Students often refer to SSLC Maths Textbook Solutions and Class 10 Maths Chapter 2 Circles and Angles Important Extra Questions and Answers Kerala State Syllabus to clear their doubts.
SSLC Maths Chapter 2 Circles and Angles Important Questions and Answers
Circles and Angles Class 10 Extra Questions Kerala Syllabus
Circles and Angles Class 10 Kerala Syllabus Extra Questions
Question 1.
In the given figure BC is the diameter of a circle and AB = AC. Then ∠ABC is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
45°
Question 2.
O is the centre of the circle and ∠ACB = 30°. What is ∠AOB?
(a) 30°
(b) 15°
(c) 60°
(d) 90°
Answer:
60°
Question 3.
O is the center of a circle. If ∠OAB = 40° and C is a point on the circle then Z ACB is
(a) 40°
(b) 50°
(c) 80°
(d) 100°
Answer:
50°
Question 4.
A and B are the points on a circle. Pisa point on the alternate segment of arc AB. If ∠ACB = 30° then what is the central angle of small arc AB.
(a) 40°
(b) 50°
(c) 30°
(d) 60°
Answer:
60°
Question 5.
In the figure AB = AC = AD, ∠BAC = 40°. What is ∠BDC?
Answer:
20°
Question 6.
In the figure radius OA is perpendicular to radius OB. Radius of the circle is 10 cm
(a) What is the measure of ∠ACB
(b) What is the length AB?
Answer:
(a) 45°
(b) 10√2
Question 7.
In the figure AB is the diameter of a circle. ∠Q is \(\frac{1}{3}\) of of ∠APB.
(a) What is the measure of ∠APB?
(b) What is the length AQB?
Answer:
(a) 90°
(b) \(\frac{1}{3}\) × 90 = 30°
Question 8.
In triangle ABC. angles are in the ratio 1: 2: 3. Angle B is the largest and A is the smallest angle.
(a) What are the angles?
(b) What is the position of A based on the circle with diameter BC?
Answer:
(a) Smallest angle = \(\frac{1}{6}\) × 180 = 30°
∠A =30°, ∠B = 90°, ∠C = 60°
(b) On the circle
Question 9.
The vertices of a quadrilateral ABCD are on a circle. ∠A = x, ∠B = 3y, ∠C = 4x, ∠D = 2y
(a) Find x, y
(b) Find the angles of ABCD
Answer:
(a) x + 4x = 180 ⇒ 5x = 180, x = 36
Angles are 36°, 144°
(b) 2y + 3y = 180 ⇒ 5y = 180,y = 36°
Angles are 72°, 108°
Question 10.
ABCD is a trapezium in which AB is parallel to CD and AD = BC.
Prove that ABCD is a cyclic quadrilateral
Answer:
AD = BC ⇒ ∠A = ∠B
Since AB is parallel to CD, ∠B + ∠C =180°
Therefore ∠A + ∠C = 180°
ABCD is cyclic
Question 11.
The comers of a trapezium are on a circle.
Prove that it is an isosceles trapezium .
Answer:
ABCD is a trapezium in which AB is parallel to CD and vertices are on the circle.
∠B + ∠C = 180°, ∠A + ∠C = 180°
That is ∠A = ∠B. Base angles of a trapezium are equal. ABCD is isosceles.
Question 12.
Can all parallelogram be cyclic ?Justify your answer
Answer:
No. For a parallelogram opposite angles are equal
If it is cyclic opposite angle sum is 180°.
That is, opposite angles are 90° each.That is it is a rectangle.
That is, cyclic parallelogram is a rectangle.
Question 13.
O is the centre of the circle, PQ is parallel to OA, PR is parallel to OB
(a) If the central angle of arc ASB is 40° then what is the measure of ∠QPR?
(b) What is the central angle of arc QSR?
Answer:
(a) ∠QPR= 40°.(Angle between two lines is equal to angle between the lines parallel to them)
(b) Central angle of arc QSR is 80°
Question 14.
Draw a right triangle having longest side 7 cm and one of the other sides is 3 cm
Answer:
See the rough diagram
Question 15.
Angles of the quadrilateral ABCD are in the ratio 1: 2: 4: 3 in the order.
a) What are the angles?
b) Is this a cyclic quadrilateral? Why?
Answer:
(a) Take the angles as x, 2x, 4x and 3x
x + 2x + 4x + 3x = 360,
10x = 360,
x = 36
Angles are 36°, 72°, 144°
(b) 36 + 144 = 360, 72 + 108 = 180.
Opposite angle sum is 180°. So it is cyclic
Question 16.
O is the centre of the circle with diametre AB. Another circle is drawn with AO as the diametre
(a) What are the measure of ∠APO, ∠ACB
(b) Outer circle has radius 5 cm and BC = 8 cm. What is the length OP?
(c) Is AP = PC? Why?
(d) What is the length of AC?
Answer:
(a) 90°. Reason ∠APO, ∠ACB are the angles in the semicircle
(b) Triangle APO, triangle ACB are similar.
\(\frac{A O}{A B}=\frac{O P}{B C}\)
\(\frac{5}{10}=\frac{O P}{8}\), OP = 4 cm
(c) AC is the chord of big circle. OP is perpendicu-lar from centre to this chord. OP bisect AC. Therefore
AP = PC
(d) AC = \(\sqrt{10-8}\) = 6 cm.
Question 17.
Sides of triangle ABC are AB = 5 cm, AC = 12 cm, BC = 13 cm
(a) What kind of triangle is this ?
(b) What is the position of A based on the circle with diametre BC?
(c) What is the position of C based on the circle with diametre ZB?
(d) What is the position of B based on the circle with diametre AC?
Answer:
(a) Right triangle
(b) On the circle
(c) Outside the circle
(d) Outside the circle
Question 18.
In the figure AB is the diametre of a semicircle.Three angles x, y, z are marked outside. on the semicircle and inside the semicircle.
(a) What is the value of y?
(b) If x, y, z are in an arithmetic sequence,then what is x + z ?
(c) If the common difference of the sequence is 50 then find x and z
Answer:
(a) y = 90°
(b) x + z = 2 × 90 = 180° (Refer the properly of arithmetic sequence)
(c) y = 90, x = 90 – 50 = 40°, z = 90 + 50 = 140°
Question 19.
Draw a circle of radius 3 cm .Construct the angles 30° and 150° with vertices on the circle using compasses and scale only.
Answer:
- Draw a circle of radius 3 cm .Mark the center of the circle as O
- Mark a point A on the circle. Draw the radius OA.
- With A as the center and OA as radius, draw an arc which cut the circle at B. Join OB, ∠AOB = 60°
- Mark a point P on the complement of the arc AB, which makes 60° at the center. ∠APB = \(\frac{1}{2}\) × 60 = 30°
- Mark a point Q on the arc AB. ∆AQB = 180 – 30 = 150°
Question 20.
In the figure ∆ABC, ∆AOC, ∆ADC are in an arithmetic sequence
(a) What is the relation between angle ABC and angle AOC
(b) What is the relation between angle ABC and ADC
(c) Find the measure of these angles
Answer:
(a) ∆AOC = 2 × ∆ABC
(b) ∆ABC + ∆ADC = 180°
(c) Let ∆ABC = x, ∆AOC = y, ∆ADC = z
x, y, z are in an arithmetic sequence. Therefore 2y = x + z
From the relations noted above
y = 2x,
x + z = 180
2y = x + z
2 × 2x = x + y = 180
4x = 180 x = 45
x = 45°,y = 90°, z = 135°
∠ABC = 45°, ∠AOC = 90°, ∠ADC = 135°
Question 21.
ABCD is a square. The diagonals AC and BD intersect at O.
(a) What is the measure of angle AOD?
(b) What is the measure of angle APD?
(c) What is the measure of angle AQD
Answer:
(a) Diagonals of a square are perpendicular to each other. ∠AOD = 90°
(b) ∠APD = 45°
(c) ∠AQD = 180 – 45 = 135°
Question 22.
In the figure O is the centre of the circle, ∠BAO = 20°, ∠BCO = 10°
(a) What is the measure of angle ∠ABC?
(b) What is the measure of angle ∠AOC?
(c) What is the measure of angle ∠ADC?
(d) Find the angles of triangle ∠AOC
(e) If the diametre of the circle is 10 cm then find the length of the chord AC
Answer:
(a) In triangle OAB. OA = OB. Angles opposite to the equal sides are equal. Similarly in the case of triangle OBC also.
∠ABC = 20 + 10 = 30°
(b) ∠AOC = 2 × 30 = 60°
(c) ∠ADC= 180 – 30 = 150°
(d) Triangle AOC, OA = OC. ∠OAC = ∠OCA = \(\frac{180-60}{2}\) = 60°
∆OAC is an equilateral triangle. Angles are 60° each.
(e) OA = AC = OC = 5cm, radius 5cm.
Question 23.
In the figure O is the centre of the circle. If angle ADC = 140°, angle 4EC= 60°then B
(a) What is the measeure of ∠APC and ∠AQC
(b) What is the measure of angle AOC?
(c) Fnd the angles of the quadrlateral PEQB
Answer:
(a) ∠APC= 180-140 = 40°, ∠AQC = 40°
(b) ∠AOC = 2 × 40 = 80°
(c) In the quadrilateral ∠AEQ = ∠AEC = 60°, ∠EPB = 180 – 40 = 140°, ∠EQB = 140° ∠PBQ = 360 – (140 + 140 + 60) = 20.
Angles are 140°, 60°, 140°, 20°
Question 24.
In the figure O is the centre of the circle, then
(a) What kind of triangle is OAC?
(b) What is the measure of angle ABC?
(c) What is the measure of angle ADC?
(d) If the radius of the circle is 6 cm then what is the length of the chord AC.
Answer:
(a) OA = OC, ∠OAC = ∠OCA = 45°, ∠AOC = 90° ∆OAC is an isosceles right triangle
(b) ∠ABC = – AOC = 45°
(c) ∠ADC = 180 – 45 = 135°
(d) AC = \(\sqrt{6^2+6^2}=6 \sqrt{2}\) cm
Question 25.
ABC is an isosceles triangle with AB = AC, ∠ABC= 50°.
(a) Name two cyclic quadrilaterals in this picture.
(b) What is the measure of angle D?
(c) What is the measure of ∠BEC?
Answer:
(a) Quadrilateral ABEC and quadrilateral DBESare cyclic.
(b) ∠ABC = ∠ACB = 50“
∴ ∠A = 180 – 100 = 80°
∠D = 80°
(c) ∠BEC = 180 – 80 = 100°
Question 26.
ABCD is a cyclic quadrilateral. AB is the diametre of the circle. AD = CD and ∠ADC = 130°.
(a) What is the measure of ∠ACB?
(b) What is the measure of ∠ABC?
(c) Find ∠DCB.
(d) What is the measure of ∠BAD?
Answer:
(a) ∠ACB = 90° (Angle in the semicircle)
(b) ∠ABC = 180 – 130 = 50°
(c) Since CD = AD,ihe angles opposite to the equal sides of triangle ADC are equal.
∠DCA = 25°, ∠DCS = 90 + 25 = 115°
(d) ∠BAD = 180 – 115 = 65°
Question 27.
Prove that any cyclic parallelogram is a rectangle.
Answer:
ABCD is a parallelogram .(Draw rough figure)
Opposite angles are equal.
∠A = ∠C, ∠B = ∠D
Sum of the opposite angles is 180°
∠A + ∠C = 180°, ∠A – ∠C
∴ ∠A = 90°, ∠C = 90°
∠B + ∠D = 180°, ∠B = ∠D
∠B = 90°, ∠D = 90°
ABCD is a square
Question 28.
In triangle ABC, AB = AC. P and Q are the mid points of the side AB and AC.
(a) Draw a rough diagram and join the points P and Q.
(b) Prove that BPQC is a cyclic quadrilateral.
(c) If ∠A in triangle ABC is 20°, find the angles of the trapezium BPQC
Answer:
(a) Figure
(b) Since AB = AC, ∠B = ∠C .
Line joining the mid points of two sides of a triangle is parallel to the third side. PQ is parallel to SC.
In PBCQ, ∠B + ∠P = 180° (co interior angles)
Since ∠C = ∠S, ∠C + ∠P = 180°
PQCB is a cyclic quadrilateral
(c) ∠A = 20°
∠B = ∠C = \(\frac{180-20}{2}\) = 80°
∠B + ∠P = 180°, ∠S = 100°, ∠Q = 100°
Angles are ∠P = 100°, ∠Q = 100°,
∠B = 80°, ∠C = 80°
Question 29.
In the figure SD = CD, ∠DSC = 25°
(a) What is the measure of ∠BDC?
(b) What is the measure of ∠BAC?
(c) What is the measure of ∠EBC?
Answer:
(a) In triangle BDC, SD = CD.
Angle opposite to these sides are equal. ∠BCD = 25°
∠BOC = 180 – (25 + 25) = 130°
(b) ∠BAC= 180 – 130 = 50°
(c) ∠EBC = ∠BAC = 50 °, ∠ESC
= 180 – (90 + 50)
= 180 – 140
= 40°
Question 30.
Two circles intersect at S and E as in the figure. The points A – B – C are along a line. Also the points D – E – F are also on a line
(a) Prove that AD is parallel to CF
(b) If AC = DF suggest a suitable name to the quadrilateral ADFC
(c) Prove that ADFC is a cyclic quadrilateral.
Answer:
(a) Draw BE. ABED is a cyclic quadrilateral.
If ∠DAB = x then ∠BED = 180 – x,
∠BEF = 180 – (180 – x) = x.
BEFC is cyclic. ∠C = 180 – x.
In quadrilateral ADFC,
∠A + ∠C = x + 180 – x = 180°
Co interior angle sum is 180°.
AD is parallel to CF
(b) ADFC is a trapezium.
Since AC = DF is an isosceles trapezium.
(c) Angles at the ends of parallel sides of an isosceles trapezium are equal.
Since ∠A = ∠D and ∠A + ∠C = 180° then ∠D + ∠C = 180°
ADFC is a cyclic quadrilateral.
Question 31.
AB is the diametre of the circle. CD is a chord of length equal to radius of the circle.
(a) What is the measure of ∠COD?
(b) What is the measure of ∠CBD?
(c) What is the measure of ∠BCF?
(d) Find the measure of ∠CPD
Answer:
(a) Draw OC, OD, OCD is an equilateral traingle.
∠COD = 60°
(b) ∠CBD = \(\frac{1}{2}\) × 60 = 30°
(c) ∠BCA = 90° (angle in the semicircle).
∴ BCP = 90°.
(d) In traingle BCP,
∠CPD = ∠CPB
= 180-(90 + 30)
= 60°