by the A(0, -3) and B(0, -5).

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q
Answer:
px + qy = p – q → (1)
qx-py = p + q → (2)
Multiply equation (1) p and equation (2) by q
p2x + pqy = p2 – pq → (3)
q2x – pqy = pq + q2 → (4)
Adding equation (3) and (4)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 13
Put x = 1 in equation (1).
p(1) + qy = p – q
qy = p – q – p
qy = – q
y = \(\frac{-q}{q}\)
y = – 1
Hence, the solution of the given pair of linear equation is x = 1, y = – 1

(ii) ax + by – c
bx + ay = 1 + c
Answer:
ax + by = c
bx + ay = 1 + c
then ax + by – c = 0 → (1)
bx + ay – (1 + c) = 0 → (2)
Solve by cross multiplication method.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 14
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 15
Hence, the solution of the given pair of linear equation is
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 16

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 17
ax + by = a2 + b2
Answer:
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 18
Hence, the solution of the given pair of equations x = a, y = b.

(iv) (a – b)x + (a + b)y = a2 – 2ab – b2
(a + b) (x + y) = a2 + b2
Answer:
(a – b)x + (a + b)y = a2 – 2ab – b2 → (1)
(a + b) (x + y) = a2 + b2
(a + b)x + (a + b)y = a2 + b2 → (2)
Subtract equation (1) and (2),
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 19
(a – b)x – (a + b)x = a2 – 2ab – b2 – a2 – b2 ax – bx – ax – bx = – 2ab – 2b2.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 20
Put x = a + b in equation (1)
(a – b) (a + b) + (a + b)y = a2 – 2ab – b2
a2 – b2 + (a + b)y = a2 – 2ab – b2
(a + b)y = a2 – 2ab – b2 – a2 + b2
(a + b)y = – 2ab
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 21
Hence, solution of the pair of linear equation is x = a + b,
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 22
(v) 152x – 378y = – 74
– 378x + 152y = – 604
Answer:
152x – 378y = – 74 → (1)
– 378x + 152y = – 604 → (2)
Adding equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 23
226x – 226y = – 678 divide by – 226
x + y = 3 → (3)
Subtract equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 24
Put x = 2 in equation (3)
2 + y = 3
y = 3 – 2
y = 1
Hence, the solution of the given pair of linear equations is x = 2, y = 1

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 1
Solution:
ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180° and
∠B + ∠D = 180°
⇒ [4y + 20] + [- 4x] = 180°
⇒ 4y – 4x + 20° -180° = 0
⇒ 4y – 4x – 160° = 0
⇒ y – x – 40° = 0 … (1)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 2

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Students can Download Class 10 Maths Chapter 2 Triangles Ex 2.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 1.
In Fig. 2.56, PS is the bisector of ∠QPR of ∆ PQR, Prove that \(\frac{\mathbf{Q S}}{\mathbf{S R}}=\frac{\mathbf{P Q}}{\mathbf{P R}}\)
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 1
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 2
Given: In the figure, PS is the bisector of

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 3

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 2.
In Fig. 2.57, D is a point on hypotenuse AD of ∆ ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that:
i) DM2 = DN.MC ii) DN2 = DM.AN
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 4

Answer:
Given: D is a point on hypotenuse AC of ∆ ABC DM ⊥ BC, DN ⊥AB and BD ⊥ AC
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 5

ii) Consider ∆le ADN And DBN
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 6

Question 3.
In the figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 – AB2 + BC2 + 2BC.BD
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 7
Solution:
∆ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
∵ In ∆ADB, ∠D = 90°
∴ Using Pythgoras Theorem, we have
AB2 = AD2 + DB2 ….. (1)
In right ∆ADC, ∠D = 90°
∴ Using Pythagoras Theorem, we have
AC2 = AD2 + DC2
= AD2 + [BD + BC]2
= AD2 + [BD2 + BC2 + 2BD.BC]
⇒ AC2 = [AD2 + DB2] + BC2 + 2BC – BD
⇒ AC2 = AB2 + BC2 + 2BC – BD [From (1)] Thus, AC2 = AB2 + BC2 + 2 BC.BD

Question 4.
In Fig. 2.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC. BD.
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 8
Answer:
Given: In figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC
To prove: AC2 = AB2 + BC2 – 2BC.BD
Proof: In right triangle ABC ∠ D = 90°
AC2 = AD2 + DC2 [By pythagoras theorem]
= AD2 + (BC – BD)2 [CD = BC – BD]
= AD2 ± BC2 + BD2 – 2BC. BD
AC2 = AB2 + BC2 – 2BC.BD
[In right angle ADB with
∠D = 90° AB2 = AD2 + BD2 By Pythagoras theorem]

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 5.
In the figure, AD is a median of triangle ABC and AM ⊥ BC. Prove that
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6 10
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 Q5
Solution:
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6 12
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6 13
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6 14

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 13

Given: In parallelogram ABCD
AB = CD & AD = BC
Construction: convert parallelogram into a rectangle and Draw AG ⊥ CD
To prove:
AC2 + BD2 = AB2 + BC22 + CD2 – AD2
Proof: consider ∆le BDF ∠D = 90°
BD2 = BF2 + FD2 = h2 + (x+d)2 — (1)
Consider ∆le AGC ∠G = 90°
AC2 = AG22 + GC2 = h2 + (x – d)2 — (2)
Adding (I) and (2)
AC2+ BD2 = h2 + (x + d)2 + h2 + (x – d)2
2h2 + x2 + 2xd + d2 + x2 – 2xd + d2
2h2 + 2x2 + 2d2
= 2x2 + 2(h2 + d2)
= 2x2 + 2y2
= x2 + y2 + x2 + AC2 + BD2 =AB2 +BC2 i-CD2 + AD2.

Question 7.
In the figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC – ∆DPB
(ii) AP.PB = CP.DP
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 Q7
Solution:
We have two chords AB and CD of a circle. AB and CD intersect at P.
(i) In ∆APC and ∆DPB,
∴ ∠APC = ∠DPB ….. (1)
[Vertically opp. angles]
∠CAP = ∠BDP …… (2)
[Angles in the same segment]
From (1) and (2) and using AA similarity, we have
∆APC ~ ∆DPB

(ii) Since, ∆APC ~ ∆DPB [As proved above]
∴ Their corresponding sides are proportional,
\(\Rightarrow \frac{A P}{D P}=\frac{C P}{B P}\)
⇒ AP.BP = CP.DP, which is the required relation.

Question 8.
In Fig. 2.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 16
Answer:
Given: In figure, two chords AB and CD of a circle intersect each other at the point P. (when produced) outside the circle.
To prove: i) ∆ PAC ~ ∆ PDB
ii )PA. PB = PC . PD
Proof: i) we know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
Therefore, for cyclic quadrilateral ABCD.
consider ABCD ∆ PAC and ∆ PBD.
∠PAC = ∠PDB → (i)
∠PCA = ∠PBD → (2)
∆ PAC ~ ∆ PDB [A A similarity criterion]

ii) A PAC ~ ∆ PDB (Proved)
\(\frac{P A}{P D}=\frac{P C}{P B}\) [Corresponding sides of the similar ∆le are proportional]
PA . PB = PC . PD

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 9.
In the figure, D is a point on side BC of ∆ABC such that \(\frac{B D}{C D}=\frac{A B}{A C}\). Prove that AD is the bisector of ∠BAC.
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 Q9
Solution:
Let us produce BA to E such that AE = AC
Join EC.
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 Q9 1
MP Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.6 24

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod ¡s 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away nad 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 2.64)?
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 22
If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 23
AC = 3m
Hence, she has 3 m string out.
Length of the string pulled in 12 seconds
at the rate of5 cm/ sec 5 × 12cm = 60 cm = O.6 m.
∴ Length of remaining string left out
= AD = 3.0 – 0.6 = 2.4m
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 24

In right angled ∆ ABD ∠B = 90°
AD2 = AB2 + BD2
BD2 = AD2 – AB2
[pythagoras theorem]
= (2.4)2 – (1.8)2
= 5.76 – 3.24 = 2.52
BD = 12.52 = 1.59 m (approx)
Hence, the horizontal distance of the fly form Nazirna after 12 seconds = 1.2 + 1.59 = 2.79 m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.6 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.6, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 1.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
Answer:
Cylinder
Diameter = 10 cm,
radius = r = \(\frac{d}{2}=\frac{10}{2}\) = 5cm
Length of wire one round of cylinder = circumference of base
= 2πr = 2π(5) = 10π cm
Cylindrical wire diameter = 3mm
= \(\frac{3}{10}\) cm
length of cylinder = h = 12 cm
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 1
Total round = 40
Now, length of wire required in one round = circumference of base of cylinder
2πr = 2π × \(\frac{10}{2}\)
2πr = 10π cm
Hence in 40round the length of wire required
= 40 × 10π
= 400π cm
= \(\frac{400 \times 22}{7}\) = 1257.14 cm
= 12.57 m.
therefore total length of wire 12.57 m
volume of wire = πr2h
= π × \(\left(\frac{0.3}{2}\right)^{2}\) × 1257.14
88.898cm3
The mass of wire of 1 cm3 = 8.885
∴ the mass of wire of 88.898 cm3
= 8.88 × 88.898
= 798.413
Hence mass of wire = 798.419 and length of wire = 12.57 m

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 2.
A right triangle, whose sides are 3cm and 4cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate).
Solution:
Let us consider the right ABAC, right angled at A such that AB = 3 cm, AC = 4 cm
∴ Hypotenuse BC = \(\sqrt{3^{2}+4^{2}}\) = 5cm
Obviously, we have obtained two cones on the same base AA’ such that radius = DA or DA’.
Now,
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 Q2
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 3

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 3.
A cistern, internally measuring 150 cm × 120 cm × no cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
Answer:
Volume of cistern =150 × 120 × 110
= 1980000 cm3
volume of water in cistern
= 129600 cm3
volume to filled in cistern
= 1980000 – 129600
= 1850400 cm3
volume of one brick = 22.5 × 7.5 × 6.5
= 1096.875 cm3
Let ‘N’ no of bricks can be placed into the cistern
Hence volume of ‘N’ no of bricks
= N × 1096.875 cm3
One bricks absorb water \(\frac{1}{17}\) of his volume hence water absorb bv 1 brick
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 6
Hence total N no. of brick absorb water
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 7
Volume to be filled in tank + water absorbs by N no. of bricks = volume of N No of bricks.
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 8
∴ therefore No. of bricks can be put in cistern without over flowing water.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 4.
In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall was approximately equivalent to the addition to the norma water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution:
Volume of three rivers = 3 {(Surface area of a river) × Depth}
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 Q4
Since 0.7236 km3 ≠ 9.728 km3
∴ The additional water in the three rivers is not equivalent to the rainfall.

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 5.
An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see Fig. 15.25).
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 12
Answer:
r1 = \(\frac{18}{2}\) = 9cm , r2 = \(\frac{8}{2}\) = 4 cm
Height h of frustum = 22 – 10 = 12 cm.
Slant height of the frustum cone =
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 13
l = 13cm
radius of cylindrical portion = \(\frac{8}{2}\)
= 4 cm.
Area of tin sheet required = CSA of cylinder + CSA of frustum portion
= 2πrh + π (r1 + r2) l
= 2π(4)(10) + π(9 + 4) 13
= 80π + π × 13 × 13 = 80π + 169π
= 249π
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 14

KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5

Question 6.
Derive the formula for the volume of the frustum of a cone.
Solution:
We have,
[Volume of the frustum RPQS] = [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 9
Since, ∆OC1Q ~ ∆OC2S [By AA similarity]
KSEEB Solutions for Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.5 Q6
From (1) and (2), we have
Volume of the frustum RPQS
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 11

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Answer:
Let P(x1, y1) be common point of both lines and divide the line segment joining A(2, – 2) and B (3, 7) in ratio K : 1
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 1

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
Let the given points be A(x, y), B( 1, 2) and C(7, 0) are collinear.
The points A, B and C will be collinear if area of ∆ABC = 0
⇒ \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
or 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0 or x + 3y – 7 = 0, which is the required relation between x and y.

Question 3.
Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Answer:
Let A → (6,- 6), B → (3,- 7) and C → (3, 3)
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 3
Squaring
(x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2
x2 – 12x + 36 +y2 +12y + 36
= x2 – 6x + 9 + y2 +14y + 49
x2 – x2 + y2 – y2 – 12x +6x + 12y – 14y + 72 – 58 = 0
– 6x – 2y = – 14 devide by – 2
3x + y = 7 → (1)
(x – 3)2 +(y + 7)2 = (x – 3)2 +(y – 3)2 (y + 7)2 = (y – 3)2
y2 +49 + 14y = y2 + 9 – 6y
y2 – y2 + 14y + 6y = 9 – 49
20y = – 40
y = – 2
Putting y = – 2 in eqn (1) 3x + y = 7
3x – 2 = 7
3x = 7 + 2
3x = 9
x = \(\frac{9}{3}\) = 3
x = 3
Thus I(x, y) → (3, – 2)
Hence, the centre of a circle is (3, – 2)

Question 4.
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
Solution:
Let we have a square ABCD such that A(-1, 2) and C(3, 2) are the opposite vertices. Let B(x, y) be an unknown vertex.
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Q4
Since, all sides of a square are equal.
∴ AB = BC ⇒ AB2 = BC22
⇒ (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2
⇒ x2 + 2x + 1 + y2 – 4y + 4
⇒ x2 – 6x + 9 + y2 – 4y + 4
⇒ 2x + 1 = -6x + 9
⇒ 8x = 8 ⇒ x = 1  …………………. (1)
Since, each angle of a square = 90°.
∴ ∆ABC is a right angled triangle.
∴ Using Pythagoras theorem, we have AB2 + BC2 = AC2
⇒ (x + 1)2 + (y – 2)2] + [(x – 3)2 + (y – 2)2]
= [(3 + 1)2 + (2 – 2)2]
⇒ [x2 + 2x + 1 + y2 – 4y + 4] + [x2 – 6x + 9 + y2 – 4y + 4]
= [42 + 02]
⇒ 2x2 + 2y2 + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
⇒ 2x2 + 2y2 – 4x – 8y + 2 = 0
⇒ x2 + y2 – 2x – 4y + 1 = 0 …………….. (2)
Substituting the value of x from (1) into (2), we have
1 + y2 – 2 – 4y + 1 = 0
⇒ y2 – 4y + 2 – 2 = 0
⇒ y2 – y = 0
⇒ y(y – 4) = 0
⇒ y = 0 or y = 4
Hence, the required other two vertices are (1, 0) and (1, 4).

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the, boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the Fig.7.4 The students are to sow seeds of flowering plants on the remaining area of the plot.
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 7
(i) Taking A as origin,find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?
Answer:
(i) Taking A as origin then AD is x – axis and AB is y – axis.
Co-ordinates of P, Q and R are
P → (4, 6), Q → (3, 2) & R → (6, 5)
area of ∆ PQR
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 8
area of ∆ PQR
= \(\frac{1}{2}\)[4(2 – 5) + 3(5 – 6) + (6 – 2)]
= \(\frac{1}{2}\)[4 × – 3 + 3 × – 1 + 6 × 4]
= \(\frac{1}{2}\)[- 12 – 3 + 24]
area of ∆ PQR
= \(\frac{1}{2}\)[- 15 + 24] = 9/2 sq units

(ii) Taking C as origin then CB is x – axis & CD is y – axis.
P(x1, y1) = (12, 2), Q(X2, y2) = (13, 6), and R(x3, y3) = (10, 3),
x1 = 12, y1 = 2, x2 = 13, y2 = 6, x3 = 10 & y3 = 3
area of ∆ PQR
= \(\frac{1}{2}\)[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= \(\frac{1}{2}\)[12 x 3 + 13 x 1 + 10 x – 4]
= \(\frac{1}{2}\)[36 + 13 – 40]
= \(\frac{1}{2}\)[49 – 40] = 9/2 Sq units
Hence we observed that area of ∆ remains same in both case.

Question 6.
The vertices of a ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\) Calculate the area of the ∆ADE and compare it with the area of ∆ABC. [Recall “The converse of basis proportionality theorem”, and “theorem of similar triangles taking their areas and corresponding sides”]
Solution:
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 3
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Q6
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 5

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 7.
Let A (4,2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2:1.
(iv) What do yo observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ ABC, find the coordinates of the centroid of the triangle.
Answer:
(i)
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 13
median AD of the triangle ABC divide the side BC into two equal parts.
Therefore D is the mid – point of side BC Co-ordinates of mid Point
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 14

(ii) Form equation AP : PD = 2 : 1
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 15
Section Formula
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 16

(iii) BQ : QE = 2 : 1
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 17
co-ordinates of
\(\frac{A D}{A B}=\frac{A E}{E C}=\frac{1}{4}\)

(iv) The co-ordinates P, Q, & R are same (11/3, 11/3) All these points represents the same point which is called centriod.

(v) Point 0 is the centroid and AD is the midian. D is the mid point of BC and point 0 divide the AD into 2 : 1
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 19

Question 8.
ABCD is a rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
We have a rectangle whose vertices are A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1).
∵ P is mid-point of AB
∴ Coordinates of P are
KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Q8
MP Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 12
We see that PQ = QR = RS = SP i.e., all sides of PQRS are equal.
∴ It can be a square or a rhombus.
But PR ≠ QS i.e., its diagonals are not equal.
∴ PQRS is a rhombus.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.4, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.3

Question 1.
Evaluate :
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 Q 1
iii) cos 48° – sin 42°
iv) cosec 31° – sec 59°
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 Q 1.1

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 2.
Show that
i) tan 48° tan 23° tan 42° tan 67° = 1
ii) cos 38° cos 52° – sin 38° sin 52° = 0.
Solution:
i) tan 48° tan 23° tan 42° tan 67° = 1
LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° × tan 23° × tan (90 – 48).tan (90 – 23)
= tan 48° × tan 23° × cot 48° × cot 23°
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 Q 2
= 1
∴ LHS = RHS

ii) cos 38° cos 52° – sin 38° sin 52° = 0
cos 38°. cos 52 – sin 38° sin 52
= cos 38° . cos52 – sin(90 – 52°) sin (90 – 38°)
= cos38° . cos52 – cos52 cos32°
= 0

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
Since, tan 1A = cot (A – 18°)
Also, tan (2A) = cot (90° – 2A) [∵ tan θ = cot (90° – θ)]
∴ A – 18° = 90° – 2A
⇒ A + 2A = 90° + 18°
⇒ 3A = 108° ⇒ A = \(\frac{108^{\circ}}{3}\) = 36°

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B
tan A = tan (90 – B)
A = 90 – B
∴ A + B = 90°.

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A = cosec (A – 20°)
Also, sec 4A = cosec (90° – 4A) [ ∵ cosec (90° – θ) = sec θ]
∴ A – 20° = 90° – 4A
⇒ A + 4A = 90° + 20°
⇒ 5A = 110° ⇒ A = \(\frac{110^{\circ}}{5}\) = 22°

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 6.
If A, B, and C are interior angles of a triangle ABC, then show that
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 Q 6
Solution:
A + B + C = 180°
B + C = 180 – A
KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 Q 6.1

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
Since, sin 67° = sin (90° – 23°) = cos 23° [ ∵ sin (90° – θ ) = cos θ]
Also, cos 75° = cos (90° – 15°) = sin 15° [∵ cos (90° – θ) = sin θ]
∴ sin 67° + cos 75° = cos 23° + sin 15°

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.3, drop a comment below and we will get back to you at the earliest.

KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Students can Download Chapter 8 Comparing Quantities Ex 8.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 1.
Convert the given fractional numbers to percents.
a) \(\frac{1}{8}\)
Solution:
Percent means per hundred. So multi-plied by 100.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 1

b) \(\frac{5}{4}\)
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 2

c) \(\frac{3}{40}\)
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 3

d) \(\frac{2}{7}\)
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 4

KSEEB Solutions

Question 2.
Convert the given decimal fractions to per cents.
a) 0.65
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 5

b) 2.1
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 6

c) 0.02
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 7

d) 12.35
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 8

KSEEB Solutions

Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 9
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 10
∴ In the first figure 25% is coloured.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 11
∴ in the second figur e 50% Is coloured,.
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 111
∴ In the third figure 37.5 % is coloured.

KSEEB Solutions

Question 4.
Find :
a) 15% of 250
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 12

b) 1% of 1 hour
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 13

c) 20% of ₹ 2500
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 14

d) 75% of ₹ 1 kg
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 15

KSEEB Solutions

Question 5.
Find the whole quantity if
a) 5% of it is 600.
Solution:
Let the whole quantity be ‘m’
∴ 5% of m = 600
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 16
∴ The whole quantity is 12,000.

b) 12% of its 1080.
Solution:
Let the whole quantity be ‘m’
∴ 12% of m = Rs. 1080
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 17
∴ The whole amount be Rs. 9.000

c) 40 % of it is 500 km
Solution:
Let the whole quantity be ‘m’
∴ 40% of m = 500 km
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 18
∴ The whole amount is 1250 kms

KSEEB Solutions

d) 70% of it is 14 minutes
Let the whole quantity be ‘m’
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 19
∴ The whole amount is 20 minutes

e) 8% of it is 40 litres.
Let the whole quantity be ‘m’
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 20
∴ The whole amount be 500 liters

KSEEB Solutions

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
Solution:
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 21

Question 7.
In a city, 30% are females, 40% are males and the remaining are children. What percent are children?
Solution:
females = 30%
males = 40%
children = 100 – (30 + 40)
remaining are children = 100 – 70 = 30%
∴ Percentage of children = 30%

KSEEB Solutions

Question 8.
Out of 15,000 votes in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Total no. of voters in a constituency } = 15,000
Percentage of voted = 60%
∴ Percentage of voters who did not vote
= 100 – 60 = 40%
Actual number of voters who did not vote} = 40% of 15,000
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 80
6000 voters who did not vote.

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let Meeta’s salary be ‘M’
10% of M is = Rs. 400
∴ M =?
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 81
∴ Her salary be Rs. 4,000/-

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Let the total matches won by them be M 25% of 20 = M
KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2 82
∴ They win 5 matches.

KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Students can Download Chapter 12 Algebraic Expressions Ex 12.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits framed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.
KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 1
Solution:
i) The number of segments required to form
5 digits of this kind
5n + 1 = 5 × 5 + 1 = 25 + 1 = 26

ii) The number of segments required to form 10 digits of this kind
5n + 1 = 5 × 10 + 1 = 50 + 1 = 51

iii) The number of segments required to form 100 digits of this kind
5n + 1 = 5 × 100 + 1 = 501

KSEEB Solutions

KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 2
Let the number of digits formed be ‘n’.
Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 3n + 1.
i) The number of segments required to form 5 digits of this kind.
3n + 1 = 3 × 5 + 1 = 15 + 1 = 16

ii) The number of segments required to form 10 digits of this kind
3n + 1 = 3 × 10 + 1 = 30 + 1 = 31

iii) The number of segments required to form 100 digits of this kind
3n + 1 = 3 × 100 + 1 = 300 + 1 = 301

KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 50
Let the number of digits formed be ‘n’. Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 5n + 2.
i) The number of segments required to form 5 digits of this kind.
5n + 2 = 5 × 5 + 2 = 25 + 2 = 27
ii) The number of segments required to form 10 digits of this kind
5n + 2 = 5 × 10 + 2 = 50 + 2 = 52

KSEEB Solutions

iii) The number of segments required to form 100 digits of this kind
5n + 2 = 5 × 100 + 2 = 500 + 2 = 502

Question 2.
Use the given algebraic expression to complete the table of number patterns.
Solution:
KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 90