Circles and Angles Questions and Answers Class 10 Maths Chapter 2 Kerala Syllabus Solutions

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SSLC Maths Chapter 2 Circles and Angles Questions and Answers

Circles and Angles Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 2 Circles and Angles Solutions

Class 10 Maths Chapter 2 Kerala Syllabus – Arcs and Angles

Textbook Page No. 36

(1) What fraction of the circle is the arc marked in the picture below?

Answer:
Given : The angle formed by the arc in its alternate arc is 40°
So the central angle of the arc = 2 × 40 = 80°
\(\frac{2}{9}\) part of the circle is the length of arc

(2) When the comer of a bent wire was placed at the centre of a circle, \(\frac{1}{10}\) of the circle was contained within it.
If the comer of this wire is placed on a point of this circle as in the second picture, what fraction of the circle would it contain? What if it is placed at a point on another circle as in the third picture.

Answer:
Angle in the bent is fixed.
When the corner of the bent is at the centre \(\frac{1}{10}\) part of the circle comes inside the wire . The angle at the bent is \(\frac{1}{10}\) × 360 = 36°
When the comer is on the circle the central angle of the arc inside the wire will be 2 × 36 =72°
It is \(\frac{72}{360}=\frac{1}{5}\) part of the circle.
Whatever be the side of circle , the part inside the bent will be \(\frac{1}{5}\), if the comer is at the circle.

Textbook Page No. 41

Question 1.
In each picture below, the central angle of an arc of a circle is shown :

In each, calculate the angles which the arc makes at the other two points
Answer:
In the first picture, central angle of arc BCD is 100°.
Therefore the measure of angle BAD is 50°.
Central angle of the arc BAD is 360 – 100 = 260°

Measure of angle BCD is \(\frac{260}{2}\) = 130°

In the second picture central angle of arc QPS is 240°
Therefore ∠QRS = 120°
Central angle of arc QRS is 360 – 240 = 120°
∠QPS = 60°

Textbook Page No. 46

Question 1.
A triangle is drawn joining the numbers 1,4 and 8 on a clock face

a) Calcualte the angles of this triangle
b) How many equilateral triangles can we make by joining the numbers on a clock face?
Answer:
a) Central angle of the arc between 1 and 4 on the clock face is 90°
Angle at 8 is 45°
Central angle of the arc between 4 and 8 is 30 × 4 = 120°
Angle at 1 is 60°
Central angle of the arc joining 8 and 1 is 5 × 30 = 150°
Angle at 4 is 75°

b) Points form an equilateral triangle are: 12 – 4 – 8, 1 – 5 – 9, 2 -6 – 10, 3 -7 -11.
Four equilateral triangles can be formed on a clock face.

Question 2.
Draw an equilateral triangle with circumradius 3.5 cm
Answer:
Draw a circle of radius 3.5 cm.
Divide the angle around the centre into three equal parts. Each angle is 120° by drawing radii Join the ends of radii. It makes a triangle with angles 60° each .

Construction:

Question 3.
Draw a triangle with circumradius 3cm and two of the angles 32\(\frac{1^0}{2}\) and 37\(\frac{1^0}{2}\)
Answer:
Draw a circle of radius 3cm.
Divide the angle around the center as 32\(\frac{1}{2}\) × 2 = 65°
and 37\(\frac{1}{2}\) × 2 = 75° by drawing radii. Third angle will be 360 – (65 + 75) = 40°
Join the ends of radii.

Construction:

Question 4.
In the picture, a semicircle is drawn with a line as diameter and a smaller semicircle with half this line as diameter.
Prove that a line joining the point where the semi-circles meet to any point on the larger semicircle is bisected by the smaller semicircle.

Answer:
AP is the chord of the circle with diameter AB with centre O.

AO is the diameter of small semicircle Draw OQ. Angle in the semicircle is 90°.
∠AQO = 90°
Perpendicular from centre to the chord bisect the chord.
That is OQ bisect AP

Question 5.
Prove that circle drawn on the equal sides of an isosceles triangle as diameters pass through the midpoint of the third side

(Hint: Consider the circles one by one)
Answer:

AB is the diameter of a circle. Angle APB = 90°.
Triangle APB is a right triangle. AB² = AP² + BP²
AC is the diameter of a circle. Angle APC = 90°,
AC² = AP² + PC²
Since AB = AC and AP is common we can write
AP² + BP² = AP² + PC² → BP² = PC² → PB = PC

Question 6.
Prove that all the circles drawn on the four sides of a rhombus as diameters pass through a common point:

Answer:

We know that the diagonals of a rhombus bisect perpendicularly
Let P be the point where the diagonals intersect △APB, △BPC, △CPD and △APD are 90° each We know that angle in the semicircle is 90°
P is the common point on the semicircles with di-ameter, the sides of the rhombus
That is, P is the point of intersection of circles.

Question 7.
In the picture below, a triangle is drawn joining the ends of the diameter of a cir-cle and another point on the semicircle; and then semicircles on the other sides of the triangle as diameters:

Prove that the sum of the areas of the blue and red crescents in the second picture is equal to the area of the triangle
(Hint: See the last problem of the lesson, Parts of Circles of Class 9 textbook).
Answer:

Since triangle ABC is a right triangle, a² = b² + c²
Sides of the triangle are the diameters of semi-circles.
Multiply both sides by π
πa² = πb² + πc²
Dividing both sides by 4 we get
\(\frac{\pi a^2}{4}=\frac{\pi b^2}{4}+\frac{\pi c^2}{4}\)
π\(\left(\frac{a}{2}\right)^2\) = π\(\left(\frac{b}{2}\right)^2\) + π\(\left(\frac{c}{2}\right)^2\)
So area of big semicircle is equal to sum of the areas of other two semicircles
Let x and y be the area of segment of circle inside the big semicircle.
Area of cresent(blue) + x + Area of cresent (red ) + y = Area of triangle + x + y
Cancelling x and y from both sides, area of triangle is equal to sum of the area of cresents.

Question 8.
In the picture, AB and CD are perpendicular chords of the circle:

Prove that the arcs APC and BQD joined together make a semicircle.
Answer:

Mark the center O
Join CB
∠AOC = 2 × ∠ABC, ∠DOB = 2 × ∠DCB
We know that in the right triangle CMB,
∠DCB + ∠ABC = 90°
∠AOC + ∠DOB = 2∠ABC + 2∠DCS
∠AOC + ∠DOB = 2 (∠ABC + ∠DCB)
= 2 × 90 = 180°
Sum of the central angles of arc AC and arc DB, which are the parts of same circle is 180°
That is, these arcs, on joining makes a semcircle.

Textbook Page No.51

Question 1.
In all the three pictures below, O is the centre of the circle and A, B, C are the points on the circle.In each, calculate all angles of triangles ABC and OBC

Answer:
In the first diagram
Join OA. In triangle OB A, ∠OAB = 20°.
In triangle OCA, ∠OAC = 30°
∠BAC= 20 + 30 = 50°
∠BOC = 100°
Since OB = OC, opposite angles are equal.
∠OBC = ∠OCB = 40°
∠ABC =20 + 40 = 60°,
∠ACB = 30 + 40=70°

In the second figure
OB = OC
∠OCB = 40°,
∠BOC= 100°, ∠A = 50°
Join OA. OA = OC → ∠OAC = 30°
∠OAB = 50 – 30 = 20°, ∠OAB = 20°
∠Z = 50°, ∠B = 60°, ∠C = 70°

In the third figure
∠ACB = \(\frac{1}{2}\) × 40 = 20°,
∠CBA = \(\frac{1}{2}\) × 70 = 35°
∠BAC= 180 – (20 + 35) = 180 – 55 = 125°

Question 2.
In each of the problem below a circle and a chord is to be drawn to split the circle into two parts. The parts must be as specified:
a) All angles in one part must be 80°
b) All angles in one part must be 110°
c) All angles in one part must be half the angles in other part
d) All angles in one part must be one and a half times the angles in the other part.
Answer:
In each of the problem below a circle and a chord is to be drawn to split the circle into two parts.
a) Draw a circle.Draw two radii such that angle between them 2 × 80°.
That is, the central angle of an arc is 160°. So angle in the alternate arc is 80°

b) Construction:

c) Sum of the angles in the alternate segments is 180°
If angle in one segment is x then angle in alter-nate segment is 2x. (Given ratio is 1 : 2)
3x = 180 → x = 60°

d) Given: one angle is 1\(\frac{1}{2}\) times the other
Ratio of these angles is 2 : 3
Angle in one segment is \(\frac{2}{5}\) × 180° = 72°.
Angle in other segment is \(\frac{3}{5}\) × 180 = 108°
Divide the angle around the centre as 144° and 216°. For this draw two radii such that angle between them is 144°. The other angle will be 360 – 144 = 216°

Class 10 Maths Kerala Syllabus Chapter 2 Solutions – Circle And Quadrilateral

Textbook Page No.55

Question 1.
Calculate the angles of the quadrilateral shown below and also the angles between its diagonals

Answer:

∠CAD = ∠CBD = 50°
∠ADB = ∠ACB = 30°
∠ACD = ∠ABD = 45°
∠BAC = ∠BDC = x.
Since ABCD is cyclic 50 + x + 45 + 30 = 180
⇒ x = 55
∠A = 55 + 50 = 105°,
∠B = 95°,
∠C = 75°,
∠D = 85°
In triangle OBC, ∠BOC = 180 – (50 + 30) = 100°, ∠AOD =100° [opposite angles are equal]
∠AOB = 180 – 100 = 80°, ∠DOC = 80° [opposite angles are equal]

Question 2.
Prove that in a cyclic quadrilateral, the outer angle at any vertex is equal to the inner angle at the opposite vertex.
Answer:
Digaram

If ∠PCD = x then ∠BCD = 180 – x
Therefore ∠A = 180 – (180 – x) = x

Question 3.
Prove that any parallelogram, which is not a rectangle, is not cyclic
Answer:
Opposite angles of a parallelogram are equal, but these are not 90°.
So the sum of opposite angles is not 180°.
That is, parallelogram which is not a rectangle is not cyclic

Question 4.
Prove that a non-isosceles trapezium is not cyclic
Answer:

ABCD is a trapezium. AB is parallell to CD, AD ≠ BC. Therefore ∠A ≠ ∠B.
Since AB parallel to CD. ∠B + ∠C = 180°. So ∠A + ∠D = 180°
ABCD is not cyclic.

Question 5.
In the first picture below, an equilateral triangle is drawn with vertices on a circle and two of its vertices are joined to a point on the circle. In the sec-ond picture,a square is drawn with vertices on a circle and two of its vertices are joined to a point on the circle

In each picture, calculate the angle marked.
Answer:
In the first figure,

triangle ABC is equilateral.
∠Z = 60°
ABDC is cyclic. ∠D = 180 – 60 = 120°

In the second figure

The diagonals of a square bisect perpendicularly.
∠APB = 45°
Since PAQB is cyclic ∠Q = 180 – 45 = 135°

Question 6.
(i) In the picture below, two circles intersect at P and Q. Lines through these points meet the circles at A, B, C, D. The lines AC and BD are not parallel. Prove that if these lines are of equal length, then ABDC is a cyclic quadrilateral

(ii) In the picture, the circles on the left and right intersect the middle circle at P, Q, R, S. Lines joining these meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.

Answer:
(i) See diagram

Join PQ. Quadrilateral APQB is cyclic.
If ∠B = x then ∠APQ is 180 – x.
Therefore ∠QPC = x, ∠D = 180 – x
∠B + ∠D = x + 180 – x = 180°.
So AB is parallel to CD. ABCD is a trapezium Also given that AC = BD. So it is an isosceles trapezium. Isosceles trapezium is cyclic.
That is ABCD is cyclic.

(ii) In the second diagram

Join PQ and RS
APQB is cyclic, PQSR is cyclic, RSDC is cyclic. If ∠B – x then ∠APQ = 180 – x
∠QPR = x, ∠QSR = 180 – x,
∠RSD = x, ∠C = 180 – x
∠B + ∠C = x + 180 – x = 180°
ABDC is cyclic

Question 7.
In the picture, the bisectors of the angles of the quadrilateral ABCD intersect at P, Q, R, S. Prove that PQRS is a cyclic quadrilateral

Prove that PQRS is a cyclic quadrilateral.
(Hint : Look at the sum of the angles of triangles PCD and RAB).
Answer:
Digaram

∠DSA = 180 – \(\left(\frac{A}{2}+\frac{D}{2}\right)\). So ∠RSP, = 180 – \(\left(\frac{A}{2}+\frac{D}{2}\right)\)
∠PQR = 180 – \(\left(\frac{C}{2}+\frac{B}{2}\right)\)
In PQRS, ∠S + ∠Q = 180 – \(\frac{A}{2}-\frac{D}{2}\) + 180 – \(\frac{C}{2}-\frac{B}{2}\)
∠S + ∠Q = 360 – \(\frac{A+B+C+D}{2}\)
= 360 – \(\frac{360}{2}\)
= 180
PQRS is cyclic.

Question 8.
In the first picture below, points P, Q, R are marked on the sides BC, CA, AB of triangle ABC and circumcircles of triangles A QR and BRP are drawn. They intersect at the point S inside the triangle :

Prove that the circumcircle of triangle CPQ also passes through S, as in the second picture.
(Hint: In the first figure, join PS, QS and RS. Then find the relations of the angles formed at the point Swith ∠A, ∠B and ∠C).
Answer:

BPSR is cyclic ∠PSR = 180 – A
∠PSQ = 360 – (180 – A + 180 – R) = A + B
In the triangle ABC, A + B + C = 180
That is ∠PSQ + ∠C = 180°
That is PSQC is cyclic
That is, circum-circle of CPQ passes through S.

Circles and Angles Class 10 Notes Pdf

Class 10 Maths Chapter 2 Circles and Angles Notes Kerala Syllabus

Introduction
Arc of a circle and its properties related to angles are the main aspects of this unit. Two points on a circle determines two arcs, one is the alternate arc of the other. An arc can make three angles: Angle on the arc itself, angle at the centre and angle in the alternate arc. These angles are related to each other.
Angle formed by an arc at the centre is two times the angle formed by that arc in its alternate arc.
Angle on a semicircle a right angle. The angles drawn on an arc are equal.
The sum of the angles on the arc and in the alternate arc is 180°. This leads to the concept of cyclic quadrilateral. If the comers of a quadrilateral are on a circle then the quadrilateral is called cyclic quadrilateral.
The unit discusses segment of a circle and its alternate segment. The angles drawn on a segment and its alternate segment make the sum 180°.

→ Two points on a circle divides the circle into two arcs. One arc is called the alternate arc of the other

→ A chord of a circle divides the circle into two segments. One segment is the alternate segment of the other.

→ An arc of a circle can make three types of angles.

1. angle on itself,
2. angle at the centre,
3. angle in the alternate arc

• Angle APB is the angle formed by the arc AB on itself
• Angle AOB is the angle formed by the arc AB at the centre
• Angle AQB is the angle formed by the arc AB in the alternate arc

→ If the ends of an arc of a circle are joined to a point on the circle, which is a point on the arc itself, then the angles so made are equal. This can be simplified as ‘angles on an arc are equal’.

→ The angle made by joining the ends of an arc of a circle to any point on the alternate arc is half the central angle of the arc.

→ If the ends of a semicircle are joined to another point on the circle, the angle made is a right angle.The angle in a semicircle is a right angle

→ In a circle, angles in the same segment are equal.

∠APB = ∠AQB

→ The sum of the angles in alternate segments is 180°.

∠APB + ∠AQB = 180°

→ If all vertices of a quadrilateral are on a circle, then the sum of its opposite angles is 180°

→ If in a quadrilateral, the sum of opposite angles is 180°, then a circle can be drawn passing through all four of its vertices.

→ If the ends of an arc of a circle are joined to a point on the circle, which is not a point on the arc itself, then the angle so made is half the central angle of the arc.

→ If the ends of a semicircle are joined to another point on the circle, the angle made is a right angle.

→ The angle in a semicircle is a right angle.

→ The angle made by joining the ends of an arc of a circle to any point on the alternate arc is half the central angle of the arc.

→ In a circle, angles in the same segment are equal, the sum of the angles in alternate segments is 180°.

→ If all vertices of a quadrilateral are on a circle, then the sum of its opposite angles is 180°.

→ If in a quadrilateral, the sum of opposite angles is 180°, then a circle can be drawn passing through all four of its vertices.

→ A quadrilateral with the property that a .circle could be drawn through all four of its vertices, is called a cyclic quadrilateral.