Kerala Syllabus 10th Standard Maths Solutions Chapter 2 Circles

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Kerala State Syllabus 10th Standard Maths Solutions Chapter 2 Circles

Circles Text Book Questions and Answers

Textbook Page No, 42

Question 1.
Suppose we draw a circle with the bottom side of the triangles in the picture as diameter. Find out whether the top corner of each triangle is inside the circle, on the circle or outside the circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 1
Answer:
Angle of the first triangle =110°
As 110° > 90° the top comer will be inside the circle
Angle of second triangle = 90°
∴ The top comer will be on the circle.
Angle of the third triangle = 70°
70° > 90°
∴ The top comer will be outside the circle

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Question 2.
For each diagonal of the quadrilateral shown, check whether the other two corners are inside, on or outside the circle with that diagonal as diameter
Kerala SSLC Class 10 Maths Chapter 2 Circles 2
Answer:
The fourth angle in ABCD
= 360 – (110 + 105 + 55) = 360
Kerala SSLC Class 10 Maths Chapter 2 Circles 3
Drawing diagonal AC and taking it as diameter of a circle As, ∠D = 90°
D will be on the circle. ∠B = 55° (90 > 55°)
∴ B will be outside the circle
Drawing diagonal BD and taking it as diameter of a circle.
∠A = 105°, ∠C = 110°
As both angles are greater than 90, they lie inside the circle.

Question 3.
If circles are drawn with each side of a triangle of sides 5 centimetres, 12 centimetres and 13 centimetres, as diametres, then with respect to each circle, where would be the third vertex?
Answer:
As the sides are 5, 12, 13 cm and also
52 + 122 = 25 + 144 = 169 = 132
∴ ABC is a right triangle
Kerala SSLC Class 10 Maths Chapter 2 Circles 4
Taking BC as diameter and drawing a circle, ∠A (<90°), A will be outside the circle.
Taking AB as diameter and drawing a circle ∠C (<90°) C will be outside the circle.
Taking AC as diameter and drawing a circle, ∠B = 90°, B will be on the circle.

Question 4.
In the picture, a circle is drawn with a line as diameter and a smaller circle with half the line as diameter. Prove that any chord of the larger circle through the point where the circles meet is bisected by the small circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 5
Answer:
∠ADO = ∠APB = 90°
(angle subtended by diameter is always 90°)
⇒ OD\\PB
Kerala SSLC Class 10 Maths Chapter 2 Circles 6
AO = OB (Radius of bigger circle)
(If in a triangle, the line drawn from midpoint of one side, is parallel to another side, then the line will bisect the third side)
Therefore AD = DP
(AB’s midpoint is ‘O’ and OD\\PB)

Question 5.
Kerala SSLC Class 10 Maths Chapter 2 Circles 7.
Use a calculator to determine up to two decimal places, the perimeter and the area of the circle in the picture.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 8

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Question 6.
The two circles in the picture cross each other at A and B. The points P and Q are the other ends of the diameters through A.
Kerala SSLC Class 10 Maths Chapter 2 Circles 9
i. Prove that P, B, Q lie on a line.
ii. Prove that PQ is parallel to the line joining the centres of the circles and is twice as long as this line.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 10
i. ∆ PAB (angle subtended by semicircle)
∠PBA = 90°
∆ ABQ be a triangle on the semi¬circle of centre D.
∴ ∠ABQ = 90°
∠PBA + ∠ABQ = 180 (Linear pair)
As AP, AQ are diameters of the circle. PQ be the line drawn through B per¬pendicularly to AB. Therefore P, B, Q lies on the same line.

ii.
Kerala SSLC Class 10 Maths Chapter 2 Circles 11
Kerala SSLC Class 10 Maths Chapter 2 Circles 12

Question 7.
Prove that the two circles drawn on the two equal sides of an isosceles triangle as diameters pass through the midpoint of the third side.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 13
∠ADB= 90° (AABD angle subtended by semicircle)
∠CDA = 90°
∴ ∠ADB +∠CDA = 180° (linear pair)
∆ ABD ∆ ADC are right angled triangles.
In ∆ ABD
BD2 = AB2 – AD2 ( AB = AC )
= AC2 – AD2 = DC2
BD = CD

Question 8.
Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point.
Kerala SSLC Class 10 Maths Chapter 2 Circles 14
Prove that this is true for any quadrilat¬eral with adjacent sides equal, as in the picture.
Kerala SSLC Class 10 Maths Chapter 2 Circles 15
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 16
ABCD is a rhombus so diameter are perpendicular bisectors.
∠AOD = 90°
O be on the circle having diameter AD.
∠AOB = 90°, therefore
O be on the circle having diameter AB.
∠BOC= 90°, therefore
O be on the circle having diameter BC
∠DOC= 90°, therefore
O be on the circle having diameter DC
O be the common point on the circle.
∠A0D = ∠AOB and
∠COD = ∠BOC and,
AD = AB,
AO be the common side.
Kerala SSLC Class 10 Maths Chapter 2 Circles 17
Δ AOD, Δ AOB are equal triangles.
OD = OB
Both the circles can passed through O.
A BCD is an isosceles triangle.
Those circles having diameters CD and BC are passing through midpoint of BD.
∴ O be common for the four circles. (Diameter)

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Question 9.
A triangle is drawn by joining a point on a semicircle to the ends of the diameter. Then semicircles are drawn with the other two sides as diameter.
Kerala SSLC Class 10 Maths Chapter 2 Circles 18
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.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 19
Area of triangle = \(\frac { 1 }{ 2 }\) × 2r × h = rh
Area of the semicircle = \(\frac{\pi r^{2}}{2}\)
Area of the rest of the figure
Kerala SSLC Class 10 Maths Chapter 2 Circles 20
The diameters of the red and blue semi-circles are the sides of the two triangles.
∴ Their areas
Kerala SSLC Class 10 Maths Chapter 2 Circles 21
Area of red and blue crescents
Kerala SSLC Class 10 Maths Chapter 2 Circles 22
= area of the triangle

Textbook Page No. 53

Question 1.
In all the pictures given below, O is the centre of the circle and A, B, C are points on it. Calculate all angles of Δ ABC and Δ OBC in each.
Kerala SSLC Class 10 Maths Chapter 2 Circles 23
Answer:
a.
Kerala SSLC Class 10 Maths Chapter 2 Circles 24
OA = OB (radii)
∴ ∠OAB = 20° ; (∠OAB = ∠OBA)
OC = OA (radii) ;
∠OAC = 30°
∠BAC = ∠OAB + ∠OAC
= 20 + 30 = 50°
∠BOC = 2x ∠BAC = 100°
OB = OC (radii)
∴ ∠OBC = ∠OCB = 40°
Kerala SSLC Class 10 Maths Chapter 2 Circles 25
Angles of triangle ABC are ∠BAC = 50° ∠ABC = 60°, ∠ACB = 70°
Angles of ∆ BOC are
∠OBC = 40°, ∠OCB = 40°, ∠BOC = 100°

b.
Kerala SSLC Class 10 Maths Chapter 2 Circles 26
Angles of ∆ ABC are ∠ABC = 50°, ∠BAC = 60°, ∠BCA = 70°
Angles of ∆ AOC are
∠OAC = 40°, ∠AOC = 100°, ∠OCA = 40°

c.
Kerala SSLC Class 10 Maths Chapter 2 Circles 27
∠ACB = 180 – 55 = 125° ; OA = OC
∠CAO = ∠ACO = 70° ; ∠OBC = ∠BCO = 55° Angles of A OBC are
∠OBC = 55° ∠COB = 70° ∠BCO = 55°
Angles of ∆ ABC
Kerala SSLC Class 10 Maths Chapter 2 Circles 28

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Question 2.
The numbers 1,4,8 on a clock’s face are joined to make a triangle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 29
Calculate the angles of this triangle.
How many equilateral triangles can we make by joining numbers on the clock’s face?
Answer:
The angle subtended by two adjacent numbers at the centre of the clock is 30°
Kerala SSLC Class 10 Maths Chapter 2 Circles 30
We can make 4 equilateral triangles by joining the numbers on the clock (1. 5, 9), (2, 6, 10), (3. 7, 11), (4, 8, 12)

Question 3.
In each problem below, draw a circle and a chord to divide it into two parts such that the parts are as specified.
i. All angles on one part 80°.
ii. All angles on one part 110°.
iii. All angles on one part half of all angles on the other.
iv. All angles on one part, one and a half times the angles on the other.
Answer:
i. ∠AOB = 160°
Therefore all angles in the are ACB arc 80°.
Kerala SSLC Class 10 Maths Chapter 2 Circles 31

ii. Draw angle as central angle 220° so angle on the small arc AB will be 110°.
Kerala SSLC Class 10 Maths Chapter 2 Circles 32

iii. Draw angle as central angle 120° or 240° All angles on one part will be 120°, and other part be 60°.
Kerala SSLC Class 10 Maths Chapter 2 Circles 33

iv. Draw a circle and draw central angle 144° All angles on the part APB will be 120° and All angles on the part AQB will be 108°.
Kerala SSLC Class 10 Maths Chapter 2 Circles 34

Question 4.
A rod bent into an angle is placed with its corner at the centre of a circle and it is found that \(\frac { 1 }{ 10 }\) of the circle lies within it. If
it is placed with its corner on another circle, what part of the circle would be within it?
Kerala SSLC Class 10 Maths Chapter 2 Circles 35
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 36

Question 5.
In the picture, O is the centre of the circle and A, B, C are points on it. Prove that
∠OAC + ∠ABC = 90°
Kerala SSLC Class 10 Maths Chapter 2 Circles 37
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 38

Question 6.
Draw a triangle of circumradius 3 centimetres and two of the angles 32\(\frac { 1° }{ 2 }\) and 37\(\frac { 1° }{ 2 }\)
Answer:
Draw a circle with radius 3 cm and central angle 65°.
Half of 65° is 32\(\frac { 1° }{ 2 }\) Similarly we can draw 75°
Join the points A, B and C Halfof75°is 373\(\frac { 1° }{ 2 }\) Complete the triangle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 39

Question 7.
In the picture, AB and CD are mutually perpendicular chords of the circle. Prove that the arcs APC and BQD joined together would make half the circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 40
Answer:
If ∠ADC = x
∠AOC =2x
(Angle subtended on the alternate arc is half of the central angle of arc )
If ∠BAD = y
∠BOD= 2y
∠AOC + ∠BOD = 2x + 2y
= 2 (x + y) = 2 × 90 = 180°
∴ The arcs APC and BQD joined together will make half the circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 41

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Question 8.
In the picture, A, B, C, D are points on a circle centred at O. The lines AC and BD are extended to meet at P. The line and BC intersect at Q. Prove that the angle which the small are AB makes at O is the sum of the angles it makes at P and Q.
Kerala SSLC Class 10 Maths Chapter 2 Circles 42
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 43

Textbook Page No. 59

Question 1.
Calcula te the angles of the quadrilateral in the picture and also the angles between their diagonals:
Kerala SSLC Class 10 Maths Chapter 2 Circles 44
Answer:
Since
∠ACD = 30°
∠ABD = 30°
(Angle in the same segment of a circle)
Since ZCBD = 45°
∠CAD = 45°
Since ZBDC = 50°
∠BAC = 50°
∠ABC + ∠ADC = 180 (cyclic quadrilateral)
∠ABC = 75°
∴ ∠ADC = 180 – 75 = 105°
Kerala SSLC Class 10 Maths Chapter 2 Circles 45
∠ADB = 105 – 50 = 55°
As, ∠BAD = 95°
∠DCB = 180 – 95 = 85°
∴ ∠ACB = 85 – 30 = 55°

Question 2.
Prove that any outer angle of a cyclic quadrilateral is equal to the inner angle at the opposite vertex.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 46
PQRS is cyclic Quadrilateral
∠PSR + ∠PQR = 180°
(sum of opposite angles)
∠PQR + ∠RQT = 180° (linear pair)
From this we get ∠PSR = ∠RQT

Question 3.
Prove that a parallelogram which is not a rectangle is not cyclic.
Answer:
PQRS is cyclic quadrilateral.
∠P + ∠R = 180
Also in a parallelogram opposite angles will be equal.
∠P + ∠R = 180°
∠P = ∠R = 90°
This means that PQRS must be a rectangle, otherwise, it is not cyclic.
Kerala SSLC Class 10 Maths Chapter 2 Circles 47

Question 4.
Prove that any non-isosceles trapezium is not cyclic.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 48
opposite angles are not supplementary.
∴ ABCD is not cyclic. Non-isosceles trapezium is not cyclic.

Question 5.
In the picture, bisectors of adjacent angles of the quadrilateral ABCD intersect at P, Q, R, S.
Kerala SSLC Class 10 Maths Chapter 2 Circles 49
Prove that PQRS is a cyclic quadrilateral.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 50
2x + 2y + 2z + 2w = 360°;
x + y + z + w = 180°
Δ DPC; ∠DPC = 180 – (w + z)
Δ ARB ; ∠ARB =180 – (x + y)
∠R + ∠P = 360 – (x + y + z + w) = 360 – 180 = 180°
In ΔBQC ZQ = m – (w + x)
In Δ ASD ∠S = 180 – (r + y)
Similarily ∠S +∠Q = 180.
PQRS is a cyclic quadrilateral.

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Question 6.
i) The two circles below intersect at P, Q and 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.
Kerala SSLC Class 10 Maths Chapter 2 Circles 51
ii) In the picture, the circle on the left and right intersect the middle circle at P, Q, R, S; the lines joining them meet the left and right circles at A, B, C, D. Prove that ABDC is a cyclic quadrilateral.
Kerala SSLC Class 10 Maths Chapter 2 Circles 52
Answer:
i.
Kerala SSLC Class 10 Maths Chapter 2 Circles 53
∴ ABCD is a cyclic quadrilateral.
Kerala SSLC Class 10 Maths Chapter 2 Circles 54
ABCD is a cyclic quadrilateral.

Question 7.
Kerala SSLC Class 10 Maths Chapter 2 Circles 55
In the picture, points P, Q, R are marked on the sides BC, CA, AB of AABC and the circumcircles of ΔAQR and ΔBRP are drawn. M is a point where these circles intersect.

Prove that the circumcircle of ΔCPQ also passes through M.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 56
Let M be a common point which three circles can passed.
Kerala SSLC Class 10 Maths Chapter 2 Circles 57
Therefore the circumcircle of ACPQ also passes through M

Textbook Page No. 67

Question 1.
In the picture, chords AB and CD of the circle are extended to meet at P.
Kerala SSLC Class 10 Maths Chapter 2 Circles 58
i. Prove that the angles of Δ APC and Δ PBD, formed by joining AC and BD, are the same.
ii. Prove that PA × PB = PC × PD.
iii. Prove that if PB = PD, then ABDC is an isosceles trapezium.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 59
i. As ABCD is a cyclic quadrilateral.
If ∠BAC = x° then
∠BDC = 180 – x ∠BDP = x°
If ∠ACD = y° then ∠PBD = y°
As ∠APC is common angle.
Angles of Δ APC and Δ PBD are same
Kerala SSLC Class 10 Maths Chapter 2 Circles 60
iii. If PB = PD ; AP= PC
In ABCD
ABDC is a cyclic quadrilateral, so their opposite angles are supplementary.
If AP = PC, in ∆ APC
∠A =∠C
Kerala SSLC Class 10 Maths Chapter 2 Circles 61
As AB = CD
AC || BD
Adjacent angles are supplementary
∴ ABCD will be an isosceles trapezium

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Question 2.
Draw a rectangle of width 5 centimetres and height 3 centimetres.
i. Draw a rectangle of the same area with width 6 centimetres.
ii. Draw a square of the same area.
Answer:
i. Draw a rectangle of length 5 cm and width 3cm.
Kerala SSLC Class 10 Maths Chapter 2 Circles 62
Extend AB up to 6cm.
(AE = 6cm) Draw an arc having radius as AE and A as centre. Extend DA and mark the point F.
Kerala SSLC Class 10 Maths Chapter 2 Circles 63
Extend BA towards left. Mark G as AD = AG
Draw Δ GFB.
Kerala SSLC Class 10 Maths Chapter 2 Circles 64
Circum circle of Δ GFB meets AD at D.
∴ AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.
Kerala SSLC Class 10 Maths Chapter 2 Circles 65

ii. Draw a rectangle of length 5cm and width 3cm. Area = 5 x 3 = 15 cm2.
Therefore side of a square will be √15.
Kerala SSLC Class 10 Maths Chapter 2 Circles 66
Draw a semicircle of diameter AH.
Extend BC, and mark the point F.
AB × BH = 5 × 3 = 15
AB × BH = BF2 ; BF = √5 cm
Area of BEGF = √15 × √15 = 15 cm2

Question 3.
Draw a square of area 15 square centimetres.
Answer:
Draw a rectangle of length 5cm and width 3cm. Area = 3 × 5 cm2 = 15 cm2. Side of the square is √15.Kerala SSLC Class 10 Maths Chapter 2 Circles 67
Draw a semicircle of diameter AH.
BC can touch the point F.
AB × BH = BF2 = √152 = 15 cm2.
Area of BEGF = 15 cm2.

Question 4.
Draw a square of area 5 square centimetres in three different ways. (Recall Pythagoras theorem)
Answer:
i. Draw a rectangle of length 5cm and width 1 cm.
Kerala SSLC Class 10 Maths Chapter 2 Circles 68
Draw a semicircle of diameter AE. Extend BC up to F. √5 is the side of the square BGHF.

ii. Draw a right-angled triangle of perpendicular sides 2 cm and 1cm.
Kerala SSLC Class 10 Maths Chapter 2 Circles 69
Hypotenuse will be y is cm. The area of the square ACDE is 5 cm2, because here we take the hypotenuse as sides of the square.

iii. Draw a right-angled triangle of hypotenuse 3 cm and one side 2cm.
Third side = \(\sqrt{3^{2}-2^{2}}=\sqrt{5} \mathrm{cm}\)
Draw a square having side BC, then area of the square BEDC will be 5cm2
Kerala SSLC Class 10 Maths Chapter 2 Circles 70

Question 5.
In the picture, a line through the centre of a circle cuts a chord into two parts:
Kerala SSLC Class 10 Maths Chapter 2 Circles 71
What is the radius of the circle?
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 72
The intersection may be with in the circle.
Chords AB & CD intersect at P i. e.,
AP × PB = CP × PD
(The intersection will be inside the circle)
AP × PB = CP × PD
4 × 6 = CP × (OP + OD)
24 = CP × (OP + OC)
24 = CP × (OP + OP + CP)
24 = CP × (5 + 5 + CP)
24 = CP × (10 + CP)
CP =2
Radius = CP + OP = 2 + 5 = 7cm

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Question 6.
In the picture, a line through the centre of a circle meets a chord of the circle:
Kerala SSLC Class 10 Maths Chapter 2 Circles 73
What are the lengths of the two pieces of the chord?
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 74
Kerala SSLC Class 10 Maths Chapter 2 Circles 75
CP= 13 – PB
Therefore equation (1)
(13 – PB)PB = 40
PB = 5, 8
If PB = 5cm then PC = 8cm
If PB = 8cm then PC = 5cm
In figure PB > PC
PB = 8 cm, PC = 5 cm

Circles Orukkam Questions & Answers

Worksheet 1

Question 1.
In triangle ABC, AB = 8cm, BC = 6cm , AC = 10cm.
1. What kind of triangle is this?
2. What is the position of B based on the circle with AC as the diameter? Why?
3. What is the position of A based on the circle with BC as the diameter? Why?
4. What is the position of the point C based on the circle with diameter AB?
Answer:
In Δ ABC
AB2 + BC2 = 82 + 62 = 64 + 36 = 100 = AC2
Δ ABC is a right angled triangle.
If we draw a circle taken in a AC as diameter, ∠B = 90°, Therefore the point B on the circle.
Δ ABC is a right-angled triangle.
If we draw a circle taking BC as diameter, ∠A < 90°, Therefore the position of point A will be outside the circle.
If we draw a circle taking AB as diameter, ∠C < 90°, Therefore the position of point C will be outside the circle.

Question 2.
Three vertices of a parallelogram are on a circle and the fourth vertex is at the center. Find the angles of the parallelogram.
Kerala SSLC Class 10 Maths Chapter 2 Circles 76
Mark a point P on the top of the figure on the circle, join AP and CP. If angle AP C = x then write? AOC
Write ABC?
Write ∠ABC + ∠APC ?
What is APC?
Find the angles
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 77
The angles of a parallelogram are 60, 120, 60 and 120.

Question 3.
In triangle ABC ,AB = AC, angle BAC = 30, BC = 5cm Find the radius of ABC
Draw the figure
Mark the center, BO and CO
Find the measure of angle BOC
Write the angles of triangle OBC
What kind of angle is triangle OBC
Write the radius of the circumcircle
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 78
∠OBC= 75 – 15 = 60 = OCB (∵OB = OC)
Angle of Δ OBC are: 60, 60, 60
ΔOBC is an equilateral triangle.
Radius of circum circle of Δ OBC = OB = OC = BC = 5 cm.

Question 4.
P QRS is cyclic.
∠P = 3x, ∠Q = y, ∠R = x, ∠ = 5 v
Find the angles
Draw circle , mark P, Q, R, S on it, complete PQRS Enter the given angles.
What is 3x + x? Find x
What is y + 5y? Find y
Find the angles
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 79
3x + x = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180″)
=> 4x = 180° => 4x = 45°
y + 5y = 180 (Sum of the opposite angles of a cyclic quadrilateral is 180°)
Kerala SSLC Class 10 Maths Chapter 2 Circles 80

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Question 5.
In the figure ABC D is a trapezium. If the vertices are on a circle, prove that it is an isosceles trapezium draw figure
What is ∠A + ∠C?
Kerala SSLC Class 10 Maths Chapter 2 Circles 81
What is ∠B + ∠C?
Write the relation between A and B. Write the conclusion.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 82
∠A + ∠C = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180°)
∠B + ∠C = 180° (AB || CD)
(Sum of the alternate angles of a cyclic quadrilateral is 180″)
∠A + ∠C = ∠B + ∠C
∴ ∠A = ∠B
∴ ABC’D is an isoceles trapezium

Question 6.
In ABC AB = AC . P is the midpoint of AB and Q is the midpoint of AC. Prove that BPQC is cyclic?
Draw figure. Mark PQ and complete BP QC?
Is PQ parallel to BC?
Note that ∠B = ∠C?
What is ∠C + ∠Q?
What is ∠B + ∠Q?
Write conclusion.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 83
PQ || BC
(The line joining midpoints of two sides of a circle will be parallel to the third side).
∠B = ∠C (1)
(∵ AB = AC are given )
∠C + ∠Q = 180° (2)
(∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°)
∠B + ∠Q = 180° .
From (1) and (2) we conclude that BPQC is an cyclic trapezium

Worksheet 2

Question 7.
Prove that ABCD given in the figure is cyclic
Kerala SSLC Class 10 Maths Chapter 2 Circles 84
Draw figure and mark PQ
If ∠BAP =xthen
what is ∠BQP?
Find ∠PQD
Find ∠PDC? Why?
What is ∠A + ∠C?
Answer:
i. Quadrilateral ABPQ is cyclic
If ∠A = x, then ∠BQP = 1 80 – x
If ∠B = y, then ∠APQ = 180 – y
Quadrilateral PQCD is cyclic, SO
∠QCD = 180 – x (∠DPQ = x )
∠PDC = 180 – x (∠PQC = y)
∴ ABCD is a cyclic quadrilateral.

Worksheet 3

Question 8.
In the figure AB, C Dare extended and intersect at P. If AB = 5, BP = 3, P D = 2 then find CD? Draw the figure.
Kerala SSLC Class 10 Maths Chapter 2 Circles 85
Write the relation between PA, PB, PC, PD
Find C D
Answer:
PA x PB = PCxPD
If CD = x, then
⇒ 8 × 3 = (2 + x)2 ⇒ 24 = 4 + 2x
⇒ 2x = 20 ⇒ x = \(\frac { 20 }{ 2 }\) = 10
∴ CD= 10

Question 9.
In the figure AB is the diameter and CD is parallel to the diameter. AB = 8cm,BD = 2cm, find CD
Kerala SSLC Class 10 Maths Chapter 2 Circles 86
Answer:
If we draw a perpendicular DP to AB from D. Then PAxPB = PD2.
Here PB = x, then PA = 8 – x.
Kerala SSLC Class 10 Maths Chapter 2 Circles 87
x(8 – x) = PD2, 22 = x2 + PD2
x(8 – x) = 4 – x2, 8x – x2 = 4 – x2
8x = 4, x = \(\frac { 1 }{ 2 }\)
Similarly, let us draw a perpendicular CQ to AB from C
AQ = ,PQ = 8– \(\left(\frac{1}{2}+\frac{1}{2}\right)\) = 7
CD = 7cm

Question 10.
Draw a rectangle of length 6cm and width 4cm. Draw another rectangle whose area equal to area of the first rectangle and one of the sides is 8cm.
Ans: Draw ABCD as in the given measurement. Mark E by extending AB 2cm more. AE = 8 will be 8cm. WithAas centre and AE radius draw an arc. This arc cut DA produced at F. Extend BA such that AD = AG and mark G. Draw triangle GF B and construct a circumcircle. The circle meets AD at H . Complete the rectangle AHIE
Kerala SSLC Class 10 Maths Chapter 2 Circles 88

Worksheet 4

Question 11.
Draw an equilateral triangle of height 3cm. What is the length of a side?
Kerala SSLC Class 10 Maths Chapter 2 Circles 89
Write the principle of construction. Students are advised to construct as in the steps given below.
Answer:
Draw a circle of radius 2cm and mark a diameter AB which is 4cm. Mark a point P from one end A is 3 cm apart on the diameter.
Draw a chord C D perpendicular to AB. Complete triangle C AD.
Using PA x PB = PD2, PD= √3 .
Now we get AD = AC = C D = 2√3
Height AP = 3cm.

Worksheet 5

Question 12.
In the figure, PA is a tangent and O is the centre of the circle. P A = 17, ∠OPA = 30° then calculate the radius of the circle and distance from centre to the point P Triangle OAP with 30°, 60°, 90° is right triangle. Using the property of this special right triangle find the radius and the distance.
Kerala SSLC Class 10 Maths Chapter 2 Circles 90
Answer:
Δ OAP is a right-angled triangle having sides 30°, 60°and 90°.
Length of side which is opposite to the angle 90°, is twice the side which is opposite to the angle 30°.
Length of side which is opposite to the
angle 60°, is √3 times of the side which is opposite to the angle 30°
That is radius of the circle , OA = \(\frac { 17 }{ √3 }\)
Distance from centre to the point P
Kerala SSLC Class 10 Maths Chapter 2 Circles 91

Worksheet 7

Question 13.
Draw a circle and construct 30°, 150° angles on it.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 92

Question 14.
Draw a circle and construct \(22 \frac{1}{2}^{0}\) on it.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 93

Question 15.
In triangle ABC the radius of the circumcircle is 6 cm, ∠A = 70°, ∠B = 80°. Construct the triangle
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 94

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Question 16.
Draw a rectangle of length 7cm, and width 5cm and construct a square whose area is same as the area of this rectangle.
Answer:
Draw a rectangle of length 7cm, and width 5cm. Area = 7 × 5 = 35 cm2.
Therefore length of one side of the square is √35.
Kerala SSLC Class 10 Maths Chapter 2 Circles 95
Draw a semicircle taking AH as diameter Extend BC, and mark a point F.
AB × BH = 7× 5 = 35
AB × BH = BF2 ;
BF = √35cm
Area of BEGF= √35 × √35 = 35 cm2

Question 17.
Draw a rectangle of one side 5cm, width 7cm. Construct another rectangle whose one side is 8cm and area equal to the area of the first rectangle.
Answer:
Draw a rectangle of length 7cm and width 5cm.
Kerala SSLC Class 10 Maths Chapter 2 Circles 96
Extend AB up to 8 cm (AE = 8cm) Draw an arc taking A as centre and AE as radius. Extend DA and mark the point F.
Kerala SSLC Class 10 Maths Chapter 2 Circles 97
Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB
Kerala SSLC Class 10 Maths Chapter 2 Circles 98
Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE and width AH.
Kerala SSLC Class 10 Maths Chapter 2 Circles 99

Question 18.
Draw a square of side 5cm and construct a rectangle having one side 7cm and area equal to area of the square.
Answer:
Draw a square of side 5cm.
Kerala SSLC Class 10 Maths Chapter 2 Circles 100
Extend AB to 7 cm
(AE = 7cm) Draw an arc taking A as centre and AE a radius. Extend DA and mark the point F.
Kerala SSLC Class 10 Maths Chapter 2 Circles 101
Elongate BA towards left, and mark G such that AD = AG.
Draw Δ GFB.
Kerala SSLC Class 10 Maths Chapter 2 Circles 102
Circumcircle of A GFB will meet side AD on H.
AG × AB = AF × AH.
That is area of the rectangle having length AB and width AD is equal to the area of rectangle having length AE a width AH.
Kerala SSLC Class 10 Maths Chapter 2 Circles 103

Question 19.
What is the position of the vertex of an equilateral triangle with opposite side as the diameter?
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 104
Angles of an equilateral triangle is 60° each. Position of the vertex of triangle with opposite side as the diameter is outside of the circle, because the angle is less than 90°.

Circles SCERT Questions & Answers

Question 20.
In the figure “ ABC is a right triangle
a. If a circle is drawn with AC as diameter find the position of B.
b. If a circle is drawn with BC as diameter, find the position of A. [ Score: 3 Time: 5 minutes]
Kerala SSLC Class 10 Maths Chapter 2 Circles 105
Answer:
a. On the circle (1)
∠B = 90°

b. Outside the circle (1)
Position of the vertex of an triangle with opposite side as the diameter is outside the circle, because. ∠A <9o°.

Question 21.
A circle is drawn with AB as diameter. Find the positions of the points C, D, E related to the circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 106 [ Score: 3 Time: 5 minute]
Answer:
C inside the circle (1)
∠C > 90°.
D on the circle (1)
∠D = 90°.
E outside the circle (i)
∠E <90°.

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Question 22.
In Δ ABC and Δ PQR, BC = QR, ∠ A = ∠P, ∠Q = 90°, QR = 5 cm, PQ = 12 cm.
Kerala SSLC Class 10 Maths Chapter 2 Circles 107
Find the diameter of the circumcircle of Δ ABC. [ Score: 4 Time: 6 minute]
Answer:
QR = BC (1)
∠A = ∠P (1)
PR Diameter of the circumcircle of Δ ABC (1)
Diameter = PR= \(\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{cm}\) .(1)

Question 23.
PQ and RS are two mutually prependicular chords of a circle. < QPR=50° find< PQS. [ Score: 3, Time: 6 minute]
Answer:
∠PRS = 90 – 50 = 40° (1)
∠PQS = 40° (1)
Kerala SSLC Class 10 Maths Chapter 2 Circles 108

Question 24.
O is the centre of the circle. If ∠BOC = 130° and ∠AOB = 110°. What is ∠AOC?
Find all angles of Δ ABC
Kerala SSLC Class 10 Maths Chapter 2 Circles 109 [ Score: 3, Time: 3 minute]
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 110

Question 25.
Find all angles of the hexagon ABCDEF
Kerala SSLC Class 10 Maths Chapter 2 Circles 111 [ Score: 4 Time:5 minute]
Answer:
∠ EFD = ∠EAD = 30°
∠ FE A = ∠FDA = 40°
∠FDE = ∠FAE = 35° (1)
∠BAC= ∠BDC = 45°
∠ABD= ∠ACD = 62°
∠ACB = ∠ADB= 35° (1)
∠A = 1480, ∠B = 100°
∠C = 97° ∠D= 155° (1)
∠E= 115° ∠F= 105° (1)

Question 26.
O is the centre of the circle and AB is a chord. AC is the bisector of ∠OAB. ∠OAB = 56°.
a. Prove that OC and AB are parallel,
b. Find ∠ABC and ∠OBE.
Kerala SSLC Class 10 Maths Chapter 2 Circles 112
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 113

Question 27.
O is the centre of the circle. AD and BC are perpendicular to XY. CB cuts the circle at E. Prove that –CE = AD.
Kerala SSLC Class 10 Maths Chapter 2 Circles 114
Answer:
∠AEB = 90° (Angle in a semi circle) ( 1)
∠AEC=90°
∴ AECD is a rectangle
∴ AD = CE (2)

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Question 28.
ABCD is a parallelogram. A, B, E, F are the points on a circle. ∠DEF = 80° Find out the angles of the quadrilateral AEFB.
Kerala SSLC Class 10 Maths Chapter 2 Circles 115. [ Score: 4, Time: 4 minute]
Answer:
∠AEF = 180 – 80 = 100° (1)
∠ABF = 180 – 100 = 80° (1)
∠A = 180 – 80 = 100° (1)
∠EFB = 180 – 100 = 80° (1)

Question 29.
Kerala SSLC Class 10 Maths Chapter 2 Circles 116
O is the centre of the circle. ? ∠OCA = x °.
a. Find ∠OAC
b. Prove that ∠OCA + ∠ABC = 90°
c. Prove that ∠ADC – ∠OCA = 90° [Score: 4, Time: 4 minute]
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 117

Circles Exam Oriented Questions & Answers

Short Answer Type Questions (Score 2)

Question 30.
In the figure AB is the diameter. PC is perpendicular to AB. PC = 6cm, PB = 3cm. Find the radius of the semi-circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 118
Answer:
AP × PB = PC2
AP × 3 = 62
AP = 36/3 = 12 .
AB = 12 + 3 = 15
ie Radius = 15/2 = 7.5 cm

Question 31.
In Δ PQR, ZP = 60°, ∠R = 30° find whether the vertex Q on the circle with PR as diameter.
Answer:
∠P + ∠R = 60 + 30 = 90°.
∴ ∠Q = 180 – 90 = 90°.
So point Q on the circle.

Question 32.
In Δ ABC, ∠A = 60°, ∠B = 70°. Find whether the vertex C is inside or outside the circle with AB as diameter.
Answer:
The vertex C is outside the circle Since ∠C = 180 – (60 + 70) = 50° ∠90°

Question 33.
In the diagram, the central angle of arc ABC is 100° and ∠OAD is 30°. Find ∠OCD.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 119
∠D = 50°
As ∠OAD
is an isosceles triangle.
∠ODA = 30°
∠ODC = 20°
∴∠OCD = 20°

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Question 34.
In the figure, find ∠PQB, O is the centre.
Kerala SSLC Class 10 Maths Chapter 2 Circles 120
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 121

Short Answer Type Questions (Score 3)

Question 35.
The central angle of arc ABC is 60°, then find out the following,
i) ∠D
ii) Central angle of arc AEC,
iii) ∠B
Kerala SSLC Class 10 Maths Chapter 2 Circles 122
Answer:
i. ∠D = 30°
ii. Then central angle of arc AEC = 300°.
iii. ∠B = 150°

Question 36.
Show that the arcs APC, BQD when joined make a semicircle?
Kerala SSLC Class 10 Maths Chapter 2 Circles 123
Answer:
∠ADC is the half of the central angle of arc APC
The central angle of are APC = 2 ∠ADC
The central angle of arc BQD = 2 ∠DAB
In triangle AOD, CD ⊥ AB, ∠AOD = 90°,
∠DAO + ∠ADO = 180 – 90° =90°,
ie, ∠DAB + ∠ADC = 90°
2( ∠DAB + ∠ADC ) = 90° × 2 = 180°

Question 37.
The central angle of the complementary arc of a circle is 40° more than 3 times the central angle of the arc. Find out the central angles of each arc?
Kerala SSLC Class 10 Maths Chapter 2 Circles 124
Answer:
x + 3x + 40 = 360
4x + 40 = 360
4x = 360 – 40 = 320
x = \(\frac { 320 }{ 4 }\) = 80
∴ central angle of arc
ABC = 80
central angle of arc ADC = 360 – 80 = 280°

Question 38.
ABCD in the diagram is a rectangle. Then find out the area of the circle.
Kerala SSLC Class 10 Maths Chapter 2 Circles 125.
Answer:
ABCD is a rectangle ∠B = ∠D = 90°
AC in the diameter of the circle
AC = \(\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10\)
∴ radius = 5cm
∴ Area of the circle = πr² = π × 5² = 25πcm²

Question 39.
In the figure, AD = 16cm, BD = 6cm, CD = 2cm. Find the length EF.
Kerala SSLC Class 10 Maths Chapter 2 Circles 126
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 127

Question 40.
In the given figure, O is the centre of the circle. If ∠OAP = 35° and ∠OBP = 40°, find the value of ∠x.
Kerala SSLC Class 10 Maths Chapter 2 Circles 128
Answer:
Join OP
Since OA = OP and
∠APO = ∠OAP=35°
Similarly, OB = OP and ∠OPB = ∠OBP = 40°
∠APB = 35°+ 40° = 75°
∠AOB = 2 × 750 = 150°

Long Answer Type Questions (Score 4)

Question 41.
In the figure find ∠APB,∠ABQ ; where O is the centre of the circle ∠OAP = 32° and  ∠OBP = 47° .
Kerala SSLC Class 10 Maths Chapter 2 Circles 129
Answer:
JoinOP.
In OAP, OA = OP = Radius
∠OAP = ∠OPA = 32°
In OPR, OB = OP = radius
∠OBP = ∠OPB =47°
∠APB = 32°+ 47°= 79°
∠AQB = 180° – 79°= 10°

Question 42.
Draw a line of √7 cm.
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 130

Question 43.
Ois the centre of the circle as shown in the figure.
Find ∠CBD.
Kerala SSLC Class 10 Maths Chapter 2 Circles 131
Answer:
Kerala SSLC Class 10 Maths Chapter 2 Circles 132
Takeapoint E on the circle, join AE and CE.
∠AEC= \(\frac { 100 }{ 2 }\) = 50°
∠AEC + ∠ABC = 180° (Opposite angles of a cyclic quadrilaterals)
∠ABC = 130°
∠ABC + ∠CBD = 180° (linearpair)
130°+ ∠CBD = 180°
∠CBD = 50°

Long Answer Type Questions (Score 5)

Question 44.
In the figure O is the centre of the circle. Central angle of arc AXB is 60°, arc CYD is 80°. Then find all the angles of ΔAPD.
Kerala SSLC Class 10 Maths Chapter 2 Circles 133
Answer:
Central angle of arc AXB = 60°
i.e., ∠AOB = 60°
i.e., ∠ADP = 30°
Central angle of arc CYD = 80°
i.e., ∠COD = 80°
i e., ∠DPA = 40°
i e., ∠APD = 180° – (30 + 40) = 110°
Angles of = 30°, 40°, 110°

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Question 45.
‘O’ is the centre of the circle ∠D = 80°, find the following measurements.
Kerala SSLC Class 10 Maths Chapter 2 Circles 134
a. ∠C
b. ∠ABC
c. ∠BAC
d. ∠F
Answer:
∠D = 80°
a. ∠C = 80° (∠D and ∠C are angled on a same arc So, both are equal)

b. ∠ABC = 90°
(AC is diameter, Angle of a hemisphere is right)

c. ∠BAC = 180 – (80 + 90)
= 180 – 170 = 100

d. ∠F= 180 – 80 = 100° (Opposite angles of a cyclic quadrilateral are equal)

Circles Memory Map

Kerala SSLC Class 10 Maths Chapter 2 Circles 135
Angle in a semicircle is right:
The angle made by any arc of a circle on the alternate arc is half the angle made at the centre.
Kerala SSLC Class 10 Maths Chapter 2 Circles 136
Kerala SSLC Class 10 Maths Chapter 2 Circles 137

All angles in an arc is equal:
Kerala SSLC Class 10 Maths Chapter 2 Circles 138
If AB, CD are two chords, then
PA × PB = PC × PD

The area of the rectangle formed of parts into which a diameter of a circle is cut by a perpendicular chord is equal to the area of the square formed by half the chord.
PA × PB = PC2
Kerala SSLC Class 10 Maths Chapter 2 Circles 139

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