Class 8 Maths Chapter 9 Circles Questions and Answers Kerala Syllabus

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SCERT Class 8 Maths Chapter 9 Solutions Circles

Class 8 Kerala Syllabus Maths Solutions Chapter 9 Circles Questions and Answers

Circles Class 8 Questions and Answers Kerala Syllabus

Diameter (Page 151)

Question 1.
Draw a circle of radius 2.75 centimetres.
Answer:
Since the radius is 2.75 cm,
the diameter will be double that length = 2.75 × 2 = 5.5 cm
Draw a line segment AB of length 5.5 cm.
Construct the perpendicular bisector of this line to find its exact midpoint O.
Place the compass needle on 0 and the pencil on A (or B) and draw the circle.
The circle will pass through both A and B.

Question 2.
Draw the pictures below:

Answer:
(i) Draw a square with a side of 4.5 cm.
Draw the perpendicular bisector of each side of the square and mark the midpoint of the side.
Draw semicircles on all 4 sides with this midpoint as the centre and half the side length as the radius.
We will get the given figure.

(ii) Draw a line of length 6 cm.
Draw the perpendicular bisector of this line and mark 3 cm above and below the point where the line and perpendicular bisector intersect.
Draw semicircles with diameters equal to half the length of these lines (3 cm), both vertically and horizontally, as shown in the figure.

Chords (Page 154-155)

Question 1.
In the picture, the horizontal line joins the centres of the circles, and the vertical line joins the points of intersection of the circles.

Write the reasons for the claims below:
(i) The perpendicular bisector of the vertical line passes through the centres of both the circles.
(ii) The horizontal line is the perpendicular bisector of the vertical line.
What general principle do we get from this?
Answer:

(i) PQ is the perpendicular bisector of the chord AB.
The perpendicular bisector of a chord passes through the centre.
The perpendicular bisector of the chord AB in the circle with centre P will pass through P.
Similarly, the perpendicular bisector of the chord AB in the circle with centre Q will pass through Q.
The perpendicular bisector of the line AB will pass through the centres P and Q of the two circles.
(ii) The line joining the centres of the two circles is the perpendicular bisector of the line joining the points where they intersect.

Question 2.
The picture shows two circles with the same centre and a line.

Prove that the portions of the line between the circles on both sides are equal.
Answer:
O is the centre of the outer circle, AB is a chord intersecting the inner circle at P and Q.
To prove AP = BQ

Draw OC perpendicular to AB.
According to the result that the perpendicular from the centre to a chord bisects the chord,
We have AC = BC …. (1)
Also PC = QC …. (2)
(1) – (2) ⇒ AC – PC = BC – QC
That is, AP = BQ
The portions of the line between the circles on both sides are equal.

Question 3.
Draw a square and the circle through all four of its vertices. Join the vertices of the square and the points where the diameters parallel to the sides of the square meet the circle, to draw another polygon.

Prove that it is a regular octagon.
Answer:
The diameters are parallel to the sides of the square.

∠OPH = ∠OPB = 90°
Consider ∆HPA and ∆BPA,
HP = BP (The perpendicular from the centre of the circle to a chord bisects the chord)
PA = PA (Common side)
∠HPA = ∠BPA = 90° (Since two sides and the included angle of a triangle)
HPA are equal to two sides and the included angle of triangle BPA; these triangles are equal.
HA = AB
From triangles BPA and BQC, we can prove that AB = BC.
Similarly, we can show that the other sides of the octagon are equal.
That is, ABCDEFGH is a regular octagon.

Equal Chords (Page 158)

Question 1.
Prove that chords of the same length in a circle are at the same distance from the centre of the circle.
Answer:

Consider a circle with centre O.
Let AB and CD be two chords of equal length (AB = CD).
Draw perpendiculars OM and ON from the centre O to the chords AB and CD, respectively.
We know that the perpendicular from the centre to a chord bisects it.
So, AM = \(\frac {1}{2}\)AB and CN = \(\frac {1}{2}\)CD.
Since AB = CD. It follows that AM = CN.
Join OA and OC (radii of the circle).
Now consider the right-angled triangles ∆OMA and ∆ONC:
OA = OC (Radii of the same circle).
AM = CN (Proved above).
∠OMA = ∠ONC = 90°.
By the RHS congruence rule (Right angle-Hypotenuse-Side),
∆OMA ≅ ∆ONC
Therefore, OM = ON (corresponding parts of congruent triangles).
Since OM and ON represent the distances from the centre, the chords are at the same distance from the centre.

Question 2.
The picture shows two chords of the same length through one end of a diameter:

Prove that the angles which the chords make with the diameter are equal.
Answer:
Let the two chords be AB and AC meeting at point A on the circle.
Let AD be the diameter passing through A.
We are given AB = AC.
Draw perpendiculars from the centre O to the chords AB and AC.
Let the feet of the perpendiculars be M and N, respectively.

Since the chords are equal in length (AB = AC), they are equidistant from the centre.
Therefore, OM = ON.
Consider the right-angled triangles ∆AMO and ∆ANO:
AO = AO (Common hypotenuse).
OM = ON (Equal chords are equidistant from the centre).
∠AMO = ∠ANO = 90°
By the RHS congruence rule,
∆AMO ≅ ∆ANO.
Therefore, ∠MAO = ∠NAO.
This means the angles the chords make with the diameter are equal.

Question 3.
In the picture, two chords of the same length in a circle are extended to meet at a point:

Prove that these lines make the same angle with the line joining the centre of the circle and the points where the lines meet.
Answer:
Let the two equal chords be AB and CD.
Let them be extended to meet at a point P outside the circle.
Join the centre O to the point P.
Draw perpendiculars OM and ON from the centre O to the chords AB and CD (extended).

Since the chords are equal (AB = CD), they are at the same distance from the centre.
So, OM = ON.
Consider the right-angled triangles ∆OMP and ∆ONP:
OP = OP (Common hypotenuse)
OM = ON (Distance to equal chords)
∠OMP = ∠ONP = 90°.
By the RHS congruence rule,
∆OMP ≅ ∆ONP.
Therefore, ∠MPO = ∠NPO.
This proves that the line joining the centre and the intersection point bisects the angle between the extended chords.

Lines and Circles (Page 162)

Question 1.
Draw triangles with sides as given below and draw the circumcircle of each.
(i) Two sides of length 4 centimetres and 5 centimetres; the angle between them is 60°.
(ii) Two sides of length 4 centimetres and Scentimetres; the angle between them is 120°.
(iii) Two sides of length 4 centimetres and 5 centimetres; the angle between them is 90°.
(Note the change in the position of the circumcentre)
Answer:
(i) Draw a triangle with given measurements.
Draw perpendicular bisectors of any two sides.
Draw a circle with the point of intersection of the perpendicular bisectors as the centre of the circumcircle.
The centre of the triangle’s circumcircle lies inside the triangle.

(ii) Draw a triangle with the given measurements.
Draw the perpendicular bisectors of any two sides.
Draw a circle with the point of intersection of the perpendicular bisectors as the centre of the circumcircle.
The centre of the triangle’s circumcircle lies outside the triangle.

(iii) Draw a triangle with the given measurements.
Draw the perpendicular bisectors of any two sides.
Draw a circle with the point of intersection of the perpendicular bisectors as the centre of the circumcircle.
The circumcentre of a right-angled triangle is the midpoint of the hypotenuse.

Class 8 Maths Chapter 9 Kerala Syllabus Circles Questions and Answers

Class 8 Maths Circles Questions and Answers

Question 1.
The radius of a circle is 10 cm. What is the length of the longest chord that can be drawn in this circle?
(A) 5 cm
(B) 10 cm
(C) 20 cm
(D) 15 cm
Answer:
(C) 20 cm
The longest chord is the diameter, which is twice the radius.

Question 2.
The perpendicular distance from the centre of a circle to a chord is 3 cm. If the radius of the circle is 5 cm, what is the length of the chord?
(A) 4 cm
(B) 8 cm
(C) 6 cm
(D) 10 cm
Answer:
(B) 8 cm
By Pythagoras theorem,
Half the chord = \(\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}\) = 4 cm
Total length = 4 + 4 = 8 cm.

Question 3.
Where does the circumcentre of a right-angled triangle lie?
(A) Inside the triangle
(B) Outside the triangle
(C) At the vertex of the right angle
(D) At the midpoint of the hypotenuse
Answer:
(D) At the midpoint of the hypotenuse

Question 4.
How many circles can be drawn passing through three non-collinear points?
(A) One
(B) Two
(C) Three
(D) Infinite
Answer:
(A) One
There is a unique circumcircle for any triangle (formed by 3 non-collinear points).

Question 5.
To find the centre of a circle using a given arc (like a piece of a bangle), we need to draw:
(A) Two diameters
(B) The perpendicular bisectors of two chords
(C) Two radii
(D) Two tangents
Answer:
(B) The perpendicular bisectors of two chords

Question 6.
Read the given statements.
Statement I: A line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.
Statement II: The perpendicular bisector of any chord passes through the centre of the circle.
Choose the correct answer:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(C) Both statements are true.

Question 7.
Read the given statements.
Statement I: Chords of equal length are at equal distances from the centre.
Statement II: As the length of a chord increases, its distance from the centre increases.
Choose the correct answer:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(A) Statement I is true, Statement II is false.
Statement II is false because as a chord gets longer (closer to being a diameter), it gets closer to the centre, so the distance decreases.

Question 8.
Read the given statements.
Statement I: It is possible to draw a circle passing through three points that lie on a straight line.
Statement II: The perpendicular bisectors of the segments joining collinear points are parallel and never intersect.
Choose the correct answer:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(B) Statement I is false, Statement II is true.
You cannot draw a circle through three collinear points because the centre cannot be found (the bisectors never meet).

Question 9.
Read the given statements.
Statement I: The circumcentre of an acute-angled triangle is inside the triangle.
Statement II: The circumcentre is the point where the perpendicular bisectors of the sides intersect.
Choose the correct answer:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true, and II explains the definition of circumcentre.
(D) Both statements are false.
Answer:
(C) Both statements are true.

Question 10.
Read the given statements.
Statement I: A diameter is a chord that passes through the centre.
Statement II: All chords of a circle are diameters.
Choose the correct answer:
(A) Statement I is true, Statement II is false.
(B) Statement I is false, Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(A) Statement I is true, Statement II is false.

Question 11.
As you move a chord away from the centre of the circle, what happens to its length?
(A) It increases
(B) It decreases
(C) It remains the same
(D) It becomes equal to the radius
Answer:
(B) It decreases

Question 12.
A line segment that represents the fixed distance from the centre to the circle is called:
(A) Chord
(B) Diameter
(C) Radius
(D) Arc
Answer:
(C) Radius

Question 13.
If two circles intersect each other at two points, the line joining their centres is the ____________ of their common chord.
(A) Parallel line
(B) Perpendicular bisector
(C) Tangent
(D) Radius
Answer:
(B) Perpendicular bisector

Question 14.
Which property is used to prove that the perpendicular from the centre to a chord bisects it?
(A) Property of Isosceles Triangles
(B) Property of Equilateral Triangles
(C) Property of Parallel Lines
(D) Area of a Circle
Answer:
(A) Property of Isosceles Triangles (The triangle formed by the centre and the two ends of the chord is isosceles).

Question 15.
How many circles can be drawn passing through exactly two given points?
(A) Only one
(B) Only two
(C) Infinite
(D) None
Answer:
(C) Infinite (Their centres will all lie on the perpendicular bisector of the line joining the two points).

Question 16.
Draw a circle of radius 3.75 centimetres.
Answer:
Since the radius is 3.75 cm,
the diameter will be double that length = 3.75 × 2 = 7.5 cm.
Draw a line segment AB of length 7.5 cm.
Construct the perpendicular bisector of this line to find its exact midpoint, O.
Place the compass needle on O and the pencil on A (or B) and draw the circle.
The circle will pass through both A and B.

Question 17.
O is the centre of the circle, and OC is perpendicular to AB.

(i) If PQ = 8 cm, find PC.
(ii) Prove that AP = BQ.
Answer:
(i) PQ = 8 cm; PC = 4 cm
(ii) The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AC = BC ……… (1)
PC = QC ……… (2)
(1) – (2) ⇒ AC – PC = BC – QC
That is, AP = BQ.

Question 18.
AB and CD are two chords of the same length in a circle, and they are perpendicular to each other.

(i) If the distance from the centre to AB is d, what is the distance from the centre to CD?
(ii) What type of quadrilateral is OQSP?
(iii) Prove that BS = DS.
Answer:
(i) Equal chords are equidistant from the centre of the circle, so OP = OQ.
∴ The distance from the centre to CD = d.
(ii) OP = OQ
Each angle is 90°.
∴ OQSP is a square.
(iii) Since the perpendicular from the centre of the circle to the chord bisects the chord.
PB = \(\frac {1}{2}\)AB
QD = \(\frac {1}{2}\)CD
Since AB = CD, PB = QD
PB = PS + BS, QD = QS + DS
∴ PS + BS = QS + DS
Since OQSP is a square, PS = QS
∴ BS = DS

Question 19.
If AB = 6 cm, ∠A = 50°, ∠B = 60°, draw triangle ABC and its circumcircle.
Answer:
First, draw triangle ABC with the given measurements.
Next, construct the perpendicular bisectors for two of its sides to find where they meet.
Mark this intersection as the circumcentre and draw a circle with radius OA.

Question 20.
Draw an equilateral triangle with a side length of 6 cm. Draw its circumcircle.
Answer:
Draw an equilateral triangle ABC with a side length of 6 cm.
Draw the perpendicular bisectors of any 2 sides.
Draw a circle with the point of intersection of the perpendicular bisectors as the circumcentre and OB as the radius.

Question 21.
In triangle ABC, ∠B = 90°, AB = 12 cm, BC = 5 cm. What is the circumradius of triangle ABC?

Answer:
AB = 12 cm, BC = 5 cm
AC² = AB² + BC²
AC² = 12² + 5²
AC = 13 cm
Circumradius = Half the hypotenuse of the right-angled triangle
= \(\frac {13}{2}\)
= 6.5 cm

Question 22.
What is the length of the chord that is 9 cm away from the centre of a circle with a radius of 15 cm?
Answer:

From the figure,
Radius, OB = 15 cm
Distance OC = 9 cm
BC² = OB² – OC²
= 15² – 9²
= 225 – 81
= 144
∴ BC = 12 cm
The length of the chord = 2 × 12 = 24 cm

Class 8 Maths Chapter 9 Notes Kerala Syllabus Circles

Diameter
Circle: It is the path traced by a point moving at a fixed distance from a fixed point.
Centre: The fixed point inside the circle.
Chord: A line joining any two points on a circle.
Radius: The fixed distance from the centre to the circle.
Diameter:

• It is double the radius.
• It is a chord that passes through the centre.
• It is the longest chord of the circle.

Radius, Diameter, and Chord

Properties of Chords

The line joining the midpoint of a chord and the centre of the circle is the perpendicular bisector of the chord.
The perpendicular bisector of a chord passes through the centre of the circle.

Here AX = BX

Example:
If a circle has a radius of 5 cm and a chord is at a distance of 3 cm from the centre, we can find the length of the chord using Pythagoras’ theorem.
Radius (hypotenuse) = 5
Distance (perpendicular) = 3
Half-chord base = \(\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}\) = 4
Total Chord Length = 4 + 4 = 8 cm

Equal Chords
Chords that are at the same distance from the centre are of equal length.
Conversely, equal chords are at equal distances from the centre.
Chords of a circle at equal distances from the centre have equal length.

Chords that subtend the same angle on both sides of the diameter have the same length.
If we draw chords of the same length from one end of a diameter, they will make the same angle with the diameter.

Constructing Circles
Through One Point: You can draw infinite circles passing through a single point.
Through Two Points: You can draw many circles passing through two fixed points. The centres of all such circles lie on the perpendicular bisector of the line joining the two points.
Through Three Points: You can draw only one circle passing through three points, provided they are not on the same straight line.

Circumcircle
A circle that passes through all three vertices of a triangle is called its circumcircle.
Circumcentre: The centre of the circumcircle is called the circumcentre.
The circumcentre is the point where the perpendicular bisectors of the sides of the triangle intersect.

Position of Circumcentre:

• In an acute triangle, it is inside.
• In a right triangle, it is the midpoint of the hypotenuse.
• In an obtuse triangle, it is outside.