Prasanna

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

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

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

Question 1.
Select the correct alternative answer for the questions given below.

i. How many midpoints does a segment have ?
(A) only one
(B) two
(C) three
(D) many
Answer:
(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
Answer:
(C) one

iii. How many lines are determined by three distinct points?
(A) two
(B) three
(C) one or three
(D) six
Answer:
(C) one or three

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
Answer:
Since, 5 > -2
∴ d(A, B) = 5 – (-2) = 5+2 = 7
(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
Answer:
d(P, R) = d(P, Q) + d(Q, R)
∴ 10 = 2 + d(Q, R)
∴ d(Q, R) = 8
(B) 8

Question 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:

Co-ordinate of the point P is 3.
Co-ordinate of the point Q is -5.
Since, 3 > -5
d(P, Q) = 3 – (-5) = 3 + 5
∴ d(P,Q) = 8
Co-ordinate of the point Q is -5.
Co-ordinate of the point R is 6.
Since, 6 > -5
d(Q, R) = 6 – (-5) = 6 + 5
∴ d(Q, R) = 11
Co-ordinate of the point P is 3.
Co-ordinate of the point R is 6.
Since, 6 > 3
d(P, R) = 6 – 3
∴ d(P, R) = 3

i. d(P, Q) + d(Q, R) = 8 + 11
= 19 …(i)
d(P, R) = 3 …(ii)
∴ d(P, Q) + d(Q, R) ≠ d(P, R) … [From (i) and (ii)]
∴ The given statement is false.

ii. d(P, R) + d(R, Q) = 3 + 11
d(P,Q) = 8 …(ii)
∴ d(P, R) + d(R, Q) + d(P, Q) …[From (i) and (ii)]
∴ The given statement is false.

iii. d(R, P) + d(P, Q) = 3 + 8
= 11 …(i)
d(R, Q) =11 . -(ii)
∴ d(R,P) + d(P,Q) = d(R,Q) ….[From (i) and (ii)]
∴ The given statement is true.

iv. d(P, Q) – d(P, R) = 8 – 3
= 5 …(i)
d(Q,R) = 11 ..(h)
∴ d(P, Q) – d(P, R) ≠ d(Q, R) …[From (i) and (ii)]
∴ The given statement is false.

Question 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. A, 5
iv. 0,-2
v. x + 3, x – 3
vi. -25, -47
vii. 80, -85
Solution:
i. Co-ordinate of first point is 3.
Co-ordinate of second point is 6.
Since, 6 > 3
∴ Distance between the points = 6 – 3 = 3

ii. Co-ordinate of first point is -9.
Co-ordinate of second point is -1.
Since, -1 > -9
∴ Distance between the points = -1 – (-9) = -1+9 = 8

iii. Co-ordinate of first point is -4.
Co-ordinate of second point is 5.
Since, 5 > -4
∴ Distance between the points = 5 – (-4)
= 5 + 4 = 9

iv. Co-ordinate of first point is 0.
Co-ordinate of second point is -2. Since,
0 > – 2
∴ Distance between the points = 0 – (-2)
= 0 + 2
= 2

v. Co-ordinate of first point is x + 3.
Co-ordinate of second point is x – 3.
Since, x + 3 > x – 3
∴ Distance between the points = x + 3 – (x – 3)
= x + 3 – x + 3 = 3 + 3
= 6

vi. Co-ordinate of first point is -25.
Co-ordinate of second point is -47.
Since, -25 > -47
∴ Distance between the points = -25 – (-47)
= -25 + 47
= 22

vii. Co-ordinate of first point is 80.
Co-ordinate of second point is -85.
Since, 80 > -85
∴ Distance between the points = 80 – (-85)
= 80 + 85
= 165

Question 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 point Q be at a distance of 8 units from P and on left side of P
Let point R be at a distance of 8 units from P and on right side of P.

i. Let the co-ordinate of point Q be x.
Co-ordinate of point P is -7.
Since, point Q is to the left of point P.
∴ -7 > x
∴ d(P, Q) = -7 -x
∴8 = -7 – x
∴ x = – 7 – 8
∴x = -15

ii. Let the co-ordinate of point R be y.
Co-ordinate of point P is -7.
Since, point R is to the right of point P.
∴ y > -7
∴ d(P, R) = 7- (-7)
∴ 8 = y + 7
∴ 8 – 7 = 7
∴ y = 1
∴ The co-ordinates of the points at a distance of 8 units from P are -15 and 1.

Question 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. Given, (A, C) = 17, d(B, C) = 6.5
d(A, C) = d(A, B) + d(B, C) …[A – B – C]
∴ 17 = d(A, B) + 6.5
∴ d(A,B)= 17 – 6.5
∴ d(A, B) = 10.5

ii. Given, d(P, Q) = 3.4, d(Q, R) = 5.7
d(P,R) = d(P,Q) + d(Q,R) …[P – Q – R]
= 34 + 5.7
∴ d(P, R) = 9.1

Question 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 point C be at a distance of 7 units from A and on left side of A
Let point B be at a distance of 7 units from A and on right side of A.

i. Let the co-ordinate of point C be x.
Co-ordinate of point A is 1.
Since, point C is to the left of point A.
∴ 1 > x
∴ d(A, C) = 1 – x
∴ 7 = 1 -x
∴x = 1 – 7
∴ x = – 6

ii. Let the co-ordinate of point B be y.
Co-ordinate of point A is 1.
Since, point B is to the right of point A.
∴y > 1
∴ d(A, B) = 7 – 1
∴ 7 = y – 1
∴ 7 + 1 = 7
∴ 7 = 8
∴ The co-ordinates of the points at a distance of 7 units from A are -6 and 8.

Question 7.
Write the following statements in conditional form.
i. Every rhombus is a square.
ii. Annies 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 rhombus, then it is a square.
ii. If iwo angles are in a linear pair, then they are supplementary.
iii. If a 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.

Question 8.
Write the converse of each of the following statements.
i. If the sum of measures of angles in a figure is 180°, then the figure is a triangle.
ii If the sum of measures of two angles is 90°, thfcn they are eomplement 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.
Answer:
i. If a figure is a triangle, then the sum of the measures of its angles is 180°.
ii. if two angles are eomplement of each other, then sum of their measures is 90°,
iii. If two lines are parallel, then the corresponding angles formed by a transversal of two lines are congruent.
iv. If a number is divisible by 3, then the sum of its digits is also divisible by 3.

Question 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.
Answer:
i. If all sides of a triangle are congruent, then its all angles are congruent.
Antecedent (Given): All the sides of the triangle are congruent.
Consequent (To prove): All the angles are congruent.

ii. The diagonals of a parallelogram bisect each other.
Conditional statement: “If a quadrilateral is a parallelogram then its diagonals bisect each other.
Antecedent (Given): Quadrilateral is a parallelogram.
Consequent (To prove): Its diagonals bisect each other.

Question 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 these two sides are congruent.
Answer:
i. Two equilateral triangles are similar.
Conditional statement: “If two triangles are equilateral, then they are similar.
Antecedent (Given): Two triangles are equailateral.
i.e. ∆ABC and ∆PQR are equilatral triangle.
Consequent (To prove): Triangles are similar
i.e. ∆ABC ∼ ∆PQR

ii. If angles in a linear pair are congruent, then each of them is a right angle.
Antecedent (Given): Angles in a linear pair are congrunent.
∠ABC and ∠ABD are angles in a linear pair i.e. ∠ABC = ∠ABD
Consequent (To prove): Each angle is a right angle.
i.e. ∠ABC – ∠ABD = 90°

iii. If the altitudes drawn on two sides of a triangle are congruent, then these two sides are congruent.
Antecedent (Given): Altitude drawn on two sides of triangle are congrunent.
In ∆ABC, AD ⊥ BC . and BE ⊥ AC. seg AD ≅ seg BE

Consequent (To prove): Two sides are congruent.
side BC ≅ side AC A

Maharashtra Board Class 9 Maths Chapter 1 Basic Concepts in Geometry Problem Set 1 Intext Questions and Activities

Question 1.
Points A, B, C are given below. Check, with a stretched thread, whether the three points are collinear or not. If they are collinear, write which one of them is between the other two. (Textbook pg. no. 4)

Answer:
Point B is between the points A and C.

Question 2.
Given below are four points P, Q, R, and S. Check which three of them are collinear and which three are non collinear. In the case of three collinear points, state which of them is between the other two. (Textbook pg. no. 4)

Answer:
Points P, R and S are collinear.
Point R is between the points P and S.

Question 3.
Students are asked to stand in a line for mass drill. How will you check whether the students standing are in a line or not ? (Textbook pg. no. 4)
Answer:
If one stands in front of the line and observes only the first student standing in the line, then all the students standing in that line are collinear i.e., standing in the same line. We can use this property of collinearity to check whether the students are standing in the same line or not.

Question 4.
How had you verified that light rays travel in a straight line? Recall an experiment in science which you have done in a previous standard. (Textbook pg. no. 4)
Answer:

The flame of the candle can be seen only when the pin holes in all cardboards are in the same straight line. We can use the set up shown in the figure above to verify that light rays travels in a straight line.

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 9 Maths Solutions covers the Practice Set 2.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 2 Parallel Lines.

9th Standard Maths 2 Practice Set 2.2 Chapter 2 Parallel Lines Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 2.2 Chapter 2 Parallel Lines Questions With Answers Maharashtra Board

Question 1.
In the given figure, y = 108° and x = 71°. Are the lines m and n parallel? Justify?

Solution:
y = 108°, x = 71° …[Given]
x + y = 71° + 108°
= 179°
∴ x + y = 180°
∴ The angles x andy are not supplementary.
∴ The angles do not satisfy the interior angles test for parallel lines
∴ line m and line n are not parallel lines.

Question 2.
In the given figure, if ∠a = ∠b then prove that line l || line m.

Given: ∠a ≅ ∠b
To prove: line l| line m
Solution:
Proof:

consider ∠c as shown in the figure ∠a ≅ ∠c …….. (i) [Vertically opposite angles]
But, ∠a ≅ ∠b I (ii) [Given]
∴ ∠b ≅ ∠c [From (i) and (ii)]
But, ∠b and ∠c are corresponding angles on lines l and m when line n is the transversal.
∴ line l || line m. [Corresponding angles test]

Question 3.
In the given figure, if ∠a ≅ ∠b and ∠x ≅ ∠y, then prove that line l | line n.

Given: ∠a ≅ ∠b and ∠x ≅ ∠y
To prove: line l | line n
Solution:
Proof:

∠a = ∠b [Given]
But, ∠a and ∠b are corresponding angles on lines l and m when line k is the transversal.
∴ line l || line m ….(i) [Corresponding angles test]
∠x ≅ ∠y [Given]
But, ∠x and ∠y are alternate angles on lines m and n when seg PQ is the transversal,

∴ line m || line n ……(ii) [Alternate angles test]
∴ From (i) and (ii),
line l || line m || line n
i.e., line l || line n

Question 4.
In the given figure, if ray BA || ray DE, ∠C = 50° and ∠D = 100°. Find the measure of ∠ABC.
(Hint: Draw a line passing through point C and parallel to line AB.)

Solution:
Draw a line FG passing through point
C and parallel to line AB
line FG || ray BA …….(i) [Construction]
Ray BA || ray DE ….(ii) [Given]
line FG || ray BA || ray DE …(iii) [From (i) and (ii)]
line FG||rayDE [From (iii)]
and seg DC is their transvensal
∴ ∠ DCF = ∠ EDC [Alternate angles]
∴ ∠ DCF = 100° [∵ ∠D = 100°]
Now, ∠ DCF = ∠ BCF + ∠ BCD [Angle addition property]
∴ 100° = ∠BCF + 50°
∴ 100° – 50° = ∠BCF
∴ ∠BCF = 50° ….(iv)
Now, line FG || ray BA and seg BC is their transversal.

∴ ∠ABC + ∠BCF = 180° [Interior angles]
∴ ∠ABC + 50° = 180° [From (iv)]
∴ ∠ABC = 180°- 50°
∴ ∠ABC = 130°

Question 5.
In the given figure, ray AE || ray BD, ray AF is the bisector of ∠EAB and ray BC is the bisector of ∠ABD. Prove that line AF || line BC.

Given: Ray AE || ray BD, and
ray AF and ray BC are the bisectors of ∠EAB and ∠ABD respectively.
To prove: line AF || line BC
Solution:
Proof:
Ray AE || ray BD and seg AB is their transversal.
∴ ∠EAB = ∠ABD ….(i) [Alternate angles]
∠FAB = \(\frac { 1 }{ 2 }\)∠EAB [Ray AF bisects ∠EAB]
∴ 2∠FAB = ∠EAB …..(ii)
∠CBA = \(\frac { 1 }{ 2 }\)∠ABD [Ray BC bisects ∠ABD]
∴ 2∠CBA = ∠ABD …(iii)
∴ 2∠FAB = 2∠CBA [From (i), (ii) and (iii)]
∴ ∠FAB = ∠CBA
But, ∠FAB and ∠ABC are alternate angles on lines AF and BC when seg AB is the transversal.

∴ line AF || line BC [Alternate angles test]

Question 6.
A transversal EF of line AB and line CD intersects the lines at points P and Q respectively. Ray PR and ray QS are parallel and bisectors of ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.

Given: Ray PR || ray QS
Ray PR and ray QS are the bisectors of ∠BPQ and ∠PQC respectively.
To prove: line AB || line CD
Solution:

Proof:
Ray PR || ray QS and seg PQ is their transversal.
∠RPQ = ∠SQP ….(i) [Alternate angles]
∠RPQ = \(\frac { 1 }{ 2 }\)∠BPQ …. (ii) [Ray PR bisects ∠BPQ]
∠SQP = \(\frac { 1 }{ 2 }\)∠PQC [Ray QS bisects ∠PQC]
∴ \(\frac { 1 }{ 2 }\)∠BPQ = \(\frac { 1 }{ 2 }\)∠PQC
∴ ∠BPQ = ∠PQC
But, ∠BPQ and ∠PQC are alternate angles on lines AB and CD when line EF is the transversal.
∴ line AB || line CD [Alternate angles test]

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Practice Set 2.2 Intext Questions and Activities

Question 1.
In the given figure, how will you decide whether line ¡ and line m are parallel or not? (Textbook pg. no. 19)

Answer:
In the figure, we observe that line I and line m are coplanar and do not intersect each other.
∴ Line l and line m are parallel lines.

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• ProblemSet 1.3 Class 9 Answers
• Practice 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 9 Maths Solutions covers the Practice Set 1.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

9th Standard Maths 2 Practice Set 1.3 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 1.3 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 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.
Answer:
i. If a quadrilateral is a parallelogram, then its opposite angles are congruent.
ii. If a quadrilateral is a rectangle, then its diagonals are congruent.
iii. If a triangle is isosceles triangle, then segment joining the vertex of a triangle and midpoint of the base is perpendicular to the base.

Question 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.
Answer:
i. If the alternate angles made by two lines and their transversal are congruent, then the two lines are parallel.
ii. If two parallel lines are intersected by a transversal, then the interior angles formed bv the transversal are supplementary.
iii. If the diagonals of a quadrilateral are congruent, then that quadrilateral is a rectangle.

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 9 Maths Solutions covers the Practice Set 1.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

9th Standard Maths 2 Practice Set 1.2 Chapter 1 Basic Concepts in Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 1.2 Chapter 1 Basic Concepts in Geometry Questions With Answers Maharashtra Board

Question 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.

i. seg DE and seg AB
ii. seg BC and seg AD
iii. seg BE and seg AD
Solution:
i. Co-ordinate of the point E is 9.
Co-ordinate of the point D is -7.
Since, 9 > -7
∴ d(D, E) = 9 – (-7) = 9 + 7 = 16
∴ l(DE) = 16 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point B is 5.
Since, 5 > -3
∴ d(A, B) = 5 – (-3) = 5 + 3 = 8
∴ l(AB) = 8 …(ii)
∴ l(DE) ≠ l(AB) …[From (i) and (ii)]
∴ seg DE and seg AB are not congruent.

ii. Co-ordinate of the point B is 5.
Co-ordinate of the point C is 2.
Since, 5 > 2
∴ d(B, C) = 5 – 2 = 3
∴ l(BC) = 3 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = -3 + 7 = 4
∴ l(AD) = 4 . ..(ii)
∴ l(BC) ≠ l(AD) … [From (i) and (ii)]
∴ seg BC and seg AD are not congruent.

iii. Co-ordinate of the point E is 9.
Co-ordinate of the point B is 5.
Since, 9 > 5
∴ d(B, E) = 9 – 5 = 4
∴ l(BE) = 4 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = 4
∴ l(AD) = 4 …(ii)
∴ l(BE) =l(AD) …[From (i) and (ii)]
∴ seg BE and seg AD are congruent.
i.e, seg BE ≅ seg AD

Question 2.
Point M is the midpoint of seg AB. If AB = 8, then find the length of AM.
Solution:
Point M is the midpoint of seg AB and l(AB) = 8. …[Given]

Question 3.
Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).
Solution:
Point P is the midpoint of seg CD and l(CP) = 2.5 …[Given]

∴ l(CD) = 2.5 x 2
∴ l(CD) = 5

Question 4.
If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.
Solution:
Given, l(AB) = 5 cm, l(BP) = 2 cm,
l(AP) = 3.4 cm … [Given]
r Since, 2 < 3.4 < 5
∴ l(BP) < l(AP) < l(AB)
i.e., seg BP < seg AP < seg AB

Question 5.
Write the answers to the following questions with reference to the figure given below:

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 ray PQ and ray 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.
Answer:
i. Ray RS or ray RT
ii. Ray PQ
iii. Line QR
iv. Ray QR, ray QS, ray QT, ray RQ, ray SQ, ray TQ
v. Ray RP and ray RS, ray RQ and ray RT vi. Ray ST, ray SR
vii. Point S

Question 6.
Answer the questions with the help of figure given below.

i. State the points which are equidistant from point B.
ii. Write a pair of points equidistant from point iii. Find d(U,V), d(P,C), d(V,B), d(U, L).
Answer:
i. Points equidistant from point B are a. A and C, because d(B, A) = d(B, C) = 2 b. D and P, because d(B, D) = d(B, P) = 4
ii. Points equidistant from point Q are a. L and U, because d(Q, L) = d(Q, U) = 1 b. P and R, because d(P, Q) = d(Q, R) = 2
iii. a. Co-ordinate of the point U is -5. Co-ordinate of the point V is 5. Since, 5 > -5
∴ d(U, V) = 5 – (-5)
= 5 + 5
∴ d(U, V) = 10

b. Co-ordinate of the point P is -2.
Co-ordinate of the point C is 4.
Since, 4 > -2
∴ d(P, C) = 4 – (-2)
= 4 + 2
∴ d(P, C) = 6

c. Co-ordinate of the point V is 5.
Co-ordinate of the point B is 2.
Since, 5 > 2
∴ d(V, B) = 5 – 2
∴ d(V, B) = 3

d. Co-ordinate of the point U is -5.
Co-ordinate of the point L is -3.
Since, -3 > -5
∴ d(U, L) = -3 – (-5)
= -3 + 5
∴ d(U, L) = 2

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 8 Maths Solutions covers the Practice Set 10.2 8th Std Maths Answers Solutions Chapter 10 Division of Polynomials.

Division of Polynomials Class 8 Maths Chapter 10 Practice Set 10.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 10.2 Chapter 10 Solutions Answers

Division of Polynomials Class 8 Practice Set 10.2 Question 1. Divide and write the quotient and the remainder.
i. (y² + 10y + 24) ÷ (y + 4)
ii. (p² + 7p – 5) ÷ (p + 3)
iii. (3x + 2x² + 4x³) ÷ (x – 4)
iv. (2m³ + m² + m + 9) ÷ (2m – 1)
v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)
vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Solution:
i. (y² + 10y + 24) ÷ (y + 4)

∴ Quotient = y + 6
Remainder = 0

ii. (p² + 7p – 5) ÷ (p + 3)

∴ Quotient = p + 4
Remainder = -17

iii. (3x + 2x² + 4x³) ÷ (x – 4)
Write the dividend in descending order of their indices.
3x + 2x² + 4x³ = 4x³ + 2x² + 3x

∴ Quotient = 4x² + 18x + 75
Remainder = 300

iv. (2m³ + m² + m + 9) ÷ (2m – 1)

∴ Quotient = m² + m + 1
Remainder = 10

v. (3x – 3x² – 12 + x⁴ + x³) ÷ (2 + x²)
Write the dividend in descending order of their indices.
(x⁴ + x³ – 3x² + 3x – 12) ÷ (x² + 2)

∴ Quotient = x² + x – 5
Remainder = x – 2

vi. (a⁴ – a³ + a² – a + 1) ÷ (a³ – 2)

∴ Quotient = a – 1
Remainder = a² + a – 1

vii. (4x⁴ – 5x³ – 7x + 1) ÷ (4x – 1)
Write the dividend in descending order of their indices.
(4x⁴ – 5x³ – 7x + 1) = (4x⁴ – 5x³ + 0x² – 7x + 1)

∴ Quotient = \(x^{3}-x^{2}-\frac{x}{4}-\frac{29}{16}\)
Remainder = \(\frac { -13 }{ 16 }\)

Std 8 Maths Digest

• Practice Set 8.1 Class 8 Answers
• Practice Set 8.2 Class 8 Answers
• Practice Set 8.3 Class 8 Answers
• Practice Set 9.1 Class 8 Answers
• Practice Set 9.2 Class 8 Answers
• Practice Set 10.1 Class 8 Answers
• Practice Set 10.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 11.3 8th Std Maths Answers Solutions Chapter 11 Statistics.

Statistics Class 8 Maths Chapter 11 Practice Set 11.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 11.3 Chapter 11 Solutions Answers

Exercise 11.3 Class 8 Question 1.
Show the following information by a percentage bar graph.

Division of standard 8	A	B	C	D
Number of students securing grade A	45	33	10	15
Total number of students	60	55	40	75

Solution:

Division of standard 8	A	B	C	D
Number of students securing grade A	45	33	10	15
Total number of students	60	55	40	75
Percentage of students securing grade A	75%	60%	25%	20%
Percentage of students not securing grade A	25%	40%	75%	80%

Statistics for Class 8 Question 2.
Observe the following graph and answer the questions.

1. State the type of the bar graph.
2. How much percent is the Tur production to total production in Ajita’s farm?
3. Compare the production of Gram in the farms of Yash and Ravi and state whose percentage of production is more and by how much?
4. Whose percentage production of Tur is the least?
5. State production percentages of Tur and Gram in Sudha’s farm.

Solution:

1. The given graph is a percentage bar graph.
2. Percent of tur production to the total production in Ajita’s farm is 60%.
3. Production of Gram in the farm of Yash = 50%
   Production of Gram in the farm of Ravi = 30%
   ∴ Difference in the production = 50% – 30% =20%
   ∴ Yash’s production of Gram is more and by 20%.
4. Sudha’s percentage production of Tur is the least.
5. Production percentages of Tur and Gram in Sudha’s farm are 40% and 60% respectively.

8th Standard Statistics Question 3.
The following data is collected in a survey of some students of 10th standard from some schools. Draw the percentage bar graph of the data.

School	1st	2nd	3rd	4th
Inclination towards science stream	90	60	25	16
Inclination towards commerce stream	60	20	25	24

Solution:

School	1st	2nd	3rd	4th
Inclination towards science stream	90	60	25	16
Inclination towards commerce stream	60	20	25	24
Total number of students	150	80	50	40
Percentage of students having inclination towards science stream	60%	75%	50%	40%
Percentage of students having inclination towards commerce stream	40%	25%	50%	60%

Maharashtra Board Class 8 Maths Chapter 11 Statistics Practice Set 11.3 Intext Questions and Activities

Statistics 8th Class Question 1.
Compare and discuss a percentage bar diagram and a subdivided bar diagram. Use it to learn the graphs in the subjects like Science, Geography. (Textbook pg, no. 74)
Solution:
[Students should attempt the above activity on their own.].

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 16 Quadrilaterals Class 6 Practice Set 37 Answers Solutions.

Quadrilaterals Class 6 Maths Chapter 16 Practice Set 37 Solutions Maharashtra Board

Std 6 Maths Practice Set 37 Solutions Answers

Question 1.
Observe the figures below and find out their names:

Solution:
i. Pentagon (5 sides)
ii. Hexagon (6 sides)
iii. Heptagon (7 sides)
iv. Octagon (8 sides)

Maharashtra Board Class 6 Maths Chapter 16 Quadrilaterals Practice Set 37 Intext Questions and Activities

Question 1.
Observe the figures given below and say which of them are quadrilaterals. (Textbook pg. no. 81)

Solution:
Is a quadrilateral: (i)

Question 2.
Draw a quadrilateral. Draw one diagonal of this quadrilateral and divided it into two triangles. Measures all the angles in the figure. Is the sum of the measures of the four angles of the quadrilateral equal to the sum of the measures of the six angles of the two triangles? Verity that this is so with other quadrilaterals. (Textbook pg. no. 84)
Solution:

m∠PQR = 104°
m∠QRP = 26°
m∠RPQ = 50°
m∠PRS = 34°
m∠RSP = 106°
m∠SPR = 40°
∴ Sum of the measures of the angles of quadrilateral = m∠PQR + m∠QRP + m∠RPQ + m∠PRS + m∠RSP + m∠SPR
= 104° + 26° + 50° + 34° + 106° + 40°
= 360°
Also, we observe that
Sum of the measures of the angles of quadrilateral = Sum of the measures of angles of the two triangles (PQR and PRS)
= (104°+ 26°+ 50°)+ (34° + 106° + 40°)
= 180° + 180°
= 360°
[Note: Students should drew different quadrilaterals and verify the property.]

Question 3.
For the pentagon shown in the figure below, answer the following: (Textbook pg. no. 84)

1. Write the names of the five vertices of the pentagon.
2. Name the sides of the pentagon.
3. Name the angles of the pentagon.
4. See if you can sometimes find players on a field forming a pentagon.

Solution:

1. The vertices of the pentagon are points A, B, C, D and E.
2. The sides of the pentagon are segments AB, BC, CD, DE and EA.
3. The angles of the pentagon are ∠ABC, ∠BCD, ∠CDE, ∠DEA and ∠EAB.
4. The players shown in the above figure form a pentagon. The players are standing on the vertices of

Question 4.
Cut out a paper in the shape of a quadrilateral. Make folds in it that join the vertices of opposite angles. What can these folds be called? (Textbook pg. no. 83)

Solution:
The folds are called diagonals of the quadrilateral.

Question 5.
Take two triangular pieces of paper such that . one side of one triangle is equal to one side of the other. Let us suppose that in ∆ABC and ∆PQR, sides AC and PQ are the equal sides. Join the triangles so that their equal sides lie B side by side. What figure do we get? (Textbook pg. no. 83)

Solution:
If we place the triangles together such that the equal sides overlap, the two triangles form a quadrilateral.

Std 6 Maths Digest

• Practice Set 35 Class 6 Answers
• Practice Set 36 Class 6 Answers
• Practice Set 37 Class 6 Answers
• Practice Set 38 Class 6 Answers
• Practice Set 39 Class 6 Answers
• Practice Set 40 Class 6 Answers
• Practice Set 41 lass 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 2 Angles Class 6 Practice Set 3 Answers Solutions.

Angles Class 6 Maths Chapter 2 Practice Set 3 Solutions Maharashtra Board

Std 6 Maths Practice Set 3 Solutions Answers

Question 1.
Use the proper geometrical instruments to construct the following angles. Use the compass and the ruler to bisect them:

1. 50°
2. 115°
3. 80°
4. 90°

Solution:

Maharashtra Board Class 6 Maths Chapter 2 Angles Practice Set 3 Intext Questions and Activities

Question 1.
Construct an angle bisector to obtain an angle of 30°. (Textbook pg. no. 11)
Solution: .
In order to get a bisected angle of a given measure, the student has to draw the angle having twice the measurement of required bisected angle.

For getting measurement of 30° (for the bisected angle), one has to make an angle of 60° (i.e. 30° × 2).

Step 1:
Draw ∠ABC of 60°.

Step 2:
Cut arcs on the rays BA and BC. Name these points as D and E respectively.

Step 3:
Place the compass point on point D and draw an arc inside the angle.
Without changing the distance of the compass, place the compass point on point E and cut the previous arc. Name the point of intersection as O

Step 4:
Draw ray BO.
Ray BO is the angle bisector of ∠ABC.
i.e. m∠ABO = m∠CBO = 30°

Question 2.
Construct an angle bisector to draw an angle of 45°. (Textbook pg. no. 11)
Solution:
For getting measurement of 45° (for the bisected angle), one has to make an angle of 90° (i.e. 45° × 2).
Step 1:
Draw ∠PQR of 90°.

Step 2:
Cut arcs on the rays QP and QR.
Name these points as M and N respectively.

Step 3:
Place the compass point on point M and draw an arc inside the angle.
Without changing the distance of the compass, place the compass point on point N and cut the

Step 4:
Draw ray QO.
Ray QO is the angle bisector of ∠PQR.
i.e. m∠PQO = m∠RQO = 45°

Question 3.
Ask three or more children to stand in a straight line. Take two long ropes. Let the child in the middle hold one end of each rope. With the help of the ropes, make the children on either side stand along a straight line. Tell them to move so as to form an acute angle, a right angle, an obtuse angle, a straight angle, a reflex angle and a full or complete angle in turn. Keeping the rope stretched will help to ensure that the children form straight lines. (Textbook pg. no. 6)
Solution:

Question 4.
Look at the pictures below and identify the different types of angles. (Textbook pg. no. 8)

Solution:
i. Complete angle
ii. Reflex and Acute angle
iii. Acute and Obtuse angle

Std 6 Maths Digest

• Practice Set 1 Class 6 Answers
• Practice Set 2 Class 6 Answers
• Practice Set 3 Class 6 Answers
• Practice Set 4 Class 6 Answers
• Practice Set 5 Class 6 Answers
• Practice Set 6 Class 6 Answers
• Practice Set 7 Class 6 Answers
• Practice Set 8 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 5 Decimal Fractions Class 6 Practice Set 15 Answers Solutions.

Decimal Fractions Class 6 Maths Chapter 5 Practice Set 15 Solutions Maharashtra Board

Std 6 Maths Practice Set 15 Solutions Answers

Question 1.
Write the proper number in the empty boxes.

Solution:

Question 2.
Convert the common fractions into decimal fractions:
i. \(\frac { 3 }{ 4 }\)
ii. \(\frac { 4 }{ 5 }\)
iii. \(\frac { 9 }{ 8 }\)
iv. \(\frac { 17 }{ 20 }\)
v. \(\frac { 36 }{ 40 }\)
vi. \(\frac { 7 }{ 25 }\)
vii. \(\frac { 19 }{ 200 }\)
Solution:
i. \(\frac { 3 }{ 4 }\)

ii. \(\frac { 4 }{ 5 }\)

iii. \(\frac { 9 }{ 8 }\)

iv. \(\frac { 17 }{ 20 }\)

v. \(\frac { 36 }{ 40 }\)

vi. \(\frac { 7 }{ 25 }\)

vii. \(\frac { 19 }{ 200 }\)

Question 3.
Convert the decimal fractions into common fractions:
i. 27.5
ii. 0.007
iii. 90.8
iv. 39.15
v. 3.12
vi. 70.400
Solution:
i. 27.5
= \(\frac { 275 }{ 10 }\)

ii. 0.007
= \(\frac { 7 }{ 1000 }\)

iii. 90.8
= \(\frac { 908 }{ 10 }\)

iv. 39.15
= \(\frac { 3915 }{ 100 }\)

v. 3.12
= \(\frac { 312 }{ 100 }\)

vi. 70.400
= 70.4
= \(\frac { 704 }{ 10 }\)

Std 6 Maths Digest

• Practice Set 9 Class 6 Answers
• Practice Set 10 Class 6 Answers
• Practice Set 11 Class 6 Answers
• Practice Set 12 Class 6 Answers
• Practice Set 13 Class 6 Answers
• Practice Set 14 Class 6 Answers
• Practice Set 15 Class 6 Answers
• Practice Set 16 Class 6 Answers
• Practice Set 17 Class 6 Answers

Balbharti Maharashtra State Board Class 6 Maths Solutions covers the Std 6 Maths Chapter 10 Equations Class 6 Practice Set 27 Answers Solutions.

Equations Class 6 Maths Chapter 10 Practice Set 27 Solutions Maharashtra Board

Std 6 Maths Practice Set 27 Solutions Answers

Question 1.
Rewrite the following using a letter:
i. The sum of a certain number and 3.
ii. The difference is obtained by subtracting 11 from another number.
iii. The product of 15 and another number.
iv. Four times a number is 24.
Solution:
i. Let the number be x.
∴ x + 3 represents the sum of a certain number x and 3.

ii. Let the number be x.
∴ x – 11 represents the number obtained by subtracting 11 from another number x.

iii. Let the number be x.
∴ 15x represents the product of 15 and another number x.

iv. Let the number be x.
∴ 4x = 24 represents four the product of a number x four times.

Question 2.
Find out which operation must be done on both sides of these equations in order to solve them:

1. x + 9 = 11
2. x – 4 = 9
3. 8x = 24
4. \(\frac { x }{ 6 }\) = 3

Solution:

1. Subtract 9 from both sides.
2. Add 4 to both sides.
3. Divide both sides by 8.
4. Multiply both sides by 6.

Question 3.
Given below are some equations and the values of the variables. Are these values the solutions to those equations?

No.	Equation	Value of the Variable	Solution (Yes/No)
i.	y – 3 = 11	y = 3	No
ii.	17 = n + 7	n = 10
iii.	30 = 5x	x = 6
iv.	\(\frac { m }{ 2 }\) = 14	m = 7

Solution:

No.	Equation	Value of the Variable	Solution (Yes/No)
i.	y – 3 = 11	y = 3	No
ii.	17 = n + 7	n = 10	Yes
iii.	30 = 5x	x = 6	Yes
iv.	\(\frac { m }{ 2 }\) = 14	m = 7	No

i. y – 3 = 11
∴ y – 3 + 3 = 11 + 3
…. (Adding 3 to both sides)
∴ y + 0 = 14
∴ y = 14

ii. 17 = n + 7
∴ 17 – 7 = n + 7 – 7
…. (Subtracting 7 from both sides)
∴ 17 + (-7) = n + 7 – 7
∴ 10 = n
∴  n = 10

iii. 30 = 5x
∴ \(\frac{30}{5}=\frac{5x}{5}\)
…. (Dividing both sides by 5)
∴  6 = 1x
∴ 6 = x
∴  x = 6

iv. \(\frac { m }{ 2 }\) = 14
∴ \(\frac { m }{ 2 }\) × 2 = 14 × 2
…. (Multiplying both sides by 2)
\(\frac { m\times2 }{ 2\times1 }\) = 28
∴ m = 28

Question 4.
Solve the following equations:
i. y – 5 = 1
ii. 8 = t + 5
iii. 4x = 52
iv. 19 = m – 4
v. \(\frac { p }{ 4 }=9\)
vi. x + 10 = 5
vi. m – 5 = -12
vii. p + 4 = -1
Solution:
i. y – 5 = 1
∴y – 5 + 5 = 1 + 5
…. (Adding 5 to both sides)
∴y + 0 = 6
∴y = 6

ii. 8 = t + 5
∴8 – 5 = t + 5 – 5
……(Subtracting 5 from both sides)
∴8 + (-5) = t + 0
∴ 3 = t
∴t = 3

iii. 4x = 52
∴\(\frac{4x}{4}=\frac{52}{4}\)
…. (Dividing both sides by 4)
∴ 1x = 13
∴ x = 13

iv. 19 = m -4
∴ 19 + 4 = m – 4 + 4
…. (Adding 4 to both sides)
∴ 23 = m + 0
∴ m = 23

v. \(\frac { p }{ 4 }\) = 9
∴ \(\frac { p }{ 4 }\) × 4 = 9 × 4 …. (Multiplying both sides by 4)
∴ \(\frac { p\times4 }{ 4\times1 }=36\)
∴ 1p = 36
∴ p = 36

vi. x + 10 = 5
∴ x + 10 – 10 = 5 – 10
…. (Subtracting 10 from both sides)
∴ x + 0 = 5 + (-10)
∴ x = -5

vii. m – 5 = -12
∴m – 5 + 5 = – 12 + 5
…. (Adding 5 to both sides)
∴m + 0 = -7
∴m = -7

viii. p + 4 = – 1
∴p + 4 – 4 = -1 – 4
…. (Subtracting 4 from both sides)
∴p + 0 = (-1) + (-4)
∴P = -5

Question 5.
Write the given information as an equation and find its solution:
i. Haraba owns some sheep. After selling 34 of them in the market, he still has 176 sheep. How many sheep did Haraba have at first?

ii. Sakshi prepared some jam at home and filled it in bottles. After giving away 7 of the bottles to her friends she still has 12 for herself. How many bottles had she made in all? If she filled 250g of jam in each bottle, what was the total weight of the jam she made?

iii. Archana bought some kilograms of wheat. She requires 12 kg per month and she got enough wheat milled for 3 months. After that, she had 14 kg left. How much wheat had Archana bought altogether?
Solution:
i. Let the number of sheep before selling be x.
∴ x – 34 = 176
∴ x – 34 + 34 = 176 + 34 ….(Adding 34 to both sides)
∴ x + 0 = 210
∴ x = 210
The number of sheep with Haraba before selling is 210.

ii. Let the total number of bottles be x.
∴ x – 7 = 12
∴ x – 7 + 7 = 12 + 7 ….(Adding 7 to both sides)
∴ x + 0 = 19
∴ x = 19
Weight of jam in each bottle = 250g
∴ Total weight of jam = 19 × 250g = 4750 g = \(\frac { 4750 }{ 1000 }\)kg = 4.75 kg
∴ The total number of bottles of jam made by Sakshi is 19, and the total weight of jam made is 4.75 kg.

iii. Let the total wheat bought by Archana be x kg.
Wheat used in 1 month = 12 kg
∴ Wheat used in 3 months = 3 × 12 = 36 kg
∴ x – 36 = 14
∴ x – 36 + 36 = 14 + 36 ….(Adding 36 to both sides)
∴ x + 0 = 50
∴ x = 50
∴ The total amount of wheat bought by Archana was 50 kg..

Std 6 Maths Digest

• Practice Set 26 Class 6 Answers
• Practice Set 27 Class 6 Answers
• Practice Set 28 Class 6 Answers
• Practice Set 29 Class 6 Answers
• Practice Set 30 Class 6 Answers
• Practice Set 31 Class 6 Answers
• Practice Set 32 Class 6 Answers
• Practice Set 33 Class 6 Answers
• Practice Set 34 Class 6 Answers