Class 10

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Problem Set 1 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Choose correct alternative for each of the following questions.

Question 1.
To draw graph of 4x + 5y = 19, find y when x = 1.
(a) 4
(b) 3
(c) 2
(d) -3
Answer:
(b)

Question 2.
For simultaneous equations in variables x and y, D_x = 49, Dy = – 63, D = 7 then what is x?
(a) 7
(b) -7
(c) \(\frac { 1 }{ 7 } \)
(d) \(\frac { -1 }{ 7 } \)
Answer:
(a)

Question 3.
Find the value of

(a) -1
(b) -41
(c) 41
(d) 1
Answer:
(d)

Question 4.
To solvex + y = 3; 3x – 2y – 4 = 0 by determinant method find D.
(a) 5
(b) 1
(c) -5
(d) -1
Answer:
(c)

Question 5.
ax + by = c and mx + n y = d and an ≠ bm then these simultaneous equations have-
(a) Only one common solution
(b) No solution
(c) Infinite number of solutions
(d) Only two solutions.
Answer:
(a)

Question 2.
Complete the following table to draw the graph of 2x – 6y = 3.
Answer:

Question 3.
Solve the following simultaneous equations graphically.
i. 2x + 3y = 12 ; x – y = 1
ii. x – 3y = 1 ; 3x – 2y + 4 = 0
iii. 5x – 6y + 30 = 0; 5x + 4y – 20 = 0
iv. 3x – y – 2 = 0 ; 2x + y = 8
v. 3x + y= 10 ; x – y = 2
Answer:
i. The given simultaneous equations are

The two lines interest at point (3,2).
∴ x = 3 and y = 2 is the solution of the simultaneous equations 2x + 3y = 12 and x – y = 1.

ii. The given simultaneous equations are

The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.

iii. The given simultaneous equations are

The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.

iv. The given simultaneous equations are


The two lines intersect at point (2, 4).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.

v. The given simultaneous equations are

The two lines intersect at point (3, 1).
∴ x = 3 and y = 1 is the solution of the simultaneous equations 3x + y = 10 and x – y = 2.

Question 4.
Find the values of each of the following determinants.

Solution:

Question 5.
Solve the following equations by Cramer’s method.

Solution:
i. The given simultaneous equations are
6x – 3y = -10 …(i)
3x + 5y – 8 = 0
∴ 3x + 5y = 8 …(ii)
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 6, b₁ = -3, c₁ = 10 and
a₂ = 3, b₂ = 5, c₂ = 8

ii. The given simultaneous equations are
4m – 2n = -4 …(i)
4m + 3n = 16 …(ii)
Equations (i) and (ii) are in am + bn = c form.
Comparing the given equations with a₁m + b₁n = c₁ and a₂m + b₂n = c₂, we get
a₁ = 4, b₁ = -2, c₁ = -4 and
a₂ = 4, b₂ = 3, c₂ = 16


∴ (m, n) = (1, 4) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

iv. The given simultaneous equations are
7x + 3y = 15 …(i)
12y – 5x = 39
i.e. -5x + 12y = 39 …(ii)
Equations (i) and (ii) are in ax + by = c form.
Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 7, b₁ = 3, c₁ = 15 and
a₂ = -5, b₂ = 12, c₂ = 39

v. The given simultaneous equations are

∴ 4(x + y – 8) = 2(3x – y)
∴ 4x + 4y – 32 = 6x – 2y
∴ 6x – 4x – 2y – 4y = -32
∴ 2x – 6y = -32
∴ x – 3y = -16 …(ii)[Dividing both sides by 2]
Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with
a₁x + b₁y = c₁ and a₂x + b₂y = c₂, we get
a₁ = 1, b₁ = -1, c₁ = -4 and
a₂ = 1, b₂ = -3, c₂ = -16

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

Question 6.
Solve the following simultaneous equations:

Answer:
i. The given simultaneous equations are

Subtracting equation (iv) from (iii), we get

∴ (x, y) = (6, – 4) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are

Adding equations (v) and (vi), we get

iii. The given simultaneous equations are


∴ (x, y) = (1, 2) is the solution of the given simultaneous equations.

iv. The given simultaneous equations are

∴ Equations (i) and (ii) become 7q – 2p = 5 …(iii)
8q + 7p = 15 …(iv)
Multiplying equation (iii) by 7, we get
49q – 14p = 35 …(v)
Multiplying equation (iv) by 2, we get
16q + 14p = 30 …(vi)
Adding equations (v) and (vi), we get

Substituting q = 1 in equation (iv), we get
8(1) + 7p = 15
∴ 8 + 7p = 15
∴ 7p = 15 – 8
∴ 7p = 7
∴ p = \(\frac { 7 }{ 7 } \) = 1
∴ (P, q) = (1,1)
Resubstituting the values of p and q, we get
1 = \(\frac { 1 }{ x } \) and 1 = \(\frac { 1 }{ y } \)
∴ x = 1 and y = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

v. The given simultaneous equations are

Resubstituting the values of p and q, we get

∴ 3x + 4y = 10 …(v)
and 2x – 3y = 1 …(vi)
Multiplying equation (v) by 3, we get
9x + 12y = 30 …(vii)
Multiplying equation (vi) by 4, we get
8x – 12y = 4 …(viii)
Adding equations (vii) and (viii), we get

Substituting x = 2 in equation (v), we get
3(2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 10 – 6
⇒ y = 4/4 = 1
∴ y = 1
∴ (x, y) = (2, 1) is the solution of the given simultaneous equations.

Question 7.
Solve the following word problems, i. A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number.
Solution:
Let the digit in unit’s place be x
and that in the ten’s place be y.

ii. Kantabai bought 1 \(\frac { 1 }{ 2 } \) kg tea and 5 kg sugar from a shop. She paid ₹ 50 as return fare for rickshaw. Total expense was ₹ 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So, next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ₹ 880 for that. Find the rate of sugar and tea per kg.
Solution:
Let the rate of tea be ₹ x per kg and that of sugar be ₹ y per kg.
According to the first condition,
cost of 1 \(\frac { 1 }{ 2 } \) kg tea + cost of 5 kg sugar + fare for rickshaw = total expense

According to the second condition,
cost of 2 kg tea + cost of 7 kg sugar = total expense
2x + 7y = 880 …(ii)
Multiplying equation (i) by 2, we get
6x + 20y = 2600 …(iii)
Multiplying equation (ii) by 3, we get
6x + 21y = 2640 …(iv)
Subtracting equation (iii) from (iv), we get

∴ The rate of tea is ₹ 300 per kg and that of sugar is ₹ 40 per kg.

iii. To find number of notes that Anushka had, complete the following activity.

Solution:
Anushka had x notes of ₹ 100 and y notes of ₹ 50.
According to the first condition,
100x + 50y = 2500
∴ 2x + y = 50 …(i) [Dividing both sides by 50]
According to the second condition,
100y + 50x = 2000
∴ 2y + x = 40 … [Dividing both sides by 50]
i.e. x + 2y = 40
∴ 2x + 4y = 80 …(ii) [Multiplying both sides by 2]
Subtracting equation (i) from (ii), we get

∴ Anushka had 20 notes of ₹ 100 and 10 notes of ₹ 50.

iv. Sum of the present ages of Manish and Savita is 31, Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Solution:
Let the present ages of Manish and Savita be x years and y years respectively.
According to the first condition,
x + y = 31 …(i)
3 years ago,
Manish’s age = (x – 3) years
Savita’s age = (y – 3) years
According to the second condition,
(x – 3) = 4 (y – 3)
∴ x – 3 = 4y – 12
∴ x – 4y = -12 + 3
∴ x – 4y = -9 …(ii)
Subtracting equation (ii) from (i), we get

∴ x = 31 – 8
∴ x = 23
The present ages of Manish and Savita are 23 years and 8 years respectively.

v. In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Solution:
Let the daily wages of skilled workers be ₹ x
that of unskilled workers be ₹ y.
According to the first condition,

∴ The daily wages of skilled workers is ₹ 450 and that of unskilled workers is ₹ 270.

vi. Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A), Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Solution:
Let the speeds of Hamid and Joseph be x km/hr andy km/hr respectively.
Distance travelled by Hamid in 20 minutes

According to the first condition,

∴ The speeds of Hamid and Joseph 50 km/hr and 40 km/hr respectively.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.4 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.
Solve the following simultaneous equations.

Solution:
i. The given simultaneous equations are

∴ Equations (i) and (ii) become
2p – 3q = 15 …(iii)
8p + 5q = 77 …(iv)
Multiplying equation (iii) by 4, we get
8p – 12q = 60 …(v)
Subtracting equation (v) from (iv), we get

ii. The given simultaneous equations are



Substituting x = 3 in equation (vi), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

∴ Equations (i) and (ii) become
27p + 31q = 85 …(iii)
31p + 27q = 89 …(iv)
Adding equations (iii) and (iv), we get

iv. The given simultaneous equations are


Substituting x = 1 in equation (vi), we get
3(1) + y = 4
∴ 3 + y = 4
∴ y = 4 – 3 = 1
∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.

Question 1.
Complete the following table. (Textbook pg. no. 16)
Solution:

Question 2.
In the above table the equations are not linear. Can you convert the equations into linear equations? (Textbook pg. no. 17)
Answer:
Yes, the above given simultaneous equations can be converted to a pair of linear equations by making suitable substitutions.

Steps for solving equations reducible to a pair of linear equations.

• Step 1: Select suitable variables other than those which are in the equations.
• Step 2: Replace the given variables with new variables such that the given equations become linear equations in two variables.
• Step 3: Solve the new simultaneous equations and find the values of the new variables.
• Step 4: By resubstituting the value(s) of the new variables, find the replaced variables which are to be determined.

Question 3.
To solve given equations fill the below boxes suitably. (Text book pg.no. 19)
Answer:

Question 4.
The examples on textbook pg. no. 17 and 18 obtained by transformation are solved by elimination method. If you solve these equations by graphical method and by Cramer’s rule will you get the same answers? Solve and check it. (Textbook pg. no. 18)

Solution:

The two lines intersect at point (1,-1).
∴ p = 1 and q = -1 is the solution of the simultaneous equations 4p + q = 3 and 2p – 3q = 5.
Re substituting the values of p and q, we get

The two lines intersect at point (0, -1).
∴ x = 0 and y = -1 is the solution of the simultaneous equations x – y = 1 and x + y = -1.
∴ (x, y) = (0, -1) is the solution of the given simultaneous equations.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.2 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

10th Maths 2 Practice Set 1.2 Question 1.
Complete the following table to draw graph of the equations.
i. x + y = 3
ii. x – y = 4
Answer:
i. x + y = 3

ii. x – y = 4

Linear Equations In Two Variables Practice Set 1.2  Question 2.
Solve the following simultaneous equations graphically.
i. x + y = 6 ; x – y = 4
ii. x + y = 5 ; x – y = 3
iii. x + y = 0 ; 2x – y = 9
iv. 3x – y = 2 ; 2x – y = 3
v. 3x – 4y = -7 ; 5x – 2y = 0
vi. 2x – 3y = 4 ; 3y – x = 4
Solution:
i. The given simultaneous equations are
x + y = 6                                                                                                        x – y = 4
∴ y = 6 – x                                                                                                     ∴ y = x – 4
The two lines intersect at point (5, 1).
∴ x = 5 and y = 1 is the solution of the simultaneous equations x + y = 6 and x – y = 4.

ii. The given simultaneous equations are


The two lines intersect at point (4, 1).
∴ x = 4 and y = 1 is the solution of the simultaneous equations x+y = 5 and x – y = 3.

iii. The given simultaneous equations are


The two lines intersect at point (3, -3).
∴ x = 3 and y = -3 is the solution of the simultaneous equations x + y = 0 and 2x – y = 9.

iv. The given simultaneous equations are


The two lines intersect at point (-1, -5).
∴ x = -1 and y = -5 is the solution of the simultaneous equations 3x- y = 2 and 2x- y = 3.

v. The given simultaneous equations are


The two lines intersect at point (1, 2.5).
∴ x = 1 and y = 2.5 is the solution of the simultaneous equations 3x – 4y = -7 and 5x – 2y = 0.

vi. The given simultaneous equations are


The two lines intersect at point (8, 4).
∴ x = 8 and y = 4 is the solution of the simultaneous equations 2x – 3y = 4 and 3y – x = 4.

10th Math Part 2 Practice Set 1.2  Question 1.
Solve the following simultaneous equations by graphical method. Complete the following tables to get ordered pairs.

i. Plot the above ordered pairs on the same co-ordinate plane.
ii. Draw graphs of the equations.
iii. Note the co-ordinates of the point of intersection of the two graphs. Write solution of these equations. (Textbook pg. no. 8)
Solution:

The two lines intersect at point (-1, -2).
∴ (x , y) = (-1, -2) is the solution of the given simultaneous equations.

Mathematics Part 1 Standard 9 Practice Set 1.2 Answer  Question 1.
Solve the above equations by method of elimination. Check your solution with the solution obtained by graphical method. (Textbook pg. no. 8)
Solution:
The given simultaneous equations are
x – y = 1 …(i)
5x – 3y = 1 …(ii)
Multiplying equation (i) by 3, we get
3x – 3y = 3 …(iii)
Subtracting equation (iii) from (ii), we get

Substituting x = -1 in equation (i), we get
-1 -y= 1
∴ -y = 1 + 1
∴ -y = 2
∴ y = -2
∴ (x,y) = (-1, -2) is the solution of the given simultaneous equations.
∴ The solution obtained by elimination method and by graphical method is the same.

1.2 Maths Class 10 Question 2.
The following table contains the values of x and y co-ordinates for ordered pairs to draw the graph of 5x – 3y = 1.

i. Is it easy to plot these points?
ii. Which precaution is to be taken to find ordered pairs so that plotting of points becomes easy? (Textbook pg. no. 8)
Solution:
i. No

The above numbers are non-terminating and recurring decimals.
∴ It is not easy to plot the given points.

ii. While finding ordered pairs, numbers should be selected in such a way that the co-ordinates obtained will be integers.

Linear Equations ¡n Two Variables Class 10 Maths Question 3.
To solve simultaneous equations x + 2y = 4; 3x + 6y = 12 graphically, following are the ordered pairs.

Plotting the above ordered pairs, graph is drawn. Observe it and find answers of the following questions.

i. Are the graphs of both the equations different or same?
ii. What are the solutions of the two equations x + 2y = 4 and 3x + 6y = 12? How many solutions are possible?
iii. What are the relations between coefficients of x, coefficients of y and constant terms in both the equations?
iv. What conclusion can you draw when two equations are given but the graph is only one line? (Textbook pg. no. 9)
Solution:
i. The graphs of both the equations are same.
ii. The solutions of the given equations are (-2, 3), (0, 2), (1, 1.5), etc.
∴ Infinite solutions are possible.
iii. Ratio of coefficients of x = \(\frac { 1 }{ 3 } \)
Ratio of coefficients of y = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
Ratio of constant terms = \(\frac { 4 }{ 12 } \) = \(\frac { 1 }{ 3 } \)
∴ Ratios of coefficients of x = ratio of coefficients of y = ratio of the constant terms
iv. When two equations are given but the graph is only one line, the equations will have infinite solutions.

Class 10 Maths Part 1 Practice Set 1.2 Question 4.
Draw graphs of x- 2y = 4, 2x – 4y = 12 on the same co-ordinate plane. Observe it. Think of the relation between the coefficients of x, coefficients ofy and the constant terms and draw the inference. (Textbook pg. no. 10)
Solution:

ii. Ratio of coefficients of x =\(\frac { 1 }{ 2 } \)
Ratio of coefficients of y = \(\frac { -2 }{ -4 } \) = \(\frac { 1 }{ 2 } \)
Ratio of constant terms = \(\frac { 4 }{ 12 } \) = \(\frac { 1 }{ 3 } \)
∴ Ratio of coefficients of x = ratio of coefficients of y ratio of constant terms
iii. If ratio of coefficients of x = ratio of coefficients of y ≠ ratio of constant terms, then the graphs of the two equations will be parallel to each other.

Condition of consistency in Equations:

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 1 Linear Equations in Two Variables.

Practice Set 1.1 Algebra 10th Std Maths Part 1 Answers Chapter 1 Linear Equations in Two Variables

Question 1.
Complete the following activity to solve the simultaneous equations.
5x + 3y = 9 …(i)
2x-3y=12 …(ii)
Solution:
5x + 3y = 9 …(i)
2x-3y=12 …(ii)
Add equations (i) and (ii).

Question 2.
Solve the following simultaneous equations.
i. 3a + 5b = 26; a + 5b = 22
ii. x + 7y = 10; 3x – 2y = 7
iii. 2x – 3y = 9; 2x + y = 13
iv. 5m – 3n = 19; m – 6n = -7
v. 5x + 2y = -3;x + 5y = 4
vi. \(\frac { 1 }{ 3 } \) x+ y = \(\frac { 10 }{ 3 } \) ; 2x + \(\frac { 1 }{ 4 } \) y = \(\frac { 11 }{ 4 } \)
vii. 99x + 101y = 499 ; 101x + 99y = 501
viii. 49x – 57y = 172; 57x – 49y = 252
Solution:
i. 3a + 5b = 26 …(i)
a + 5b = 22 …(ii)
Subtracting equation (ii) from (i), we get

Substituting a = 2 in equation (ii), we get
2 + 5b = 22
∴ 5b = 22 – 2
∴ 5b = 20
∴ b = \(\frac { 20 }{ 5 } \) =4
∴ (a, b) = (2, 4) is the solution of the given simultaneous equations.

ii. x + 7y = 10
∴ x = 10 – 7y …(i)
3x – 2y = 7 …1(ii)
Substituting x = 10 – ly in equation (ii), we get
3 (10 – 7y) – 2y = 7
∴ 30 – 21y – 2y = 7
∴ -23y = 7 – 30
∴ -23y = -23
∴ y = \(\frac { -23 }{ -23 } \)
Substituting y = 1 in equation (i), we get
x = 10 – 7 (1)
= 10 – 7 = 3
∴ (x, y) = (3, 1) is the solution of the given simultaneous equations.

iii. 2x – 3y = 9 …(i)
2x + y = 13 …(ii)
Subtracting equation (ii) from (i), we get

∴ (x, y) = (6, 1) is the solution of the given simultaneous equations.

iv. 5m – 3n = 19 …(i)
m – 6n = -7
∴ m = 6n – 7 …(ii)
Substituting m = 6n – 7 in equation (i), we get
5(6n – 7) – 3n = 19
∴ 30n – 35 – 3n = 19
∴ 27n = 19 + 35
∴ 27n = 54
∴ n = \(\frac { 54 }{ 27 } \) = 2
Substituting n = 2 in equation (ii), we get
m = 6(2) – 7
= 12 – 7 = 5
∴ (m, n) = (5, 2) is the solution of the given simultaneous equations.

v. 5x + 2y = -3 …(i)
x + 5y = 4
∴ x = 4 – 5y …(ii)
Substituting x = 4 – 5y in equation (i), we get
5(4 – 5y) + 2y = -3
∴ 20 – 25y + 2y = -3
∴ -23y = -3 – 20
∴ -23y = -23
∴ y = \(\frac { -23 }{ -23 } \) = 1
Substituting y = 1 in equation (ii), we get
x = 4 – 5(1)
= 4 – 5 = -1
∴ (x, y) = (-1, 1) is the solution of the given simultaneous equations.

Substituting y = 3 in equation (i), we get
x = 10 – 3(3)
= 10 – 9 = 1
∴ (x, y) = (1, 3) is the solution of the given simultaneous equations.

vii. 99x + 101 y = 499 …(i)
101 x + 99y = 501 …(ii)
Adding equations (i) and (ii), we get

Substituting x = 3 in equation (iii), we get
3 + y = 5
∴ y = 5 – 3 = 2
∴ (x, y) = (3, 2) is the solution of the given simultaneous equations.

viii. 49x – 57y = 172 …(i)
57x – 49y = 252 …(ii)
Adding equations (i) and (ii), we get

Substituting x = 7 in equation (iv), we get
7 + y = 10
∴ y = 10 – 7 = 3
∴ (x, y) = (7, 3) is the solution of the given simultaneous equations.

Complete the following table. (Textbook pg. no. 1)

Question 1.
Solve: 3x+ 2y = 29; 5x – y = 18 (Textbook pg. no. 3)
Solution:
3x + 2y = 29 …(i)
and 5x- y = 18 …(ii)
Let’s solve the equations by eliminating ‘y’.
Fill suitably the boxes below.
Multiplying equation (ii) by 2, we get

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Practice Set 7.3 Geometry 10th Std Maths Part 2 Answers Chapter 7 Mensuration

Practice Set 7.3 Geometry Class 10 Question 1.
Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)
Given : Radius (r) = 10 cm,
Measure of the arc (θ) = 54°
To find : Area of the sector.
Solution:
Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\)
= \(\frac { 54 }{ 360 } \) × 3.14 × (10)²
= \(\frac { 3 }{ 20 } \) × 3.14 × 100
= 3 × 3.14 × 5
= 15 × 3.14
= 47.1 cm²
∴ The area of the sector is 47.1 cm².

Mensuration Practice Set 7.3 Question 2.
Measure of an arc of a circle is 80° and its radius is 18 cm. Find the length of the arc. (π = 3.14)
Given: Radius (r) = 18 cm,
Measure of the arc (θ) = 80°
To find: Length of the arc.
Solution:
Length of arc = \(\frac{\theta}{360} \times 2 \pi r\) × 2πr
= \(\frac { 8 }{ 360 } \) × 2 × 3.14 × 18
= \(\frac { 2 }{ 9 } \) × 2 × 3.14 × 18
= 2 × 2 × 3.14 × 2 = 25.12 cm
∴ The length of the arc is 25.12 cm.

Practice Set 7.3 Geometry Question 3.
Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.
Solution:
Given: Radius (r) = 3.5 cm,
length of arc (l) = 2.2 cm
To find: Area of the sector.
Solution:
Area of sector = \(\frac{l \times \mathrm{r}}{2}\)
= \(\frac{2.2 \times 3.5}{2}\)
= 1.1 × 3.5 = 3.85 cm²
∴ The area of the sector is 3.85 cm².

Question 4.
Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector, (π = 3.14)
Given: Radius (r) = 10 cm,
area of minor sector =100 cm²
To find: Area of maj or sector.
Solution:
Area of circle = πr²
= 3.14 × (10)²
= 3.14 × 100 = 314 cm²
Now, area of major sector
= area of circle – area of minor sector
= 314 – 100
= 214 cm²
∴ The area of the corresponding major sector is 214 cm².

Question 5.
Area of a sector of a circle of radius 15 cm is 30 cm². Find the length of the arc of the sector.
Given: Radius (r) =15 cm,
area of sector = 30 cm²
To find: Length of the arc (l).
Solution:

∴ The length of the arc is 4 cm.

Practice Set 7.3 Question 6.
In the adjoining figure, radius of the circle is 7 cm and m (arc MBN) = 60°, find
i. Area of the circle.
ii. A(O-MBN).
iii. A(O-MCN).

Given: radius (r) = 7 cm,
m(arc MBN) = θ = 60°
Solution:
i. Area of circle = πr²
= \(\frac { 22 }{ 7 } \) × (7)²
= 22 × 7
= 154 cm²
∴ The area of the circle is 154 cm²

ii. Central angle (θ) = ∠MON = 60°
Area of sector = \(\frac{\theta}{360} \times \pi r^{2}\)
∴ A(O – MBN) = \(\frac { 60 }{ 360 } \) × \(\frac { 22 }{ 7 } \) × (7)²
\(\frac { 1 }{ 6 } \) × 22 × 7
= 25.67 cm²
= 25.7 cm²
∴ A(O-MBN) = 25.7 cm²

iii. Area of major sector = area of circle – area of minor sector
∴ A(O-MCN) = Area of circle – A(O-MBN)
= 154 – 25.7
∴ A(O-MCN) = 128.3 cm²

Question 7.
In the adjoining figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC).
Given: Radius (r) = 3.4 cm,
perimeter of sector 12.8 cm
To find: A(P-ABC)

Solution:
Perimeter of sector
= Iength of arc ABC + AP + CP

∴ 12.8 = l + 3.4 + 3.4
∴ 12.8 = l + 6.8
∴ l = 12.8 – 6.8 = 6cm

∴ A(P-ABC) = 10.2 cm²

7.3 Class 10 Question 8.
In the adjoining figure, O is the centre of the sector. ∠ROQ = ∠MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and (π = \(\frac { 22 }{ 7 } \))

Given: ∠ROQ = ∠MON = 60°,
radius (r) = OR = 7 cm, radius (R) = OM = 21 cm
To find: Lengths of arc RXQ and arc MYN.
Solution:
i. Length of arc RXQ = \(\frac{\theta}{360} \times 2 \pi r\)
= \(\frac { 60 }{ 2 } \) × 2 × \(\frac { 22 }{ 7 } \) × 7
= \(\frac { 1 }{ 6 } \) × 2 × 22
= 7.33 cm
ii. Length of arc MYN = \(\frac{\theta}{360} \times 2 \pi R\)
= \(\frac { 60 }{ 2 } \) × 2 × \(\frac { 22 }{ 7 } \) × 21
= \(\frac { 1 }{ 6 } \) × 2 × 22 × 3
= 22 cm
∴ The lengths of arc RXQ and arc MYN are 7.33 cm and 22 cm respectively.

Question 9.
In the adjoining figure, if A(P-ABC) = 154 cm², radius of the circle is 14 cm, find
i. ∠APC,
ii. l(arc ABC).

Given: A(P – ABC) = 154 cm²,
radius (r) = 14 cm
Solution:
i. Let ∠APC = θ

Std 10 Geometry Mensuration Question 10.
Radius of a sector of a circle is 7 cm. If measure of arc of the sector is
i. 30°
ii. 210°
iii. three right angles, find the area of the sector in each case.
Given: Radius (r) = 7 cm
To find: Area of the sector.
Solution:
i. Measure of the arc (θ) = 30°

∴ Area of the sector is 12.83 cm².
ii. Measure of the arc (θ) = 210°

∴ Area of the sector is 89.83 cm².
iii. Measure of the arc (θ) = 3 right angle
= 3 × 90° = 270°

∴ Area of the sector is 115.50 cm².

Mensuration Practice Question 11.
The area of a minor sector of a circle is 3.85 cm² and the measure of its central angle is 36°. Find the radius of the circle.
Given: Area of minor sector = 3.85 cm²,
central angle (θ) = 36°
To find: Radius of the circle (r).
Solution:
Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\)

∴ The radius of the circle ¡s 3.5 cm.

10th Geometry Practice Set 7.3 Question 12.
In the given figure, ꠸PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Given: In rectangle PQRS,
PQ = 14 cm, QR = 21 cm
To find: Areas of the parts x, y and z.
Solution:

∠Q = ∠R = θ = 90° …[Angles of a rectangle]

For the sector (Q – PA),
PQ = QA …[Radii of the same circle]
∴ QA = 14 cm
Now, QR = QA + AR … [Q – A – R]
∴ 21 = 14 + AR
∴ AR = 7 cm

Area of rectangle = length × breadth
area of ꠸PQRS = PQ × QR
= 14 × 21
= 294 cm²
Area of part z = area of ꠸PQRS
– area of part x – area of part y
= 294 – 154 – 38.5
= 101.5 cm²
∴ The area of part x is 154 cm², the area of part y is 38,5 cm² and the area of part z is 101.5 cm².

Question 13.
∆ALMN is an equilat triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm. Find,

i. A (∆ LMN).
ii. Area of any one of the sectors.
iii. Total area of all the three sectors.
iv. Area of the shaded region. (\(\sqrt { 3 }\) = 1.732 )
Given: In equilateral triangle LMN, LM =14 cm,
radius of sectors (r) = 7 cm
Solution:
i. ∆LMN is an equilateral triangle.

ii. Central angle (θ) = 60° …[Angle of an equilateral triangle]

∴ Area of one sector = 25.67 cm²
iii. Total area of all three sectors
= 3 × Area of one sector
= 3 × 25.67
= 77.01 cm²
∴ Total area of all three sectors = 77.01 cm²
iv. Area of shaded region
= A(∆LMN) – total area of all three sectors
= 84.87 – 77.01
= 7.86 cm²
∴ Area of shaded region = 7.86 cm

Maharashtra Board Class 10 Maths Chapter 7 Mensuration Intext Questions and Activities

Mensuration Practice Set 7.3 Question 1.
Complete the following table with the help of given figure. (Textbook pg. no. 149)


Solution:

Question 2.
Observe the figures below. Radii of all circles are equal. Observe the areas of the shaded regions and complete the following table. (Textbook pg. no. 150)

Solution:

Thus, if measure of an arc of a circle is θ, then
Area of sector (A) = \(\frac{\theta}{360}\) × Area of circle
∴ Area of sector (A) = \(\frac{\theta}{360}\) × πr²

Mensuration In Maths Question 3.
In the following figures, radii of all circles are equal. Observe the length of arc in each figure and complete the table. (Textbook pg. no. 151)

Solution:

Thus, if the measure of an arc of a circle is 0, then
Length of arc (l) = \(\frac{\theta}{360}\) × circumference of circle
∴ Length of arc (l) = \(\frac{\theta}{360}\) × 2πr

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Practice Set 7.2 Geometry 10th Std Maths Part 2 Answers Chapter 7 Mensuration

Question 1.
The radii of two circular ends of frustum shaped bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm³)
Given: Radii (r₁) = 14 cm, and (r₂) = 7 cm,
height (h) = 30 cm
To find: Amount of water the bucket can hold.
Solution:
Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r₁² + r₂² + r₁ × r₂)

∴ The bucket can hold 10.78 litres of water.

Question 2.
The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its
i. curved surface area,
ii. total surface area,
iii. volume, (π = 3.14)
Given: Radii (r₁) = 14 cm, and (r₂) = 6 cm,
height (h) = 6 cm
Solution:

i. Curved surface area of frustum
= πl (r₁ + r₂)
= 3.14 × 10(14 + 6)
= 3.14 × 10 × 20 = 628 cm²
∴ The curved surface area of the frustum is 628 cm².

ii. Total surface area of frustum
= πl (r₁+ r₂) + πr₁² + πr₂²
= 628 + 3.14 × (14)² + 3.14 × (6)²
= 628 + 3.14 × 196 + 3.14 × 36
= 628 + 3.14(196 + 36)
= 628 + 3.14 × 232
= 628 + 728.48
= 1356.48 cm²
∴ The total surface area of the frustum is 1356.48 cm².

iii. Volume of frustum
= \(\frac { 1 }{ 3 } \) πth(r₁² +r₂² + r₁ × r₂)
= \(\frac { 1 }{ 3 } \) × 3.14 × 6(14² + 6² + 14 × 6)
= 3.14 × 2(196 + 36 + 84)
= 3.14 × 2 × 316
= 1984.48 cm³
∴ The volume of the frustum is 1984.48 cm³.

Question 3.
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of frustum, complete the following activity. (π = \(\frac { 22 }{ 7 } \))
Solution:
Circumference1 = 27πr₁ = 132 cm

Curved surface area of frustum = π (r₁ + r₂) l
= π (21 + 14) × 25
=π × 35 × 35
= \(\frac { 22 }{ 7 } \) × 35 × 25
= 2750 cm²

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.1  Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Practice Set 7.1 Geometry 10th Std Maths Part 2 Answers Chapter 7 Mensuration

Practice Set 7.1 Geometry 10th Question 1. Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.
Given: For the cone,
radius (r) = 1.5 cm,
perpendicular height (h) = 5 cm
To find: Volume of the cone.
Solution:
Volume of cone = \(\frac { 1 }{ 3 } \) πr²h

∴ The volume of the cone is 11.79 cm³.

Mensuration Practice Set 7.1 Question 2. Find the volume of a sphere of diameter 6 cm. [π = 3.14]
Given: For the sphere, diameter (d) = 6 cm
To find: Volume of the sphere.
Solution:
Radius (r) = \(\frac { d }{ 2 } \) = \(\frac { 6 }{ 2 } \) = 3 cm
Volume of sphere = \(\frac { 4 }{ 3 } \) πr²
= \(\frac { 4 }{ 3 } \) × 3.14 × (3)³
= 4 × 3.14 × 3 × 3
= 113.04 cm³
∴ The volume of the sphere is 113.04 cm³.

Practice Set 7.1 Geometry Class 10 Question 3. Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm. [π = 3.14]
Given: For the cylinder,
radius (r) = 5 cm,
height (h) = 40 cm
To find: Total surface area of the cylinder.
Solution:
Total surface area of cylinder = 2πr (r + h)
= 2 × 3.14 × 5 (5 + 40)
= 2 × 3.14 × 5 × 45
= 1413 cm²
The total surface area of the cylinder is 1413 cm².

Practice Set 7.1 Geometry Question 4. Find the surface area of a sphere of radius 7 cm.
Given: For the sphere, radius (r) = 7 cm
To find: Surface area of the sphere.
Solution:
Surface area of sphere = Aπr²
= 4 × \(\frac { 22 }{ 7 } \) × (7)²
= 88 × 7
= 616 cm²
∴ The surface area of the sphere is 616 cm².

Practice Set 7.1 Question 5. The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.
Given: For the cuboid,
length (l) = 44 cm, breadth (b) = 21 cm,
height (h) = 12 cm
For the cone, height (H) = 24 cm
To find: Radius of base of the cone (r).
Solution:
Volume of cuboid = l × b × h
= 44 × 21 × 12 cm³
Volume of cone = \(\frac { 1 }{ 3 } \) πr²H
= \(\frac { 1 }{ 3 } \) × \(\frac { 22 }{ 7 } \) × r² × 24 cm³
Since the cuboid is melted to form a cone,
∴ volume of cuboid = volume of cone

∴ r² = 21 × 21
∴ r = 21 cm …[Taking square root of both sides]
∴ The radius of the base of the cone is 21 cm.

10th Class Geometry Practice Set 7.1 Question 6. Observe the measures of pots in the given figures. How many jugs of water can the cylindrical pot hold?

Given: For the conical water jug,
radius (r) = 3.5 cm, height (h) = 10 cm
For the cylindrical water pot,
radius (R) = 7 cm, height (H) = 10 cm
To find: Number of jugs of water the cylindrical pot can hold.
Solution:
Volume of conical jug = \(\frac { 1 }{ 3 } \) πr²h

∴ The cylindrical pot can hold 12 jugs of water.

Mensuration Class 10 Practice Set 7.1 Question 7. A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm². The cone is placed up on the cylinder. Volume of the solid figure so formed is 500 cm³. Find the total height of the figure

Given: For the cylindrical part,
height (h) = 3 cm,
area of the base (πr²)= 100 cm²
Volume of the entire figure = 500 cm³
To find: Total height of the figure.
Solution:
A cylinder and a cone have equal bases.
∴ they have equal radii.
radius of cylinder = radius of cone = r
Area of base = 100 cm²
∴ πr² =100 …(i)
Let the height of the conical part be H.
Volume of the entire figure
= Volume of the entire + Volume of cone

∴ Height of conical part (H) =6 cm
Total height of the figure = h + H
= 3 + 6
= 9 cm
∴ The total height of the figure is 9 cm.

10th Geometry Practice Set 7.1 Question 8. In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

Given: For the conical Part,
height (h) = 4 cm, radius (r) = 3 cm
For the cylindrical part,
height (H) = 40 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Total area of the toy.
Solution:
Slant height of cone (l) = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}\)
= \(\sqrt{\mathrm{4}^{2}+\mathrm{3}^{2}}\)
= \(\sqrt { 16+9 }\)
= \(\sqrt { 25 }\) = 5 cm
Curved surface area of cone = πrl
= π × 3 × 5
= 15π cm²
Curved surface area of cylinder = 2πrH
= 2 × π × 3 × 40
= 240π cm²
Curved surface area of hemisphere = 2πr²
= 2 × π × 3²
= 18π cm²
Total area of the toy
= Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere
= 15π + 240π + 18π
= 2737 π cm²
∴ The total area of the toy is 273π cm².

7.1.8 Practice Question 9. In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

Given: For the cylindrical tablets,
radius (r) = 7 mm,
thickness = height(h) = 5 mm
For the cylindrical wrapper,
diameter (D) = 14 mm, height (H) = 10 cm
To find: Number of tablets that can be wrapped.
Solution:
Radius of wrapper (R) = \(\frac { Diameter }{ 2 } \)
= \(\frac { 14 }{ 2 } \) = 7 mm
Height of wrapper (H) = 10 cm
= 10 × 10 mm
= 100 mm
Volume of a cylindrical wrapper = πR²H
= π(7)² × 100
= 4900π mm³
Volume of a cylindrical tablet = πr²h
= π(7)² × 5
= 245 π mm³
No. of tablets that can be wrapped

∴ 20 tables can be wrapped in the wrapper

Class 10 Maths 7.1 Question 10. The given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure.
(π = 3.14)
Given: For the conical part,
height (h) = 4 cm, radius (r) = 3 cm
For the hemispherical part,
radius (r) = 3 cm
To find: Volume and surface area of the toy.
Solution:

Now, volume of the toy
= Volume of cone + volume of hemisphere
= 12π + 18π
= 30π
= 30 × 3.14
= 94.20 cm³
Also, surface area of the toy
= Curved surface area of cone + Curved surface area of hemisphere
= 15π + 18π
= 33π
= 33 × 3.14
= 103.62 cm²
∴ The volume and surface area of the toy are 94.20 cm³ and 103.62 cm² respectively.

Question 11.
Find the surface area and the volume of a beach ball shown in the figure.

Given: For the spherical ball,
diameter (d) = 42 cm
To find: Surface area and volume of the beach ball.
Solution:
Radius (r) = \(\frac { d }{ 2 } \) = \(\frac { 42 }{ 2 } \) = 21 cm
Surface area of sphere= 4πr²
= 4 × 3.14 × (21)²
= 4 × 3.14 × 21 × 21
= 5538.96 cm²
Volume of sphere = \(\frac { 4 }{ 3 } \) πr³
= \(\frac { 4 }{ 3 } \) × 3.14 × (21)³
= 4 × 3.14 × 7 × 21 × 21
= 38772.72 cm³
∴ The surface area and the volume of the beach ball are 5538.96 cm² and 38772.72 cm³ respectively.

Question 12.
As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.

Given: For the metal sphere,
diameter (d) = 2 cm
For the cylindrical glass, diameter (D) =14 cm,
height of water in the glass (H) = 30 cm
To find: Volume of water in the glass.
Solution:
Let the radii of the sphere and glass be r and R respectively.

Volume of water with sphere in it = πR²H
= π × (7)² × 30
= 1470π cm³
Volume of water in the glass
= Volume of water with sphere in it – Volume of sphere

∴ The volume of the water in the glass is 1468.67 π cm³ (i.e. 4615.80 cm³).

Maharashtra Board Class 10 Maths Chapter 7 Mensuration Intext Questions and Activities

Question 1.
The length, breadth and height of an oil can are 20 cm, 20 cm and 30 cm respectively as shown in the adjacent figure. How much oil will it contain? (1 litre = 1000 cm³) (Textbook pg. no.141)

Given: For the cuboidal can,
length (l) = 20 cm,
breadth (b) = 20 cm,
height (h) = 30 cm
To find: Oil that can be contained in the can.
Solution:
Volume of cuboid = l × b × h
= 20 × 20 × 30
= 12000 cm³
= \(\frac { 12000 }{ 1000 } \) litres
= 12 litres
∴ The oil can will contain 12 litres of oil.

Question 2.
The adjoining figure shows the measures of a Joker’s dap. How much cloth is needed to make such a cap? (Textbook pg. no. 141)

Given: For the conical cap,
radius (r) = 10 cm,
slant height (l) = 21 cm
To find: Cloth required to make the cap.
Solution:
Cloth required to make the cap
= Curved surface area of the conical cap
= πrl = \(\frac { 22 }{ 7 } \) × 10 × 21
=22 × 10 × 3
= 660 cm²
∴ 660 cm² of cloth will be required to make the cap.

Question 3.
As shown in the adjacent figure, a sphere is placed in a cylinder. It touches the top, bottom and the curved surface of the cylinder. If radius of the base of the cylinder is ‘r’,

i. what is the ratio of the radii of the sphere and the cylinder ?
ii. what is the ratio of the curved surface area of the cylinder and the surface area of the sphere?
iii. what is the ratio of the volumes of the cylinder and the sphere? (Textbook pg. no. 141)
Solution:
Radius of base of cylinder = radius of sphere
∴ Radius of sphere = r
Also, height of cylinder = diameter of sphere
∴ h = d
∴ h = 2r …(i)

∴ radius of sphere : radius of cylinder = 1 : 1.

∴ curved surface area of cylinder : surface area of sphere = 1:1.

∴ volume of cylinder : volume of sphere = 3 : 2.

Question 4.
Finding volume of a sphere using cylindrical beaker and water. (Textbook, pg. no. 142)

i. Take a ball and a beaker of the same radius.
ii. Cut a strip of paper of length equal to the diameter of the beaker.
iii. Draw two lines on the strip dividing it into three equal parts.
iv. Stick this strip on the beaker straight up from the bottom.
v. Fill the water in the beaker upto the first mark of the strip from bottom.
vi. Push the ball in the beaker so that it touches the bottom.
Observe how much water level rises.
You will notice that the water level has risen exactly upto the total height of the strip. Try to obtain the formula for volume of sphere using the volume of the cylindrical beaker.
Solution:
Suppose volume of beaker upto height 2r is V.
V = πr² h
∴ V = πr²(2r) …[∵ h = 2r]
∴ V = 2πr³
But, V = volume of the ball + volume of water in the beaker
∴ 2πr³ = Volume of the ball + \(\frac { 1 }{ 3 } \) × 2πr³
∴ Volume of the ball = 2πr³ – \(\frac { 2 }{ 3 } \) πr³
= \(\frac{6 \pi r^{3}-2 \pi r^{3}}{3}\)
∴ Volume of the ball = \(\frac { 4 }{ 3 } \) πr³
∴ Volume of a sphere = \(\frac { 4 }{ 3 } \) πr³

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Problem Set 6 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 6 Trigonometry.

Problem Set 6 Geometry 10th Std Maths Part 2 Answers Chapter 6 Trigonometry

Question 1.
Choose the correct alternative answer for the following questions.

i. sin θ.cosec θ = ?
(A) 1
(B) 0
(C) \(\frac { 1 }{ 2 } \)
(D) \(\sqrt { 2 }\)
Answer:
(A)

ii. cosec 45° = ?
(A) \(\frac{1}{\sqrt{2}}\)
(B) \(\sqrt { 2 }\)
(C) \(\frac{\sqrt{3}}{2}\)
(D) \(\frac{1}{\sqrt{3}}\)
Answer:
(B)

iii. 1 + tan² θ = ?
(A) cot² θ
(B) cosec² θ
(C) sec² θ
(D) tan² θ
Answer:
(C)

iv. When we see at a higher level, from the horizontal line, angle formed is ______
(A) angle of elevation.
(B) angle of depression.
(C) 0
(D) straight angle.
Answer:
(A)

Question 2.
If sin θ = \(\frac { 11 }{ 61 } \), find the value of cos θ using trigonometric identity.
Solution:
sin θ = \(\frac { 11 }{ 61 } \) … [Given]
We know that,
sin² θ + cos² θ = 1


…[Taking square root of both sides]

Question 3.
If tan θ = 2, find the values of other trigonometric ratios.
Solution:
tan θ = 2 …[Given]
We know that,
1 + tan² θ = sec⁷ θ
∴ 1 + (2)⁷ = sec⁷ θ
∴ 1 + 4 = sec⁷ θ
∴ sec⁷ θ = 5
∴ sec θ = \(\sqrt { 5 }\) …[Taking square root of both sides]

Question 4.
If sec θ = \(\frac { 13 }{ 12 } \), find the values of other trigonometric ratios.
Solution:
sec θ = \(\frac { 13 }{ 12 } \) … [Given]
We know that,
1 + tan² θ = sec² θ


∴ sin θ = \(\frac { 5 }{ 13 } \), cos θ = \(\frac { 12 }{ 13 } \), tan θ = \(\frac { 5 }{ 12 } \), cot θ = \(\frac { 12 }{ 5 } \), cosec θ = \(\frac { 13 }{ 5 } \)

Question 5.
Prove the following:
i. sec θ (1 – sin θ) (sec θ + tan θ) = 1
ii. (sec θ + tan θ) (1 – sin θ) = cos θ
iii. sec² θ + cosec² θ = sec² θ × cosec² θ
iv. cot² θ – tan² θ = cosec² θ – sec² θ
v. tan⁴ θ + tan² θ = sec⁴ θ – sec² θ
vi. \(\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\) = 2 sec2 θ
vii. sec⁶ x – tan⁶ x = 1 + 3 sec² x × tan² x

Proof:
i. L.H.S. = sec θ (1 – sin θ) (sec θ + tan θ)

∴ sec θ (1 – sin θ) (sec θ + tan θ) = 1

ii. L.H.S. = (sec θ + tan θ) (1 – sin θ)

∴ (sec θ + tan θ) (1 – sin θ) = cos θ

iii. L.H.S. = sec² θ + cosec² θ

∴ sec² θ + cosec² θ = sec² θ × cosec² θ

iv. L.H.S. = cot² θ – tan² θ
= (cosec² θ – 1) – (sec² θ – 1)
[∵ tan² θ = sec² θ – 1,
cot² θ = cosec² θ – 1]
= cosec² θ – 1 – sec² θ + 1
cosec² θ – sec² θ
= R.H.S.
∴ cot² θ – tan² θ = cosec² θ – sec² θ

v. L.H.S. = tan⁴ θ + tan² θ
= tan² θ (tan² θ + 1)
= tan² θ. sec² θ
…[∵ 1 + tan² θ = sec² θ]
= (sec² θ – 1) sec² θ
…[∵ tan² θ = sec² θ – 1]
= sec⁴ θ – sec² θ
= R.H.S.
∴ tan⁴ θ + tan² θ = sec⁴ θ – sec² θ

vii. L.H.S. = sec⁶ x – tan⁶ x
= (sec² x)³ – tan⁶ x
= (1 + tan² x)³ – tan⁶ x …[∵ 1 + tan² θ = sec² θ]
= 1 + 3tan² x + 3(tan² x)² + (tan² x)³ – tan⁶ x …[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1 + 3 tan² x (1 + tan² x) + tan⁶ x – tan⁶ x
= 1 + 3 tan² x sec² x …[∵ 1 + tan² θ = sec² θ]
= R.H.S.
∴ sec³x – tan⁶x = 1 + 3sec²x.tan²x


x. We know that,
sin² θ + cos² θ = 1
∴ 1 – sin² θ = cos² θ
∴ (1 – sin θ) (1 + sin θ) = cos θ. cos θ

Question 6.
A boy standing at a distance of 48 metres from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Solution:
Let AB represent the height of the building and point C represent the position of the boy.
Angle of elevation = ∠ACB = 30°
BC = 48 m

In right angled ∆ABC,
tan 30° = \(\frac { AB }{ BC } \) … [By definition]

∴ The height of the building is 16\(\sqrt { 3 }\) m.

Question 7.
From the top of the lighthouse, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the lighthouse is 100 metres, then find how far the ship is from the lighthouse.
Solution:
Let AB represent the height of lighthouse and point C represent the position of the ship.

Angle of depression ∠PAC 30°
AB = 100m.
Now, ray AP || seg BC
∴ ∠ACB = ∠PAC … [Alternate angles]
∴ ∠ACB = 30°
AB = 100m
In right angled ∆ABC,
tan 30° = \(\frac { AB }{ BC } \) …[By definition]
∴ \(\frac{1}{\sqrt{3}}=\frac{100}{\mathrm{BC}}\)
∴ BC = 100\(\sqrt { 3 }\)m
∴ The ship is 1oo\(\sqrt { 3 }\)m far from the lighthouse.

Question 8.
Two buildings are in front of each other on a road of width 15 metres. From the top of the first building, having a height of 12 metre, the angle of elevation of the top of the second building is 30°. What is the height of the second building?
Solution:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road.

Draw seg AM ⊥ seg CD
Angle of elevation = ∠CAM = 30°
AB = 12m
BD = 15m
In ꠸ ABDM,
∠B = ∠D = 90°
∠M 90° …[segAM ⊥ segCD]
∠A 90° …[Remaining angle of ꠸ABDM]
꠸ABDM is a rectangle …[Each angle is 90°]

∴ The height of the second building is 20.65 m.

Question 9.
A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
Solution:
Let AB represent the length of the ladder and AE represent the height of the platform.

Draw seg AC ⊥ seg BD.
Angle of elevation = ∠BAC = 70°
AB = 20 m
AE = 2m
In right angled ∆ABC,
sin 70° = \(\frac { BC }{ AB } \) …..[By definition]
∴ 0.94 = \(\frac { BC }{ 20 } \)
∴ BC = 0.94 × 20
= 18.80 m
In ꠸ACDE,
∠E = ∠D = 90°
∠C = 90° … [seg AC ⊥ seg BD]
∴ ∠A = 90° … [Remaining angle of ꠸ACDE]
∴ ꠸ACDE is a rectangle. … [Each angle is 90°]
∴ CD = AE = 2 m … [Opposite sides of a rectangle]
Now, BD = BC + CD … [B – C – D]
= 18.80 + 2
= 20.80 m
∴ The maximum height from the ground upto which the ladder can reach is 20.80 metres.

Question 10.
While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing, (sin 20° = 0.342)
Solution:
Let AC represent the initial height and point A represent the initial position of the plane.
Let point B represent the position where plane lands.
Angle of depression = ∠EAB = 20°

Now, seg AE || seg BC
∴ ∠ABC = ∠EAB … [Alternate angles]
∴ ∠ABC = 20°
Speed of the plane = 200 km/hr
= 200 × \(\frac { 1000 }{ 3600 } \) m/sec
= \(\frac { 500 }{ 9 } \) m/sec
∴ Distance travelled in 54 sec = speed × time
= \(\frac { 500 }{ 9 } \) × 54
= 3000 m
∴ AB = 3000 m
In right angled ∆ABC,
sin 20° = \(\frac { AC }{ AB } \) ….[By definition]
∴ 0.342 = \(\frac { AC }{ 3000 } \)
∴ AC = 0.342 × 3000
= 1026 m
∴ The plane was at a height of 1026 m when it started landing.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.3 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

Practice Set 5.3 Geometry 10th Std Maths Part 2 Answers Chapter 5 Co-ordinate Geometry

Practice Set 5.3 Geometry Class 10 Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
i. 45°
ii. 60°
iii. 90°
Solution:
i. Angle made with the positive direction of
X-axis (θ) = 45°
Slope of the line (m) = tan θ
∴ m = tan 45° = 1
∴ The slope of the line is 1.

ii. Angle made with the positive direction of X-axis (θ) = 60°
Slope of the line (m) = tan θ
∴ m = tan 60° = \(\sqrt { 3 }\)
∴ The slope of the line is \(\sqrt { 3 }\).

iii. Angle made with the positive direction of
X-axis (θ) = 90°
Slope of the line (m) = tan θ
∴ m = tan 90°
But, the value of tan 90° is not defined.
∴ The slope of the line cannot be determined.

Practice Set 5.3 Geometry Question 2. Find the slopes of the lines passing through the given points.
i. A (2, 3), B (4, 7)
ii. P(-3, 1), Q (5, -2)
iii. C (5, -2), D (7, 3)
iv. L (-2, -3), M (-6, -8)
v. E (-4, -2), F (6, 3)
vi. T (0, -3), s (0,4)
Solution:
i. A (x₁, y₁) = A (2, 3) and B (x₂, y₂) = B (4, 7)
Here, x₁ = 2, x₂ = 4, y₁ = 3, y₂ = 7

∴ The slope of line AB is 2.

ii. P (x₁, y₁) = P (-3, 1) and Q (x₂, y₂) = Q (5, -2)
Here, x₁ = -3, x₂ = 5, y₁ = 1, y₂ = -2

∴ The slope of line PQ is \(\frac { -3 }{ 8 } \)

iii. C (x₁, y₁) = C (5, -2) and D (x₂, y₂) = D (7, 3)
Here, x₁ = 5, x₂ = 7, y₁ = -2, y₂ = 3

∴ The slope of line CD is \(\frac { 5 }{ 2 } \)

iv. L (x₁, y₁) = L (-2, -3) and M (x₂,y₂) = M (-6, -8)
Here, x₁ = -2, x₂ = – 6, y₁ = – 3, y₂ = – 8

∴ The slope of line LM is \(\frac { 5 }{ 4 } \)

v. E (x₁, y₁) = E (-4, -2) and F (x₂, y₂) = F (6, 3)
Here,x₁ = -4, x₂ = 6, y₁ = -2, y₂ = 3

∴ The slope of line EF is \(\frac { 1 }{ 2 } \).

vi. T (x₁, y₁) = T (0, -3) and S (x₂, y₂) = S (0, 4)
Here, x₁ = 0, x₂ = 0, y₁ = -3, y₂ = 4

∴ The slope of line TS cannot be determined.

5.3.5 Practice Question 3. Determine whether the following points are collinear.
i. A (-1, -1), B (0, 1), C (1, 3)
ii. D (- 2, -3), E (1, 0), F (2, 1)
iii. L (2, 5), M (3, 3), N (5, 1)
iv. P (2, -5), Q (1, -3), R (-2, 3)
v. R (1, -4), S (-2, 2), T (-3,4)
vi. A(-4,4),K[-2,\(\frac { 5 }{ 2 } \)], N (4,-2)
Solution:


∴ slope of line AB = slope of line BC
∴ line AB || line BC
Also, point B is common to both the lines.
∴ Both lines are the same.
∴ Points A, B and C are collinear.

∴ slope of line DE = slope of line EF
∴ line DE || line EF
Also, point E is common to both the lines.
∴ Both lines are the same.
∴ Points D, E and F are collinear.

∴ slope of line LM ≠ slope of line MN
∴ Points L, M and N are not collinear.

∴ slope of line PQ = slope of line QR
∴ line PQ || line QR
Also, point Q is common to both the lines.
∴ Both lines are the same.
∴ Points P, Q and R are collinear.

∴ slope of line RS = slope of line ST
∴ line RS || line ST
Also, point S is common to both the lines.
∴ Both lines are the same.
∴ Points R, S and T are collinear.

∴ slope of line AK = slope of line KN
∴ line AK || line KN
Also, point K is common to both the lines.
∴ Both lines are the same.
∴ Points A, K and N are collinear.

Practice Set 5.3 Geometry 9th Standard Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Solution:

∴ The slopes of the sides AB, BC and AC are -5, \(\frac { 1 }{ 5 } \) and \(\frac { -2 }{ 3 } \) respectively.

Geometry 5.3 Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Proof:


∴ Slope of side AB = Slope of side CD … [From (i) and (iii)]
∴ side AB || side CD
Slope of side BC = Slope of side AD … [From (ii) and (iv)]
∴ side BC || side AD
Both the pairs of opposite sides of ꠸ABCD are parallel.
꠸ABCD is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.

Question 6.
Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Solution:
R(x₁, y₁) = R (1, -1), S (x₂, y₂) = S (-2, k)
Here, x₁ = 1, x₂ = -2, y₁ = -1, y₂ = k

But, slope of line RS is -2. … [Given]
∴ -2 = \(\frac { k+1 }{ -3 } \)
∴ k + 1 = 6
∴ k = 6 – 1
∴ k = 5

5.3 Class 10 Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Solution:
B(x₁, y₁) = B (k, -5), C (x₂, y₂) = C (1, 2)
Here, x₁ = k, x₂ = 1, y₁ = -5, y₂ = 2

But, slope of line BC is 7. …[Given]
∴ 7 = \(\frac { 7 }{ 1-k } \)
∴ 7(1 – k) = 7
∴ 1 – k = \(\frac { 7 }{ 7 } \)
∴ 1 – k = 1
∴ k = 0

Question 8.
Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Solution:

But, line PQ || line RS … [Given]
∴ Slope of line PQ = Slope of line RS
∴ 2 = \(\frac { k-1 }{ 2 } \)
∴ 4 = k – 1
∴ k = 4 + 1
∴ k = 5