## Maharashtra State Board Class 9 Maths Solutions Chapter 1 Basic Concepts in Geometry Practice Set 1.1

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.