NCERT Solutions for Class 9 Maths Chapter 9 Circles Ex 9.3
Ex 9.3 Class 9 Maths Question 1.
In figure, A, B and C are three
points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other
than the arc ABC, find ∠ ADC.
Solution:
Given a circle with centre O, such that ∠AOB = 60° and ∠BOC = 30°
∵ ∠AOB + ∠BOC = ∠AOC
∴ ∠AOC = 60° + 30° = 90°
Since, the angle subtended by an arc at the circle is half the angle subtended by it at the centre.
∴ ∠ADC = ½ (∠AOC) = ½ (90°) = 45°
Ex 9.3 Class 9 Maths Question 2.
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.
Solution:
We have a circle having a chord AB
equal to radius of the circle.
⇒ ∆AOB is an equilateral triangle.
Since, each angle of an equilateral triangle is 60°.
⇒ ∠AOB = 60°
Since, the arc ACB makes reflex ∠AOB = 360° – 60° = 300° at the centre of the circle and ∠ACB at a point on the minor arc of the circle.
Therefore,
∠ACB = ½ [reflex ∠AOB]
= ½ [300°] = 150°
Hence,
the angle subtended by the chord on the minor arc = 150°
Similarly, ∠ADB = ½ [∠AOB] = ½ × 60° = 30°
Hence, the angle subtended by the chord on the
major arc = 30°
Ex 9.3 Class 9 Maths Question 3.
In figure, ∠PQR = 100°, where P, Q and R are
points on a circle with centre O. Find ∠OPR.
The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
∴ reflex ∠POR = 2∠PQR
But ∠PQR = 100°
∴ reflex ∠POR = 2 × 100° = 200°
Since, ∠POR + reflex ∠POR = 360°
⇒ ∠POR = 360° – 200°
⇒ ∠POR = 160°
Since, OP = OR [Radii of the same circle]
∴ In ∆POR, ∠OPR = ∠ORP
[Angles opposite to equal sides of a triangle are equal]
Also, ∠OPR + ∠ORP + ∠POR = 180°
[Sum of the angles of a triangle is 180°]
⇒ ∠OPR + ∠ORP + 160° = 180°
⇒ 2∠OPR = 180° – 160° = 20° [∠OPR = ∠ORP]
⇒ ∠OPR = 20°/2 = 10°
Ex 9.3 Class 9 Maths Question 4.
In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.
Solution:
In ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°
⇒ 69° + 31° + ∠BAC = 180°
⇒ ∠BAC = 180° – 100° = 80°
Since, angles in the same segment are equal.
∴ ∠BDC = ∠BAC
⇒ ∠BDC = 80°
Ex 9.3 Class 9 Maths Question 5.
In figure, 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.
Solution:
∠BEC = ∠EDC + ∠ECD
[Exterior angle is equal to the sum of the opposite interior angles]
⇒ 130° = ∠EDC + 20°
⇒ ∠EDC = 130° – 20° = 110°
⇒ ∠BDC = 110°
Since, angles in the same segment are equal.
∴ ∠BAC = ∠BDC
⇒ ∠BAC = 110°
Ex 9.3 Class 9 Maths Question 6.
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.
Solution:
Since angles in the same segment of a circle are
equal.
∴ ∠BAC = ∠BDC
⇒ ∠BDC = 30°
We have ∠DBC = 70° [Given]
In ∆BCD, we have
∠BCD + ∠DBC + ∠CDB = 180° [Sum of
angles of a triangle is 180°]
⇒ ∠BCD + 70° + 30° = 180°
⇒ ∠BCD = 180° – 100° = 80°
Now, in ∆ABC,
AB = BC [Given]
∴ ∠BCA = ∠BAC [Angles opposite to
equal sides of a triangle are equal]
⇒ ∠BCA = 30° [∵ ∠BAC = 30°]
Now, ∠BCA + ∠ECD = ∠BCD
⇒ 30° + ∠ECD = 80°
⇒ ∠ECD = 80° – 30° = 50°
Ex 9.3 Class 9 Maths Question 7.
If diagonals of a cyclic quadrilateral are diameters
of the circle through the vertices of the quadrilateral, prove that it is a
rectangle.
Solution:
Since AC and BD are diameters.
⇒ AC = BD ….…(i)
[All diameters of a circle
are equal]
Also,
∠BAD = 90° [Angle formed in a
semicircle is 90°]
Similarly, ∠ABC = 90°, ∠BCD = 90° and ∠CDA = 90°
Now, in ∆ABC and ∆BAD, we have
AC = BD [From (i)]
AB = BA [Common side]
∠ABC = ∠BAD [Each equal to 90°]
∴ ∆ABC ≅ ∆BAD [By RHS congruence rule]
⇒ BC = AD [C.P.C.T.]
Similarly, AB = DC
Thus, the cyclic quadrilateral ABCD is such that
its opposite sides are equal and each of its angle is a right angle.
∴ ABCD is a rectangle.
Ex 9.3 Class 9 Maths Question 8.
If the non-parallel sides of a
trapezium are equal, prove that it is cyclic.
Solution:
We have a trapezium ABCD such that
AB ॥ CD and AD = BC.
Let us draw BE ॥ AD such that ABED is a parallelogram.
∵ The opposite angles and opposite
sides of a parallelogram are equal.
∴ ∠BAD = ∠BED ….…(i)
and AD = BE ….…(ii)
But AD = BC [Given] ….…(iii)
∴ From (ii) and (iii), we have BE =
BC
⇒ ∠BCE = ∠BEC …… (iv) [Angles opposite to equal sides of a
triangle are equal]
Now, ∠BED + ∠BEC = 180° [Linear pair]
⇒ ∠BAD + ∠BCE = 180° [Using (i) and (iv)]
i.e., A pair of opposite angles of a
quadrilateral ABCD is 180°.
∴ ABCD is cyclic.
⇒ The trapezium ABCD is cyclic.
Ex 9.3 Class 9 Maths Question 9.
Two circles intersect at two points B and C. Through
B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and
P, Q respectively (see figure). Prove that ∠ACP = ∠QCD.
Solution:
Since, angles in the same segment of a circle are equal.
∴ ∠ACP = ∠ABP …..…(i)
Similarly, ∠QCD = ∠QBD ….…(ii)
Since, ∠ABP = ∠QBD ….…(iii) [Vertically opposite angles]
∴ From (i), (ii) and (iii), we have
∠ACP = ∠QCD
Ex 9.3 Class 9 Maths Question 10.
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.
Solution:
We have ∆ABC, and two circles described with
diameter as AB and AC respectively. They intersect at a point D, other than A.
Let us join A and D.
∵ AB is a diameter.
∴ ∠ADB is an angle formed in a semicircle.
⇒ ∠ADB = 90° ….…(i)
Similarly, ∠ADC = 90° ..….(ii)
Adding (i) and (ii), we have
∠ADB + ∠ADC = 90° + 90° = 180°
i.e., B, D and C are collinear points.
⇒ BC is a straight line. Thus, D lies on BC.
Ex 9.3 Class 9 Maths Question 11.
ABC and ADC are two right-angled triangles with
common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
We have ∆ABC and ∆ADC such that they are having AC
as their common hypotenuse and ∠ADC = ∠ABC = 90°
∴ Both the triangles are in semi-circle.
Case I:
If both the triangles are in the same semi-circle.
⇒ A, B, C and D are concyclic. Join BD. DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.
⇒ ∠CAD = ∠CBD
Case II:
If both the triangles are not in the same semi-circle.
⇒ A, B, C and D are concyclic. Join BD. DC is a chord.
∴ ∠CAD and ∠CBD are formed in the same segment.
⇒ ∠CAD = ∠CBD
Ex 9.3 Class 9 Maths Question 12.
Prove that a cyclic parallelogram
is a rectangle.
Solution:
We have a cyclic parallelogram
ABCD. Since, ABCD is a cyclic quadrilateral.
∴ Sum of its opposite angles is
180°.
⇒ ∠A + ∠C = 180° …..…(i)
But ∠A = ∠C …..…(ii)
[Opposite angles of a parallelogram are equal]
From (i) and (ii), we have
∠A = ∠C = 90°
Similarly, ∠B = ∠D = 90°
⇒ Each angle of the parallelogram
ABCD is 90°.
Thus, ABCD is a rectangle.