NCERT Class 9 Maths Solutions Chapter - 10 Circles, Ex - 10.5

Ex - 10.5

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


Solution

We may observe that
    AOC = AOB + BOC
        = 60^o + 30^o
        = 90^o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

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

In OAB
    AB = OA = OB = radius
OAB is an equilateral triangle.

So, each interior angle of this triangle will be of 60^o

AOB = 60^o

Now,  

In cyclic quadrilateral ACBD
ACB + ADB = 180^o        (Opposite angle in cyclic quadrilateral)
ADB = 180^o - 30^o = 150^o
So, angle subtended by this chord at a point on major arc and minor arc are 30^o and 150^o respectively.

Question 3.  In the given figure, PQR = 100^o, where P, Q and R are points on a circle with centre O. Find OPR.

Solution

Consider PR as a chord of circle.
Take any point S on major arc of circle.
Now PQRS is a cyclic quadrilateral.

PQR + PSR = 180^o                    (Opposite angles of cyclic quadrilateral)
PSR = 180^o - 100^o = 80^o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.
 POR = 2PSR = 2 (80^o) = 160^o

In POR
OP = OR                                        (radii of same circle) 

 OPR = ORP                         (Angles opposite equal sides of a triangle)

OPR + ORP + POR = 180^o    (Angle sum property of a triangle)

2 OPR + 160^o= 180^o
2 OPR = 180^o - 160^o = 20^o

OPR = 10^o 

Question 4.  In the given figure, ABC = 69^o, ACB = 31^o, find BDC.

Solution

In ABC

BAC + ABC + ACB = 180^o     (Angle sum property of a triangle)
BAC + 69^o + 31^o = 180^o

BAC = 180^o - 100º
BAC = 80^o

BDC = BAC = 80^o                    (Angles in same segment of circle are equal)

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

Solution

In CDE
CDE + DCE = CEB        (Exterior angle)
CDE + 20^o = 130^o

CDE = 110^o
But BAC = CDE               (Angles in same segment of circle)
BAC = 110^o

^{Question 6.}  ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If DBC = 70^o, BAC is 30^o, find BCD. Further, if AB = BC, find ECD.

Solution

For chord CD

CBD = CAD                    (Angles in same segment)

CAD = 70^o

BAD = BAC + CAD = 30^o + 70^o = 100^o
BCD + BAD = 180^o        (Opposite angles of a cyclic quadrilateral)
BCD + 100^o = 180^o
BCD = 80^o
In ABC

AB = BC                               (given)
 BCA = CAB               (Angles opposite to equal sides of a triangle)
BCA = 30^o
We have BCD = 80^o
BCA + ACD = 80^o
30^o + ACD = 80^o

ACD = 50^o

ECD = 50^o

^{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

Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.

        (Consider BD as a chord)

    BCD + BAD = 180^o            (Cyclic quadrilateral)
    BCD = 180^o - 90^o = 90^o

            (Considering AC as a chord)

   ADC + ABC = 180^o            (Cyclic quadrilateral)
    90^o + ABC = 180^o
    ABC = 90^o
    Here, each interior angle of cyclic quadrilateral is of 90^o. Hence it is a rectangle.

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

Solution

Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM  CD and BN  CD
In AMD and BNC
 AD = BC                                 (Given)
AMD = BNC                      (By construction each is 90^o)
AM = BM    (Perpendicular distance between two parallel lines is same)
AMD  BNC              (RHS congruence rule) 

ADC = BCD                   (CPCT)    ... (1)

BAD and ADC are on same side of transversal AD

BAD + ADC = 180^o                ... (2)  
BAD + BCD = 180^o          [Using equation (1)]
This equation shows that the opposite angles are supplementary.
So, ABCD is a cyclic quadrilateral.

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 the given figure). Prove that ACP = QCD

Solution

    Join chords AP and DQ 
    For chord AP 
    PBA = ACP         (Angles in same segment)        ... (1)  
    For chord DQ
    DBQ = QCD         (Angles in same segment)        ... (2)   
    ABD and PBQ are line segments intersecting at B.
    PBA = DBQ         (Vertically opposite angles)        ... (3)     
    From equations (1), (2) and (3), we have
    ACP = QCD

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

Consider a ABC
Two circles are drawn while taking AB and AC as diameter.
 Let they intersect each other at D and let D does not lie on BC.
 Join AD 
    ADB = 90^o            (Angle subtend by semicircle)
    ADC = 90^o            (Angle subtend by semicircle)
    BDC = ADB + ADC = 90^o + 90^o = 180^o 
 Hence BDC is straight line and our assumption was wrong.
 Thus, Point D lies on third side BC of ABC

Question 11.  ABC and ADC are two right triangle with common hypotenuse AC. Prove that CAD = CBD.

Solution

In ABC

ABC + BCA + CAB = 180^o    (Angle sum property of a triangle)
 90^o + BCA + CAB = 180^o
 BCA + CAB = 90^o        ... (1)
In ADC

CDA + ACD + DAC = 180^o    (Angle sum property of a triangle)
 90^o + ACD + DAC = 180^o
 ACD + DAC = 90^o        ... (2)
Adding equations (1) and (2), we have

BCA + CAB + ACD + DAC = 180^o

 (BCA + ACD) + (CAB + DAC) = 180^o BCD + DAB = 180^o        ... (3)
    But it is given that
B + D = 90^o + 90^o = 180^o        ... (4)
From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180^o.
So, it is a cyclic quadrilateral.
Consider chord CD.
Now, CAD = CBD                      (Angles in same segment)

Question 12.  Prove that a cyclic parallelogram is a rectangle.

Solution

Let ABCD be a cyclic parallelogram.
    A + C = 180^o     (Opposite angle of cyclic quadrilateral)    ... (1)
    We know that opposite angles of a parallelogram are equal
    A = C and B = D 
    From equation (1)
    A + C = 180^o
A + A = 180^o
2 A = 180^o
A = 90^o 
Parallelogram ABCD is having its one of interior angles as 90^o, so, it is a rectangle.

Ex-10.4

Ex-10.6