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Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

9th Standard Maths 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 1.1 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 1.
Find the distances with the help of the number line given below.

i. d(B, E)
ii. d (J, J)
iii. d(P, C)
iv. d(J, H)
v. d(K, O)
vi. d(O, E)
vii. d(P, J)
viii. d(Q, B)
Solution:
i. Co-ordinate of the point B is 2.
Co-ordinate of the point E is 5.
Since, 5 > 2
∴ d(B, E) = 5 – 2
∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.
Co-ordinate of the point A is 1.
Since, 1 > -2
∴ d(J, A) = 1 – (-2)
= 1 + 2
∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.
Co-ordinate of the point C is 3.
Since, 3 > -4
∴ d(P,C) = 3 – (-4)
= 3 + 4
∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.
Co-ordinate of the point H is -1.
Since, -1 > -2
∴ d(J,H) = – 1 – (-2)
= -1 + 2
∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.
Co-ordinate of the point O is 0.
Since,0 > -3
∴ d(K, O) = 0 – (-3)
= 0 + 3
∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0.
∴ Co-ordinate of the point E is 5.
Since, 5 > 0
∴ d(O, E) = 5 – 0
∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4.
Co-ordinate of the point J is -2.
Since -2 > -4
∴ d(P, J) = -2 – (-4)
= – 2+ 4
∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5.
Co-ordinate of the point B is 2.
Since,2 > -5
∴ d(Q,B) = 2 – (-5)
= 2 + 5
∴ d(Q, B) = 7

Question 2.
If the co-ordinate of A is x and that of B is . y, find d(A, B).
i. x = 1, y = 7
ii. x = 6, y = -2
iii. x = -3, y = 7
iv. x = -4, y = -5
v. x = -3, y = -6
vi. x = 4, y = -8
Solution:
i. Co-ordinate of point A is x = 1.
Co-ordinate of point B is y = 7
Since, 7 > 1
∴ d(A, B) = 7 – 1
∴ d(A, B) = 6

ii. Co-ordinate of point A is x = 6.
Co-ordinate of point B is y = -2.
Since, 6 > -2
∴ d(A, B) = 6 – ( -2) = 6 + 2
∴ d(A, B) = 8

iii. Co-ordinate of point A is x = -3.
Co-ordinate of point B is y = 7.
Since, 7 > -3
∴ d(A, B) = 7 – (-3) = 7 + 3
∴ d(A, B) = 10

iv. Co-ordinate of point A is x = -4.
Co-ordinate of point B is y = -5.
Since, -4 > -5
∴ d(A, B) = -4 – (-5)
= -4 + 5
∴ d(A, B) = 1

v. Co-ordinate of point A is x =-3.
Co-ordinate of point B is y = -6.
Since, -3 > -6
∴ d(A, B) = -3 – (-6)
= -3 + 6
∴ d(A, B) = 3

vi. Co-ordinate of point A is x = 4.
Co-ordinate of point B is y = -8.
Since, 4 > -8
∴ d(A, B) = 4 – (-8)
= 4 + 8
∴d(A, B) = 12

Question 3.
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
i. d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
ii. d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
iii. d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
iv. d(L, M) =11, d(M, N) = 12, d(N, L) = 8
v. d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
vi. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Solution:
i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, Q) = 10 …(i)
d(P, R) + d(Q, R) = 7 + 3 = 10 .. .(ii)
∴ d(P, Q) = d(P, R) + d(Q, R) …[From (i) and (ii)]
∴ Point R is between the points P and Q
i. e., P – R – Q or Q – R – P.
∴ Points P, R, Q are collinear.

ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
d(R, S) = 8 …(i)
d(S, T) + d(R, T) = 6 + 4 = 10 …(h)
∴ d(R, S) ≠ d(S, T) + d(R, T) … [From (i) and (ii)]
∴ The given points are not collinear.

iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(A, B) = 16 …(i)
d(C, A) + d(B, C) = 9 + 7 = 16 …(ii)
∴ d(A, B) = d(C, A) + d(B, C) …[From(i) and (ii)]
∴ Point C is between the points A and B.
i. e., A – C – B or B – C – A.
∴ Points A, C, B are collinear

iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
d(M, N) = 12 …(i)
d(L, M) + d(N, L) = 11 + 8 = 19 …(ii)
∴d(M, N) + d(L, M) + d(N, L) … [From (i) and (ii)]
∴ The given points are not collinear.

v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Y) = 15 …(i)
d(X,Z) + d(Y, Z) = 8 + 7= 15 …(ii)
∴ d(X, Y) = d(X, Z) + d(Y, Z) …[From (i) and (ii)]
∴ Point Z is between the points X and Y
i. e.,X – Z – Y or Y – Z – X.
∴ Points X, Z, Y are collinear.

vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
d(E, F) = 8 …(i)
d(D, E) + d(D, F) = 5 + 6 = 11 …(ii)
∴ d(E, F) ≠ d(D, E) + d(D, F) … [From (i) and (ii)]
∴ The given points are not collinear.

Question 4.
On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Solution:
Given, d(A, C) = 10, d(C, B) = 8.

Case I: Points A, B, C are such that, A – B – C.

∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.

∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.

From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.

Question 5.
Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Solution:
Given,d(X, Y) = 17, d(Y, Z) = 8
Case I: Points X, Y, Z are such that, X – Y – Z.

∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.

∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.

From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Question 6.
Sketch proper figure and write the answers of the following questions. [2 Marks each]
i. If A – B – C and l(AC) = 11,
l(BC) = 6.5, then l(AB) = ?
ii. If R – S – T and l(ST) = 3.7,
l(RS) = 2.5, then l(RT) = ?
iii. If X – Y – Z and l(XZ) = 3√7,
l(XY) = √7, then l(YZ) = ?
Solution:
i. Given, l(AC) =11, l(BC) = 6.5

l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5

l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,

l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

Question 7.
Which figure is formed by three non-collinear points?
Solution:
Three non-collinear points form a triangle.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Problem Set 2 Class 9 Answers

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.

10th Standard Maths 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.3 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

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

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.2 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
Find the co-ordinates of point P if P divides the line segment joining the points A (-1, 7) and B (4, -3) in the ratio 2:3.
Solution:
Let the co-ordinates of point P be (x, y) and A (x₁, y₁) B (x₂, y₂) be the given points.
Here, x₁ = -1, y₁ = 7, x₂ = 4, y₂ = -3, m = 2, n = 3
∴ By section formula,

∴ The co-ordinates of point P are (1,3).

Question 2.
In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
i. P (-3, 7), Q (1, -4), a : b = 2 : 1
ii. P (-2, -5), Q (4, 3), a : b = 3 : 4
iii. P (2, 6), Q (-4, 1), a : b = 1 : 2
Solution:
Let the co-ordinates of point A be (x, y).
i. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -3, y₁ = 7, x₂ = 1, y₂ = -4, a = 2, b = 1
∴ By section formula,

∴ The co-ordinates of point A are (\(\frac { -1 }{ 3 } \),\(\frac { -1 }{ 3 } \)).

ii. Let P (x₁,y₁), Q (x₂, y₂) be the given points.
Here, x₁ = -2, y₁ = -5, x₂ = 4, y₂ = 3, a = 3, b = 4
By section formula,

∴ The co-ordinates of point A are (\(\frac { 4 }{ 7 } \),\(\frac { -11 }{ 7 } \))

iii. Let P (x₁, y₁), Q (x₂, y₂) be the given points.
Here,x₁ = 2,y₁ = 6, x₂ = -4, y₂ = 1, a = 1,b = 2
∴ By section formula,

∴ The co-ordinates of point A are (0,\(\frac { 13 }{ 3 } \))

Question 3.
Find the ratio in which point T (-1, 6) divides the line segment joining the points P (-3,10) and Q (6, -8).
Solution:
Let P (x₁, y₁), Q (x₂, y₂) and T (x, y) be the given points.
Here, x₁ = -3, y₁ = 10, x₂ = 6, y₂ = -8, x = -1, y = 6
∴ By section formula,

∴ Point T divides seg PQ in the ratio 2 : 7.

Question 4.
Point P is the centre of the circle and AB is a diameter. Find the co-ordinates of point B if co-ordinates of point A and P are (2, -3) and (-2,0) respectively.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 2, y₁ =-3,
x = -2, y = 0

Point P is the midpoint of seg AB.
∴ By midpoint formula,

∴ The co-ordinates of point B are (-6,3).

Question 5.
Find the ratio in which point P (k, 7) divides the segment joining A (8, 9) and B (1,2). Also find k.
Solution:
Let A (x₁, y₁), B (x₂, y₂) and P (x, y) be the given points.
Here, x₁ = 8, y₁ = 9, x₂ = 1, y₂ = 2, x = k, y = 7
∴ By section formula,

∴ Point P divides seg AB in the ratio 2 : 5, and the value of k is 6.

Question 6.
Find the co-ordinates of midpoint of the segment joining the points (22, 20) and (0,16).
Solution:
Let A (x₁, y₁) = A (22, 20),
B (x₂,y₂) = B (0, 16)
Let the co-ordinates of the midpoint be P (x,y).
∴ By midpoint formula,

The co-ordinates of the midpoint of the segment joining (22, 20) and (0, 16) are (11,18).

Question 7.
Find the centroids of the triangles whose vertices are given below.
i. (-7, 6), (2,-2), (8, 5)
ii. (3, -5), (4, 3), (11,-4)
iii. (4, 7), (8, 4), (7, 11)
Solution:
i. Let A (x₁, y₁) = A (-7, 6),
B (x₂, y₂) = B (2, -2),
C (x₃, y₃) = C(8, 5)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (1,3).

ii. Let A (x₁ y₁) = A (3, -5),
B (x₂, y₂) = B (4, 3),
C(x₃, y₃) = C(11,-4)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (6, -2).

iii. Let A (x₁, y₁) = A (4, 7),
B (x₂, y₂) = B (8,4),
C (x₃, y₃) = C(7,11)
∴ By centroid formula,

∴ The co-ordinates of the centroid are (\(\frac { 19 }{ 3 } \),\(\frac { 22 }{ 3 } \))

Question 8.
In ∆ABC, G (-4, -7) is the centroid. If A (-14, -19) and B (3, 5), then find the co-ordinates of C.
Solution:
G (x, y) = G (-4, -7),
A (x₁, y₁) = A (-14, -19),
B(x₂, y₂) = B(3,5)
Let the co-ordinates of point C be (x₃, y₃).
G is the centroid.
By centroid formula,

∴ The co-ordinates of point C are (-1, – 7).

Question 9.
A (h, -6), B (2, 3) and C (-6, k) are the co-ordinates of vertices of a triangle whose centroid is G (1,5). Find h and k.
Solution:
A(x₁,y₁) = A(h, -6),
B (x₂, y₂) = B(2, 3),
C (x₃, y₃) = C (-6, k)
∴ centroid G (x, y) = G (1, 5)
G is the centroid.
By centroid formula,

∴ 3 = h – 4
∴ h = 3 + 4
∴ h = 7

∴ 15 = -3 + k
∴ k = 15 + 3
∴ k = 18
∴ h = 7 and k = 18

Question 10.
Find the co-ordinates of the points of trisection of the line segment AB with A (2,7) and B (-4, -8).
Solution:
A (2, 7), B H,-8)
Suppose the points P and Q trisect seg AB.
∴ AP = PQ = QB

∴ Point P divides seg AB in the ratio 1:2.
∴ By section formula,

Co-ordinates of P are (0, 2).
Point Q is the midpoint of PB.
By midpoint formula,

Co-ordinates of Q are (-2, -3).
∴ The co-ordinates of the points of trisection seg AB are (0,2) and (-2, -3).

Question 11.
If A (-14, -10), B (6, -2) are given, find the co-ordinates of the points which divide segment AB into four equal parts.
Solution:
Let the points C, D and E divide seg AB in four equal parts.

Point D is the midpoint of seg AB.
∴ By midpoint formula,

∴ Co-ordinates of D are (-4, -6).
Point C is the midpoint of seg AD.
∴ By midpoint formula,

∴ Co-ordinates of C are (-9, -8).
Point E is the midpoint of seg DB.
∴ By midpoint formula,


∴ Co-ordinates of E are (1,-4).
∴ The co-ordinates of the points dividing seg AB in four equal parts are C(-9, -8), D(-4, -6) and E(1, – 4).

Question 12.
If A (20, 10), B (0, 20) are given, find the co-ordinates of the points which divide segment AB into five congruent parts.
Solution:
Suppose the points C, D, E and F divide seg AB in five congruent parts.
∴ AC = CD = DE = EF = FB

∴ co-ordinates of C are (16, 12).
E is the midpoint of seg CB.
By midpoint formula,

∴ co-ordinates of E are (8, 16).
D is the midpoint of seg CE.

∴ co-ordinates of F are (4, 18).
∴ The co-ordinates of the points dividing seg AB in five congruent parts are C (16, 12), D (12, 14), E (8, 16) and F (4, 18).

Maharashtra Board Class 10 Maths Chapter 5 Co-ordinate Geometry Intext Questions and Activities

Question 1.
A (15, 5), B (9, 20) and A-P-B. Find the ratio in which point P (11, 15) divides segment AB. Find the ratio using x and y co-ordinates. Write the conclusion. (Textbook pg. no. 113)
Solution:
Suppose point P (11,15) divides segment AB in the ratio m : n.
By section formula,

∴ Point P divides seg AB in the ratio 2 : 1.
The ratio obtained by using x and y co-ordinates is the same.

Question 2.
External division: (Textbook pg. no. 115)
Suppose point R divides seg PQ externally in the ratio 3:1.

Let the common multiple be k.
Let PR = 3k and QR = k
Now, PR = PQ + QR … [P – Q – R]
∴ 3k = PQ + k
∴ \(\frac { PQ }{ QR } \) = \(\frac { 2k }{ k } \) = \(\frac { 2 }{ 1 } \)
∴ Point Q divides seg PR in the ratio 2 : 1 internally.
Thus, we can find the co-ordinates of point R, when co-ordinates of points P and Q are given.

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 5.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 5 Co-ordinate Geometry.

10th Standard Maths 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 5.1 Chapter 5 Co-ordinate Geometry Questions With Answers Maharashtra Board

Practice Set 5.1 Geometry Class 10 Question 1. Find the distance between each of the following pairs of points.
i. A (2, 3), B (4,1)
ii. P (-5, 7), Q (-1, 3)
iii. R (0, -3), S (0,\(\frac { 5 }{ 2 } \))
iv. L (5, -8), M (-7, -3)
v. T (-3, 6), R (9, -10)
vi. W(\(\frac { -7 }{ 2 } \),4), X(11, 4)
Solution:
i. Let A (x₁, y₁) and B (x₂, y₂) be the given points.
∴ x₁ = 2, y₁ = 3, x₂ = 4, y₂ = 1
By distance formula,

∴ d(A, B) = 2\(\sqrt { 2 }\) units
∴ The distance between the points A and B is 2\(\sqrt { 2 }\) units.

ii. Let P (x₁, y₁ ) and Q (x₂, y₂) be the given points.
∴ x₁ = -5, y₁ = 7, x₂ = -1, y₂ = 3
By distance formula,

∴ d(P, Q) = 4\(\sqrt { 2 }\) units
∴ The distance between the points P and Q is 4\(\sqrt { 2 }\) units.

iii. Let R (x₁, y₁) and S (x₂, y₂) be the given points.
∴ x₁ = 0, y₁ = -3, x₂ = 0, y₂ = \(\frac { 5 }{ 2 } \)
By distance formula,

∴ d(R, S) = \(\frac { 11 }{ 2 } \) units
∴ The distance between the points R and S is \(\frac { 11 }{ 2 } \) units.

iv. Let L (x₁, y₁) and M (x₂, y₂) be the given points.
∴ x₁ = 5, y₁ = -8, x₂ = -7, y₂ = -3
By distance formula,

∴ d(L, M) = 13 units
∴ The distance between the points L and M is 13 units.

v. Let T (x₁,y₁) and R (x₂, y₂) be the given points.
∴ x₁ = -3, y₁ = 6,x₂ = 9,y₂ = -10
By distance formula,

∴ d(T, R) = 20 units
∴ The distance between the points T and R 20 units.

vi. Let W (x₁, y₁) and X (x₂, y₂) be the given points.

∴ d(W, X) = \(\frac { 29 }{ 2 } \) units
∴ The distance between the points W and X is \(\frac { 29 }{ 2 } \) units.

Practice Set 5.1 Geometry 10th Question 2. Determine whether the points are collinear.
i. A (1, -3), B (2, -5), C (-4, 7)
ii. L (-2, 3), M (1, -3), N (5, 4)
iii. R (0, 3), D (2, 1), S (3, -1)
iv. P (-2, 3), Q (1, 2), R (4, 1)
Solution:
i. By distance formula,


∴ d(A, B) = \(\sqrt { 5 }\) …(i)
On adding (i) and (iii),
d(A, B) + d(A, C)= \(\sqrt { 5 }\) + 5\(\sqrt { 5 }\) = 6\(\sqrt { 5 }\)
∴ d(A, B) + d(A, C) = d(B, C) … [From (ii)]
∴ Points A, B and C are collinear.

ii. By distance formula,

On adding (i) and (iii),
d(L, M) + d(L, N) = 3\(\sqrt { 5 }\) + 5\(\sqrt { 2 }\) ≠ \(\sqrt { 65 }\)
∴ d(L, M) + d(L, N) ≠ d(M, N) … [From (ii)]
∴ Points L, M and N are not collinear.

iii. By distance formula,

On adding (i) and (ii),
∴ d(R, D) + d(D, S) = \(\sqrt { 8 }\) + \(\sqrt { 5 }\) ≠ 5
∴ d(R, D) + d(D, S) ≠ d(R, S) … [From (iii)]
∴ Points R, D and S are not collinear.

iv. By distance formula,

On adding (i) and (ii),
d(P, Q) + d(Q, R) = \(\sqrt { 10 }\) + \(\sqrt { 10 }\) = 2\(\sqrt { 10 }\)
∴ d(P, Q) + d(Q, R) = d(P, R) … [From (iii)]
∴ Points P, Q and R are collinear.

Coordinate Geometry Class 10 Practice Set 5.1 Question 3. Find the point on the X-axis which is equidistant from A (-3,4) and B (1, -4).
Solution:
Let point C be on the X-axis which is equidistant from points A and B.
Point C lies on X-axis.
∴ its y co-ordinate is 0.
Let C = (x, 0)
C is equidistant from points A and B.
∴ AC = BC

∴ (x + 3)² + (-4)² = (x- 1)² + 4²
∴ x² + 6x + 9 + 16 = x² – 2x + 1 + 16
∴ 8x = – 8
∴ x = – \(\frac { 8 }{ 8 } \) = -1
∴ The point on X-axis which is equidistant from points A and B is (-1,0).

10th Geometry Practice Set 5.1 Question 4. Verify that points P (-2, 2), Q (2, 2) and R (2, 7) are vertices of a right angled triangle.
Solution:
Distance between two points

Consider, PQ² + QR² = 42 + 52 = 16 + 25 = 41 … [From (i) and (ii)]
∴ PR² = PQ² + QR² … [From (iii)]
∴ ∆PQR is a right angled triangle. … [Converse of Pythagoras theorem]
∴ Points P, Q and R are the vertices of a right angled triangle.

Question 5.
Show that points P (2, -2), Q (7, 3), R (11, -1) and S (6, -6) are vertices of a parallelogram.
Proof:
Distance between two points

PQ = RS … [From (i) and (iii)]
QR = PS … [From (ii) and (iv)]
A quadrilateral is a parallelogram, if both the pairs of its opposite sides are congruent.
∴ □ PQRS is a parallelogram.
∴ Points P, Q, R and S are the vertices of a parallelogram.

Question 6.
Show that points A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are vertices of rhombus ABCD.
Proof:
Distance between two points


∴ AB = BC = CD = AD …[From (i), (ii), (iii) and (iv)]
In a quadrilateral, if all the sides are equal, then it is a rhombus.
∴ □ ABCD is a rhombus.
∴ Points A, B, C and D are the vertices of rhombus ABCD.

Practice Set 5.1 Question 7. Find x if distance between points L (x, 7) and M (1,15) is 10.
Solution:
X₁ = x, y₁ = 7, x₂ = 1, y₂ = 15
By distance formula,

∴ 1 – x = ± 6
∴ 1 – x = 6 or l – x = -6
∴ x = – 5 or x = 7
∴ The value of x is – 5 or 7.

Geometry 5.1 Question 8. Show that the points A (1, 2), B (1, 6), C (1 + 2\(\sqrt { 3 }\), 4) are vertices of an equilateral triangle.
Proof:
Distance between two points

∴ AB = BC = AC … [From (i), (ii) and (iii)]
∴ ∆ABC is an equilateral triangle.
∴ Points A, B and C are the vertices of an equilateral triangle.

Maharashtra Board Class 10 Maths Chapter 5 Coordinate Geometry Intext Questions and Activities

Question 1.
In the figure, seg AB || Y-axis and seg CB || X-axis. Co-ordinates of points A and C are given. To find AC, fill in the boxes given below. (Textbook pa. no. 102)

Solution:
In ∆ABC, ∠B = 900
∴ (AB)² + (BC)² = [(Ac)² …(i) … [Pythagoras theorem]
seg CB || X-axis
∴ y co-ordinate of B = 2
seg BA || Y-axis
∴ x co-ordinate of B = 2
∴ co-ordinate of B is (2, 2) = (x₁,y₁)
co-ordinate of A is (2, 3) = (x₂, Y₂)
Since, AB || to Y-axis,
d(A, B) = Y₂ – Y₁
d(A,B) = 3 – 2 = 1
co-ordinate of C is (-2,2) = (x₁,y₁)
co-ordinate of B is (2, 2) = (x₂, y₂)
Since, BC || to X-axis,
d(B, C) = x₂ – x₁
d(B,C) = 2 – -2 = 4
∴ AC² = 12 + 42 …[From (i)]
= 1 + 16 = 17
∴ AC = \(\sqrt { 17 }\) units …[Taking square root of both sides]

Class 10 Maths Digest

• Practice Set 4.1 Class 10 Answers
• Practice Set 4.2 Class 10 Answers
• Problem Set 4 Class 10 Answers
• Practice Set 5.1 Class 10 Answers
• Practice Set 5.2 Class 10 Answers
• Practice Set 5.3 Class 10 Answers
• Problem Set 5 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.2 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.2 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.2 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Given below are some triangles and lengths of line segments. Identify in which figures, ray PM is the bisector of ∠QPR.

Solution:
In ∆ PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 7 }{ 3 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.5 }{ 1.5 } \) = \(\frac { 35 }{ 15 } \) = \(\frac { 7 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR. [Converse of angle bisector theorem]

ii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 10 }{ 7 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \) (ii)
∴ \(\frac { PQ }{ PR } \) ≠ \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is not the bisector of ∠QPR

iii. In ∆PQR,
\(\frac { PQ }{ PR } \) = \(\frac { 9 }{ 10 } \) (i)
\(\frac { QM }{ RM } \) = \(\frac { 3.6 }{ 4 } \) = \(\frac { 36 }{ 40 } \) = \(\frac { 9 }{ 10 } \) (ii)
∴ \(\frac { PQ }{ PR } \) = \(\frac { QM }{ RM } \) [From (i) and (ii)]
∴ Ray PM is the bisector of ∠QPR [Converse of angle bisector theorem]

Question 2.
In ∆PQR PM = 15, PQ = 25, PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Solution:
PN + NR = PR [P – N – R]
∴ PN + 8 = 20
∴ PN = 20 – 8 = 12
Also, PM + MQ = PQ [P – M – Q]
∴ 15 + MQ = 25

∴ line NM || side RQ [Converse of basic proportionality theorem]

Question 3.
In ∆MNP, NQ is a bisector of ∠N. If MN = 5, PN = 7, MQ = 2.5, then find QP.

Solution:
In ∆MNP, NQ is the bisector of ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 7 }{ 5 } \) = \(\frac { QP }{ 2.5 } \)
∴ QP = \(\frac { 7\times 2.5 }{ 5 } \)
∴ QP = 3.5 units

Question 4.
Measures of some angles in the figure are given. Prove that \(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \)

Solution:
Proof
∠APQ = ∠ABC = 60° [Given]
∴ ∠APQ ≅ ∠ABC
∴ side PQ || side BC (i) [Corresponding angles test]
In ∆ABC,
sidePQ || sideBC [From (i)]
∴\(\frac { AP }{ PB } \) = \(\frac { AQ }{ QC } \) [Basic proportionality theorem]

Question 5.
In trapezium ABCD, side AB || side PQ || side DC, AP = 15, PD = 12, QC = 14, find BQ.

Solution:
side AB || side PQ || side DC [Given]
∴\(\frac { AP }{ PD } \) = \(\frac { BQ }{ QC } \) [Property of three parallel lines and their transversals]
∴\(\frac { 15 }{ 12 } \) = \(\frac { BQ }{ 14 } \)
∴ BQ = \(\frac { 15\times 14 }{ 12 } \)
∴ BQ = 17.5 units

Question 6.
Find QP using given information in the figure.

Solution:
In ∆MNP, seg NQ bisects ∠N. [Given]
∴\(\frac { PN }{ MN } \) = \(\frac { QP }{ MQ } \) [Property of angle bisector of a triangle]
∴\(\frac { 40 }{ 25 } \) = \(\frac { QP }{ 14 } \)
∴ QP = \(\frac { 40\times 14 }{ 25 } \)
∴ QP = 22.4 units

Question 7.
In the adjoining figure, if AB || CD || FE, then find x and AE.

Solution:
line AB || line CD || line FE [Given]
∴\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \) [Property of three parallel lines and their transversals]
∴\(\frac { 8 }{ 4 } \) = \(\frac { 12 }{ X } \)
∴ X = \(\frac { 12\times 4 }{ 8 } \)
∴ X = 6 units
Now, AE AC + CE [A – C – E]
= 12 + x
= 12 + 6
= 18 units
∴ x = 6 units and AE = 18 units

Question 8.
In ∆LMN, ray MT bisects ∠LMN. If LM = 6, MN = 10, TN = 8, then find LT.

Solution:
In ∆LMN, ray MT bisects ∠LMN. [Given]
∴\(\frac { LM }{ MN } \) = \(\frac { LT }{ TN } \) [Property of angle bisector of a triangle]
∴\(\frac { 6 }{ 10 } \) = \(\frac { LT }{ 8 } \)
∴ LT = \(\frac { 6\times 8 }{ 10 } \)
∴ LT = 4.8 units

Question 9.
In ∆ABC,seg BD bisects ∠ABC. If AB = x,BC x+ 5, AD = x – 2, DC = x + 2, then find the value of x.

Solution:
In ∆ABC, seg BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AD }{ CD } \) [Property of angle bisector of a triangle]
∴\(\frac { x }{ x+5 } \) = \(\frac { x-2 }{ x+2 } \)
∴ x(x + 2) = (x – 2)(x + 5)
∴ x2 + 2x = x2 + 5x – 2x – 10
∴ 2x = 3x – 10
∴ 10 = 3x – 2x
∴ x = 10

Question 10.
In the adjoining figure, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

Solution:

Question 11.
In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Solution:
In ∆ABC, ray BD bisects ∠ABC. [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (i) [Property of angle bisector of a triangle]
Also, in ∆ABC, ray CE bisects ∠ACB. [Given]
∴\(\frac { AC }{ BC } \) = \(\frac { AE }{ EB } \) (ii) [Property of angle bisector of a triangle]
But, seg AB = seg AC (iii) [Given]
∴\(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (iv) [From (ii) and (iii)]
∴\(\frac { AD }{ DC } \) = \(\frac { AE }{ EB } \) [From (i) and (iv)]
∴ ED || BC [Converse of basic proportionality theorem]

Question 1.
i. Draw a ∆ABC.
ii. Bisect ∠B and name the point of intersection of AC and the angle bisector as D.
iii. Measure the sides.

iv. Find ratios \(\frac { AB }{ BC } \) and \(\frac { AD }{ DC } \)
v. You will find that both the ratios are almost equal.
vi. Bisect remaining angles of the triangle and find the ratios as above. Verify that the ratios are equal. (Textbook pg. no. 8)
Solution:

Note: Students should bisect the remaining angles and verify that the ratios are equal.

Question 2.
Write another proof of the above theorem (property of an angle bisector of a triangle). Use the following properties and write the proof.
i. The areas of two triangles of equal height are proportional to their bases.
ii. Every point on the bisector of an angle is equidistant from the sides of the angle. (Textbook pg. no. 9)

Given: In ∆CAB, ray AD bisects ∠A.
To prove: \(\frac { AB }{ AC } \) = \(\frac { BD }{ DC } \)
Construction: Draw seg DM ⊥ seg AB A – M – B and seg DN ⊥ seg AC, A – N – C.
Solution:
Proof:
In ∆ABC,
Point D is on angle bisector of ∠A. [Given]
∴DM = DN [Every point on the bisector of an angle is equidistant from the sides of the angle]
\(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B \times D M}{A C \times D N}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{A(\Delta A B D)}{A(\Delta A C D)}=\frac{A B}{A C}\) (ii) [From (i)]
Also, ∆ABD and ∆ACD have equal height.
∴ \(\frac{\mathrm{A}(\Delta \mathrm{ABD})}{\mathrm{A}(\Delta \mathrm{ACD})}=\frac{\mathrm{BD}}{\mathrm{CD}}\) (iii) [Triangles having equal height]
∴\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\) [From (ii) and (iii)]

Question 3.
i. Draw three parallel lines.
ii. Label them as l, m, n.
iii. Draw transversals t₁ and t₂.
iv. AB and BC are intercepts on transversal t₁.
v. PQ and QR are intercepts on transversal t₂.
vi. Find ratios \(\frac { AB }{ BC } \) and \(\frac { PQ }{ QR } \). You will find that they are almost equal. Verify that they are equal.(Textbook pg, no. 10)
Solution:

(Students should draw figures similar to the ones given and verify the properties.)

Question 4.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.(Textbook pg, no. 12)

Solution:

Question 5.
In ∆ABC, ray BD bisects ∠ABC. A – D – C, side DE || side BC, A – E – B, then prove that \(\frac { AB }{ BC } \) = \(\frac { AE }{ EB } \) (Textbook pg, no. 13)

Solution:

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b₁, h₁, and A₁ respectively.
Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)

Solution:
∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).

Solution:
Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.

Solution:
i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)

Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 Marathi Solutions Kumarbharti Chapter 8 वाट पाहताना Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 10 Marathi Kumarbharti Chapter 8 वाट पाहताना

Marathi Kumarbharti Std 10 Digest Chapter 8 वाट पाहताना Textbook Questions and Answers

कृति

कतिपत्रिकेतील प्रश्न १ (अ) आणि (आ) यांसाठी…

प्रश्न 1.
आकृत्या पूर्ण करा.
(i)
(ii)
(iiii)
(iv)
उत्तर:
(i)
(ii)
(iii)
(iv)

प्रश्न 2.
कारणे शोधा
(अ) आवाजाची वाट पाहण्याचं सार्थक व्हायचं, कारण ……………………………………
(आ) म्हातारीच्या तोंडावर समाधान पसरायचं, कारण ……………………………………
(इ) पुस्तक वाचण्याची वाट पाहण्यात उन्हाळ्याच्या आधीचा काळ लेखिकेला वेड लावायचा, कारण ……………………………………
(ई) पोस्टमन मनानंच कोरं पत्र वाचतो, कारण ……………………………………
उत्तर:
(अ) आवाजाची वाट पाहण्याचं सार्थक व्हायचं, कारण पहाटे कुहुकुहु ऐकू यावा, ही रात्री झोपताना बाळगलेली इच्छा पहाटे पहाटे पूर्ण होई.
(आ) म्हातारीच्या तोंडावर समाधान पसरायचं; कारण दूर परगावी राहणारा आपला मुलगा आपली आठवण काढतो, आपल्याला तो त्याच्याकडे नेणार आहे, या कल्पने- तेचे मन सुखायचे,
(इ) पुस्तक वाचण्याची वाट पाहण्यात उन्हाळ्याच्या आधीचा काळ लेखिकेला वेड लावायचा, कारण पुस्तकांतून भाषेची शक्ती, लेखकांच्या प्रतिभेची शक्ती समजू लागली होती.
(ई) पोस्टमन मनानंच कोरं पत्र वाचतो; कारण त्या म्हातारीला पुत्रभेटीचा आनंद मिळावा आणि तिचे शेवटचे दिवस समाधानात जावेत, अशी पोस्टमनची इच्छा होती.

प्रश्न 3.
तुलना करा.

व्यक्तीशी मैत्री	कवितेशी मैत्री
……………………..	……………………..
……………………..	……………………..
……………………..	……………………..

उत्तर:

व्यक्तीशी मैत्री	कवितेशी मैत्री
आपण त्या व्यक्तीला हाक मारतो. तिच्याकडे धावतो. मनसोक्त गप्पा मारतो. ती व्यक्ती प्रतिसादही देते. व्यक्ती हवी तेव्हा भेटू शकते.	कविता तिच्याकडे धाव घेऊनही भेटत नसे. मात्र ती प्रसन्न झाली तर कधीही धावत येऊन भेटे. कविता मात्र खूप वाट पाहायला लावते.

प्रश्न 4.
‘वाट पाहणे’ या प्रक्रियेबाबत पुढील मुद्द्यांना अनुसरून लेखिकेचे मत लिहून तक्ता पूर्ण करा.

वाट पाहणे प्रक्रियेतील समाविष्ट गोष्टी	वाट पाहणे प्रक्रियेतून माणसाने शिकायच्या गोष्टी	वाट पाहण्याचे फायदे
……………………..	……………………..	……………………..
……………………..	……………………..	……………………..
……………………..	……………………..	……………………..

उत्तर:

प्रश्न 5.
स्वमत.
(अ) पाठाच्या शीर्षकाची समर्पकता तुमच्या शब्दांत सांगा.
उत्तर :
अरुणा ढेरे यांचा ‘वाट पाहताना’ हा अत्यंत हृदय ललित लेख आहे. जीवनातील एक मूलभूत महत्त्वाचे तत्त्व या लेखात त्या उलगडून दाखवतात. तसे पाहिले तर माणूस वाट पाहत पाहतच वाटचाल करीत असतो. प्रत्येक पावलावर त्याच्या मनात ‘नंतर काय होईल?’, ‘माझ्या स्वप्नांप्रमाणे, कल्पनेप्रमाणे घडेल ना?’ अशी तगमग असते. हीच तगमग त्याला पुढे जायला, जीवन जगायला लावते. हे तत्व लेखिकांनी अनेक उदाहरणांच्या साहाय्याने स्पष्ट केले आहे.

सुट्टीतल्या सगळ्या गोष्टी जगायला मिळतील या आशेने लेखिका सुट्टीची वाट पाहत. अनेक अनोळखी प्रदेश, माणसे, प्रसंग यांचा सहवास घडवणाऱ्या पुस्तकांची वाट पाहणे अत्यंत रमणीय होते. उंबराच्या झाडावर बसणाऱ्या पोपटांच्या थव्यामुळे हिरवेगार बनलेले ते झाड पाहून लेखिकांचे मन हळवे, कोमल होऊन जाते. त्यातच त्यांच्या कवितांची मुळे रुजतात. वाट पाहण्याने त्यांची निर्मितिशीलता जागृत होते. आत्याची वाट पाहताना त्यांचे मन अस्वस्थता आणि अनामिक भीती यांनी भरून जाते. या सर्वात जगण्याचाच अनुभव होता. अस्वस्थता, हुरहुर, दुःख, तगमग, शंकाकुलता हे सारे भाव पोस्टमनला भेटलेली म्हातारी, तसेच शेतकरी, वारकरी भक्त यांच्या चेहऱ्यांवर लेखिकांना गवसतात. अशा प्रकारे जगण्याच्या मुळाशीच वाट पाहण्याची भावना असल्याचे भान लेखिका या लेखातून वाचकांना देतात. म्हणून वाट पाहताना’ हे शीर्षक अत्यंत समर्पक आहे.

(आ) म्हातारीचं वाट पाहणं सुखाचं करण्यासाठी पोस्टमनने केलेल्या युक्तीबाबत तुमचे मत लिहा.
उत्तर:
एखादी निर्जीव वस्तू पोहोचती करावी, त्याप्रमाणे तो पोस्टमन पत्रे देत नसे. कारण पत्रे ही निर्जीव वस्तू नसतात. ती माणसांच्या सुखदुःखांनी, आशा-आकांक्षांनी भरलेली असतात. त्यात माणसांचे मन असते, हृदय असते. पत्रांचे हे स्वरूप चित्रपटातल्या त्या पोस्टमनने जाणले होते. म्हणून तो अंध म्हातारीला मुलाचे काल्पनिक पत्र वाचून दाखवतो. ते पत्र खोटे असते. मजकूर खोटा असतो. त्या अंध म्हातारीच्या मुलाचा स्पर्शसुद्धा त्या पत्राला झालेला नसतो. पण म्हातारी सुखावते. तिचे उरलेले दिवस आनंदात जातात. या विपरीत स्थितीने पोस्टमनचे मन कळवळते. पण म्हातारी सुखावणे हे अधिक मूल्ययुक्त होते. आपल्या मुलालाही तो पोस्टमन हीच उदात्त शिकवण देतो. मुलातला माणूस जागा करतो. माणसाशी माणसासारखे वागण्याची ही महान शिकवण होती. प्रत्येक आई-वडिलांनी आपल्या मुलांना असे माणूसपण शिकवले पाहिजे; तरच मानवी समाजाला भविष्य आहे.

(इ) ‘वाट पाहणे एरवी सुखाची गोष्ट नसली तरी अनेक गोष्टींचे मोल जाणवून देणारी आहे’. या विधानाची सत्यता पटवून दया.
उत्तर:
‘वाट पाहताना’ या पाठात लेखिकांनी जीवनाचा एक सुखमंत्रच सांगितला आहे. वाट पाहणे हा तो मंत्र होय. कोणत्याही गोष्टीसाठी पाहायला शिकले पाहिजे, असे त्यांचे सांगणे आहे. वाद पाहणे हे तसे कधीच सुखाचे नसते. आपल्या आशा-आकांक्षा पूर्ण करण्यासाठी, एखादी गोष्ट मिळवण्यासाठी आपले मन अधीर झालेले असते. मन शंकेने व्याकुळ होते. हवी ती गोष्ट आपल्याला मिळेल का? असा प्रश्न मचात काहूर माजवतो.

एखादी गोष्ट वाट न पाहता, चटकन मिळाली, तर ती गोष्ट आपली जिवाभावाची आहे की वरवरची आहे, हे कळायला मार्ग राहत नाही. इच्छा तत्काळ पूर्ण झाल्यास आपल्याला आनंद मिळेल, हे खरे आहे.

पण आपण कदाचित वरवरच्या गोष्टींमध्ये बुडून जाण्याची शक्यता असते. अधिकाधिक वाट पाहिल्यामुळे आपली खरी ओढ कुठे आहे, हे कळते. म्हणजेच आपल्याला खरोखर काय हवे आहे, नेमकी कशाची गरज आहे, हे कळून चुकते. जे आपल्या दृष्टीने मोलाचे आहे, हे शोधण्याची दृष्टी या वाट पाहण्यातून मिळते. आपल्या दृष्टीने मोलाच्या असलेल्या गोष्टी मिळाल्या तर आपले जीवन समृद्ध होते. समृद्घ जीवन जगणे हेच तर प्रत्येक माणसाचे ध्येय असते. म्हणून वाट पाहणे त्रासाचे असले तरी अनेक गोष्टींचे मोल ओळखण्यासाठी ते उपयोगी ठरते, हे खरे आहे.

भाषासौंदर्य
मराठी भाषेतील शब्दसामर्थ्य शब्दातीत आहे. ‘वाट’ या एकाच शब्दाचा वापर विविध अर्थानी करून एक अर्थपूर्ण मनोगत तयार झाले आहे.

नमस्कार,
तू वाट दाखवणार,
म्हणून काल तुझ्या पत्राची वाट पाहत होतो.
त्या वाटेने पत्र आलेच नाही.
नेहमी त्या वाटेवरून धावणारी
पोस्टमन दादाची सायकलही त्या दिवशी धावली नाही.
शेवटी सगळा दिवस वाट पाहण्यात गेला,
साऱ्या दिवसाचीच वाट लागली
आणि मी माझ्या घरच्या वाटेने माघारी फिरलो
मनात आले आपण पत्राचीच वाट पाहत होतो
आता कशाचीच वाट पाहू नये
आपणच आपली वाट निर्माण करावी
जी वाट नवनिर्मितीची ठरेल.

वरील मनोगताचा अभ्यास करा व त्यातील भाषिक सामर्थ्य जाणून घ्या. एका शब्दाचे वेगवेगळ्या संदर्भात वेगवेगळे अर्थ असणारे इतर काही शब्द वापरून तुम्हांलाही असे मनोगत लिहिता येईल.

आपल्या भाषिक क्षमता वाढवण्यासाठी याचा सराव करा.

उतारा क्र. १
प्रश्न. पुढील उतारा वाचा आणि दिलेल्या
सूचनांनुसार कृती करा :

कृती १: (आकलन)

प्रश्न 1.
आकृत्या पूर्ण करा :




उत्तर:

कृती २ : (आकलन)

प्रश्न 1.
अंगणात मोकळ्या वातावरणात झोपायला मिळण्यापर्यंतचा घटनाक्रम :
(i) होळीनंतर थंडी झपाट्याने कमी होत जायची आणि आंब्याचा मोहोर नुसता घमघमत असायचा.
(ii) ………………………..
(iii) ………………………..
उत्तर:
(i) होळीनंतर थंडी झपाट्याने कमी होत जायची आणि आंब्याचा मोहोर नुसता घमघमत असायचा.
(ii) मार्च-एप्रिलमध्ये मुलांना गॅलरीत झोपायला मिळे.
(iii) सुट्टी लागल्यावर अंगणात अंथरुणे पडत.

प्रश्न 2.
लेखिकांचा वाट पाहण्याचा पहिला अनुभव :
(i) ………………………..
(ii) ………………………..
(iii) ………………………..
(iv) ………………………..
उत्तर:
(i) अंगणात रात्रीच्या थंड वातावरणात हळूहळू झोप येई.
(ii) उदया कुहुकुहु ‘ ऐकू येईल का, ही हुरहुर लागे.
(iii) पहाटे पहाटे झोपेत असतानाच कुहुकुहु ऐकू येई.
(iv) त्या आवाजाची वाट पाहिली, याची धन्यता वाटे.

कृती ३ : (व्याकरण)

प्रश्न 1.
भांडे ‘ या शब्दातील पहिल्या अक्षरावरील अनुस्वार काढला की ‘भाडे’ हा शब्द मिळतो. दोन्ही अर्थपूर्ण शब्द आहेत. असे उताऱ्यातून दोन शब्द शोधा आणि अनुस्वारसहित व अनुस्वारविरहित असे प्रत्येकी दोन्ही शब्द लिहा.
उत्तर:
पाठातील शब्द : थंडी. दोन शब्द : थंडी, थडी. तोंड. दोन शब्द : तोंड, तोड.

प्रश्न 2.
कंसात दिलेला प्रत्यय जोडून प्रत्ययासहितचे पूर्णरूप लिहा :
(i) झपाटा (ने)
(ii) झोप (चा)
(iii) भाषा (ला)
(iv) पुस्तके (त)
उत्तर:
(i) झपाट्याने
(ii) झोपेचा
(iii) भाषेला
(iv) पुस्तकांत.

प्रश्न 3.
अधोरेखित नामांच्या जागी अन्य योग्य नामे लिहून वाक्य पुन्हा लिहा :
आमच्या भल्यामोठ्या वाड्यात पुष्कळ बिहाडे होती.
उत्तर:
आमच्या भल्यामोठ्या इमारतीत पुष्कळ कुटुंबे होती.

प्रश्न 4.
घमघमाट’ यासारखे तुम्हांला ठाऊक असलेले चार शब्द लिहा.
उत्तर:
चमचमाट, दणदणाट, ठणठणाट, फडफडाट.

Marathi Kumarbharti Class 10 Textbook Solutions Chapter 8 वाट पाहताना Additional Important Questions and Answers
प्रश्न. पुढील उतारा वाचा आणि दिलेल्या सूचनांनुसार कृती करा :

कृती १ : (आकलन)

प्रश्न 1.
कारणे लिहा :
(i) पण तेव्हा जीव नुसता फुटून जायचा; कारण ……………………………….
(ii) पोस्टमनचे काम वाटते तितके सोपे नव्हते; कारण ……………………………….
उत्तर:
(i) पण तेव्हा जीव नुसता फुटून जायचा; कारण आत्याला घरी यायला रात्र होई म्हणून लेखिकांचे मन अनामिक भीतीने व्यापून जायचे.
(ii) पोस्टमनचे काम वाटते तितके सोपे नव्हते; कारण पत्रांचा थैला पाठीवर घेऊन वाहनांची सोय नसलेल्या वाड्या वस्त्यांवर पायी चालत जावे लागे.

प्रश्न 2.
अर्थ स्पष्ट करा :
(i) तो नुसता पत्र पोहोचवणारा सरकारी नोकर नाही. तो माणूस आहे.
(ii) पावसाची वाट पाहणाऱ्या शेतकऱ्याचे डोळे आठवा जरा.
उत्तर:
(i) तो पोस्टमन एक वस्तू नेऊन दुसऱ्याला दयावी, इतक्या कोरडेपणाने काम करणारा हमाल नव्हता. तो त्या पत्रात दडलेल्या माणसांच्या भावभावना ओळखू शकत होता, त्या माणसांशी तो मनाने जोडला जायचा.
(ii) पाऊस पडण्याचे दिवस आले की शेतकरी आतुरतेने पावसाची वाट पाहतो. त्या वेळी त्याच्या मनात पाऊस पडेल की नाही, पडला तर पुरेसा पडेल की नाही, ही धाकधुकी असते. आणि पडलाच नाही तर? ही जिवाची तडफड करणारी भीतीही असते. हे सर्व भाव शेतकऱ्यांच्या डोळ्यांत दिसतात.

कृती ३ : (व्याकरण)
प्रश्न 1.
‘रडू गळ्याशी दाटून येणे’ या वाक्प्रचारात ‘गळा’ या अवयवाचा उपयोग केलेला आहे, असे शरीराच्या अवयवांवर आधारित आणखी चार वाक्प्रचार लिहा.
उत्तर:
(i) राग नाकावर असणे,
(ii) पाऊल वाकडे पडणे.
(iii) छाती पिटणे.
(iv) नाकातोंडात पाणी जाणे.

प्रश्न 2.
‘वाड्यावस्त्या’ यासारखे आणखी चार जोडशब्द लिहा.
उत्तर:
(i) गल्लीबोळ
(ii) बाजारहाट
(iii) नदीनाले
(iv) झाडेझुडपे.

प्रश्न 3.
अधोरेखित सर्वनाम कोणाला उद्देशून योजले आहे, ते लिहा :
(i) पोस्टमन आल्याचे तिला बरोबर समजते.
(ii) त्याच्या येण्याची वाट पाहत आहे.
(iii) तो त्याला माणसे दाखवतो.
उत्तर:
(i) तिला – अंध म्हातारी.
(ii) त्याच्या – म्हातारीचा मुलगा.
(iii) तो – पोस्टमन, त्याला – पोस्टमनचा मुलगा.

व्याकरण व भाषाभ्यास

कृतिपत्रिकेतील प्रश्न ४ (अ) आणि (आ) यांसाठी…
व्याकरण घटकांवर आधारित कृती :

१. समास:
विग्रहावरून सामासिक शब्द लिहा :
विग्रह – सामासिक शब्द
(i) कानापर्यंत
(ii) राजाचा वाडा
(iii) सात सागरांचा समूह
(iv) दहा किंवा बारा
उत्तर:
विग्रह – सामासिक शब्द
(i) कानापर्यंत – आकर्ण
(ii) राजाचा वाडा – राजवाडा
(iii) सात सागरांचा समूह – सप्तसिंधू
(iv) दहा किंवा बारा – दहाबारा

२. अलंकार :

प्रश्न 1.
पुढील ओळींमधील अलंकार ओळखा व स्पष्टीकरण दया :
‘कुटुंबवत्सल इथे फणस हा।
कटिखांदयावर घेऊनि बाळे।।’
उत्तर :
अलंकार → चेतनगुणोक्ती
स्पष्टीकरण : फणसाच्या झाडाला लगडलेली फळे म्हणजे फणसाची लेकरे आहेत, अशा मानवी भावनांचे आरोपण फणसाच्या निर्जीव झाडावर केल्यामुळे हा चेतनगुणोक्ती अलंकार आहे.

प्रश्न 2.
पुढील वैशिष्ट्यावरून अलंकार ओळखा व समर्पक उदाहरण दया : (सराव कृतिपत्रिका -३)
(i) उपमेय व उपमान या दोघात भेद नाही.
(ii) उपमेय हे उपमानच आहे.
(अ) अलंकाराचे नाव → [ ]
(आ) अलंकाराचे उदाहरण → [ ]
उत्तर :
(अ) अलंकाराचे नाव → [रूपक]
(आ) अलंकाराचे उदाहरण → [वारणेचा ढाण्या वाघ बाहेर पडला]

३. वृत्त :
पुढील ओळींचे गण पाडून वृत्त ओळखा :
तदितर खग भेणे वेगळाले पळाले
उपवन जल केली जे कराया मिळाले
उत्तर :
वृत्त : हे मालिनी वृत्त आहे.

४. शब्दसिद्धी :

प्रश्न 1.
पुढील शब्दांना ‘खोर’ हा प्रत्यय लावून शब्द तयार करा :
(i) भांडण –
(i) चुगली –
उत्तर:
(i) भांडखोर
(ii) चुगलखोर

प्रश्न 2.
पुढील शब्दांच्या आधी ‘अव’ हा उपसर्ग लावून शब्द तयार करा :
(i) गुण – (ii) लक्षण –
उत्तर:
(i) अवगुण
(ii) अवलक्षण

प्रश्न 3.
वर्गीकरण करा : (सराव कृतिपत्रिका -१).
शब्द : सामाजिक, अभिनंदन, नम्रता, अपयश.
प्रत्ययघटित – उपसर्गघटित
(i) ……………………………
(ii) ……………………………
उत्तर:
प्रत्ययघटित – उपसर्गघटित
(i) सामाजिक – (ii) नम्रता
(i) अभिनंदन – (ii) अपयश

५. सामान्यरूप :
पुढील शब्दांची सामान्यरूपे लिहा :
(i) रात्रीचे –
(ii) पंखांनी –
(iii) आंब्यावर –
(iv) म्हातारीला –
(v) संगीताने –
(vi) हाताला –
उत्तरे :
(i) रात्रीचे – रात्री
(ii) पंखांनी – पंखां
(iii) आंब्यावर – आंब्या
(iv) म्हातारीला – म्हातारी
(v) संगीताने – संगीता
(vi) हाताला – हाता

६. वाक्प्रचार :

प्रश्न 1.
जोड्या जुळवा :
वाक्प्रचार – अर्थ
(i) चाहूल येणे – (अ) चौकशी करणे
(ii) सार्थक होणे – (आ) गुंग होणे
(iii) थक्क होणे – (इ) अंदाज येणे
(iv) विचारपूस करणे – (ई) धन्य वाटणे
(v) भान विसरणे – (उ) चकित होणे
उत्तरे :
(i) चाहूल येणे – अंदाज येणे
(ii) सार्थक होणे – धन्य वाटणे
(iii) थक्क होणे – चकित होणे
(iv) विचारपूस करणे – चौकशी करणे
(v) भान विसरणे – गुंग होणे.

प्रश्न 2.
दिलेल्या वाक्यांत योग्य वाक्प्रचारांचा उपयोग करून वाक्ये पुन्हा लिहा : (कपाळाला आठी पडणे, सहीसलामत बाहेर पडणे, भान विसरणे) (सराव कृतिपत्रिका -१)
(i) त्सुनामीच्या तडाख्यात सापडलेल्या लोकांना भारतीय जवानांनी सुखरूप बाहेर काढले.
(ii) दिवाळीसाठी आणलेले नवीन कपडे नमिताला न आवडल्यामुळे तिने नाराजी व्यक्त केली.
उत्तर:
(i) त्सुनामीच्या तडाख्यात सापडलेले लोक भारतीय … जवानांच्या मदतीने सहीसलामत बाहेर पडले.
(ii) दिवाळीसाठी आणलेले नवीन कपडे पाहून नमिताच्या कपाळालां आठी पडली.

भाषिक घटकांवर आधारित कृती:

१. शब्दसंपत्ती :

प्रश्न 1.
गटात न बसणारा शब्द लिहा :
(i) कोकीळ, पोपट, कावळा, गाय, मोर,
(ii) कुरड्या, पापड्या, शेवया, चकल्या, वाळवण.
उत्तर:
(i) गाय
(ii) वाळवण,

प्रश्न 2.
पुढील पक्ष्यांसमोर त्यांची घरे लिहा :
जसे : कोकिळा – घरटे; तसे
(i) पोपट – …………………….
(ii) कोंबडा – …………………….
उत्तर:
(i) पोपट – ढोली
(ii) कोंबडा – खुराडे.

प्रश्न 3.
जसे : पोपटांचा – थवा; तसे –
(i) गुरांचा – …………………….
(ii) फुलांचा – …………………….
उत्तर:
(i) गुरांचा – कळप
(ii) फुलांचा – गुच्छ,

प्रश्न 4.
पुढील शब्दांचे भिन्न अर्थ लिहा :
← माळा →
← गार →
उत्तर:
मजला ← माळा → हार
थंड ← गार → गारगोटी

प्रश्न 5.
गटात न बसणारा शब्द शोधा : (सराव कृतिपत्रिका-३)
(i) खाणे, जेवणे, जेवण, करणे →
(ii) मधुर, स्वस्त, पाणी, स्वच्छ →
उत्तर:
(i) जेवण
(ii) पाणी

प्रश्न 6.
विरुद्धार्थी शब्द लिहा :
(i) मऊ x …………………..
(ii) गार x …………………..
(iii) धाकटा x …………………..
(iv) अलीकडे x …………………..
(v) अंध x …………………..
(vi) दूर x …………………..
(vii) पक्की x …………………..
(viii) शहर x …………………..
उत्तर:
(i) मऊ x टणक
(ii) गार x गरम
(iii) धाकटा x थोरला
(iv) अलीकडे x पलीकडे
(v) अंघ x डोळस
(vi) दूर x जवळ
(vii) पक्की x कच्ची
(viii) शहर x खेडे

वाट पाहताना शब्दार्थ

• घमघमणे – सुगंध दाटून येऊन पसरणे.
• हजारी मोगरा – अनेक फुलांचा गुच्छ येणारे मोगऱ्याचे झाड.
• गराडा – गर्दी करून घातलेला वेढा.
• प्रतिमा – नवनवीन कल्पना योजून निर्मिती करण्याची क्षमता.
• दिंडी दरवाजा – (दिंडी = मोठ्या दरवाजात असलेला लहान दरवाजा.) दिंडी असलेला मोठा दरवाजा.
• शोष – कोरडेपणा, सुकलेपणा, घशास पडलेली कोरड.
• भला – चांगला, सज्जन.
• डोळस – डोळे-दृष्टी शाबूत असलेला, आंधळेपणाने विश्वास न ठेवणारा.
• धीर धरणे – अधीरता, उत्सुकता दाबून ठेवून संयम बाळगणे.

वाट पाहताना वाक्प्रचार व त्यांचे अर्थ

• तोंडावर येणे : (एखादी भावी घटना) नजीक येऊन ठेपणे.
• सार्थक होणे : धन्यता वाटणे, परिपूर्ती होणे.
• मन आतून फुलून येणे : मनातल्या मनात अमाप आनंद होणे.
• जीव फुटून जाणे : अतोनात कासावीस होणे, भयभीत होणे.
• नाटक चालू ठेवणे : सोंग, बतावणी चालू ठेवणे.

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Balbharti Maharashtra State Board Class 10 English Solutions Unit 1.3 On Wings of Courage Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 1.3 Question Answer Maharashtra Board

On Wings of Courage Poem 10th Std Question Answer

Question 1.
The ranks of officers in Indian Army, Navy and Air Force are jumbled up. Discuss with your group and put them in the appropriate boxes.

Commander, Brigadier, Wing-Commander, Vice-Admiral, Squadron-Leader, Major, Colonel, Field Marshal, Air Marshal, Admiral of Fleet, Lieutenant-General, Flying Officer, Commodore, Rear Admiral, Air-Commodore.

ARMY	NAVY	AIR FORCE

Answer:

Army	Navy	Air Force
Brigadier,	Commander,	Wing-
Major, Colonel,	Vice-Admiral,	Commander,
Field Marshal,	Admiral	Squadron-
Lieutenant-	of Fleet,	Leader, Air
General	Commodore,	Marshal, Flying
	Rear Admiral	Officer, Air-Commodore

Question 2.
Homophones/ Homographs
(A) Make sentences to bring out the difference between-
(1) (a) wear ……………………………………..
(b) ware ……………………………………..
(2) (a) here ……………………………………..
(b) hear ……………………………………..
(3) (a) there ……………………………………..
(b) their ……………………………………..
(4) (a) cell ……………………………………..
(b) sell ……………………………………..
Answer:
(1) (a) wear: The little girl wanted to wear a pink, frilly dress.
(b) ware: The silver ware laid out on the King’s table was exquisite.

(2) (a) here: “You must sit here,” said the man to his guest.
(b) hear: The children could hear the sound of the planes quite clearly.

(3) (a) there: “I had kept my bag there,” said the woman to the policeman.
(b) their: The girls picked up their bags and went home.

(4) (a) cell: The prisoner sat in the dark cell without talking.
(b) sell: The hawker wanted to sell all his wares before evening.

(B) Write what the underlined Homographs in the following sentences mean.
(1) (a) A bear is an omnivorous animal. ……………………………………..
(b) She could not bear the injustice. ……………………………………..
(2) (a) A bat is the only bird which is a mammal. ……………………………………..
(b) His bat broke as it struck the ball. ……………………………………..
(3) (a) He had to pay a fine for breaking the traffic signal. ……………………………………..
(b) Use a fine cloth for the baby’s clothes. ……………………………………..
(4) (a) We enjoyed a lot at the temple fair. ……………………………………..
(b) She has a fair complexion. ……………………………………..
Answer:
(1) (a) A bear is an omnivorous animal.
bear – a large, heavy animal
(b) She could not bear the injustice,
bear – to tolerate

(2) (a) A bat is the only bird which is a mammal.
bat – a mammal that flies
(b) His bat broke as it struck the ball.
bat – a wooden implement used for hitting the ball in many games.

(3) (a) He had to pay a fine for breaking the traffic signal.
fine – penalty
(b) Use a fine cloth for the baby’s clothes,
fine – delicate, soft

(4) (a) We enjoyed a lot at the temple fair.
fair – a gathering of stalls and amusements for public entertainment
(b) She has a fair complexion, fair – light, not dark

Maharashtra Board Class 10 English Kumarbharati Unit 1.3 Questions and Answers

Question 1.
Read the text and fill in the flow chart of the promotions received by Arjan Singh.

Answer:

Question 2.
With the help of facts given in the text prepare a Fact file of Air Marshal Arjan Singh.
(a) Date of Birth
(b) Place of Birth
(c) Education
(d) First Assignments
(e) Important posts held
(a) In Air Force
(b) After retirement
(f) Awards
(g) Most outstanding contribution in IAF
(h) Retirement
Answer:
(a) Date of birth: April 15, 1919
(b) Place of birth: Lyalpur
(c) Education: at Montgomery; Empire Pilot Training Course at RAF (Cranwell)
(d) First Assignment: to fly Westland Wapiti biplanes in the North-Western Frontier Province as a member of the No. 1 RIAF Squadron
(e) Important posts held:
(1) In Air Force: Member of No. 1. RIAF, Flying Officer, Squadron Leader, Wing Commander, Group Captain, Air Commodore, Air Officer Commanding, Air Vice Marshal, Air Officer Commanding-in-Chief, Deputy Chief of Air Staff, Vice Chief of Air Staff, Chief of Air Staff, Air Chief Marshal.
(2) After retirement: Ambassador to Switzerland Lieutenant Governor of Delhi
(f) Awards: Distinguished Flying Cross (1944); Padma Vibhushan
(g) Most outstanding contribution in IAF: Transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(h) Retirement: in August 1969.

Question 3.
Fill in the web.

Answer:
(1) Singh had successfully led a young IAF during the 1965 Indo-Pak war.
(2) Singh played a major role in transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(3) Singh was honoured with the rank of Marshal on the Republic Day in 2002.
(4) Singh’s contribution was most outstanding during the 1965 Indo-Pak war.

Question 4.
Say what actions preceded the following promotions of Arjan Singh in his career in the IAF.
(a) Selected for Empire Pilot training course at RAF
(b) Promoted to Squadron Leader
(c) Leader of a flypast of over 100 aircraft at Red Fort, Delhi
(d) Awarded Padma Vibhushan
(e) First Air Chief Marshal of Indian Air Force
Answer:
(a) The authorities selected Singh for the Empire Pilot training course.
(b) He flew against the tribal forces and moved back to No. 1 Squadron as a Flying Officer to fly the Hawker Hurricane.
(c) On 15th August 1947, Arjan Singh achieved the unique honour of leading a fly-past of over a hundred IAF aircraft over the Red Fort in Delhi.
(d) He was awarded the Padma Vibhushan for his astute leadership of the Air Force and for inspiring the IAF to victory in the 1965 Indo-Pak war.
(e) He was a source of inspiration to all the personnel of the Armed Forces through the years.

Question 5.
Replace the underlined words/phrases with the appropriate ones, to retain the proper meaning.
(be the epitome of, gear up, a brief stint, play a major role, in recognition of, take over reins)
(a) He contributed notably in bringing up the school.
(b) Our school cricket team got ready for the final match against P. Q. R. High School.
(c) After a short period of working as a lecturer, Ravi took up an important post in a multi-national company.
(d) Our class monitor is a perfect symbol of duty and discipline.
(e) Accepting the great value of his research; they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the control of Scotland.
Answer:
(a) He played a major role in bringing up the school.
(b) Our school cricket team geared up for the final match against P.Q.R.High School.
(c) After a brief stint as a lecturer, Ravi took up an important post in a multinational company.
(d) Our class monitor is the epitome of duty and discipline.
(e) In recognition of his research, they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the reins of Scotland.

Question 6.
Build the word wall with the words related to ‘Military’.

Answer:

Question 7.
(A) State the different meanings of the following pairs of Homophones and make sentences of your own with each of them.

Word	Meaning	Sentence
(a) led lead(b) role roll(c) air heir(d) feat feet(e) reign rein rain	………………………….. ………………………………………………………… ……………………………………………………… ………………………………………………………. ………………………………………………………. …………………………..	………………………….. ………………………………………………………. ………………………………………………………. ………………………………………………………. ………………………………………………………. …………………………..

Answer:

Word	Meaning	Sentence
(a) led	past participle of lead (to guide or conduct)	The captain led his team to safety.
lead	graphite used as part of a pencil	Do you have a lead pencil?
(b) role	a part (in a play, film, etc.)	Marie got the leading role in the new movie.
roll	move in a particular direction by turning over and over	The boy wanted to roll in the mud while playing.
(c) air	the invisible gaseous substance surrounding the earth	There Is a lot of humidity in the air during the monsoon.
heir	successor or inheritor	The family did not know who the heir to the property was.
(d) feat	a great achievement	Climbing Mt. Everest is a feat.
feet	a unit of measurement	The girl saw to her shock that the lion was only a few feet away.
(e) reign	rule as king or queen	Queen Elizabeth’s reign has been a long one.
rein	a restraining influence	The new manager kept a tight rein on her employees.
rain	water that falls In drops from clouds in the sky	Children love to play in the rain.

(B) The following Homographs have the same spelling and pronunciation but can have different meanings. Make sentences of your own to show the difference.

Answer:
(a) firm: (i) My neighbour recently Joined an electronics firm as Sales Executive.
(ii) Many people feel that they must be firm with their children when they are growing.

(b) train: (i) The train left from platform 2 at seven p.m. sharp.
(ii) You must always train your pets to obey you.

(c) type: (i) The man asked his secretary to type the letter immediately.
(ii) Cows eat only a particular type of grass.

(d) post: (i) My aunt quit her job because she felt that the post was not suitable for her.
(ii) The little boy ran to the post office to post the letter to Santa Claus.

(e) current : (i) The minister was disturbed when he read about the current situation of unrest In the country.
(ii) It is a difficult task to row against the current in a river.

Question 8.
Glance through the text and prepare notes from the information that you get. Take only relevant points. Don’t use sentences. Arrange the points in the same order. You may use symbols or short forms. Present the points sequentially. Use highlighting techniques.
Answer:
Air Force Marshal Arjart Singh—Icon of India’s Military History

1. Date of Birth: 15 April, 1919
2. Qualifications: Empire Pilot Training Course at RAF (Cranwell)
3. Responsibilities:

• first assignment to fly Westland Wapiti biplanes in No.l RIAF Squadron
• brief stint in No.2 RIAF Squadron; moved back to No. 1 RIAF Squadron as Flying Officer
• overall commander of ‘Shiksha’
• led the IAF during the 1965 Indo-Pak war
• led a squadron against the Japanese during the Arakan Campaign; assisted the advance of Allied Forces to Yangoon
• led a fly-past on August 15, 1947
• commanded Ambala in the rank of Group Captain; took over as AOC of an operational command
• took over reins of the IAF
• ambassador to Switzerland; Lieutenant Governor of Delhi

(4) Achievements:

• selected for the Empire Pilot Training Course at RAF (Cranwell) in 1938, at age 19
• promoted to the rank of Squadron Leader in 1944
• led a fly-past over the Red Fort on August 15, 1947
• promoted to the rank of Wing Commander; promoted to the rank of Air Commodore in 1949
• longest tenure as AOC (1949-1952 and 19571961)
• appointed as Deputy Chief of Air Staff at the end of the 1962 war; appointed as Vice Chief of Air Staff in 1963
• rank of Air Marshal in August 1964; took over reins of IAF
• successfully led the IAF in 1965 Indo-Pak war
• promoted as Air Vice Marshal; appointed as AOC-in-C of an operational command
• first Air Chief to keep his flying currency till his CAS rank; has flown more than 60 different types of * aircraft
• first and only Air Chief Marshal of the IAF

(5) Awards:

– Distinguished Flying Cross (1944)
– Padma Vibhushan

(6) After retirement: Ambassador to Switzerland; Lieutenant Governor of Delhi
(Students can put these points attractively in boxes and use highlighting techniques.)

Question 9.
Develop a story suitable to the conclusion/end given below. Suggest a suitable title.
………………………………………………….. (Title)
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
………………………………………………………….. and so, with tears of joy and pride, the 10 year old Sanyogita More received the National Bravery Award from the Prime Minister.
Answer:
A WONDERFUL ACT OF BRAVERY
It was the 26th of July in Mumbai. It was raining cats and dogs. Ten-year-old Sanyogita More stood at the door of her hut. The street was flooded with water. Sanyogita was frightened. Her parents had not returned from work and she was all alone.

Suddenly, she saw two little boys, Rohan and Sohan, come out from the neighbouring hut to play in the water. As Sanyogita watched, there came a sudden gush of water and the boys were dragged towards an open manhole, which had been marked with a pole. They caught hold of the pole, but the pole began to tilt. It would soon fall—and the boys would go down the manhole!

Sanyogita ran as fast as she could towards the boys. Pulling a rope from a nearby door, she looped it around a large stone. She held onto the rope and extended her hand towards the boys. “Catch my hand, Sohan, Rohan,” she shouted. “Catch! Catch soon!”

The boys were in a panic but they did as they were told. Sohan held Rohan’s leg, Rohan held Sanyogita’s hand, and Sanyogita held onto the rope.

“Help! Help! she shouted, knowing that if the rope broke or the stone was dislodged, they would all go into the manhole.

She stood there shivering, her arms numb, for nearly 15 minutes before help arrived. Sanyogita collapsed after the incident. The news of her brave deed spread far and wide, and reached the ears of 1 the Prime Minister, who decided to honour her with an award. And so, with tears of joy and pride, the 10- I year-old Sanyogita More received the National Bravery ‘ Award from the Prime Minister.

Question 10.
You wish to join any one of the Indian Armed Forces. Fill in the following application form.
To
The Advertiser
N/AF Recruitment Service
Purangaon – 456 789

Affix recent
passport size
photograph

Application For Recruitment
Rect notice No 1234

1. Post applied for
2. Name and surname of Candidate (in Block letters)
3. Father’s Name ………………………………… Mother’s Name …………………………………
4. Date of Birth
5. Contact details :
Tel. No. (Res) ………………….. . Mobile No.
Email ID ………………….. .
6. Permanent Address :
House No./Street/Village ………………….. .
Post Office ………………….. .
District ………………….. State ………………….. .
Pincode ………………….. .
7. Educational Qualifications :

Serial Number	Qualification	Name of School/College	Name of Board/University	Percentage obtained

8. Whether registered at any employment exchange Yes/ No ………………….. (If yes, mention registration number and the name of the Employment Exchange.)

9. Outstanding achievements in extra-curricular activities/ sports/ games, etc.
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

10. Why you wish to join Armed Forces. …………………………………………………………………
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

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