Maharashtra Board Class 9 Math Part 2 Chapter 1 Basic Concept in Geometry Solution

Maharashtra Board Class 9 Math Part 2 Solution Chapter 1 – Basic Concept in Geometry

Balbharati Maharashtra Board Class 9 Math Part 2 Solution Chapter 1: Basic Concept in Geometry. Marathi or English Medium Students of Class 9 get here Basic Concept in Geometry full Exercise Solution.

Std

Maharashtra Class 9

Subject

Math Part 2 Solution

Chapter

Basic Concept in Geometry

 

Practice – 1.1

 

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

(i) d (B,E) = 5-2 = 3

(ii) d (J, A) = 1 – (-2) = 1+2 = 3

(iii) d (P, C) = 3 – (-4) = 3 + 4 = 7

(iv) d (J, H) = – 1 – (-2) = – 1 + 2 = 1

(v) d(K, O) = 0 – (-3) = 0 + 3 = 3

(vi) d(O, E) = 5 – 0 = 5

(vii) d (P, J) = – 2 – (-4) = – 2 + 4 = 2

(viii) d (Q, B) = 2 – (-4) = 2 + 5 = 7

 

(2) If the co-ordinate of A is x and that of B is y, find d(A, B).

(i) x = 1, y = 7

= d (A, B) = d (1, 7) = 7 – 1 = 6

 

(ii) x = 6, y = – 2

= d (A, B) = d (6, – 2) = 6 – (-2) = 8

 

(iii) x = – 3, y = 7

= d (A, B) = d (-3, 7) = 7 – (-3) = 7 + 3 = 10

 

(iv) x = – 4, y = – 5

= d (A, B) = d (-4, -5) = – 4 – (-5) = – 4 + 5 = 1

 

(v) x = – 3, y = – 6

= d (A, B) = d (-3, -6) = – 3 – (-6) = – 3 + 6 = 3

 

(vi) x = 4, y = – 8

= d (A, B) = d (4, -8) = 4 – (-8) = 4 + 8 = 12

 

(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

= d (P, R) + d (Q, R) = 7 + 3 = 10 = d (P,Q)

∴ R is in between P and Q.

 

(ii) d(R, S) = 8,

d(S, T) = 6,

d(R, T) = 4

= The points are non-collinear.

 

(iii) d(A, B) = 16,

d(C, A) = 9,

d(B, C) = 7

= d (C, A) + d (B, C) = 9 + 7 = 16 = d (A, B)

∴ C is point between A & B

 

(iv) d(L, M) = 11,

d(M, N) = 12,

d(N, L) = 8

= Points are non-collinear.

 

(v) d(X, Y) = 15,

d(Y, Z) = 7,

d(X, Z) = 8

= d (X, Z) + d (Y, Z) = 8 + 7 = 15 = d (X, Y)

∴ Z is a point between X and y.

 

(vi) d (D, E) = 5,

d(E, F) = 8,

d(D, F) = 6

= Points are non-collinear.

 

(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.

= d (A, C) = 10 & d (C, B) = 8

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

Or

d (A, C) – d (C, B)  = 10 – 8 = 2

∴ d (A, B) = 2 or 18

 

(5) Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).

Solution:

d (X, y) = 17 , d (Y, Z) = 8

d (X, z) = d (X, Y) + d (Y, Z)

= 17 + 8

= 25

Or

d (X, Y) – d (Y, Z)

= 17 – 8

= 9

 

(6) Sketch proper figure and write the answers of the following questions.

(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) l(AC) = 11, l(BC) = 6.5

l(AB) = l(AC) – l(BC) = 11 – 6.5 = 4.5

 

 

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

l(RT) = l(ST) + l(RS)

= 37 + 2.5

= 6.2

 

(iii) l(XZ) = 3√7,

l(XY) = √7

l(YZ) = l(XZ) – l(XY)

= 3√7 – √7

= 2√7

 

(7) Which figure is formed by three non-collinear points?

Ans: Triangle.

 

Practice Set – 1.2

 

(1) The following table shows points on a number line and their co-ordinates. Decide whether the pair of segments given below the table are congruent or not.

Point A B C D E
Co-ordinate -3 5 2 -7 9

 (i) seg DE and seg AB

(ii) seg BC and seg AD

(iii) seg BE and seg AD

Solution:

(i) Seg DE and Seg AB

l (DE) = 9 – (-7) = 9 + 7 = 16

l (AB) = 5 – (-3) = 5 + 3 = 8

∴ l(DE) >  l (AB)

∴ The segments are not congruent

 

(ii) seg BC and seg AD

l (BC) = 5 – 2 = 3

l (AD) = – 3 – (-7) = – 3 + 7 = 4

∴ l (AD) > l (BC)

∴ Segments are not congruent

 

(iii) seg BE and seg AD

l (BE) = 9 – 5 = 4

l (AD) = 4

∴ l (BE) = l (AD)

∴ Segments are not congruent

 

(2) Point M is the midpoint of seg AB. If AB = 8 then find the length of AM.  

Solution:

M is midpoint of AB

∴ l (AM) = l (MB) or l (AB) = l (AM) × 8

∴ l (AM) = ½ × 8 = 4

 

(3) Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).  

Solution:

∵ P is midpoint of CD

∴ l (CD) = 2 × l (CP)

= 2×2.5

= 5

 

(4) If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments

Solution:

l (AB) = 5, l (BP) = 2, l(AP) = 3.4

∴ 5> 3.4> 2 => l (AB) > l (AB) > l (BP)

∴ Deg AB > Deg AP> Deg BP

 

(5) Write the answers to the following questions with reference to figure 1.13.

(i) Write the name of the opposite ray of ray RP

(ii) Write the intersection set of ray PQ and ray RP.

(iii) Write the union set of seg PQ and seg QR.

(iv) State the rays of which seg QR is a subset.

(v) Write the pair of opposite rays with common end point R.

(vi) Write any two rays with common end point S.

(vii) Write the intersection set of ray SP and ray ST.

Solution:

(i) RT or RS

(ii) Ray PQ

(iii) Ray QR

(iv) Ray RQ

(v) Ray RQ and Ray RT

(vi)Ray ST and Ray SQ

(vii) Point S

 

(6) Answer the questions with the help of figure 1.14. (PHOTO)

(i) State the points which are equidistant from point B.

(ii) Write a pair of points equidistant from point Q.

(iii) Find d (U, V), d (P, C), d (V,B), d (U, L).

Solution:

(i) Clearly A and C are equidistant from B.

Now, l (BD) = 6 – 2 = 4

l (BP) = 2 – (-2) = 4

∴ l (BD) = l (BP)

∴ D and P are equidistant from B.

 

(ii) l (QR) = – 4 – (-6) = 2

l (QP) = – 2 – (-4) = 2

∴ P and R are equidistant from Q

also, U and L are equidistant from Q.

 

(iii) d (U, V) = 5 – (-5) = 5 + 5 = 10

d (P, C) = 4 – (-2) = 4 + 2 = 6

d (V, B) = 5 – 2 = 3

d (U, L) = – 3 – (-5) = – 3 + 5 = 2

 

Practice Set – 1.3

 

(1) Write the following statements in ‘if-then’ form.

(i) The opposite angles of a parallelogram are congruent.

(ii) The diagonals of a rectangle are congruent.

(iii) In an isosceles triangle, the segment joining the vertex and the midpoint of the base is perpendicular to the base.

Solution:

(i) If a quadrilateral is a parallelogram, then the opposite angle of that quadrilateral are congruent.

(ii) If a quadrilateral is a rectangle, then the diagonals of that quadrilateral are congruent.

(iii) If a triangle is an isosceles, then the segment joining the vertex and midpoint of the base is perpendicular to the base.

 

(2) Write converses of the following statements.

(i) The alternate angles formed by two parallel lines and their transversal are congruent

(ii) If a pair of the interior angles made by a transversal of two lines are supplementary then the lines are parallel

(iii) The diagonals of a rectangle are congruent

Solution:

(i) If the alternate angles formed by the transversal with two lines are congruent then the lines are parallel.

(ii) The pair of interior angles formed by two parallel lines and their transversal are supplementary.

(iii) If diagonals of a quadrilateral are congruent then it is a rectangle.

 

Problem Set – 1

 

(1) Select the correct alternative from the answers of the questions given below.

(i) How many mid points does a segment have?

(A) Only one

(B) Two

(C) three

(D) many

Solution: (A) Only one.

 

(ii) How many points are there in the intersection of two distinct lines?

(A) infinite

(B) Two

(C) One

(D) Not a single

Solution: (C) One.

 

(iii) How many lines are determined by three distinct points?

(A) two

(B) Three

(C) One or three

(D) six

Solution: (C) Three or one.

 

(iv) Find d (A, B), if co-ordinates of A and B are – 2 and 5 respectively.

(A) – 2

(B) 5

(C) 7

(D) 3

Solution: (C) 7.

 

(v) If P – Q – R and d (P,Q) = 2, d (P,R) = 10, then find d (Q,R).

(A) 12

(B) 8

(C) √96

(D) 20

Solution:

d (P, R) = d (P, Q) + d (Q, R)

= d (Q, R) = d (P, R) – d (P, Q) = 10 – 2 = 8

(B) 8.

 

(2) On a number line, co-ordinates of P, Q, R are 3, – 5 and 6 respectively. State with reason whether the following statements are true or false.

(i) d(P,Q) + d(Q,R) = d(P,R)

(ii) d(P,R) + d(R,Q) = d(P,Q)

(iii) d(R,P) + d(P,Q) = d(R,Q)

(iv) d(P,Q) – d(P,R) = d(Q,R)  

Solution:

d (P, Q) = 3 + 5 = 8

d (P, R) = 6 – 3 = 3

d (Q, R) = 6 + 5 = 11

 

(i) d (P, Q) + d (Q, R)

= 8 + 11 = 19 ≠ d (P, R)

= False.

 

(ii) d (P, R) + d (R, Q)

= 3 + 11 = 14 ≠ d (P, Q)

= False

 

(iii) d (R, P) + d (P, Q)

= 3 + 8 = 11 = d (Q, R)

= True

 

(iv) d (P, Q) – d (P, R)

= 8 – 3 = 5 ≠ d (Q, R)

= False.

 

(3) Co-ordinates of some pairs of points are given below. Hence find the distance between each pair.

(i) 3, 6

(ii) – 9, – 1

(iii) – 4, 5

(iv) 0, – 2

(v) x + 3, x- 3

(vi) – 25, – 47

(vii) 80, – 85

Solution:

(i) d (3, 6) = 6 – 3 = 3

(ii) d (-9, -1) = -1 + 9 = 8

(iii) d (-4, 5,) = 5 – (-4) = 9

(iv) d (0, -2) = 0 – (-2) = 2

(v) d (x + 3, x – 3) = x + 3 – x + 3 = 6

(vi) d (-25, -47) = -25 – (-47) = 22

(vii) d (80, -85) = 80 – (-85) = 165

 

(4) Co-ordinate of point P on a number line is – 7. Find the co-ordinates of points on the number line which are at a distance of 8 units from point P.

Solution:

Let, the points be Q and R.

∴ Coordinates of Q = – 7 – 8

= – 15

Coordinates of R = – 7 + 8

= – 1

 

(5) Answer the following questions.

(i) If A – B – C and d (A, C) = 17, d (B, C) = 6.5 then d (A, B) =?

(ii) If P – Q – R and d (P, Q) = 3.4, d (Q, R) = 5.7 then d(P, R) =?

Solution:

(i) d (A, C) = 17,

d (B, C) = 65 Then d (A, B) = ?

d (A, C) = d (A, B) + d (B, C) => d (A, B) = 17 – 6.5

= 10.5

 

(ii) d (P, Q) = 3.4,

d (Q, R) = 5.7 Then d (P, R) =?

d (P, R) = d (P, Q) + d (Q, R)

= 3.4 + 57 = 9.1

 

(6) Co-ordinate of point A on a number line is 1. What are the co-ordinates of points on the number line which are at a distance of 7 units from A?  

Solution:

Let, B and C be the points

Coordinates of B = 1.7 = – 6

Coordinates of C = 1 + 7 = 8

 

(7) Write the following statements in conditional form.

(i) Every rhombus is a square.

(ii) Angles in a linear pair are supplementary.

(iii) A triangle is a figure formed by three segments.

(iv) A number having only two divisors is called a prime number.

Solution:

(i) If a quadrilateral is a square then it is a rhombus

(ii) If the angles from a linear pair then they are supplementary

(iii) If given figure is a triangle, then it is formed by three segments.

(iv) If a number has only two divisors, then it is a prime number.

 

(8) Write the converse of each of the following statements.

(i) If the sum of measures of angles in a figure is 1800, then the figure is a triangle.

(ii) If the sum of measures of two angles is 900 then they are complement of each other.

(iii) If the corresponding angles formed by a transversal of two lines are congruent then the two lines are parallel.

(iv) If the sum of the digits of a number is divisible by 3 then the number is divisible by 3.

Solution:

If the figure is a triangle, then the sum of measures of angles is 180°.

(ii) IF the two angles are complement of each other, then the sum of measures of angles is 90°

(iii) If two lines are parallel then the corresponding angles formed by the transversal are congruent.

(iv) If a number is divisible by 3 then the sum of their digits is divisible by 3.

 

(9) Write the antecedent (given part) and the consequent (part to be proved) in the following statements.

(i) If all sides of a triangle are congruent then its all angles are congruent.

(ii) The diagonals of a parallelogram bisect each other.

Solution:

(i) Antecedent: All sides wides of a triangle are congruent

Congruent: All angles are congruent.

(ii) Antecedent: Quadrilateral is parallelogram

Consequent: Diagonals bisect each other

 

(10) Draw a labelled figure showing information in each of the following statements and write the antecedent and the consequent.

(i) Two equilateral triangles are similar.

(ii) If angles in a linear pair are congruent then each of them is a right angle.

(iii) If the altitudes drawn on two sides of a triangle are congruent then those two sides are congruent

Solution:

 

 

Here is your solution of Maharashtra Board Class 9 Math Part 2 Chapter 1 Basic Concept in Geometry Solution

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Updated: August 28, 2021 — 4:35 pm

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