Class 9

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 7 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

9th Standard Maths 2 Problem Set 7 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Problem Set 7 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

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

i. What is the form of co-ordinates of a point on the X-axis?
(A) (b,b)
(B) (0, b)
(C) (a, 0)
(D) (a, a)
Answer:
(C) (a, 0)

ii. Any point on the line y = x is of the form _____.
(A) (a, a)
(B) (0, a)
(C) (a, 0)
(D) (a, -a)
Answer:
(A) (a, a)

iii. What is the equation of the X-axis ?
(A) x = 0
(B) y = 0
(C) x + y = 0
(D) x = y
Answer:
(B) y = 0

iv. In which quadrant does the point (-4, -3) lie ?
(A) First
(B) Second
(C) Third
(D) Fourth
Answer:
(C) Third

v. What is the nature of the line which includes the points (-5, 5), (6, 5), (-3, 5), (0, 5)?
(A) Passes through the origin
(B) Parallel to Y-axis
(C) Parallel to X-axis
(D) None of these
Answer:
The y co-ordinate of all the points is the same.
∴ The line which passes through the given points is parallel to X-axis.
(C) Parallel to X-axis

vi. Which of the points P(-1, 1), Q(3, -4), R( -1, -1), S(-2, -3), T (-4, 4) lie in the fourth quadrant?
(A) P and T
(B) Q and R
(C) only S
(D) P and R
Answer:
(B) Q and R

Question 2.
Some points are shown in the adjoining figure. With the help of it answer the following questions :

i. Write the co-ordinates of the points Q and R.
ii. Write the co-ordinates of the points T and M.
iii. Which point lies in the third quadrant ?
iv. Which are the points whose x and y co-ordinates are equal ?
Solution:
i. Q(-2, 2) and R(4, -1)
ii. T(0, -1) and M(3, 0)
iii. Point S lies in the third quadrant.
iv. The x and y co-ordinates of point O are equal.

Question 3.
Without plotting the points on a graph, state in which quadrant or on which axis do the following points lie.
i. (5, -3)
ii. (-7, -12)
iii. (-23, 4)
iv. (-9, 5)
v. (0, -3)
vi. (-6, 0)
Solution:

Question 4.
Plot the following points on one and the same co-ordinate system.
A(1, 3), B(-3, -1), C(1, -4), D(-2, 3), E(0, -8), F(1, 0)
Solution:

Question 5.
In the graph alongside, line LM is parallel to the Y-axis.

i. What is the distance of line LM from the Y-axis?
ii. Write the co-ordinates of the points P, Q and R.
iii. What is the difference between the x co-ordinates of the points L and M?
Solution:
i. Distance of line LM from the Y-axis is 3 units.
ii. P(3, 2), Q (3, -1), R(3, 0)
iii. x co-ordinate of point L = 3
x co-ordinate of point M = 3
∴ Difference between the x co-ordinates of the points L and M = 3 – 3
= 0

Question 6.
How many lines are there which are parallel to X-axis and having a distance 5 units?
Solution:
The equation of a line parallel to the X-axis is y = b.
There are 2 lines which are parallel to X-axis and at a distance of 5 units.
Their equations are y = 5 and y = -5.

Question 7.
If ‘a’ is a real number, what is the distance between the Y-axis and the line x = a?
Solution:
Equation of Y-axis is x = 0.
Since, ‘a’ is a real number, there are two possibilities.
Case I: a > 0
Case II: a < 0 ∴ Distance between the Y-axis and the line x = a = a-0 = a Since, |a| = a, a > 0
= – a, a < 0
∴ Distance between the Y-axis and the line x = a is |a|.

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Problem Set 7 Intext Questions and Activities

Question 1.
As shown in the adjoining figure, ask girls to sit in lines so as to form the X-axis and Y-axis.
i. Ask some boys to sit at the positions marked by the coloured dots in the four quadrants.
i. Now, call the students turn by turn using the initial letter of each student’s name. As his or her initial is called, the student stands and gives his or her own co-ordinates. For example Rajendra (2, 2) and Kirti (-1, 0)
iii. Even as they have fun during this field activity, the students will leam how to state the position of a point in a plane. (Textbook pg. no. 92)

Class 9 Maths Digest

• Practice Set 7.1 Class 9 Answers
• Practice Set 7.2 Class 9 Answers
• Problem Set 7 Class 9 Answers
• Practice Set 8.1 Class 9 Answers
• Practice Set 8.2 Class 9 Answers
• Problem Set 8 Class 9 Answers
• Practice Set 9.1 Class 9 Answers
• Practice Set 9.2 Class 9 Answers
• Practice Set 9.3 Class 9 Answers
• Problem Set 9 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

9th Standard Maths 2 Practice Set 7.2 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 7.2 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
On a graph paper plot the points A(3, 0), B(3, 3), C(0, 3). Join A, B and B, C. What is the figure formed?
Soiution:

d(O, A) = 3 cm, d(A, B) = 3 cm, d(B, C) = 3 cm, d(O, C) = 3 cm and each angle of □ OABC is 90°
∴ □ OABC is a square.

Question 2.
Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Since, the line is at a distance of 7 units to the left of Y-axis,
∴ a = -7
∴ x = -1 is the equation of the required line.

Question 3.
Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis.
Solution:
The equation of a line parallel to the X-axis is y = b.
Since, the line is at a distance of 5 units below the X-axis.
∴ b = -5
∴ y = -5 is the equation of the required line.

Question 4.
The point Q( -3, -2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph.
Solution:
The equation of a line parallel to the Y-axis is x = a.
Here, a = -3
∴ x = -3 is the equation of the required line.

Question 5.
Y-axis and line x = – 4 are parallel lines. What is the distance between them?
Solution:
Equation of Y-axis is x = 0.
Equation of the line parallel to the Y-axis is x = – 4. … [Given]
∴ Distance between the Y-axis and the line x = – 4 is 0 – (- 4) … [0 > -4]
= 0 + 4 = 4 units
∴ The distance between the Y-axis and the line x = – 4 is 4 units.
[Note: The question is modified as X-axis cannot be parallel to the line x = – 4.]

Question 6.
Which of the equations given below have graphs parallel to the X-axis, and which ones have graphs parallel to the Y-axis? [1 Mark each]
i. x = 3
ii. y – 2 = 0
iii. x + 6 = 0
iv. y = -5
Solution:
i. The equation of a line parallel to the Y-axis is x = a.
∴ The line x = 3 is parallel to the Y-axis.

ii. y – 2 = 0
∴ y = 2
The equation of a line parallel to the X-axis is y = b.
∴ The line y – 2 = 0 is parallel to the X-axis.

iii. x + 6 = 0
∴ x = -6
The equation of a line parallel to the Y-axis is x = a.
∴ The line x + 6 = 0 is parallel to the Y-axis.

iv. The equation of a line parallel to the X-axis is y = b.
∴ The line y = – 5 is parallel to the X-axis.

Question 7.
On a graph paper, plot the points A(2, 3), B(6, -1) and C(0, 5). If these points are collinear, then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis.
Solution:

From the graph, the line drawn intersects the X-axis at D(5, 0) and the Y-axis at C(0, 5).

Question 8.
Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection.
x + 4 = 0,
y – 1 = 0,
2x + 3 = 0,
3y – 15 = 0
Solution:
i. x + 4 = 0
∴ x = – 4

ii. y – 1 = 0
∴ y = 1

iii. 2x + 3 = 0
∴2x = -3
∴ x = \(\frac { -3 }{ 2 }\)
∴ x = -1.5

iv. 3y- 15 = 0
3y = 15
y = \(\frac { 15 }{ 3 }\)
∴ y = 5

The co-ordinates of the point of intersection of x + 4 = 0 and y – 1 = 0 are A(-4, 1).
The co-ordinates of the point of intersection ofy – 1 = 0 and 2x + 3 = 0 are B(-1.5, 1).
The co-ordinates of the point of intersection of 3y – 15 = 0 and 2x + 3 = 0 are C(-1.5, 5).
The co-ordinates of the point of intersection of x + 4 = 0 and 3y – 15 = 0 are D(-4, 5).

Question 9.
Draw the graphs of the equations given below.
i. x + y = 2
ii. 3x – y = 0
iii. 2x + y = 1
Solution:
i. x + y = 2
∴ y = 2 – x
When x = 0,
y = 2 – x
= 2 – 0
= 2
When x = 1,
y = 2 – x
= 2 – 1
= 1
When x = 2,
y = 2 – x
= 0

ii. 3x – y = 0
∴ y = 3x
When x = 0,
y = 3x
= 3(0)
= 0

When x = 1,
y = 3x
= 3(1)
= 3

When x = -1,
y = 3x
= 3(-1)
= -3

iii. 2x + y = 1
∴ y = 1 – 2x
When x = 0,
y = 1 – 2x
= 1 – 2(0)
= 1 – o
When x = 1,
y = 1 – 2x
= 1- 2(1)
= 1 – 2
= -1
When x = -1,
y = 1 – 2x
= 1 – 2(-1)
= 1 + 2
= 3

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.2 Intext Questions and Activities

Question 1.
i. Can we draw a line parallel to the X-axis at a distance of 6 unIts from It and below the X-axis?
ii. Will all of the points (-3,-6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) be on that line?
iii. What would be the equation of this line?(Textbook pg. no. 94)
Solution:

i. Yes.
This line will pass through the point (0,-6).

ii. Yes.
Here, y co-ordinate of the points (-3, -6), (10,-6), ( \(\frac { 1 }{ 2 }\), -6) is the same, which is -6.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 6 units below the X-axis.
∴ b = -6
∴ Equation of the line is y = -6.

Question 2.
i. Can we draw a line parallel to the Y – axis at a distance of 2 units from ¡t and to its right?
ii. Will all of the points (2, 10), (2, 8), (2, -) be on that line?
iii. What would be the equation of this line? (Textbook pg. no. 95)
Solution:

i. Yes.
(2, 10)
This line will pass through the point (2, 0).
(2,8)
ii. Yes.
Here, x co-ordinate of the points (2, 10), (2, 8), (2,-\(\frac { 1 }{ 2 }\) ) is the same, which is 2.
∴ All the above points lie on the same line.

iii. Since, the line is at a distance of 2 units to the right of Y-axis.
a = 2
∴ Equation of the line is x = 2.

Question 3.
On a graph paper, plot the points (0, 1), (1, 3), (2, 5). Are they collinear? If so, draw the line that passes through them.
i. Through which quadrants does this line pass ?
ii. Write the co-ordinates of the point at which it intersects the Y-axis.
iii. Show any point in the third quadrant which lies on this line. Write the co-ordinates of the point. (Textbook pg. no. 96)
Solution:
i. The line passes through the quadrants I, II and III.
ii. The line intersects the Y-axis at (0, 1).
iii. (-1,-1)

Class 9 Maths Digest

• Practice Set 7.1 Class 9 Answers
• Practice Set 7.2 Class 9 Answers
• Problem Set 7 Class 9 Answers
• Practice Set 8.1 Class 9 Answers
• Practice Set 8.2 Class 9 Answers
• Problem Set 8 Class 9 Answers
• Practice Set 9.1 Class 9 Answers
• Practice Set 9.2 Class 9 Answers
• Practice Set 9.3 Class 9 Answers
• Problem Set 9 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 7.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 7 Co-ordinate Geometry.

9th Standard Maths 2 Practice Set 7.1 Chapter 7 Co-ordinate Geometry Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 7.1 Chapter 7 Co-ordinate Geometry Questions With Answers Maharashtra Board

Question 1.
State in which quadrant or on which axis do the following points lie.
i. A(-3, 2)
ii. B(-5, -2)
iii. K(3.5, 1.5)
iv. D(2, 10)
V. E(37, 35)
vi. F(15, -18)
vii. G(3, -7)
viii. H(0, -5)
ix. M(12, 0)
x. N(0, 9)
xi. P(0, 2.5)
xii. Q(-7, -3)
Solution:

Question 2.
In which quadrant are the following points?
i. whose both co-ordinates are positive.
ii. whose both co-ordinates are negative.
iii. whose x co-ordinate is positive and the y co-ordinate is negative.
iv. whose x co-ordinate is negative and y co-ordinate is positive.
Solution:
i. Quadrant I
ii. Quadrant III
iii. Quadrant IV
iv. Quadrant II

Question 3.
Draw the co-ordinate system on a plane and plot the following points.
L(-2, 4), M(5, 6), N(-3, -4), P(2, -3), Q(6, -5), S(7, 0), T(0, -5)
Solution:

Maharashtra Board Class 9 Maths Chapter 7 Co-ordinate Geometry Practice Set 7.1 Intext Questions and Activities

Question 1.
Plot the points R(-3,-4), S(3,-l) on the same co-ordinate system. (Textbook pg. no. 93)
Steps for plotting the points:
i. Draw X-axis and Y-axis on the plane. Show the origin.
ii. Draw a line parallel to Y-axis at a distance of 3 units in the -ve direction of X-axis.
iii. Draw another line parallel to X-axis at a distance of 4 units in the -ve direction of Y-axis.
iv. Intersection of these lines is the point R (-3, -4).
v. The point S can be plotted in the same manner.

Class 9 Maths Digest

• Practice Set 7.1 Class 9 Answers
• Practice Set 7.2 Class 9 Answers
• Problem Set 7 Class 9 Answers
• Practice Set 8.1 Class 9 Answers
• Practice Set 8.2 Class 9 Answers
• Problem Set 8 Class 9 Answers
• Practice Set 9.1 Class 9 Answers
• Practice Set 9.2 Class 9 Answers
• Practice Set 9.3 Class 9 Answers
• Problem Set 9 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 6 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

9th Standard Maths 2 Problem Set 6 Chapter 6 Circle Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Problem Set 6 Chapter 6 Circle Questions With Answers Maharashtra Board

Question 1.
Choose correct alternative answer and fill in the blanks.

i. Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence, the length of the chord is ____.
(A) 16 cm
(B) 8 cm
(C) 12 cm
(D) 32 cm
Answer:

∴ OA² = AC² + OC²
∴ 10² = AC² + 6²
∴ AC² = 64
∴ AC = 8 cm
∴ AB = 2(AC)= 16 cm
(A) 16 cm

ii. The point of concurrence of all angle bisectors of a triangle is called the ____.
(A) centroid
(B) circumcentre
(C) incentre
(D) orthocentre
Answer:
(C) incentre

iii. The circle which passes through all the vertices of a triangle is called ____.
(A) circumcircle
(B) incircle
(C) congruent circle
(D) concentric circle
Answer:
(A) circumcircle

iv. Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ____.
(A) 12 cm
(B) 13 cm
(C) 14 cm
(D) 15 cm
Answer:

OA² = AC² + OC²
∴ OA² = 12² + 5²
∴ OA² = 169
∴ OA = 13 cm
(B) 13 cm

v. The length of the longest chord of the circle with radius 2.9 cm is ____.
(A) 3.5 cm
(B) 7 cm
(C) 10 cm
(D) 5.8 cm
Answer:
Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm
(D) 5.8 cm

vi. Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____.
(A) on the centre
(B) inside the circle
(C) outside the circle
(D) on the circle
Answer:
l(OP) > radius
∴Point P lies in the exterior of the circle.
(C) outside the circle

vii. The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is _____.
(A) 2 cm
(B) 1 cm
(C) 8 cm
(D) 7 cm
Answer:

PQ = 8 cm, MN = 6 cm
∴ AQ = 4 cm, BN = 3 cm
∴ OQ² = OA² + AQ²
∴ 5² = OA² + 4²
∴ OA² = 25 – 16 = 9
∴ OA = 3 cm
Also, ON² = OB² + BN²
∴ 5² = OB² + 3²
∴ OB = 4 cm
Now, AB = OA + OB = 3 + 4 = 7 cm

Question 2.
Construct incircle and circumcircle of an equilateral ADSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.
Solution:


Steps of construction:
i. Construct ∆DPS of the given measurement.
ii. Draw the perpendicular bisectors of side DP and side PS of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. With C as centre and CM as radius, draw a circle which touches all the three sides of the triangle.
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.

Radius of incircle = 2.2 cm and Radius of circumcircle = 4.4 cm

Question 3.
Construct ∆NTS where NT = 5.7 cm. TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it.
Solution:


Steps of construction:
For incircle:
i. Construct ∆NTS of the given measurement.
ii. Draw the bisectors of ∠T and ∠S. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side TS. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
For circumcircle:
i. Draw the perpendicular bisectors of side NT and side TS of the triangle.
ii. Name the point of intersection of the perpendicular bisectors as point C.
iii. Join seg CN
iv. With C as centre and CN as radius, draw a circle which passes through the three vertices of the triangle.

Question 4.
In the adjoining figure, C is the centre of the circle, seg QT is a diameter, CT = 13, CP = 5. Find the length of chord RS.

Given: In a circle with centre C, QT is a diameter, CT = 13 units, CP = 5 units
To find: Length of chord RS
Construction: Join points R and C.
Solution:

i. CR = CT= 13 units …..(i) [Radii of the same circle]
In ∆CPR, ∠CPR = 90°
∴ CR² = CP² + RP² [Pythagoras theorem]
∴ 13² = 5² + RP² [From (i)]
∴ 169 = 25 + RP² [From (i)]
∴ RP² = 169 – 25 = 144
∴ RP = \(\sqrt { 144 }\) [Taking square root on both sides]
∴ RP = 12 cm ….(ii)

ii. Now, seg CP _L chord RS [Given]
∴ RP = \(\frac { 1 }{ 2 }\) RS [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \(\frac { 1 }{ 2 }\) RS [From (ii)]
∴ RS = 2 x 12 = 24
∴ The length of chord RS is 24 units.

Question 5.
In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD.

Given: P is the centre of the circle.
Chord AB and chord CD intersect on the diameter at the point E. ∠AEP ≅ ∠DEP
To prove: AB = CD
Construction: Draw seg PM ⊥ chord AB, A-M-B
seg PN ⊥ chord CD, C-N-D
Proof:

∠AEP ≅ ∠DEP [Given]
∴ Seg ES is the bisector of ∠AED.
PoInt P is on the bisector of ∠AED.
∴ PM = PN [Every point on the bisector of an angle is equidistant from the sides of the angle.]
∴ chord AB ≅ chord CD [Chords which are equidistant from the centre are congruent.]
∴ AB = CD [Length of congruent segments]

Question 6.
In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.

Given: O is the centre of the circle.
diameter CD ⊥ chord AB, A-E-B
To prove: ∆ABC is an isosceles triangle.
Proof:
diameter CD ⊥ chord AB [Given]
∴ seg OE ⊥ chord AB [C-O-E, O-E-D]
∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
In ∆CEA and ∆CEB,
∠CEA ≅ ∠CEB [Each is of 90°]
seg AE ≅ seg BE [From (i)]
seg CE ≅ seg CE [Common side]
∴ ∆CEA ≅ ∆CEB [SAS test]
∴ seg AC ≅ seg BC [c. s. c. t.]
∴ ∆ABC is an isosceles triangle.

Maharashtra Board Class 9 Maths Chapter 6 Circle Problem Set 6 Intext Questions and Activities

Question 1.
Every student in the group should do this activity. Draw a circle in your notebook. Draw any chord of that circle. Draw perpendicular to the chord through the centre of the circle. Measure the lengths of the two parts of the chord. Group leader should prepare a table as shown below and ask other students to write their observations in it. Write the property which you have observed. (Textbook pg. no. 77)


Answer:
On completing the above table, you will observe that the perpendicular drawn from the centre of a circle on its chord bisects the chord.

Question 2.
Every student from the group should do this activity. Draw a circle in your notebook. Draw a chord of the circle. Join the midpoint of the chord and centre of the circle. Measure the angles made by the segment with the chord.
Discuss about the measures of the angles with your friends. Which property do the observations suggest ? (Textbook pg. no. 77)

Answer:
The meausure of the angles made by the drawn segment with the chord is 90°. Thus, we can conclude that, the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.

Question 3.
Draw circles of convenient radii. Draw two equal chords in each circle. Draw perpendicular to each chord from the centre. Measure the distance of each chord from the centre. What do you observe? (Textbook pg. no. 79)
Answer:
Congruent chords of a circle are equidistant from the centre.

Question 4.
Measure the lengths of the perpendiculars on chords in the following figures.

Did you find OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii)?
Write the property which you have noticed from this activity. (Textbook pg. no. 80)
Answer:
In each figure, the chords are equidistant from the centre. Also, we can see that the measures of the chords in each circle are equal.
Thus, we can conclude that chords of a circle equidistant from the centre of a circle are congruent.

Question 5.
Draw different triangles of different measures and draw in circles and circumcircles of them. Complete the table of observations and discuss. (Textbook pg. no. 85)
Answer:

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

9th Standard Maths 2 Practice Set 6.3 Chapter 6 Circle Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 6.3 Chapter 6 Circle Questions With Answers Maharashtra Board

Question 1.
Construct ∆ABC such that ∠B =100°, BC = 6.4 cm, ∠C = 50° and construct its incircle.
Solution:


Steps of construction:
i. Construct ∆ABC of the given measurement.
ii. Draw the bisectors of ∠B and ∠C. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side BC. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.

Question 2.
Construct ∆PQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and construct its circumcircle.
Solution:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° … [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° – 50°
∴ m∠Q = 60°


Steps of construction:
i. Construct A PQR of the given measurement.
ii. Draw the perpendicular bisectors of side PQ and side QR of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CP
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.

Question 3.
Construct ∆XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.
Solution:


Steps of construction:
i. Construct ∆XYZ of the given measurement
ii. Draw the bisectors of ∠X and ∠Z. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side XZ. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.

Question 4.
In ∆LMN, LM = 7.2 cm, ∠M = 105°, MN = 6.4 cm, then draw ∆LMN and construct its circumcircle.
Solution:


Steps of construction:
i. Construct ∆LMN of the given measurement.
ii. Draw the perpendicular bisectors of side MN and side ML of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CM
v. With C as centre and CM as radius, draw a circle which passes through the three vertices of the triangle.

Question 5.
Construct ∆DEF such that DE = EF = 6 cm. ∠F = 45° and construct its circumcircle.
Solution:


Steps of construction:
i. Construct ∆DEF of the given measurement.
ii. Draw the perpendicular bisectors of side DE and side EF of the triangle.
iii. Name the point of intersection of perpendicular bisectors as point C.
iv. Join seg CE
v. With C as centre and CE as radius, draw a circle which passes through the three vertices of the triangle.

Maharashtra Board Class 9 Maths Chapter 6 Circle Practice Set 6.3 Intext Questions and Activities

Question 1.
Draw any equilateral triangle. Draw incircle and circumcircle of it. What did you observe while doing this activity? (Textbook pg. no. 85)
i. While drawing incircle and circumcircle, do the angle bisectors and perpendicular bisectors coincide with each other?
ii. Do the incentre and circumcenter coincide with each other? If so, what can be the reason of it?
iii. Measure the radii of incircle and circumcircle and write their ratio.
Solution:


Steps of construction:
i. Construct equilateral ∆XYZ of any measurement.
ii. Draw the perpendicular bisectors of side XY and side YZ of the triangle.
iii. Draw the bisectors of ∠X and ∠Z.
iv. Name the point of intersection of the perpendicular bisectors and angle bisectors as point I.
v. With I as centre and IM as radïus, draw a circle which touches all the three sides of the triangle.
vi. With I as centre and IZ as radius, draw a circle which passes through the three vertices of the triangle.
[Note: Here, point of intersection of perpendicular bisector and angle bisector is same.]

i. Yes.
ii. Yes.
The angle bisectors of the angles and the perpendicular bisectors of the sides of an equilateral triangle are coincedent. Hence, its incentre and circumcentre coincide.
iii. Radius of circumcircle = 3.6 cm,
Radius of incircle = 1.8 cm
\(\text { Ratio }=\frac{\text { Radius of circumcircle }}{\text { Radius of incircle }}=\frac{3.6}{1.8}=\frac{2}{1}=2 : 1\)

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

9th Standard Maths 2 Problem Set 5 Chapter 5 Quadrilaterals Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Problem Set 5 Chapter 5 Quadrilaterals Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer and fill in the blanks.

i. If all pairs of adjacent sides of a quadrilateral are congruent, then it is called ____.
(A) rectangle
(B) parallelogram
(C) trapezium
(D) rhombus
Answer:
(D) rhombus

ii. If the diagonal of a square is 22√2 cm, then the perimeter of square is ____.
(A) 24 cm
(B) 24√2 cm
(C) 48 cm
(D) 48√2 cm
Answer:
In ∆ABC,
AC² = AB² + BC²

∴ (12² √2 )² = AB² + AB²
∴ \( A B^{2}=\frac{12^{2} \times 2}{2}=12^{2}\)
∴ AB = 12 cm
∴ Perimeter of □ABCD = 4 x 12 = 48 cm
(C) 48 cm

iii. If opposite angles of a rhombus are (2x)° and (3x – 40)°, then the value of x is ____.
(A) 100°
(B) 80°
(C) 160°
(D) 40°
Answer:
2x = 3x – 40 … [Pythagoras theorem]
∴ x = 40°
(D) 40°

Question 2.
Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
Solution:

Let □ABCD be the rectangle.
AB = 7 cm, BC = 24 cm
In ∆ABC, ∠B = 90° [Angle of a rectangle]
AC² = AB² + BC² [Pythagoras theorem]
= 7² + 24²
= 49 + 576
= 625
AC = √625 [Taking square root of both sides]
= 25 cm
∴ The length of the diagonal of the rectangle is 25 cm.

Question 3.
If diagonal of a square is 13 cm, then find its side.
Solution:

Let □PQRS be the square of side x cm.
∴ PQ = QR = x cm …..(i) [Sides of a square]
∴ In ∆PQR, ∠Q = 90° [Angle of a square]
∴ PR² = PQ² + QR² [Pythagoras theorem]
∴ 13 = x + x [From (i)]
∴ 169 = 2x²

The length of the side of the square is 6.5√2 cm.

Question 4.
Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.
Solution:

Let □STUV be the parallelogram.
Ratio of two adjacent sides of a parallelogram is 3 : 4.
Let the common multiple be x.
ST = 3x cm and TU = 4x cm
∴ ST = UV = 3x cm
TU = SV = 4x cm …..(i) [Opposite sides of a parallelogram]
Perimeter of □STUV = 112 [Given]
∴ ST + TU + UV + SV = 112
∴ 3x + 4x + 3x + 4x = 112 [From (i)]
∴ 14x = 112
∴ x = \(\frac { 112 }{ 14 }\)
∴ x = 8
∴ ST = UV = 3x = 3 x 8 = 24 cm
∴ TU = SV = 4x = 4 x 8 = 32 cm [From (i)]
∴ The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm.

Question 5.
Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
Solution:

□PQRS is a rhombus. [Given]
PR = 20 cm and QS = 48 cm [Given]
∴ PT = \(\frac { 1 }{ 2 }\) PR [Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) x 20 = 10 cm
Also, QT = \(\frac { 1 }{ 2 }\) QS [Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) x 48 = 24 cm

ii. In ∆PQT, ∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]
∴ PQ² = PT² + QT² [Pythagoras- theorem]
= 10² + 24²
= 100 + 576
∴ PQ² = 676
∴ PQ = \(\sqrt {676 }\) [Taking square root of both sides]
= 26 cm
∴ The length of side PQ is 26 cm.

Question 6.
Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50°, then find the measure of ∠MPS.
Solution:

□PQRS is a rectangle.
∴ PM = \(\frac { 1 }{ 2 }\) PR …(i)
MS = \(\frac { 1 }{ 2 }\) QS …(ii) [Diagonals of a rectangle bisect each other]
Also, PR = QS …..(iii) [Diagonals of a rectangle are congruent]
∴ PM = MS ….(iv) [From (i), (ii) and (iii)]
In ∆PMS,
PM = MS [From (iv)]
∴ ∠MSP = ∠MPS = x° …..(v) [Isosceles triangle theorem]
∠PMS = ∠QMR = 50° ……(vi) [Vertically opposite angles]
In ∆MPS,
∠PMS + ∠MPS + ∠MSP = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 50° +x + x = 180° [From (v) and (vi)]
∴ 50° + 2x= 180
∴ 2x= 180-50
∴ 2x= 130
∴ x = \(\frac { 130 }{ 2 }\) = 65°
∴ ∠MPS = 65° [From (v)]

Question 7.
In the adjoining figure, if seg AB || seg PQ , seg AB ≅ seg PQ, seg AC || seg PR, seg AC ≅ seg PR, then prove that seg BC || seg QR and seg BC ≅ seg QR.

Solution:
Given: seg AB || seg PQ , seg AB ≅ seg PQ,
seg AC || seg PR, seg AC ≅ seg PR
To prove: seg BC || seg QR, seg BC ≅ seg QR
Proof:
Consider □ABQP,
seg AB || seg PQ [Given]
seg AB ≅ seg PQ [Given]
∴ □ABQP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ segAP || segBQ …..(i)
∴ seg AP ≅ seg BQ …..(ii) [Opposite sides of a parallelogram]
Consider □ACRP,
seg AC || seg PR [Given]
seg AC ≅ seg PR [Given]
∴ □ACRP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg AP || seg CR …(iii)
∴ seg AP ≅ seg CR …….(iv) [Opposite sides of a parallelogram]
Consider □BCRQ,
seg BQ || seg CR
seg BQ ≅ seg CR
∴ □BCRQ is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg BC || seg QR
∴ seg BC ≅ seg QR [Opposite sides of a parallelogram]

Question 8.
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that PQ || AB and PQ = \(\frac { 1 }{ 2 }\) ( AB + DC).

Given : □ ABCD is a trapezium.
To prove:
Construction: Join points A and Q. Extend seg AQ and let it meet produced DC at R.
Proof:

seg AB || seg DC [Given]
and seg BC is their transversal.
∴ ∠ABC ≅ ∠RCB [Alternate angles]
∴ ∠ABQ ≅ ∠RCQ ….(i) [B-Q-C]
In ∆ABQ and ∆RCQ,
∠ABQ ≅∠RCQ [From (i)]

seg BQ ≅ seg CQ [Q is the midpoint of seg BC]
∠BQA ≅ ∠CQR [Vertically opposite angles]
∴ ∆ABQ ≅ ∆RCQ [ASA test]
seg AB ≅ seg CR …(ii) [c. s. c. t.]
seg AQ ≅ seg RQ [c. s. c. t.]
∴ Q is the midpoint of seg AR. ….(iii)

In ∆ADR,
Points P and Q are the midpoints of seg AD and seg AR respectively. [Given and from (iii)]

∴ seg PQ || seg DR [Midpoint theorem]
i.e. seg PQ || seg DC ……..(iv) [D-C-R]
But, seg AB || seg DC …….(v) [Given]
∴ seg PQ || seg AB [From (iv) and (v)]
In ∆ADR,

Question 9.
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively, then prove that MN || AB.

Solution:
Given: □ABCD is a trapezium. AB || DC.
Points M and N are midpoints of diagonals AC and DB respectively.
To prove: MN || AB
Construction: Join D and M. Extend seg DM to meet seg AB at point E such that A-E-B.
Proof:
seg AB || seg DC and seg AC is their transversal. [Given]
∴ ∠CAB ≅ ∠ACD [Alternate angles]
∴ ∠MAE ≅ ∠MCD ….(i) [C-M-A, A-E-B]
In ∆AME and ∆CMD,

∠AME ≅ ∠CMD [Vertically opposite angles]
seg AM ≅ seg CM [M is the midpoint of seg AC]
∠MAE ≅∠MCD [From (i)]
∴ ∆AME ≅ ∆CMD [ASA test]
∴ seg ME ≅ seg MD [c.s.c.t]
∴ Point M is the midpoint of seg DE. …(ii)
In ∆DEB,

Points M and N are the midpoints of seg DE and seg DB respectively. [Given and from (ii)]
∴ seg MN || seg EB [Midpoint theorem]
∴ seg MN || seg AB [A-E-B]

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Problem Set 5 Intext Questions and Activities

Question 1.
Draw five parallelograms by taking various measures of lengths and angles. (Textbook page no. 59)

Question 2.
Draw a parallelogram PQRS. Draw diagonals PR and QS. Denote the intersection of diagonals by letter O. Compare the two parts of each diagonal with a divider. What do you find? (Textbook page no. 60)

Answer:
seg OP = seg OR, and seg OQ = seg OS
Thus we can conclude that, point O divides the diagonals PR and QS in two equal parts.

Question 3.
To verify the different properties of quadrilaterals.

Material: A piece of plywood measuring about 15 cm x 10 cm, 15 thin screws, twine, scissor.
Note: On the plywood sheet, fix five screws in a horizontal row keeping a distance of 2 cm between any two adjacent screws. Similarly make two more rows of screws exactly below the first one. Take care that the vertical distance between any two adjacent screws is also 2 cm.
With the help of the screws, make different types of quadrilaterals of twine. Verify the properties of sides and angles of the quadrilaterals. (Textbook page no. 75)

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

9th Standard Maths 2 Practice Set 6.2 Chapter 6 Circle Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 6.2 Chapter 6 Circle Questions With Answers Maharashtra Board

Question 1.
Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?
Given: In a circle with centre O,
OR and OP are radii and RS and PQ are its congruent chords.
PQ = RS= 16 cm,
OR = OP = 10 cm
seg OU ⊥ chord PQ, P-U-Q
seg OT ⊥ chord RS, R-T-S
To find: Distance of chords from centre of the circle.
Solution:

i. PU = \(\frac { 1 }{ 2 }\)(PQ) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ PU= \(\frac { 1 }{ 2 }\) x 16 = 8 cm …(i)

ii. In ∆OUP, ∠OUP = 90°
∴ OP² = OU² + PU² [Pythagoras theorem]
∴ 10² = OU² + 8² [From (i)]
∴ 100 = OU² + 64
∴ OU² = 100 – 64 = 36
∴ OU = √36 [Taking square root on both sides]
∴ OU = 6 cm

iii. Now, OT = OU [Congruent chords of a circle are equidistant from the centre.]
∴ OT = OU = 6cm
∴ The distance of the chords from the centre of the circle is 6 cm.

Question 2.
In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of chords.
Given: In a circle with cente O,
OA and OC are the radii and AB and CD are its congruent chords,
OA = OC = 13cm
0E = OF = 5 cm
seg 0E ⊥ chord CD, C-E-D
seg OF ⊥ chord AB. A-F-B
To find: length of the chords
Solution:

i. In ∆AFO, ∠AFO = 90°
∴ AO² = AF² + FO² [Pythagoras theorem]
∴ 13² = AF² + 5²
∴ 169 = AF² + 25
∴ AF² = 169-25
∴ AF² = 144
∴ AF = \(\sqrt { 144 }\) [Taking square root on both sides]
∴ AF = 12 cm …..(i)

ii. Now AF = \(\frac { 1 }{ 2 }\)AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \(\frac { 1 }{ 2 }\) (AB) [From (i)]
∴ AB = 12 x 2 = 24 cm
∴ CD = AB = 24 cm [chord AB ≅ chord CD]
∴ The lengths of the two chords are 24 cm each.

Question 3.
Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of ∠NPM.
Given: Point C is the centre of the circle.
chord PM ≅ chord PN
To prove: Ray PC is the bisector of ∠NPM.
Construction: Draw seg CR ⊥ chord PN, P-R-N
seg CQ ⊥ chord PM, P-Q-M
Proof:

chord PM chord PN [Given]
seg CR ⊥ chord PN
seg CQ ⊥ chord PM [Construction]
∴ segCR ≅ segCQ ….(i) [Congruent chords are equidistant from the centre]
In ∆PRC and ∆PQC,
∠PRC ≅ ∠PQC [Each is of 90°]
segCR ≅ segCQ [From (i)]
seg PC ≅ seg PC [Common side]
∴ ∆PRC ≅ ∆PQC [Hypotenuse side test]
∴ ∠RPC ≅ ∠QPC [c. a. c. t.]
∴ ∠NPC ≅ ∠MPC [N- R-P, M-Q-P]
∴ Ray PC is the bisector of ∠NPM.

Maharashtra Board Class 9 Maths Chapter 6 Circle Practice Set 6.2 Intext Questions and Activities

Question 1.
Prove the following two theorems for two congruent circles. (Textbook pg. no. 81)
i. Congruent chords in congruent circles are equidistant from their respective centres.
ii. Chords of congruent circles which are equidistant from their respective centres are congruent.
Write ‘Given’. ‘To prove’ and the proofs of these theorems.
Solution:
(i) Congruent chords in congruent circles are equidistant from their respective centres.
Given: Point P and point Q are the centres of congruent circles.
chord AB ≅ chord CD
seg PM ⊥ chord AB, A-M-B
seg QN ⊥ chord CD, C-N-D
To prove: PM = QN

Construction: Draw seg PA and seg QC.
Proof:
seg PM ⊥ chord AB, seg QN ⊥ chord CD [Given]
∴ AM = \(\frac { 1 }{ 2 }\)(AB) ………(i) [Perpendicular drawn from the centre of the circle to the
∴ CN = \(\frac { 1 }{ 2 }\)(CD) ……..(ii) chord bisects the chord.]
But, AB = CD ………(iii) [Given]
∴ AM = CN [From (i), (ii) and (iii)]
i.e., segAM ≅ segCN ….(iv) [Segments of equal lengths]
In ∆PMA and ∆QNC,
∠PMA ≅ ∠QNC [Each is of 90°]
hypotenuse PA ≅ hypotenuse QC [Radii of congruent circles]
seg AM ≅ seg CN [From (iv)]
∴ ∆PMA ≅ ∆QNC [Hypotenuse side test]
∴ segPM ≅ segQN [c. s. c. t.]
∴ PM ≅ QN [Length of congruent segments]

(ii) Chords of congruent circles which are equidistant from their respective centres are congruent.

Given: Point P and point Q are the centres of congruent circles.
seg PM ⊥ chord AB, A-M-B
seg QN ⊥ chord CD, C-N-D
PM = QN
To prove: chord AB ≅ chord CD
Construction: Draw seg PA and seg QC.
Proof:
In ∆PMA and ∆QNC,
∴ ∠PMA ≅ ∠QNC [Each is of 90°]
seg PM ≅ seg QN [Given]
hypotenuse PA ≅ hypotenuse QC [Radii of the congruent circles]
∴ ∆PMA ≅ ∆QNC [Hypotenuse side test]
∴ seg AM ≅ seg CN [c. s. c. t.]
∴ AM = CN ….(i) [Length of congruent segments]
Now, seg PM ⊥ chord AB, and seg QN ⊥ chord CD
∴ AM = \(\frac { 1 }{ 2 }\)(AB) …(ii)
∴ CN = \(\frac { 1 }{ 2 }\) (CD) ..(iii) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AB = CD [From (i), (ii) and (ii)]
∴ chord AB ≅ chord CD [Segments of equal lengths]

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

9th Standard Maths 2 Practice Set 5.5 Chapter 5 Quadrilaterals Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 5.5 Chapter 5 Quadrilaterals Questions With Answers Maharashtra Board

Question 1.
In the adjoining figure, points X, Y, Z are the midpoints of of ∆ABC respectively, cm. Find the lengths of side AB, side BC and side AC AB = 5 cm, AC = 9 cm and BC = 11c.m. Find the lengths of XY, YZ, XZ.

Solution:
i. AC = 9 cm [Given]
Points X and Y are the midpoints of sides AB and BC respectively. [Given]
∴ XY = \(\frac { 1 }{ 2 }\) AC [Midpoint tfyeprem]
= \(\frac { 1 }{ 2 }\) x 9 = 4.5 cm

ii. AB = 5 cm [Given]
Points Y and Z are the midpoints of sides BC and AC respectively. [Given]
∴ YZ = \(\frac { 1 }{ 2 }\) AB [Midpoint theorem]
= \(\frac { 1 }{ 2 }\) x 5 = 2.5 cm

iii. BC = 11 cm [Given]
Points X and Z are the midpoints of sides AB and AC respectively. [Given]
∴ XZ = \(\frac { 1 }{ 2 }\) BC [Midpoint theorem]
= \(\frac { 1 }{ 2 }\) x 11 = 5.5 cm
l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

Question 2.
In the adjoining figure, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR, then prove that,
i. SL = LR
ii. LN = \(\frac { 1 }{ 2 }\) SQ.

Given: □PQRS and □MNRL are rectangles. M is the midpoint of side PR.
Solution:
Toprove:
i. SL = LR
ii. LN = \(\frac { 1 }{ 2 }\) (SQ)
Proof:
i. □PQRS and □MNRL are rectangles. [Given]
∴ ∠S = ∠L = 90° [Angles of rectangles]
∠S and ∠L form a pair of corresponding angles on sides SP and LM when SR is their transversal.

∴eg ML || seg PS …(i) [Corresponding angles test]
In ∆PRS,
Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]
∴ Point L is the midpoint of seg SR. ……(ii) [Converse of midpoint theorem]
∴ SL = LR

ii. Similarly for ∆PRQ, we can prove that,

Point N is the midpoint of seg QR. ….(iii)
In ∆RSQ,
Points L and N are the midpoints of seg SR and seg QR respectively. [From (ii) and (iii)]
∴ LN = \(\frac { 1 }{ 2 }\)SQ [Midpoint theorem]

Question 3.
In the adjoining figure, ∆ABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ∆FED is an equilateral triangle.

Given: ∆ABC is an equilateral triangle.
Points F, D and E are midpoints of side AB, side BC, side AC respectively.
To prove: ∆FED is an equilateral triangle.
Solution:
Proof:
∆ABC is an equilateral triangle. [Given]
∴ AB = BC = AC ….(i) [Sides of an equilateral triangle]
Points F, D and E are midpoints of side AB and BC respectively.

∴ FD = \(\frac { 1 }{ 2 }\)AC …..(ii) [Midpoint theorem]
Points D and E are the midpoints of sides BC and AC respectively.

∴ DE = \(\frac { 1 }{ 2 }\)AB …..(iii) [Midpoint theorem]
Points F and E are the midpoints of sides AB and AC respectively.
∴ FE = \(\frac { 1 }{ 2 }\)BC
∴ FD = DE = FE [From (i), (ii), (iii) and (iv) ]
∴ ∆FED is an equilateral triangle.

Question 4.
In the adjoining figure, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\). [Hint: Draw DN || QM]

Solution:
Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD.
To Prove: \(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\)
Construction: Draw seg DN ||seg QM such that P-M-N and M-N-R.
Proof:
In ∆PDN,
Point T is the midpoint of seg PD and seg TM || seg DN [Given]
∴ Point M is the midpoint of seg PN. [Construction and Q-T-M]
∴ PM = MN [Converse of midpoint theorem]
In ∆QMR,
Point D is the midpoint of seg QR and seg DN || seg QM [Construction]

∴ Point N is the midpoint of seg MR. [Converse of midpoint theorem]
∴ RN = MN …..(ii)
∴ PM = MN = RN …..(iii) [From (i) and (ii)]
Now, PR = PM + MN + RN [ P-M-R-Q-T-M]
∴ PR = PM + PM + PM [From (iii) ]
∴ PR = 3PM
\(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\)

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

9th Standard Maths 2 Practice Set 6.1 Chapter 6 Circle Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 6.1 Chapter 6 Circle Questions With Answers Maharashtra Board

Question 1.
Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of a circle.
Given: In a circle with centre O,
OA is radius and AB is its chord,
seg OP ⊥ chord AB, A-P-B
AB = 12 cm, OP =8 cm
To Find: Diameter of the circle
Solution:

i. AP = \(\frac { 1 }{ 2 }\) AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
∴ AP = \(\frac { 1 }{ 2 }\) x 12 = 6 cm ….(i)

ii. In ∆OPA, ∠OPA = 90°
∴ OA² = OP² + AP² [Pythagoras theorem]
= 8² + 6² [From (i)]
= 64 + 36
∴ OA² = 100
∴ OA = \(\sqrt { 100 }\) [Taking square root on both sides]
= 10 cm

iii. Radius (r) = 10 cm
∴ Diameter = 2r = 2 x 10 = 20 cm
∴ The diameter of the circle is 20 cm.

Question 2.
Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.
Solution:

Given: In a circle with centre O,
PO is radius and PQ is its chord,
seg OR ⊥ chord PQ, P-R-Q
PQ = 24 cm, diameter (d) = 26 cm
To Find: Distance of the chord from the centre (OR)
Solution:
Radius (OP) = \(\frac { d }{ 2 }\) = \(\frac { 26 }{ 2 }\) = 13 cm ……(i)
∴ PR = \(\frac { 1 }{ 2 }\) PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= \(\frac { 1 }{ 2 }\) x 24 = 12 cm …..(ii)

ii. In ∆ORP, ∠ORP = 90°
∴ OP²= OR² + PR² [Pythagoras theorem]
∴ 13² = OR² + 12² [From (i) and (ii)]
∴ 169 = OR² + 144
∴ OR² = 169 – 144
∴ OR² = 25
∴ OR = √25 = 5 cm [Taking square root on both sides]
∴ The distance of the chord from the centre of the circle is 5 cm.

Question 3.
Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.
Given: in a circle with centre A,
PA is radius and PQ is chord,
seg AM ⊥ chord PQ, P-M-Q
AP = 34 cm, AM = 30 cm
To Find: Length of the chord (PQ)
Solution:

I. In ∆AMP, ∠AMP = 90°
∴ AP² = AM² + PM² [Pythagoras theorem]
34² = 30² + PM²
∴ PM² = 34² – 30²
∴ PM² (34 – 30)(34 + 30) [a² – b² = (a – b)(a + b)]
= 4 x 64
∴ PM = \(\sqrt { 4\times64 }\) ………(i) [Taking square root on both sides]
= 2 x 8 = 16cm

ii. Now, PM = \(\frac { 1 }{ 2 }\)(PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
16 = \(\frac { 1 }{ 2 }\)(PQ) [From (i)]
∴ PQ = 16 x 2
= 32cm
∴ The length of the chord of the circle is 32cm.

Question 4.
Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.
Given: In a circle with centre O,
OP is radius and PQ is its chord,
seg OM ⊥ chord PQ, P-M-Q
OP = 41 units, PQ = 80 units,
To Find: Distance of the chord from the centre of the circle(OM)
Solution:

i. \(\frac { 1 }{ 2 }\)PM = (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= \(\frac { 1 }{ 2 }\)(80) = 40 Units ….(i)

ii. In ∆OMP, ∠OMP = 90°
∴ OP² = OM² + PM² [Pythagoras theorem]
∴ 41² = OM² + 40² [From (i)]
∴ OM² = 41² – 40²
= (41 -40) (41 +40) [a² – b² = (a – b) (a + b)]
= (1)(81)
∴ OM² = 81 OM = √81 = 9 units [Taking square root on both sides] [From (i)]
∴ The distance of the chord from the centre of the circle is 9 units.

Question 5.
In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.

Given: Two concentric circles having centre O.
To prove: AP = BQ
Construction: Draw seg OM ⊥ chord AB, A-M-B
Solution:
Proof:
For smaller circle,
seg OM ⊥ chord PQ [Construction, A-P-M, M-Q-B]
∴ PM = MQ …..(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
For bigger circle,
seg OM ⊥ chord AB [Construction]
∴ AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AP + PM = MQ + QB [A-P-M, M-Q-B]
∴ AP + MQ = MQ + QB [From (i)]
∴ AP = BQ

Question 6.
Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.
Solution:

Given: O is the centre of the circle.
seg PQ is the diameter.
Diameter PQ bisects the chords AB and CD in points M and N respectively.
To prove: chord AB || chord CD.
Proof:
Diameter PQ bisects the chord AB in point M [Given]
∴ seg AM ≅ seg BM
∴ seg OM ⊥ chord AB [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
∴ ∠OMA = 90° …..(i)
Also, diameter PQ bisects the chord CD in point N [Given]
∴ seg CN ≅ seg DN
seg ON ⊥ chord CD [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
∴ ∠ONC = 90° …..(ii)
Now, ∠OMA + ∠ONC = 90° + 90° [From (i) and (ii)]
= 180°
But, ∠OMA and ∠ONC form a pair of interior angles on lines AB and CD when seg MN is their transversal.
∴ chord AB || chord CD [Interior angles test]

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.4 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

9th Standard Maths 2 Practice Set 5.4 Chapter 5 Quadrilaterals Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 5.4 Chapter 5 Quadrilaterals Questions With Answers Maharashtra Board

Question 1.
In □IJKL, side IJ || side KL, ∠I = 108° and ∠K = 53°, then find the measures of ∠J and ∠L.
Solution:

i. ∠I = 108° [Given]
side IJ || side KL and side IL is their transveral. [Given]
∴ ∠I + ∠L = 180° [Interior angles]
∴ 108° + ∠L = 180°
∴ ∠L = 180° – 108° = 72°

ii. ∠K = 53° [Given]
side IJ || side KL and side JK is their transveral. [Given]
∴ ∠J + ∠K = 180° [Interior angles]
∴ ∠J + 53° = 180°
∴ ∠J= 180°- 53° = 127°
∴ ∠L = 72°, ∠J = 127°

Question 2.
In □ABCD, side BC || side AD, side AB ≅ side DC. If ∠A = 72°, then find the measures of ∠B and ∠D.
Construction: Draw seg BP ⊥ side AD, A – P – D, seg CQ ⊥ side AD, A – Q – D.
Solution:

i. ∠A = 72° [Given]
In □ABCD, side BC || side AD and side AB is their transversal. [Given]
∴ ∠A + ∠B = 180° [Interior angles]
∴ 72° +∠B = 180°
∴ ∠B = 180° – 72° = 108°

ii. In ∆BPA and ∆CQD,
∠BPA ≅ ∠CQD [Each angle is of measure 90°]
Hypotenuse AB ≅ Hypotenuse DC [Given]
seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]
∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]
∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]
∴ ∠A = ∠D
∴ ∠D = 72°
∴ ∠B = 108°, ∠D = 72°

Question 3.
In □ABCD, side BC < side AD, side BC || side AD and if side BA ≅ side CD, then prove that ∠ABC = ∠DCB.

Given: side BC < side AD, side BC || side AD, side BA = side CD
To prove: ∠ABC ≅ ∠DCB
Construction: Draw seg BP ⊥ side AD, A – P – D
seg CQ ⊥ side AD, A – Q – D
Solution:
Proof:
In ∆BPA and ∆CQD,
∠BPA ≅ ∠CQB [Each angle is of measure 90°]
Hypotenuse BA ≅ Hypotenuse CD [Given]
seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]
∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]
∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]
∴ ∠A = ∠D ….(i)
Now, side BC || side AD and side AB is their transversal. [Given]
∴ ∠A + ∠B = 180°…..(ii) [Interior angles]
Also, side BC || side AD and side CD is their transversal. [Given]
∴ ∠C + ∠D = 180° …..(iii) [Interior angles]
∴ ∠A + ∠B = ∠C + ∠D [From (ii) and (iii)]
∴ ∠A + ∠B = ∠C + ∠A [From (i)]
∴ ∠B = ∠C
∴ ∠ABC ≅ ∠DCB

Class 9 Maths Digest

• Practice Set 5.1 Class 9 Answers
• Practice Set 5.2 Class 9 Answers
• Practice Set 5.3 Class 9 Answers
• Practice Set 5.4 Class 9 Answers
• Practice Set 5.5 Class 9 Answers
• Problem Set 5 Class 9 Answers
• Practice Set 6.1 Class 9 Answers
• Practice Set 6.2 Class 9 Answers
• Practice Set 6.3 Class 9 Answers
• Problem Set 6 Class 9 Answers