Mathematics Part I Solutions Solutions for Class 10 Maths Chapter 2 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 10 students for Maths Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Question 1:

In ΔABC, we have A = 60° and B = 70°. Is the vertex C inside or outside the circle with diameter AB?

Given: In ΔABC, A = 60° and B = 70°

Using angle sum property:

A + B + C = 180°

60° + 70° + C = 180°

130° + C = 180°

⇒ ∠C = 180° 130°

= 50°

We know that angle in a semi circle is a right angle.

If AB is the diametre of the circle then C should be equal to 90°.

However, here C is less than 90°. Therefore, vertex C will lie outside the circle.

#### Question 2:

Prove that if a pair of opposite angles of a quadrilateral are right, then a circle can be drawn through all four of its vertices.

Consider the figure below.

Here, ABCD is a quadrilateral in which A = C = 90°.

We know that angle in a semi circle is a right angle.

Suppose a circle is drawn such that it passes through vertex A.

As BAD = 90°, BD acts like the diametre of this circle.

Similarly, suppose a circle is drawn such that it passes through vertex C.

As BCD = 90°, BD acts like the diametre of this circle.

Thus, BD is the diametre of the circle passing through vertices A and C.

As the end points of the diametre lie on the circle, all the four vertices of quadrilateral ABCD lie on the circle.

Hence, if a pair of opposite angles of a quadrilateral are right then a circle can be drawn through all four of its vertices.

#### Question 3:

In the quadrilateral ABCD, we have AB = 3 cm, BC = 4 cm, AC = 5 cm, A = 120°, C = 70°. If we draw the circle with AC as diameter, which of the four vertices of ABCD would be inside the circle? Which of them would be outside this circle? Is any vertex on the circle? What about the circle with BD as diameter.

Given: AB = 3 cm, BC = 4 cm, AC = 5 cm, A = 120° and C = 70°

Consider the following:

AB2 = 32 = 9

BC2 = 42 = 16

AC2 = 52 = 25

AC2 = AB2 + BC2

Therefore, by the converse of Pythagoras theorem, we get that B = 90°.

Using angle sum property in quadrilateral ABCD:

A + B + C + D = 360°

120° + 90° + 70° + D = 360°

280° + D = 360°

⇒ ∠D = 360° 280° = 80°

We know that angle in a semi circle is a right angle.

If we draw a circle with AC as the diametre then vertex B will lie on the circle.

As the measure of D is less than 90°, vertex D will lie outside the circle.

Thus, vertices A, B and C will lie on the circle.

Now, let us draw a circle with BD as the diametre.

As A = 120° and C = 70°, vertex A will lie inside the circle and vertex C will lie outside the circle.

Only vertices B and D will lie on the circle.

#### Question 1:

In this figure, what fraction of the circumference of the circle is the length of the arc ADB?

Construction: Join points A and B with centre O of the circle.

Given: ACB = 60°

We know that angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

We know that measure of an arc is equal to the measure of the corresponding central angle.

Now,

Thus, the length of the arc ADB is of the circumference of the circle.

#### Question 1:

What is the radius of the circle shown below?

Given: ΔABC, with B = 90° and C = 45°

Using angle sum property:

A + B + C = 180°

⇒ ∠A + 90° + 45° = 180°

⇒ ∠A + 135° = 180°

⇒ ∠A = 180° 135°

= 45°

A = C

BC = AB = 3 cm (Sides opposite to equal angles are equal in length.)

Using Pythagoras theorem in ΔABC:

As ABC = 90° and we know that angle in a semi circle is a right angle, AC is the diametre of the circle.

#### Question 2:

What is the area of the circle shown below?

Let the given rectangle be ABCD.

Construction: Join BD.

We know that each angle of a rectangle measures 90°.

We know that angle in a semi circle is a right angle.

Therefore, BD is the diametre of the circle.

Using Pythagoras theorem in ΔABD:

Radius of the circle, r =

Area of the circle = πr2

#### Question 3:

How do we draw a triangle with two of the angle 40° and 120° and circumradius 3 centimetres?

To construct the required triangle, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

The steps of construction are as follows:

(1) Draw a circle with centre O and radius 3 cm.

(2) Take a point B on this circle and join OB.

(3) Taking O as the centre and OB as the segment, draw an angle of measure 80° such that it intersects the circle at point C.

(Since we need BAC = 40° , we have drawn BOC = 80°.)

(4) Taking O as the centre and OB as the segment, draw an angle of measure 240° on the other side of BOC such that it intersects the circle at point A.

(Since we need ACB = 120°, we have drawn AOB = 240°.)

(5) Join AC, AB and BC.

ΔABC is the required triangle.

#### Question 4:

How do we draw aangle?

To construct the required angle, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

The steps of construction are as follows:

(1) Draw a circle of any radius with O as the centre.

(2) Take a point B on the circle and join OB.

(3) Taking O as the centre and OB as the segment, draw an angle of measure 45° such that it intersects the circle at point A.

(Since we need ACB =we have drawn AOB = 45°.)

(4) Join CA and CB.

ACB is the required angle.

#### Question 5:

In each of the pictures below, draw aangle, according to the specifications:

(i) At the point A

(ii) At the point A with one side along OA

(iii) At the point A one side along AB

(i)

To construct the required angle at point A, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

The steps of construction are as follows:

(1) Draw a circle of any radius with O as the centre.

(2) Take a point B on the circle and join OB.

(3) Taking O as the centre and OB as the segment, draw an angle of measure 45° such that it intersects the circle at point C.

(Since we need CAB =we have drawn COB = 45°.)

(4) Join CA and AB.

CAB is the required angle.

(ii)

To construct the required angle at point A with one side along OA, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

(1) Draw a circle of any radius with O as the centre.

(2) Join the diametre AB of the circle.

(3) Taking O as the centre and OB as the segment, draw an angle of measure 45° such that it intersects the circle at point C.

(Since we need CAB =, we have drawn COB = 45°.)

(4) Join AC.

CAB is the required angle.

(iii)

To construct the required angle at point A with one side along AB, we will use the following property.

Angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

(1) Draw a circle of any radius with O as the centre.

(2) Take two points A and B on the upper side of the centre and join AB.

(3) Join OB.

(4) Taking O as the centre and OB as the segment, draw an angle of measure 45° such that it intersects the circle at point C.

(Since we need CAB =, we have drawn COB = 45°.)

(5) Join AC.

CAB is the required angle.

#### Question 1:

In the picture below, O is the centre of the circle and the line OD is parallel to the line CA.

Prove that OD bisects AOB.

Given: CA is parallel to OD.

CB acts as the transversal.

∴ ∠ACO = DOB (Corresponding angles) … (1)

We know that angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

∴ ∠AOB = 2ACB

Using equation (1):

AOB = 2DOB

OD acts as the bisector of AOB.

Therefore, OD bisects AOB.

Yes, we can use this concept to draw the bisector of a given angle.

Suppose, in the given figure, we are given only AOB and CB as the diametre of the circle.

Let us join one end point, C of the diametre CB with point A of AOB.

We get that 2ACB = AOB.

If we draw a line segment passing through point O and parallel to AC such that it intersects the circle at point D, then OD acts as the bisector of AOB.

#### Question 1:

In the figure below, O is the centre of the circle.

Prove that x + y = 90°

We know that angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.

∴ ∠BOC = 2BAC

⇒ ∠BOC = 2y

Now, OB = OC (Radii of the same circle)

∴ ∠OBC = OCB = x (Angles opposite to equal sides are equal in measure.)

Using angle sum property in ΔOBC:

BOC + OCB + CBO = 180°

2y + x + x = 180°

2y + 2x = 180°

2(y + x) = 180°

y + x =

x + y = 90°

#### Question 1:

In the figure below, A, B, C, D are points on the circle.

Compute the angles of the quadrilateral ABCD and the angles between its diagonals.

We know that angles in a same segment are equal.

Here, CBD and CAD are the angles in the same segment.

⇒ ∠CBD = CAD = 30°

Now, BAD = BAC + CAD = 35° + 30° = 65°

ADB and ACB are the angles in a same segment.

⇒ ∠ADB = ACB = 50°

BDC and BAC are the angles in a same segment.

⇒ ∠BDC = BAC = 35°

Now, ADC = ADB + BDC = 50° + 35° = 85°

We know that angles in the alternate segments are supplementary.

∴ ∠BAD + BCD = 180°

65° + BCD = 180°

⇒ ∠BCD = 180° 65° = 115°

Also, ABC + ADC = 180°

⇒ ∠ABC + 85° = 180°

⇒ ∠ABC = 180° 85° = 95°

∴ ∠BAD = 65°, ABC = 95°, BCD = 115° and ADC = 85°

Using angle sum property in ΔPBC:

PBC + BCP + CPB = 180°

30° + 50° + CPB = 180°

80° + CPB = 180°

⇒ ∠CPB = 180° 80° = 100°

APD = BPC = 100° (Vertically opposite angles)

We know that sum of angles forming a linear pair is 180°.

∴ ∠BPC + BPA = 180°

100° + BPA = 180°

⇒ ∠BPA = 180° 100° = 80°

CPD = BPA = 80° (Vertically opposite angles)

Hence, the angles between the diagonals are CPD = BPA = 80° and APD = BPC = 100°.

#### Question 2:

In the figure below, ΔABC is equilateral and O is its circumcentre.

Prove that the length of AD is equal to the radius of the circle.

Construction: Join AO.

Given: ΔABC is an equilateral triangle.

⇒ ∠ABC = 60°

We know that angles in the same segment are equal.

⇒ ∠ADC = ABC = 60°

Now, OD = OA (Radii of the same circle)

∴ ∠OAD = ADO = 60° (Angle opposite to equal sides are equal in measure.)

Using angle sum property in ΔAOD:

60° + 60° + AOD = 180°

120° + AOD = 180°

⇒ ∠AOD = 180° 120° = 60°

Therefore, ΔAOD is an equilateral triangle.

Thus, the length of AD is equal to the radius of the circle.

#### Question 3:

In the picture below, ΔPQR is right angled. Also, A = P and BC = QR.

Prove that the diameter of the circumcircle of ΔABC is equal to the length of PQ.

Given: PRQ = 90°

We know that angle in a semi circle is a right angle.

If we draw a circle passing through vertex R of ΔPRQ then PQ acts as the diametre of the circle.

We have A = P and BC = QR.

We know that angles in the same segment are equal.

BC and QR are equal segments having equal angles, A and P. So, points A, B and C also lie on the same circle on which points P, Q and R lie.

As PQ is the diametre of the circumcircle of ΔPQR, PQ is the diametre of the circumcircle of ΔABC.

#### Question 1:

Prove that in a cyclic quadrilateral, the exterior angle at any vertex is equal to the interior angle at the opposite vertex.

Consider the cyclic quadrilateral ABCD whose one side DC is extended to point E.

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠DAB + DCB = 180°

⇒ ∠DCB = 180° DAB … (1)

We know that sum of the angles forming a linear pair is 180°.

∴ ∠BCE + BCD = 180°

⇒ ∠BCD = 180° BCE … (2)

From equation (1) and equation (2), we get:

180° DAB = 180° BCE

⇒ −∠DAB = −∠BCE

⇒ ∠DAB = BCE

Hence, in a cyclic quadrilateral, the exterior angle at any vertex is equal to the interior angle at the opposite vertex.

#### Question 2:

Prove that the non-rectangular parallelogram is not a cyclic quadrilateral.

Consider the non-rectangular parallelogram, ABCD.

We know that opposite angles of a parallelogram are equal.

∴ ∠A = C … (1)

Let us suppose that ABCD is a cyclic quadrilateral.

We know that opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠A + C = 180°

⇒ ∠A + A = 180° (From equation (1))

2A = 180°

⇒ ∠C = 90°

This is a contradiction to the given condition that ABCD is a non-rectangular parallelogram.

Thus, our supposition was wrong.

Hence, a non- rectangular parallelogram is not a cyclic quadrilateral.

#### Question 3:

Prove that non-isosceles trapeziums are not cyclic.

Consider the non-isosceles trapezium, ABCD.

Let us suppose that ABCD is a cyclic quadrilateral.

We know that opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠B + D = 180°

⇒ ∠B = 180° D … (1)

ABCD is a non-isosceles trapezium and AB is parallel to CD

∴ ∠A + D = 180° (Sum of the interior angles on the same side of the transversal is 180°.)

⇒ ∠A = 180° D … (2)

From equation (1) and equation (2), we get:

A = B

This is not possible as ABCD is a non-isosceles trapezium.

Thus, our supposition was wrong.

Hence, non-isosceles trapeziums are not cyclic.

#### Question 4:

In the figure below, ABCD is a square.

How much is APB?

Given: ABCD is a square.

Construction: Join AC.

We know that the diagonals of a square bisect the angles.

(Each angle of a square measures 90°)

Now, APBC is a cyclic quadrilateral as all its vertices lie on the circle.

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠APB + ACB = 180°

⇒ ∠APB + 45° = 180°

⇒ ∠APB = 180° 45°

⇒ ∠APB = 135°

#### Question 5:

Prove that in a cyclic hexagon ABCDEF as shown below, A + C + E = B + D + F.

Given: A cyclic hexagon ABCDEF.

Construction: Join CF.

ABCF is a cyclic quadrilateral as all its vertices lie on the circle.

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠B + CFA = 180° … (1)

A + BCF = 180° … (2)

From equation (1) and (2), we have:

B + CFA = A + BCF … (3)

CDEF is a cyclic quadrilateral as all its vertices lie on the circle.

∴ ∠D + EFC = 180° … (4)

E + DCF = 180° … (5)

From equation (4) and (5), we have:

D + EFC = E + DCF … (6)

Adding equation (3) and equation (6):

B + CFA + D + EFC = A + BCF + E + DCF

⇒ ∠B + D + EFC + CFA = A + E + DCF + BCF

⇒ ∠B + D + F = A + E + C

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