These NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Exercise 10.5
In Fig. 10.36, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D In a point on the circle other than the arc ABC, find ∠ADC.
We have given,
∠BOC = 30° and ∠AOB = 60°
∠AQC = 30° + 60°
∴ ∠AOC = 90° …….(i)
Again, ∠AOC and ∠ADC subtended by same arc AC.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠AOC = 2∠ADC
⇒ 90° = 2∠ADC (From (i))
⇒ ∠ADC = 45°
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
In circle whose centre is O. AB is a chord equal to its radius.
∴ OAB is a an equilateral triangle.
We know that each angle of an equilateral triangle is 60°.
∴ AOB = 60°
Now, ∠AOB = 2∠AQB
(The angle subtended by an arc at the centre is double the angle subtanded by it at any point on the remaining part of the circle)
⇒ 60° = 2∠AQB
⇒ ∠AQB = 30°
Again, QAPB is a cyclic quadrilateral.
∴ ∠AQB + ∠APB = 180° (Sum of opposite angles, of a cyclic quadrilateral is 180°)
⇒ 30 + ∠APB = 180°
⇒ ∠APB = 180° – 30° = 150°
Therefore, angle subtended by the chord at a point on the minor arc is 150° and at a point on major arc is 30°.
In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
We have given ∠PQR = 100°.
Since, angle subtended by an arc at the centre is double tire angle subtended by it at any point on the remaining part of the circle.
∴ ∠1 = 2∠PQR
⇒ ∠1 = 2 × 100°
⇒ ∠1 = 200°
Again, ∠1 + ∠2 = 360°
⇒ 200° + ∠2 = 360° (∵ ∠1 = 200°)
⇒ ∠2 = 160°
Now, In ∆POR
OP = OR (Radii of circle)
∴ ∠OPR = ∠ORP (Angle opposite to equal sides of ∆ are equal)
∠OPR + ∠ORP + ∠2 = 180 (Angle sum property of triangle)
⇒ ∠OPR + ∠OPR + 160 = 180
⇒ 2∠OPR = 20
⇒ ∠OPR = 10°
In Fig. 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
∠ABC = 69° and ∠ACB = 31°
∴ ∠ABC + ∠ACB + ∠BAC = 180° (Angle sum property of ∆)
⇒ 69°+ 31° + ∠BAC = 180°
⇒ ∠BAC = 80°
We know that angles in the same segment of a circle are equal.
∴ ∠BAC = ∠BDC
∠BDC = 80° (∵ ∠BAC = 80°)
In Fig. 10.39, A, B, C and D are four points on a Circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
We have, ∠BEC = 130° and ∠ECD = 20°
Again, ∠BEC + ∠CED = 180° (linear pair)
⇒ 130° + ∠CED = 180°
⇒ ∠CED = 50°
Now, In triangle CDE
∠CED + ∠ECD + ∠CDE = 180° (Sum of all angles of a ∆)
⇒ 50° + 20° + ∠CDE = 180
⇒ ∠CDE = 110°
We know that angles on the same segments of a circle are equal.
∴ ∠CDB = ∠BAC
or, ∠BAC = 110° (∵ ∠CDB = 110°)
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
We know that angles on the same segment of a circle are equal.
∴ ∠DAC = ∠DBC
or, ∠DAC = 70° (∵ ∠DBC = 70°)
Again, ABCD is a cyclic quadrilateral.
∴ ∠DAB + ∠BCD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
100° + ∠BCD = 180° (∵∠DAB = ∠DAC + ∠BAC)
⇒ ∠BCD = 80°
Now, in triangle ABC,
AB = BC (Given)
∴ ∠BAC = ∠BCA (Angle opposite to equal sides are equal)
or, ∠BCA = 30°
Again, ∠BCD = ∠BCA + ∠ACD
⇒ 80° = 30° + ∠ACD
⇒ ∠ACD = 50°
∴ ∠ECD = 50°
If diagonals of a cyclic quadrilateral are diameter of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Given: ABCD is a cyclic quadrilateral in which diagonal AC and BD are diameter of circle.
To prove that: ABCD is a rectangle.
Proof: In ABCD,
AC = BD (Diameter of circle are equal)
Again, ∠ABC = 90° (Angle in a semicircle is a right angle)
Similarly, ∠BAD = ∠ADC = ∠BCD = 90°
Now, we know that, if a quadrilateral has both diagonals are equal and each angle is 90° then the quadrilateral is a rectangle.
∴ ABCD is a rectangle.
If the non parallel sides of a trapezium are equal, prove that it is cyclic.
Given: A trapezium ABCD in which AB || CD and AD = BC.
To prove that: ABCD is a cyclic trapezium.
Construction: Draw DE ⊥ AB and CF ⊥ AB.
Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that ∠B + ∠D = 180°.
In triangles DEA and CFB, we have
AD = BC (Given)
∠DEA = ∠CFB (Each equal to 90°)
DE = CF (Distance between two parallel lines is always same)
So, by S- A-S congruency condition
∆DEA ≅ ∆CFB
∴ ∠A = ∠B and ∠ADE = ∠BCF (By CPCT)
⇒ 90° + ∠ADE = 90° + ∠BCF (Add 90° both side)
⇒ ∠EDC + ∠ADE = ∠FCD + ∠BCF (∵ ∠EDC = 90° and ∠CD = 90°)
⇒ ∠ADC = ∠BCD
⇒ ∠D = ∠C
Thus, ∠A = ∠B and ∠C = ∠D
∴ ∠A + ∠B + ∠C + ∠D = 360° (Sum of the angles of a equal is 360°)
⇒ 2∠B + 2∠D = 360°
⇒ ∠B + ∠D = 180°
Hence ABCD is a cyclic quadrilateral.
Two circles intersect at two points B and C. Through B, two line segment ABD and PBQ are drawn to intersect the circle at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.
We know that angles in the same segment of a circle are equal.
∠D = ∠Q and ∠P = ∠A
Again, In ∆PQC,
∠P + ∠Q + ∠PCQ = 180° ……(i)
(Sum of all three angles of the triangle is 180°)
Now, In ∆ADC,
∠A + ∠D + ∠ACD = 180° …….(ii)
(Sum of all three angles of a triangle is 180°)
From equation (i) and (ii)
∠P + ∠Q + ∠PCQ = ∠A + ∠D + ∠ACD
⇒ ∠P + ∠Q + ∠PCQ = ∠P + ∠Q + ∠ACD (∵ ∠A = ∠P and ∠D = ∠Q prove above)
∴ ∠PCQ = ∠ACD
⇒ ∠PCQ – ∠PCD = ∠ACD – ∠PCD (Subtract ∠PCD both side)
⇒ ∠QCD = ∠ACP
⇒ ∠ACP = ∠QCD
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Since angle in a semicircle is right angle.
∠ADB = 90° and ∠ADC = 90°
⇒ ∠ADB + ∠ADC = 90° + 90°
⇒ ∠ADB + ∠ADC = 180°
⇒ ∠BDC = 180°
Therefore, BDC is a straight line or, D must be lie on BC.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Given: ABC and DBC are two right angle triangles on same base AC.
To prove that: ∠CAD = ∠CBD
Construction: Take AC as a diameter and draw a circle and join BD.
Proof: We know that angle in a semicircle is a right angle.
Therefore, vertex B of ∆ABC and vertex D of ∆ADC must be lie on the circumference of the circle.
Again we know that angles in the same segment of a circle are equal.
∴ ∠CAD = ∠CBD.
Prove that a cyclic parallelogram is a rectangle.
Let ABCD be acyclic parallelogram. In order to prove that it is a rectangle, it is sufficient to show that one of the angles of parallelogram ABCD is a right angle.
Now ABCD is a parallelogram.
∠B = ∠D ……(i) [∵ opposite angle of a || gm are equal]
Also, ABCD is a cyclic quadrilateral
∴ ∠B + ∠D = 180° ……(ii) (Sum of opposite angles of cyclic quadrilateral)
⇒ ∠B + ∠B = 180° [∵ ∠B = ∠D from (i)]
⇒ 2∠B = 180°
⇒ ∠B = 90°
Hence ABCD is a rectangle.