NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

# NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 10 Circles Ex 10.5.

Ex 10.5 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 10.5 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.

AO = BO = AB
∆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 10.5 Class 9 Maths Question 3.
In figure, PQR = 100°, where P, Q and R are points on a circle with centre O. Find OPR.

Solution:
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 = 2PQR
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°

2OPR = 180° – 160° = 20°        [OPR = ORP]
OPR = 2/2 = 10°

Ex 10.5 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 10.5 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 10.5 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 10.5 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]
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 10.5 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.
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 10.5 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 10.5 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.
Similarly,
Adding (i) and (ii), we have
i.e., B, D and C are collinear points.
BC is a straight line. Thus, D lies on BC.

Ex 10.5 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.

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.

Ex 10.5 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.

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