Tag: Class 10 Maths RS Aggarwal Solutions

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Solve each of the following systems of equations by using the method of cross multiplication:
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

x = 15, y= 5

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

MCQ
Question 1.
Solution:
(a)

Question 2.
Solution:
(d)

Question 3.
Solution:
(a)

Question 4.
Solution:
(d)

Short-Answer Questions
Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:
Let first number = x
and second number = y
According to the conditions, x – y = 26 …(i)
and x = 3y …..(ii)
From (i),
3y – y = 26
⇒ 2y = 26
⇒ y = 13
and x = 3 x 13 = 39
Numbers are 39 and 13

Short-Answer Questions (3 marks)
Question 11.
Solution:
23x + 29y = 98 …..(i)
29x + 23y = 110 …..(ii)
Adding, we get 52x + 52y = 208
x + y = 4 …..(iii) (Dividing by 52)
and subtracting,
-6x + 6y = -12
x – y = 2. …..(iv) (Dividing by -6)
Adding (iii) and (iv),
2x = 6 ⇒ x = 3
Subtracting,
2x = 2 ⇒ y = 1
Hence, x = 3, y = 1

Question 12.
Solution:


x = 1, y = \(\frac { 3 }{ 2 }\)

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to the condition,
5x + 7y = 195 …(i)
7x + 5y= 153 …(ii)
Adding, (i) and (ii)
12x + 12y = 348
x + y = 29 ….(iii) (Dividing by 12)
and subtracting,
-2x + 2y = 42
-x + y = 21 …..(iv) (Dividing by -2)
Now, Adding (iii) and (iv),
2y = 50 ⇒ y = 25
and from (iv),
-x + 25 = 21 ⇒ -x = 21 – 25 = -4
x = 4
Cost of one pencil = ₹ 4
and cost of one pen = ₹ 25

Question 16.
Solution:
2x – 3y = 1, 4x – 3y + 1 = 0
2x – 3y = 1
2x = 1 + 3y
x = \(\frac { 1 + 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below

Now plot the points (2, 1), (5, 3) and (-1, -1) on the graph and join them to get a line.
Similarly,
4x – 3y + 1 = 0
⇒ 4x = 3y – 1
⇒ x = \(\frac { 3y – 1 }{ 4 }\)

Now plot the points (-1, -1), (-4, -5) and (2, 3) on the graph and join them to get another line which intersects the first line at the point (-1, -1).
Hence, x = -1, y = -1

Long-Answer Questions
Question 17.
Solution:
We know that opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
Now, ∠A = 4x° + 20°, ∠B = 3x° – 5°, ∠C = 4y° and ∠D = 7y° + 5°
But ∠A + ∠C = 180°
4x + 20° + 4y° = 180°
⇒ 4x + 4y = 180° – 20 = 160°
x + y = 40° …(i) (Dividing by 4)
and ∠B + ∠D = 180°
⇒ 3x – 5 + 7y + 5 = 180°
⇒ 3x + 7y = 180° …(ii)
From (i), x = 40° – y
Substituting the value of x in (ii),
3(40° – y) + 7y = 180°
⇒ 120° – 3y + 7y = 180°
⇒ 4y = 180°- 120° = 60°
y = 15°
and x = 40° – y = 40° – 15° = 25°
∠A = 4x + 20 = 4 x 25 + 20 = 100 + 20= 120°
∠B = 3x – 5 = 3 x 25 – 5 = 75 – 5 = 70°
∠C = 4y = 4 x 15 = 60°
∠D = 7y + 5 = 7 x 15 + 5 = 105 + 5 = 110°

Question 18.
Solution:

Question 19.
Solution:
Let numerator of a fraction = x
and denominator = y
Fraction = \(\frac { x }{ y }\)
According to the conditions,

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Choose the correct answer in each of the following questions.
Question 1.
Solution:
(c)

Question 2.
Solution:
(c)

Question 3.
Solution:
(a)

Question 4.
Solution:
(d)

Question 5.
Solution:
(a)

Question 6.
Solution:
(b)

Question 7.
Solution:
(c) 4x + 6y = 3xy, 8x + 9y = 5xy
Dividing each term by xy,

Question 8.
Solution:
(a)

Question 9.
Solution:
(c)

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(b)

Question 13.
Solution:
(a)

Question 14.
Solution:
(d)

Question 15.
Solution:
(d)

Question 16.
Solution:
(d)

Question 17.
Solution:
(d)

Question 18.
Solution:
(d) The system of equations is consistent then their graph lines will be either intersecting or coincident.

Question 19.
Solution:
(a) The pair of lines of equation is inconsistent then the system will not have no solution i.e., their lines will be parallel.

Question 20.
Solution:
(b)

Question 21.
Solution:
(b) ABCD is a cyclic quadrilateral
∠A = (x + y + 10)°, ∠B = (y + 20)°, ∠C = (x + y – 30)° and ∠D = (x + y)°
∠A + ∠C = 180°
Now, x + y + 10°+ x + y – 30° = 180°
⇒ 2x + 2y – 20 = 180°
⇒ 2x + 2y = 180° + 20° = 200°
⇒ x + y = 100° …(i)
and ∠B + ∠D = 180°
⇒ y + 20° + x + y = 180°
⇒ x + 2y = 180° – 20° = 160° …(ii)
Subtracting,
-y = -60° ⇒ y = 60°
and x + 60° = 100°
⇒ x = 100° – 60° = 40°
Now, ∠B = y + 20° = 60° + 20° = 80°

Question 22.
Solution:
(d) Let one’s digit of a two digit number = x
and ten’s digit = y
Number = x + 10y
By interchanging the digits,
One’s digit = y
and ten’s digit = x
Number = y + 10x
According to the conditions,
x + y = 15 …(i)
y + 10x = x + 10y + 9
⇒ y + 10x – x – 10y = 9
⇒ 9x – 9y = 9
⇒ x – y = 1 …(ii)
Adding (i) and (ii),
2x = 16 ⇒ x = 8
and x + y = 15
⇒ 8 + y = 15
⇒ y = 15 – 8 = 7
Number = x + 10y = 8 + 10 x 7 = 8 + 70 = 78

Question 23.
Solution:
(b) Let the numerator of a fractions = x
and denominator = y

Question 24.
Solution:
(d) Let present age of man = x years
and age of his son = y years
5 years hence,
Age of man = (x + 5) years
and age of son = y + 5 years
(x + 5) = 3 (y + 5)
⇒ x + 5 = 3y + 15
x = 3y + 15 – 5
x = 3y + 10 ……(i)
and 5 years earlier
Age of man = x – 5 years
and age of son = y – 5 years
x – 5 = 7 (y – 5)
x – 5 = 7y – 35
⇒ x = 7y – 35 + 5
x = 7y – 30 ……….(ii)
From (i) and (ii),
7y – 30 = 3y + 10
⇒ 7y – 3y = 10 + 30
⇒ 4y = 40
y = 10
x = 3y + 10 = 3 x 10 + 10 = 30 + 10 = 40
Present age of father = 40 years

Question 25.
Solution:
(b)

Question 26.
Solution:
(c)

Question 27.
Solution:
(a)

The system has infinitely many solutions.
The lines are coincident.

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Very-Short and Short-Answer Questions
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Let first, number = x
and second number = y
x – y = 5

Question 8.
Solution:
Let cost of one pen = ₹ x
and cost of one pencil = ₹ y
According to the conditions,
5x + 8y = 120 …(i)
8x + 5y = 153 …(ii)
Adding, we get
13x + 13y = 273
x + y = 21 …(iii) (Dividing by 13)
and subtracting (i) from (ii),
3x – 3y = 33
⇒ x – y = 11 …….(iv) (Dividing by 3)
Again adding (iii) and (iv),
2x = 32 ⇒ x = 16
Subtracting,
2y = 10 ⇒ y = 5
Cost of 1 pen = ₹ 16
and cost of 1 pencil = ₹ 5

Question 9.
Solution:
Let first number = x
and second number = y
According to the conditions,

and x + y = 80
⇒ x + 15 = 80
x = 80 – 15 = 65
Numbers are : 65, 15

Question 10.
Solution:
Let one’s digit of a two digits number = x
and ten’s digit = y
Number = x + 10y
By reversing its digits One’s digit = y
and ten’s digit = x
Then number = y + 10x
According to the conditions,
x + y = 10 …(i)
x + 10y – 18 = y + 10x
x+ 10y – y – 10x = 18
⇒ -9x + 9y = 18
⇒ x – y = -2 (Dividing by -9) …..(ii)
Adding (i) and (ii),
2x = 8 ⇒ x = 4
and by subtracting,
2y = 12 ⇒ y = 6
Number = x + 10y = 4 + 10 x 6 = 4 + 60 = 64

Question 11.
Solution:
Let number of stamps of 20p = x
and stamps of 25 p = y
According to the conditions,
x + y = 47 …..(i)
20x + 25y = 1000
4x + 5y = 200 …(ii)
From (i), x = 47 – y
Substituting the value of x in (ii),
4 (47 – y) + 5y = 200
188 – 4y + 5y = 200
⇒ y = 200 – 188 = 12
and x + y = 47
⇒ x + 12 = 47
⇒ x = 47 – 12 = 35
Hence, number of stamps of 20 p = 35
and number of stamps of 25 p = 12

Question 12.
Solution:
Let number of hens = x
and number of cows = y
According to the conditions,
x + y = 48 …..(i)
x x 2 + y x 4 = 140
⇒ 2x + 4y = 140
⇒ x + 2y = 70 ……(ii)
Subtracting (i) from (ii),
y = 22
and x + y = 48
⇒ x + 22 = 48
⇒ x = 48 – 22 = 26
Number of hens = 26
and number of cows = 22

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
12x + 17y = 53 …(i)
17x + 12y = 63 …(ii)
Adding, 29x + 29y = 116
Dividing by 29,
x + y = 4 …(iii)
Subtracting,
-5x + 5y = -10
⇒ x – y = 2 …(iv) (Dividing by -5)
Adding (iii) and (iv)
2x = 6 ⇒ x = 3
Subtracting,
2y = 2 ⇒ y = 1
x = 3, y = 1
x + y = 3 + 1 = 4

Question 16.
Solution:

Question 17.
Solution:
kx – y = 2
6x – 2y = 3

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 3 Linear equations in two variables Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A.

RS Aggarwal Solutions Class 10 Chapter 3

• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3C
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3D
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3E
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3F
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables MCQS
• RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Test Yourself

Solve each of the following systems of equations graphically.
Question 1.
Solution:
2x + 3y = 2 …..(i)
x – 2y = 8 …(ii)
From Eq. (i),
⇒ 2x = 2 – 3y
⇒x = \(\frac { 2 – 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (1, 0), (-2, 2) and (4, -2) on the graph and join them to get a line.
Similarly x – 2y = 8 ⇒ x = 8 + 2y

Now plot the points (6, -1), (4, -2) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at point(4, -2).
x = 4, y = -2

Question 2.
Solution:
3x + 2y = 4
⇒ 3x = 4 – 2y
⇒ x = \(\frac { 4 – 2y }{ 3 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now plot the points (0, 2), (2, -1) and (4, -4) on the graph and join them to get a line. Similarly,
2x – 3y = 7
2x = 3y+ 7
x = \(\frac { 3y+ 7 }{ 2 }\)

Now plot the points on the graph and join them to get another line.
We see that these two lines intersect each other at point (2, -1).
x = 2, y = -1

Question 3.
Solution:
2x + 3y = 8
⇒ 2x = 8 – 3y
x = \(\frac { 8 – 3y }{ 2 }\)
Now, giving some different values to y, we get corresponding values of x as given below

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
x – 2y + 3 = 0
x = 2y – 3

Now, plot the points (-3, 0), (-1, 1) and (1, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (1, 2).
x = 1, y = 2

Question 4.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = \(\frac { 5y – 4 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-2, 0) (3, 2) and (8,4) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
⇒ y = 8 – 2x

Plot the points (1, 6), (2, 1) and (3, 2) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (3, 2).
x = 3, y = 2

Question 5.
Solution:
3x + 2y = 12
⇒ 3x = 12 – 2y
x = \(\frac { 12 – 2y }{ 2 }\)
Giving some different values to y, we get corresponding the values of x as given below:

Now, we plot the points (4,0), (2,3) and (0, 6) on the graph and join them to get a line.
Similarly,
5x – 2y = 4
⇒ 5x = 4 + 2 y
⇒ x = \(\frac { 4 + 2 y }{ 5 }\)

Plot the points (0, -2), (2, 3) and (4, 8) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, 3).
x = 2, y = 3

Question 6.
Solution:
3x + y + 1 = 0 ⇒y = -3x – 1
Giving some different values to x, we get corresponding values of y as given below:

Now, plot the points (0, -1), (-1, 2) and (-2, 5) on the graph and join them to get a line.
Similarly,
2x – 3y + 8 = 0
2x = 3y – 8
x = \(\frac { 3y – 8 }{ 2 }\)

Now, plot the points (-4, 0), (-1, 2) and (2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the points (-1, 2).
x = -1, y = 2

Question 7.
Solution:
2x + 3y + 5 = 0
2x = -3y – 5
x = \(\frac { -3y – 5 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-4, 1), (-1, -1) and (2, -3) on the graph and join them to get a line. Similarly,
3x – 2y – 12 = 0
⇒ 3x = 2y + 12
x = \(\frac { 2y + 12 }{ 3 }\)

Plot the points (4, 0), (0, -6) and (2, -3) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -3).
x = 2, y = -3

Question 8.
Solution:
2x – 3y + 13 = 0
⇒ 2x = 3y – 13
⇒ x = \(\frac { 3y – 13 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Now, plot the points (-5, 1), (-2, 3) and (1, 5) on the graph and join them to get a line.
Similarly,
3x – 2y + 12 = 0
3x = 2y – 12
x = \(\frac { 2y – 12 }{ 3 }\)

Now, plot the points (-4, 0), (-2, 3) and (0, 6) on the graph and join them to get another line.
We see that these lines intersect each other at the point (-2, 3).
x = -2, y = 3

Question 9.
Solution:
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
⇒ x = \(\frac { 4 – 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0) (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
3x – y + 5 = 0
⇒ -y = -5 – 3x
⇒ y = 5 + 3x

Plot the points (0, 5), (-1, 2) and (-2, -1) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (-1, 2).
x = -1, y = 2

Question 10.
Solution:
x + 2y + 2 = 0
⇒ x = – (2y + 2)
Giving some different values to y, we get corresponding the values of x as given below:

Plot the points (-2,0), (0, -1) and (2, -2) on the graph and join them to get a line.
Similarly,
3x + 2y – 2 = 0
⇒ 3x = 2 – 2y
x = \(\frac { 2 – 2y }{ 3 }\)

Plot the points (0, 1), (2, -2) and (-2, 4) on the graph and join them to get another line.
We see that these two lines intersect each other at the point (2, -2).
x = 2, y = -2

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the x-axis:
Question 11.
Solution:
x – y + 3 = 0
⇒ x = y – 3
Giving some different value to y, we get corresponding values of x as given below:

Plot the points (-3, 0), (-1, 2) and (0, 3) on the graph and join them to get a line.
Similarly,
2x + 3y – 4 = 0
⇒ 2x = 4 – 3y
x = \(\frac { 4 – 3y }{ 2 }\)

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get another line.
We see that these line intersect each other at (-1, 2) and x-axis at A (-3, 0) and D (2, 0).

Area of ∆BAD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AD x BL
= \(\frac { 1 }{ 2 }\) x 5 x 2 = 5 sq.units

Question 12.
Solution:
2x – 3y + 4 = 0
⇒ 2x = 3y – 4
x = \(\frac { 3y – 4 }{ 2 }\)
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-2, 0), (1, 2) and (4, 4) on the graph and join them to get a line.
Similarly,
x + 2y – 5 = 0
x = 5 – 2y

Now, plot the points (5, 0), (3, 1) and (1, 2) on the graph and join them to get another line.
We see that there two lines intersect each other at the point B (1, 2) and intersect x- axis at A (-2, 0) and D (5, 0) respectively.

Now, area of ∆BAD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AD x BL
= \(\frac { 1 }{ 2 }\) x 7 x 2
= 7 sq. units

Question 13.
Solution:
4x – 3y + 4 = 0
⇒ 4x = 3y – 4
x = \(\frac { 3y – 4 }{ 4 }\)
Giving some different values to y, we get corresponding value of x as given below:

Plot the points (-1, 0), (2, 4) and (-4, -4) on the graph and join them, to get a line.
Similarly,
4x + 3y – 20 = 0
4x = 20 – 3y
x = \(\frac { 20 – 3y }{ 4 }\)

Plot the points (5, 0), (2, 4) and (-1, 8) on the graph and join them to get a line.

We see that these two lines intersect each other at the point B (2, 4) and intersect x-axis at A (-1, 0) and D (5, 0).
Area ∆BAD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AD x BL
= \(\frac { 1 }{ 2 }\) x 6 x 4 = 12 sq. units

Question 14.
Solution:
x – y + 1 = 0 ⇒ x = y – 1
Giving some different values toy, we get the values of x as given below:

Plot the points (-1, 0), (0, 1) and (1, 2) on the graph and join them to get a line.
Similarly,
3x + 2y – 12 = 0
⇒ 3x = 12 – 2y
x = \(\frac { 12 – 2y }{ 3 }\)

Plot the points (4, 0), (2, 3) and (0, 6) on the graph and join them to get another line.

We see that these two lines intersect each ohter at the point E (2, 3) and intersect x- axis at A (-1, 0) and D (4, 0).
Area of ∆EAD = \(\frac { 1 }{ 2 }\) x base x altitude 1
= \(\frac { 1 }{ 2 }\) x AD x EL
= \(\frac { 1 }{ 2 }\) x 5 x 3
= \(\frac { 15 }{ 2 }\)
= 7.5 sq. units

Question 15.
Solution:
x – 2y + 2 = 0 ⇒ x = 2y – 2
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (0, 1), (2, 2) on the graph and join them to get a line.
Similarly,
2x + y – 6 = 0
⇒ y = 6 – 2x

Plot the points on the graph and join them to get a line.

We see that these two lines intersect each other at the point C (2, 2) and intersect the x-axis at A (-2, 0) and F (3, 0).
Now, area of ∆CAF = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AF x CL
= \(\frac { 1 }{ 2 }\) x 5 x 2 = 5 sq. units

Solve each of the following given systems of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis:
Question 16.
Solution:
2x – 3y + 6 = 0
⇒ 2x = 3y – 6
x = \(\frac { 3y – 6 }{ 2 }\)
Giving some different values to y, we get corresponding values of x, as given below:

Now, plot the points (-3, 0), (0, 2), (3, 4) on the graph and join them to get a line.
Similarly,
2x + 3y – 18 = 0
2x = 18 – 3y
x = \(\frac { 18 – 3y }{ 2 }\)

Plot the points (0, 6), (3, 4) and (6, 2) on the graph and join them to get a line.
We see that these two lines intersect each other at C (3, 4) and intersect y-axis at B (0, 2) and D (0, 6).

Now, area of ∆CBD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x BD x CL
= \(\frac { 1 }{ 2 }\) x 4 x 3 sq. units = 6 sq. units

Question 17.
Solution:
4x – y – 4 = 0
⇒ 4x = y + 4
x = \(\frac { y + 4 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (1, 0), (0, -4) and (2, 4) on the graph and join them to get a line.
Similarly,
3x + 2y – 14 = 0
2y = 14 – 3x
y = \(\frac { 14 – 3x }{ 2 }\)

Now, plot the points (0, 7), (2, 4) and (4, 1) on the graph and join them to get another line.
We see that these two lines intersect each other at C (2, 4) and y-axis at B (0, -4) and D (0, 7) respectively.

Area of ∆CBD = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x BD x CL
= \(\frac { 1 }{ 2 }\) x 11 x 2 sq. units = 11 sq. units

Question 18.
Solution:
x – y – 5 = 0 ⇒ x = y + 5
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (5, 0), (0, -5) and (1, -4) on the graph and join these to get a line.
Similarly,
3x + 5y – 15 = 0
3x = 15 – 5y
x = \(\frac { 15 – 5y }{ 3 }\)

Plot the points (5, 0), (0, 3) and (-5, 6) on the graph and join them to get a line.
We see that these two lines intersect each other at A (5, 0) and y-axis at B (0, -5) and E (0, 3) respectively.

Now area of ∆ABE = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x BE x AL
= \(\frac { 1 }{ 2 }\) x 8 x 5 sq. units = 20 sq. units

Question 19.
Solution:
2x – 5y + 4 = 0
⇒ 2x = 5y – 4
x = \(\frac { 5y – 4 }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (-2, 0), (3, 2) and (-7, -2) on the graph and join them to get a line.
Similarly,
2x + y – 8 = 0
2x = 8 – y
x = \(\frac { 8 – y }{ 2 }\)

Plot the points (4, 0), (3, 2) and (2, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at B (3, 2) and y-axis at G (0, 1) and H (0, 8) respectively.

Now area of ∆EGH = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x GH x EL
= \(\frac { 1 }{ 2 }\) x 7 x 3 sq. units
= \(\frac { 21 }{ 2 }\) = 10.5 sq. units

Question 20.
Solution:
5x – y – 7 = 0 ⇒ y = 5x – 7
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, -7), (1, -2), (2, 3) on the graph and join them to get a line.
Similarly,
x – y + 1 = 0 ⇒ x = y – 1

Plot the points (-1, 0), (1, 2) and (2, 3) on the graph and join them to get a line.

We see that these two lines intersect each other at C (2, 3) and intersect y-axis at A (0, -7) and F (0, 1) respectively.
Now, area of ∆CAF = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x AF x CL
= \(\frac { 1 }{ 2 }\) x 8 x 2 = 8 sq.units

Question 21.
Solution:
2x – 3y – 12 ⇒ 2x = 12 + 3y
x = \(\frac { 12 + 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (3, -2) and (0, -4) on the graph and join them to get a line.
Similarly,
x + 3y = 6 ⇒ x = 6 – 3y

Plot the points (6, 0), (0, 2) and (-6, 4) on the graph and join them to get a line.
We see that these two lines intersect each other at the points A (6, 0) and intersect the y-axis at C (0, -4) and E (0, 2) respectively.

Now area of ∆ACE = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x CE x AO
= \(\frac { 1 }{ 2 }\) x 6 x 6 sq. units = 18 sq.units

Show graphically that each of the following given systems of equations has infinitely many solutions:
Question 22.
Solution:
2x + 3y = 6
2x = 6 – 3y
x = \(\frac { 6 – 3y }{ 2 }\)
Giving some different values to y, we get the corresponding val ues of x as given below:

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get a line.
Similarly,
4x + 6y = 12
4x = 12 – 6y
x = \(\frac { 12 – 6y }{ 2 }\)

on the graph and join them to get a line.
We see that all the points lie on the same straight line.
This system has infinite many solutions.

Question 23.
Solution:
3x – y = 5
⇒ y = 3x – 5
Giving some different values to x, we get corresponding values of y as shown below:

Now, plot the points (0, -5), (1, -2), (2, 1) on the graph and join them to get a line:
Similarly,
6x – 2y = 10
⇒ 6x = 10 + 2y
x = \(\frac { 10 + 2y }{ 6 }\)

Plot the points (2, 1), (4, 7), (3, 4) on the graph and join them to get another line.
We see that these lines coincide each other.
This system has infinite many solutions.

Question 24.
Solution:
2x + y = 6
y = 6 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 6), (2, 2), (4, -2) on the graph and join them to get a line.
Similarly,
6x + 3y = 18
⇒ 6x = 18 – 3y
⇒ x = \(\frac { 18 – 3y }{ 2 }\)

Now, plot the points (3, 0), (1, 4) and (5, -4) on the graph and join them to get another line.
We see that these two lines coincide each other.
This system has infinitely many solutions

Question 25.
Solution:
x – 2y = 5 ⇒ x = 5 + 2y
Giving some different values to y, we get corresponding values of x as shown below:

Plot the points (5, 0), (3, -1), (1, -2) on the graph and join them to get a line.
similarly,
3x – 6y = 15
⇒ 3x = 15 + 6y
x = \(\frac { 15 + 6y }{ 3 }\)

Plot the points (7, 1), (-1, -3) and (-3, -4) on the graph and join them to get another line.
We see that all the points lie on the same line.
Lines coincide each other.
Hence, the system has infinite many solutions.

Show graphically that each of the following given systems of equations is inconsistent, i.e., has no solution:
Question 26.
Solution:
x – 2y = 6 ⇒ x = 6 + 2y
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (6, 0), (4, -1) and (0, -3) on the graph and join them to get a line.
Similarly,
3x – 6y = 0
⇒ 3x = 6y
⇒ x = 2y

Plot the points (0, 0), (2, 1), (4, 2) on the graph and join them to get a line.

We see that these two lines are parallel i.e., do not intersect each other.
This system has no solution.

Question 27.
Solution:
2x + 3y = 4
⇒ 2x = 4 – 3y
⇒ x = \(\frac { 4 – 3y }{ 2 }\)
Giving some different values to y, we get corresponding values of x as given below:

Plot the points (2, 0), (-1, 2) and (5, -2) on the graph and join them to get a line.
Similarly,
4x + 6y = 12 ⇒ 4x = 12 – 6y
x = \(\frac { 12 – 6y }{ 4 }\)

Plot the points (3, 0), (0, 2) and (-3, 4) on the graph and join them to get another line.
We see that two lines are parallel i.e., these do not intersect each other at any point.
Therefore the system has no solution.

Question 28.
Solution:
2x + y = 6, y = 6 – 2x
Giving some different values to x, we get corresponding values ofy as given below:

Plot the points (0, 6), (1, 4) and (3, 0) on the graph and join them to get a line.
Similarly,
6x + 3y = 20 ⇒ 6x = 20 – 3y

We see that these two lines are parallel and do not intersect each other.
Therefore this system has no solution.

Question 29.
Solution:
2x + y = 2 ⇒ y = 2 – 2x
Giving some different values to x, we get corresponding values of y as given below:

Plot the points (0, 2), (1, 0) and (-2, 6) on the graph and join them to get a line.
Similarly,
2x + y = 6 ⇒ y = 6 – 2x

Plot the points (0, 6), (2, 2) and (3, 0) on the graph and join them to get another line.
ABFD is the trapezium whose vertices are A (0, 2), B (1, 0), F (3, 0), D (0, 6).

Area of trapezium ABFD = Area ∆DOF – Area ∆AOB
= \(\frac { 1 }{ 2 }\) (DO x OF) – \(\frac { 1 }{ 2 }\) (AO x OB)
= \(\frac { 1 }{ 2 }\) (6 x 3) – \(\frac { 1 }{ 2 }\) (2 x 1) sq.units
= 9 – 1 = 8 sq.units

Hope given RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 12 Circles Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B
• RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS
• RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself

MCQ
Question 1.
Solution:
In the given figure,
PT is the tangent and PQ is the chord of the circle with centre O.
∠OPT = 50°

OP is radius and PT is the tangent.
OP ⊥ PT ⇒ ∠OPT = 90°
∠OPQ + ∠QPT = 90°
⇒ ∠OPQ + 50° = 90°
⇒ ∠OPQ = 90° – 50° = 40°
In ∆OPQ,
OP = OQ (radii of the same circle)
∠OQP = ∠OPQ = 40°
In ∆OPQ,
∠POQ = 180° – (∠OPQ + ∠OQP)
= 180° – (40° + 40°) = 180° – 80°
= 100° (b)

Question 2.
Solution:
Angle between two radii of a circle = 130°

Then ∠APB = 180° – ∠AOB
= 180°- 130° = 50° (c)

Question 3.
Solution:
In the given figure,
PA and PB are the tangents drawn from P to the circle with centre O
∠APB = 80°

OA is radius of the circle and AP is the tangent
OA ⊥ AP ⇒ ∠OAP = 90°
OP bisects ∠APB,
∠APO = \(\frac { 1 }{ 2 }\) x 80 = 40°
∠POA = 180° – (90° + 40°)
= 180° – 130° = 50° (b)

Question 4.
Solution:
In the given figure, AD and AE are tangents to the circle with centre O.
BC is the tangent at F which meets AD at C and AE at B
AE = 5 cm

AE and AD are the tangents to the circle
AE = AD = 5 cm
Tangents from an external point drawn to the circle are equal
CD = CF and BE = BF
Now, perimeter of ∆ABC = AB + AC + BC
= AB + AC + BF + CF (BE = BF and CF = CD)
= AB + AC + BE + CD
= AB + BE + AC + CD
= AE + AD
= 5 + 5 = 10 cm (b)

Short-Answer Questions
Question 5.
Solution:
In the given figure, a quadrilateral ABCD is circumscribed a circle touching its sides at P, Q, R and S respectively.
AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm

A circle touches the sides of a quadrilateral ABCD.
AB + CD = BC + AD …(i)
Now, AP and AS are tangents to the circle
AP = AS = 5 cm …(ii)
Similarly, CQ = CR = 3 cm
BP = BQ = x – 5 = 4
BQ = BC – CQ = 7 – 3 = 4 cm
x – 5 = 4
⇒ x = 4 + 5 = 9cm

Question 6.
Solution:
In the given figure, PA and PB are the tangents drawn from P to the circle.
OA and OB are the radii of the circle and AP and BP are the tangents.
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
In quad. AOBP
∠A + ∠B = 90° + 90° = 180°
But these are opposite angles of a quadrilateral
AOBP is a cyclic quadrilateral
A, O, B, P are concyclic

Question 7.
Solution:
In the given figure, PA and PB are two tangents to the circle with centre O from an external point P.
∠PBA = 65°,
To find : ∠OAB and ∠APB
In ∆APB
AP = BP (Tangents from P to the circle)
∠PAB = ∠PBA = 65°
∠APB = 180° – (∠PAB + ∠PBA)
= 180° – (65° + 65°) = 180° – 130° = 50°
OA is radius and AP is tangent
OA ⊥ AP
∠OAP = 90°
∠OAB = ∠OAP – ∠PAB = 90° – 65° = 25°
Hence, ∠OAB = 25° and ∠APB = 50°

Question 8.
Solution:
Given : In the figure,
BC and BD are the tangents drawn from B
to the circle with centre O.
∠CBD = 120°
To prove : OB = 2BC
Construction : Join OB.
Proof: OB bisects ∠CBD (OC = OD and BC = BD)

Question 9.
Solution:
(i) A line intersecting a circle in two distinct points is called a secant.
(ii) A circle can have two parallel tangents at the most.
(iii) The common point of a tangent to a circle and the circle is called the point of contact.
(iv) A circle can have infinitely many tangents.

Question 10.
Solution:
Given : In a circle, from an external point P, PA and PB are the tangents drawn to the circle with centre O.
To prove : PA = PB
Construction : Join OA, OB and OP.
Proof : OA and OB are the radii of the circle and AP and BP are tangents.

OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
PA = PB (c.p.c.t.)
Hence proved.

Short-Answer Questions
Question 11.
Solution:
Given : In a circle with centre O and AB is its diameter.
From A and B, PQ and RS are the tangents drawn to the circle

To prove : PQ || RS
Proof : OA is radius and PAQ is the tangent
OA ⊥ PQ
∠PAO = 90° …(i)
Similarly, OB is the radius and RBS is tangent
∠OBS = 90° …(ii)
From (i) and (ii)
∠PAO = ∠OBS
But there are alternate angles
PQ || RS

Question 12.
Solution:
Given : In the given figure,
In ∆ABC,
AB = AC.
A circle is inscribed the triangle which touches it at D, E and F
To prove : BE = CE
Proof: AD and AF are the tangents drawn from A to the circle
AD = AF
But, AB = AC
AB – AD = AC -AF
⇒ BD = CF
But BD = BE and CF = CE (tangent drawn to the circle)
But BD = CF
BE = CE
Hence proved.

Question 13.
Solution:
Given : In a circle from an external point P, PA and PB are the tangents to the circle
OP, OA and OB are joined.

To prove: ∠POA = ∠POB
Proof: OA and OB are the radii of the circle and PA and PB are the tangents to the circle
OA ⊥ AP and OB ⊥ BP
∠OAP = ∠OBP = 90°
Now, in right ∆OAP and ∆OBP,
Hyp. OP = OP (common)
Side OA = OB (radii of the same circle)
∆OAP = ∆OBP (RHS axiom)
∠POA = ∠POB (c.p.c.t.)
Hence proved.

Question 14.
Solution:
Given : A circle with centre O, PA and PB are the tangents drawn from A and B which meets at P.
AB is chord of the circle
To prove : ∠PAB = ∠PBA
Construction : Join OA, OB and OP

Proof: OA is radius and AP is tangent
OA ⊥ AP ⇒ ∠OAP = 90°
Similarly, OB ⊥ BP ⇒ ∠OBP = 90°
In ∆OAB, OA = OB (radii of the circle)
∠OAB = ∠OBA
⇒ ∠OAP – ∠OAB = ∠OBP – ∠OBA
⇒ ∠PAB = ∠PBA
Hence proved.

Question 15.
Solution:
Given : A parallelogram ABCD is circumscribed a circle.
To prove : ABCD is a rhombus.
Proof: In a parallelogram ABCD.
Opposite sides are equal and parallel.

AB = CD and AD = BC
Tangents drawn from an external point of a circle to the circle are equal.
AP = AS BP = BQ
CQ = CR and DR = DS
Adding, we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC [AB = CD and AD = BC]
⇒ AB + AB = BC + BC
⇒ 2AB = 2BC
⇒ AB = BC
But AB = CD and BC = AD
AB = BC = CD = AD
Hence || gm ABCD is a rhombus.

Question 16.
Solution:
Given : O is the centre of two concentric circles with radii 5 cm and 3 cm respectively.
AB is the chord of the larger circle which touches the smaller circle at P.

OP and OA are joined.
To find : Length of AB
Proof: OP is the radius of the smaller circle and touches the smaller circle at P
OP ⊥ AB and also bisects AB at P
AP = PB = \(\frac { 1 }{ 2 }\) AB
Now, in right ∆OAP,
OA² = OP² + AP² (Pythagoras Theorem)
⇒ (5)² = (3)² + AP²
⇒ 25 = 9 + AP²
⇒ AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
Hence AB = 2 x AP = 2 x 4 = 8 cm

Long-Answer Questions
Question 17.
Solution:
In the figure, quad. ABCD is circumscribed about a circle which touches its sides at P, Q, R and S respectively

To prove : AB + CD = AD + BC
Proof: Tangents drawn from an external point to a circle are equal
AP = AS
BP = BQ
CR = CQ
DR = DS
Adding, we get,
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence AB + CD = AD + BC

Question 18.
Solution:
Given : A quad. ABCD circumscribe a circle with centre O and touches at P, Q, R and S respectively
OA, OB, OC and OD are joined forming angles AOB, BOC, COD and DOA
To prove : ∠AOB + ∠COD = 180°
and ∠BOC + ∠AOD = 180°
Construction : Join OP, OQ, OR and OS

Proof: In right ∆AOP and ∆AOS,
Side OP = OS (radii of the same circle)
Hyp. OA = OA (common)
∆AOP = ∆AOS (RHS axiom)
∠1 = ∠2 (c.p.c.t.)
Similarly, we can prove that
∠4 = ∠3
∠5 = ∠6
∠8 = ∠7
Adding, we get
∠1 + ∠4 + ∠5 + ∠8 = ∠2 + ∠3 + ∠6 + ∠7
⇒ (∠1 + ∠8) + (∠4 + ∠5) = (∠2 + ∠3) + (∠6 + ∠7)
⇒ ∠AOB + ∠COD = ∠AOD + ∠BOC
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° (angles at a point)
∠AOB + ∠COD = ∠AOD + ∠BOC = 180°
Hence proved

Question 19.
Solution:
Given : From an external point P, PA and PB are the tangents drawn to the circle,
OA and OB are joined.
To prove : ∠APB + ∠AOB = 180°
Construction : Join OP.

Proof : Now, in ∆POA and ∆PBO,
OP = OP (common)
PA = PB (Tangents from P to the circle)
OA = OB (Radii of the same circle)
∆POA = ∆PBO (SSS axiom)
∠APO = ∠BPO (c.p.c.t.)
and ∠AOP = ∠BOP (c.p.c.t.)
OA and OB are the radii and PA and PB are the tangents
OA ⊥ AP and OB ⊥ BP
⇒ ∠OAP = 90° and ∠OBP = 90°
In ∆POA,
∠OAP = 90°
∠APO + ∠AOP = 90°
Similarly, ∠BPO + ∠BOP = 90°
Adding, we get
(∠APO + ∠BPO) + (∠AOP + ∠BOP) = 90° + 90°
⇒ ∠APB + ∠AOB = 180°.
Hence proved.

Question 20.
Solution:
Given : PQ is chord of a circle with centre O.
TP and TQ are tangents to the circle
Radius of the circle = 10 cm
i.e. OP = OQ = 10 cm and PQ = 16 cm

To find : The length of TP.
OT bisects the chord PQ at M at right angle.
PM = MQ = \(\frac { 16 }{ 2 }\) = 8 cm
In right ∆PMO,
OP² = PM² + MO² (Pythagoras Theorem)
⇒ (10)² = (8)² + MO²
⇒ 100 = 64 + MO²
⇒ MO² = 100 – 64 = 36 = (6)²
MO = 6 cm
Let TP = x and TM = y
In right ∆TPM,
TP² = TM² + PM²
⇒ x² = y² + 8²
⇒ x² = y² + 64 …(i)
and in right ∆TPM
OT² = TP² + OP²
⇒ (y + 6)² = x² + 10²
⇒ y² + 12y + 36 = x² + 100
⇒ y² + 12y + 36 = y2 + 64 + 100 {From (i)}
⇒ 12y = 64 + 100 – 36 = 128

Hope given RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Choose the correct answer in each of the following questions.
Question 1.
Solution:
Distance between P (-6, 8) and origin O (0, 0)

Question 2.
Solution:
Distance of the point (-3, 4) from x-axis = 4 (c)

Question 3.
Solution:
Let point P (x, 0) is on x-axis which is equidistant from points A (-1, 0) and B (5, 0)

Question 4.
Solution:
R (5, 6) is the midpoint of the line segment joining the points A (6, 5) and B (4, y)

Question 5.
Solution:
Point C (k, 4) divides the join of points A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 6.
Solution:
Vertices of ∆ABC are A (0, 4), B (0, 0) and C (3, 0)

Question 7.
Solution:
A (1, 3), B (-1, 2), C (2, 5) and D (x, 4) are the vertices of a ||gm ABCD.
AB = CD

Question 8.
Solution:
Points A (x, 2), B (-3, -4) and C (7, -5) are collinear

Question 9.
Solution:
Area of ∆ABC whose vertices are A (5, 0), B (8, 0) and C (8, 4)

Question 10.
Solution:
Area of ∆ABC with vertices A (a, 0), O (0, 0) and B (0, b)

Question 11.
Solution:
P( \(\frac { a }{ 2 }\), 4) is midpoint of the line segment joining the points A (-6, 5) and B (-2, 3)

Question 12.
Solution:
ABCD is a rectangle whose three vertices are B (4, 0), C (4, 3) and D (0, 3)

Question 13.
Solution:
Let coordinates of P be (x, 7)
P divides the line segment joining the points A (1, 3) and B (4, 6) in the ratio 2 : 1

Question 14.
Solution:
Let coordinates of one end of diameter of a circle are A (2, 3) coordinates of centre are (-2, 5)
Let coordinates of other end B of the diameter be (x, y)

Question 15.
Solution:
In the given figure, P (5, -3) and Q (3, y) are the points of trisection of the line segment joining A (7, -2) and B (1, -5)

Question 16.
Solution:
Midpoint of AB is P (0, 4) coordinates of B are (-2, 3)

Question 17.
Solution:
Let point P (x, y) divides AB with vertices A(2, -5) and B (5, 2) in the ratio 2 : 3

Question 18.
Solution:
A (-6, 7) and B (-1, -5) two points, then

Question 19.
Solution:
Let point P (x, 0) on x-axis is equidistant from A (7, 6) and B (-3, 4)

Question 20.
Solution:
Distance of P (3, 4) from the x-axis = 4 units. (b)

Question 21.
Solution:
Let a point P (x, 0) on x-axis divides the line segment.
AB joining the points A (2, -3) and B (5, 6) in the ratio m : n.

Question 22.
Solution:
Let a point A (0, y) on y-axis divides the line segment PQ joining the points P (-4, 2) and Q (8, 3) in the ratio of m : n.

Question 23.
Solution:
P (-1, 1) is the midpoint of line segment joining the points A (-3, b) and B (1, b + 4)

Question 24.
Solution:
Let the point P (x, y) divides the line segment joining the points A (2, -2) and B (3, 7) in the ratio k : 1, then

Question 25.
Solution:
AD is the median of ∆ABC with vertices A (4, 2), B (6, 5) and C (1, 4)

Question 26.
Solution:
A (-1, 0), B (5, -2) and C (8, 2) are the vertices of a ∆ABC then centroid

Question 27.
Solution:
Two vertices A (-1, 4), B (5, 2) of a ∆ABC and its centroid G is (0, -3)
Let (x, y) be the co-ordinates of vertex C, then

Question 28.
Solution:
Points A (-4, 0), B (4, 0) and C (0, 3) are vertices of a ∆ABC.

Question 29.
Solution:
Points P (0, 6), Q (-5, 3) and R (3, 1) are the vertices of a ∆PQR

Question 30.
Solution:
Points A (2, 3), B (5, k) and C (6, 7) are collinear
Area of ∆ABC = 0

Question 31.
Solution:
Points A (1, 2), B (0, 0) and C (a, b) are collinear
Area of ∆ABC = 0

Question 32.
Solution:
A (3, 0), B (7, 0) and C (8, 4)
Area ∆ABC

Question 33.
Solution:
AOBC is a rectangle with vertices A (0, 3), O (0, 0) and B (5, 0) and each diagonal of rectangle are equal.

Question 34.
Solution:
Points are A (4, p) and B (1, 0)
Distance = 5 units

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) Let P (x, y) be the required point which divides the line joining the points A (-1, 7) and B (4, -3) in the ratio 2 : 3.

Question 2.
Solution:
Let (7, -2) and B (1, -5) be the given points and P (x, y) and Q (x’, y’) are the points of trisection.

Question 3.
Solution:
Let coordinates of P be (x, y) which divides the line segment A (-2, -2) and B (2, -4) in

Question 4.
Solution:
Let the coordinates of A be (x, y) which lies on line joining P (6, -6) and Q (-4, -1)

Question 5.
Solution:
Points P, Q, R and S divides a line segment joining the points A (1, 2) and B (6, 7) in 5 equal parts.

Question 6.
Solution:
Points P, Q and R in order divide a line segment joining the points A (1, 6) and B (5, -2) in 4 equal parts.

P divides AB in the ratio of 1 : 3 Let coordinates of P be (x, y), then

Question 7.
Solution:
The line segment joining the point A (3, -4) and B (1, 2) is trisected by the points P (p, -2) and Q(\(\frac { 1 }{ 2 }\), q).

Question 8.
Solution:
Mid point of the line segment joining the points A (3, 0) and B (-5, 4)

Question 9.
Solution:
(2, p) is the mid point of the line segment joining the points A (6, -5), B (-2, 11)

Question 10.
Solution:
Mid point of the line segment joining the points A (2a, 4) and B (-2, 3b) is C (1, 2a + 1)

Question 11.
Solution:
The line segment joining the points A (-2, 9) and B (6, 3) is a diameter of a circle with centre C.

Question 12.
Solution:
AB is diameter of a circle with centre C.
Coordinates of C (2, -3) and of B (1, 4)

Question 13.
Solution:
Let P (2, 5) divides the line segment joining the points A (8,2) and B (-6, 9) in the ratio m : n

Question 14.
Solution:

Question 15.
Solution:
Let P (m, 6) divides the join of A (-4, 3) and B (2, 8) in the ratio k : 1
Then coordinates of P will be

Question 16.
Solution:
Let point P (-3, k) divides the join of A (-5, -4) and B (-2, 3) in the ratio m : n, then

Question 17.
Solution:
Let point P on the x-axis divides the line segment joining the points A (2, -3) and B (5, 6) the ratio m : n
Let P is the point on x-axis whose coordinates are (x, 0)

Question 18.
Solution:
Let a point P on y-axis divides the line segment joining the points A (-2, -3) and B (3, 7) in the ratio m : n
Let the coordinates of P be (0, y)

Question 19.
Solution:
Let a point P (x, y) on the given line x – y – 2 = 0 divides the line segment joining the points A (3, -1) and B (8, 9) in the ratio m : n, then

Question 20.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)
Let AD, BE and CF are the medians of sides BC, CA and AB respectively, then
Coordinates of D will be =

Question 21.
Solution:
Centroid of ∆ABC where coordinates of A are (-1, 0), of B are (5, -2) and of C are (8, 2)

Question 22.
Solution:
G (-2, 1) is the centroid of ∆ABC whose vertex A is (1, -6) and B is (-5, 2)
Let the vertex C be (x, y), then

Question 23.
Solution:
O (0, 0) is the centroid of ∆ABC in which B is (-3, 1), C is (0, -2)
Let A be (x, y), then

Question 24.
Solution:
Points are A (3, 1), B (0, -2), C (1, 1) and D (4, 4)

Question 25.
Solution:
Points P (a, -11), Q (5, b), R (2, 15) and S (1, 1) are the vertices of a parallelogram PQRS.
Diagonals of a parallelogram bisect each other.
O is mid point of PR and QS.

Question 26.
Solution:
Three vertices of a parallelogram ABCD are A (1, -2), B (3, 6), C (5, 10).
Let fourth vertices D be (x, y)

Question 27.
Solution:
Let a point P (0, y) on 7-axis, divides the line segment joining the points (-4, 7) and (3, -7) in the ratio m : n

Question 28.
Solution:

Question 29.
Solution:
Let a point P (x, 0) divides the line segment joining the points A (3, -3) and B (-2, 2) in the ratio m : n

Question 30.
Solution:
Base QR of an equilateral APQR lies on x- axis is O (0, 0) is mid point PQR and coordinate of Q are (-4, 0).
Coordinate of R will be (4, 0)

Question 31.
Solution:
Base BC of an equilateral triangle ABC lies on y-axis in such a way that origin O (0, 0) lies is the middle of BC and coordinates of C are (0, -3).
Coordinates of B will be (0, 3)

Question 32.
Solution:
Let the points P (-1, y) lying on the line segment joining points A (-3, 10) and B (6, -8) divides it in the ratio m : n.

Question 33.
Solution:
In rectangle ABCD, A (-1, -1), B (-1, 4), C (5, 4), D (5, -1)
P, Q, R and S are the mid points of AB, BC, CD and DA respectively.

Question 34.
Solution:
P is mid point of line segment joining the points A (-10, 4) and B (-2, 0)

Question 35.
Solution:
Let coordinates of P and Q be (0, y) and (x, 0) respectively.
Let M (2, -5) be the mid-point of PQ.
By midpoint formula

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:
Let the other two vertices be (h, k) and (m, n).
Hence, the vertices in order are (3, 2), (-1, 0), (h, k) and (m, n).
It is to be kept in mind that the diagonals of a parallelogram bisect each other.
Hence, the point of intersection (2, -5) is the midpoint of the diagonal whose ends are (3, 2) and (h, k). Then

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B are helpful to complete your math homework.

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RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16C

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) In ∆ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -A)

Question 2.
Solution:
Vertices of quadrilateral ABCD are A (3,-1), B (9, -5), C (14, 0) and D (9, 19)
Join diagonal AC.

Question 3.
Solution:
PQRS is a quadrilateral whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2)
Join PR which forms two triangles PQR and PSR.

Question 4.
Solution:
ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4)
Join AC which terms two triangles ABC and ADC

Question 5.
Solution:

Question 6.
Solution:
D, E and F are the midpoints of the sides CB, CA and AB respectively of ∆ABC.
Vertices are A (2, 1), B (4, 3) and C (2, 5)

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)
D is midpoint of BC


Similarly area of ∆ACD = 5 sq. units (10 – 5 = 5 sq. units)
Hence, median divides the triangle into two triangles of equal in area.

Question 8.
Solution:
In ∆ABC, coordinates of A are (1, -4) and let C and E are the midpoints of AB and AC respectively.
Coordinates of F are (2, -1) and of E are (0, -1)
Let coordinates of B be (x1, y1) and of C be (x2, y2)

Question 9.
Solution:
A (6, 1), B (8, 2) and C (9, 4) are the three vertices of a parallelogram ABCD.
E is the midpoint of DC.
Join AE, AC and BD which intersects at O.
O is midpoint of AC

Question 10.
Solution:
(i) Vertices of a ∆ABC are A (1, -3), B (4, p) and C (-9, 7) and area = 15 sq. units

Question 11.
Solution:
Vertices of ∆ABC are A (k + 1, 1), B (4, -3) and C (7, -k) and area = 6 sq. units.

Question 12.
Solution:
Let the vertices of a triangle ABC are A (-2, 5), B (k, -4) and C (2k + 1, 10) and area = 53 sq. units

Question 13.
Solution:
(j) Points are A (2, -2), B (-3, 8) and C (-1, 4)

Question 14.
Solution:
Points are A (x, 2), B (-3, -4) and C (7, -5)

Question 15.
Solution:
Points are given A (-3, 12), B (7, 6) and C (x, 9)

Question 16.
Solution:
Points are given P (1, 4), Q (3, y) and R (-3, 16)

Question 17.
Solution:
The given points are A (-3, 9), B (2, y) and C (4, -5)

Question 18.
Solution:
The points are given A (8, 1), B (3, -2k) and C (k, -5)

Question 19.
Solution:
The points are given A (2, 1), B (x, y) and C (7, 5)

Question 20.
Solution:
The points are given A (x, y), B (-5, 7) and C (-4, 5)

Question 21.
Solution:
Points are given A (a, 0), B (0, b) and C (1,1)
Points are collinear.
Area of ∆ABC = 0

Question 22.
Solution:
The points are given P (-3, 9), Q (a, ti) and R (4, -5)
Points are collinear.
Area of ∆PQR = 0

Question 23.
Solution:
Vertices of ∆ABC are A (0, -1), B (2, 1) and C (0, 3)

Question 24.
Solution:
Let A (a, a²), B (b, b²) and C (0, 0)
For the points A, B and C to collinear area of ∆ABC must be zero.
Now, area of ∆ABC

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16A

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Question 1.
Solution:
(i) A (9, 3) and B (15, 11)

Question 2.
Solution:
Distance from origin O (0, 0) and the given points (x, y) = √(x² + y²)

Question 3.
Solution:
Points A (x, -1), B (5, 3) and AB = 5 units

Question 4.
Solution:
Points A (2, -3), B (10, y) and AB = 10

Question 5.
Solution:
Points P (x, 4), Q (9, 10) and PQ = 10

Question 6.
Solution:
Point A (x, 2) is equidistant from B (8, -2 and C (2, -2)

Question 7.
Solution:
A (0, 2) is equidistant from B (3, p) and C ip, 5)
Then AB = AC

Question 8.
Solution:
Let point P (x, 0) is on x-axis and P is equidistant from A (2, -5) and B (-2, 9)

Question 9.
Solution:
Let the points on x-axis be P (x,, 0) and Q (x2, 0) and A (11, -8)

or x – 5 = 0, then x = 5
Points are (17, 0) and (5, 0)

Question 10.
Solution:
Let point P (0, y) is on the y-axis, then

Question 11.
Solution:
P (x, y) is equidistant from A (5, 1) and B (-1, 5)

Question 12.
Solution:
P (x, y) is equidistant from A (6, -1) and B (2, 3)

Question 13.
Solution:
Let the coordinates of the points be O (x, y)

Question 14.
Solution:
Points A (4, 3) and B (x, 5) lie on a circle with centre O (2, 3)

Question 15.
Solution:
Point C (-2, 3) is equidistant from points A (3, -1) and B (x, 8)

Question 16.
Solution:
Point P (2, 2) is equidistant from the two points A (-2, k) and B (-2k, -3)

Question 17.
Solution:
(i) Let point P (x, y) is equidistant from A (a + b, b – a) and B (a – b, a + b), then

Question 18.
Solution:
We know that if the sum of any two of these distances is equal to the distance of the third, then the points are collinear.
Now, (i) Let the points are A (1, -1), B (5, 2), C (9, 5)

Question 19.
Solution:
The points are A (7, 10), B (-2, 5) apd C (3, -4)

Question 20.
Solution:
Points are A (3, 0), B (6, 4) and C (-1, 3)

Question 21.
Solution:
Points are A (5, 2), B (2, -2) and C (-2, t)

Question 22.
Solution:
Points are A (2, 4), B (2, 6) and C (2 + √3, 5)

Question 23.
Solution:
Let the points are A (-3, -3), B (3, 3), C (-3√3, 3√3)

Question 24.
Solution:
Points are A (-5, 6), B (3, 0), C (9, 8)

Question 25.
Solution:
Points are O (0, 0), A (3, √3) and B (3, -√3)

Question 26.
Solution:
(i) Points are A (3, 2), B (0, 5), C (-3, 2), D (0, -1)

Question 27.
Solution:
Points are A (-3, 2), B (-5, -5), C (2, -3), D (4, 4)

Question 28.
Solution:
Points are A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)

Question 29.
Solution:
Points are A (6, 1), B (8, 2), C (9, 4) and D (7, 3)

Question 30.
Solution:
Points are A (2, 1), B (5, 2), C (6, 4) and D (3, 3)

Question 31.
Solution:
Points are A (1, 2), B (4, 3), C (6, 6) and D (3, 5)

Question 32.
Solution:
(i) Points are A (-4, -1), B (-2, -4), C (4, 0) and D (2, 3)

Question 33.
Solution:

Question 34.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 10 Solutions Chapter 16 Co-ordinate Geometry Ex 16D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16A
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16B
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16C
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D
• RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry MCQS

Very-Short-Answer Questions
Question 1.
Solution:
Points are A (-1, y), B (5, 7) and centre O (2, -3y).
Points A and B lie on the circle with centre O.

or y + 1 = 0, then y = -1
y = -1, 7

Question 2.
Solution:
Point A (0, 2) is equidistant from B (3, p) and also from C (p, 5)

Question 3.
Solution:
Three vertices of a rectangle ABCD are B (4, 0), C (4, 3) and D (0, 3)
Its diagonal are equal.

Question 4.
Solution:
Point P (k – 1, 2) is equidistant from two points A (3, k) and B (k, 5)

Question 5.
Solution:
Let P (x, 2) divides the join of A (12, 5) and B (4, -3) in the ration m : n.

Question 6.
Solution:

Question 7.
Solution:
Vertices of ∆ABC are A (7, -3), B (5, 3) and C (3, -1)

Question 8.
Solution:
Point C (k, 4) divides the join of A (2, 6) and B (5, 1) in the ratio 2 : 3

Question 9.
Solution:
Let a point P (x, 0) on x-axis is equidistant from two points A (-1, 0) and B (5, 0)

Question 10.
Solution:
Distance between two points

Question 11.
Solution:
The points (3, a) lies on the line 2x – 3y = 5
It will satisfy it.
2 x 3 – 3 x a = 5
6 – 3a = 5 => 3a = 6 – 5 = 1
a = \(\frac { 1 }{ 3 }\)

Question 12.
Solution:
Points A (4, 3) and B (x, 5) lie on the circle with centre O (2, 3)
OA = OB

Question 13.
Solution:
P (x, y) is equidistant from the point A (7, 1) and B (3, 5)

Question 14.
Solution:
O (0, 0) is the centroid of ∆ABC whose vertices are A (a, b), B (b, c) and C (c, a)

Question 15.
Solution:
Coordinates of centroid

Question 16.
Solution:
Let point P (4, 5) divides the join of A (2, 3) and B (7, 8) in the ratio m : n

Question 17.
Solution:
Points are given A (2, 3), B (4, k) and C (6, -3)
Points are collinear.

Hope given RS Aggarwal Solutions Class 10 Chapter 16 Co-ordinate Geometry Ex 16D are helpful to complete your math homework.

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