Circle : Exercise 10.5 (Mathematics NCERT Class 9th)


Q. 1     In figure A, B and C are three points on a circle with centre O such that \angle BOC = 30º and \angle AOB = 60º. If D is a point on the circle other than the arc ABC, find \angle ADC.

11Sol.

Since are ABC makes \angle AOC = \angle AOB + \angle BOC = 60º + 30º = 90º at the centre of the circle and \angle ADC at a point on the remaining part of the circle.
Therefore \angle ADC = {1 \over 2}\left( {\angle AOC} \right) = {1 \over 2} \times 90^o = 45^o


Q. 2     A chord of a circle is equal to the radius of the circle. Find th eangle subtended by the chord at a point on the minor arc and also at a point on the major are.
Sol.

Let PQ be chord. Join OP and OQ.
It is given that PQ = OP = OQ (Since Chord = radius)
Therefore \Delta OPQ is equilateral.
 \Rightarrow \angle POQ = 60º13Since are PBQ makes reflex POQ = 360º – 60º = 300º at centre of the circle and \angle PBQ at a point in the minor arc of the circle.
Therefore \angle PBQ = {1 \over 2}\left( {reflex\,\angle POQ} \right)
 = {1 \over 2} \times 300^o = 150^o
Similarly, \angle PAQ = {1 \over 2}\left( {\angle POQ} \right) = {1 \over 2} (60º) = 30º
Hence, angle subtended by the chord on the minor are 150º and on the major chord = 30º.


Q.3      In figure \angle PQR = 100º, there P, Q and R are points on a circle with centre O. Find \angle OPR.
12Sol.

Since the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same are at a point on the circumference.
Therefore Reflex \angle POR = 2\angle PQR
 \Rightarrow Reflex \angle POR = 2 \times 100^\circ = 200^\circ
 \Rightarrow \angle POR = 360º – 200º = 160º
In \Delta OPR, OP = OR [Radii of the same circle]
 \Rightarrow \angle OPR = \angle ORP
[Angles opp. to equal sides are equal]
and \angle POR = 160º ......... (1) [Proved above]
In \Delta OPR,
\angle OPR + \angle ORP + \angle PQR = 180^\circ [Angle sum property]
 \Rightarrow
 \Rightarrow
 \Rightarrow 2\angle OPR = 20^\circ
 \Rightarrow \angle OPR = 10^\circ


Q.4      In figure \angle ABC = 69º, \angle ACB = 31º, Find \angle BDC.

14Sol.

In \Delta ABC,
\angle BAC + \angle ABC + \angle BCA = 180^o
 \Rightarrow \angle BAC + {69^o} + {31^o} = 180^\circ
 \Rightarrow \angle BAC = 180^o - \left( {69^o + 31^o} \right)
 = 180^o - 100^o = 80^o
Since angles in the same segment are equal
Therefore \angle BDC = \angle BAC = 80^o


Q.5      In figure A, B, C and D are four points on a circle. AC and BD intersect at a point E such that \angle BEC = 130º and \angle ECD = 20º.  Find \angle BAC.
remainingSol.

\angle CED + \angle CEB = 180º [Linear pair]
 \Rightarrow \angle CED + 130º = 180º
 \Rightarrow \angle CED = 180º – 130º
= 50º
In \Delta ECD, \angle EDC + \angle CED + \angle ECD = 180º
 \Rightarrow \angle EDC + 50º + 20º = 180º
 \Rightarrow \angle EDC = 180º – 50º – 20º
= 110º
 \Rightarrow \angle BDC = \angle EDC = 110º
Since angles in the same segment are equal
Therefore \angle BAC = \angle BDC = 110º.


Q.6     ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \angle DBC = 70º, \angle BAC is 30º, find \angle BCDFurther, if AB = BC, find \angle ECD.
Sol.

\angle BDC = \angle BAC [Angles in the same segment]
 \Rightarrow \angle BDC = 30º [Since \angle BAC = 30º (given]

16In \Delta BCD, we have
\angle BDC + \angle DBC + \angle BCD = 180º [Sum of \angle s of a \Delta ]
 \Rightarrow 30º + 70º + \angle BCD = 180º [Since \angle DBC = 70º, \angle BDC = 30º]
 \Rightarrow \angle BCD = 180º – 30º – 70º = 80º
If AB = BC, than \angle BCA = \angle BAC = 30º [Angles opp. to equal sides in a \Delta are equal]
Now , \angle ECD = \angle BCD - \angle BCE
= 80º – 30º = 50º
[Since \angle BCD = 80º (found above) and \angle BCE = \angle BCA = 30º]
Hence, \angle BCD = 80º and \angle ECD = 50º


Q.7     If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Sol.

Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quad. ABCD.
To prove : Quadrilateral ABCD is a rectangle.

15Solution : - Since all the radii of the same circle are equal
Therefore OA = OB = OC = OD

 \Rightarrow OA = OC = {1 \over 2}AC
and OB = OD = {1 \over 2}BD
 \Rightarrow AC = BD
Therefore the diagonals of the quadrilateral ABCD are equal and bisect each other.
 \Rightarrow Quadrilateral ABCD is a rectangle.


Q.8     If the non- parallel sides of a trapezium are equal prove that it is cyclic.
Sol.

Given : Non - parallel sides AD and BC of a trapezium are equal.
To prove : ABCD is a cyclic trapezium.
Construction : Draw DE  \bot AB and CF  \bot AB.

17Proof : In order to prove that ABCD is a cyclic trapezium it is sufficient to prove that \angle B + \angle D = 180º.
In \Delta s DEA and CFB, we have
AD = BC [Given]
\angle DEA = \angle CFB [Each = 90º]
and DE = CF [Distance between two|| lines is always equal]
Therefore by RHS criterion of congruence, we have
DEA \cong CFB
 \Rightarrow \angle A = \angle B\,\,and\,\,\angle ADE = \angle BCF
(Corresponding parts of congruent triangles are equal)

Now, \angle ADE = \angle BCF
 \Rightarrow 90º + \angle ADE = 90º + \angle BCF
 \Rightarrow \angle EDC + \angle ADE = \angle FCD + \angle BCF
[Since \angle EDC = 90º and \angle FCD = 90º]
 \Rightarrow \angle ADC = \angle BCD
 \Rightarrow \angle D = \angle C
Thus , \angle A = \angle B\,\,and\,\,\angle C = \angle D
Therefore , \angle A + \angle B + \angle C + \angle D = 360º
[Since sum of the angles of a quad. is 360º]
 \Rightarrow 2\angle B + 2\angle D = 360º
 \Rightarrow \angle B + \angle D = = 180º
Hence, ABCD is a cyclic trapezium.


Q.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 \angle ACP = \angle QCD

18Sol.

Since angles in the same segment are equal.
Therefore \angle ACP = \angle ABP ... (1)
and \angle QCD = \angle QBD ... (2)
Also \angle ABP = \angle QBD ... (3) [Vertically opp. angles]
Therefore From (1), (2) and (3) , we have
\angle ACP = \angle QCD


Q.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.
Sol.

Given : Two circles are drawn with sides AB and AC of \Delta ABC as diameters. The circle intersect at D.

20To prove : D lies on BC.
Construction : Join A and D.
Proof : Since AB and AC are the diameters of the two circles. [Given]
Therefore \angle ADB = 90º [Angles in a semi- circle]
and, \angle ADC = 90º [Angles in a semi-circle]
Adding we get \angle ADB + \angle ADC = 90º + 90º = 180º
 \Rightarrow BDC is a straight line.
Hence, D lies on BC.


Q.11    ABC and ADC are two right triangles with common hypotenuse AC. Prove that \angle CAD = \angle CBD.
Sol.

\Delta ABC and ADC are right \angle d with common hypotenuse AC. Draw a circle with AC as diameter passing through B and D. Join BD.
21Clearly, \angle CAD = \angle CBD [Since Angles in the same segment are equal]


Q.12    Prove that a cyclic parallelogram is a rectangle.
Sol.

Given : ABCD is a parallelogram inscribed in circle.
To prove : ABCD is a rectangle.
Proof : Since ABCD is a cyclic parallelogram.
22Therefore \angle A + \angle C = 180º ... (1)
But \angle A = \angle C ... (2)
From (1) and (2), we have
\angle A = \angle C = 90º
Similarly, \angle B = \angle D = 90º
Therefore Each angle of ABCD is of 90º
Hence, ABCD is a rectangle.



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