NCERT Solutions for Class 9 Maths Exercise 10.5

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NCERT Solutions for Class 9 Maths Exercise 10.5 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Circles Download as PDF

NCERT Solutions for Class 9 Maths Exercise 10.5

NCERT Solutions for Class 9 Mathematics Circles

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

Ans. AOC = AOB + BOC

AOC =

Now AOC = 2ADC

[Angled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]

ADC = AOC

ADC = =


2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord on a point on the minor arc and also at a point on the major arc.

Ans. Let AB be the minor arc of circle.

Chord AB = Radius OA = Radius OB

AOB is an equilateral triangle.

AOB =

Now

AOB + BOA =

+ BOA =

BOA =

D is a point in the minor arc.

= 2BDA

BOA = 2BDA

BDA = BOA =

BDA =

Thus angle subtended by major arc, at any point D in the minor arc is

Let E be a point in the major arc

= 2AEB

AOB = 2AEB

AEB = AOB

AEB = =


3. In figure, PQR = where P, Q, R are points on a circle with centre O. Find OPR.

Ans. In the figure, Q is a point in the minor arc

= 2PQR

ROP = 2PQR

ROP = =

Now

POR + ROP =

POR + =

POR = = …..(i)

Now OPR is an isosceles triangle.

OP = OR [radii of the circle]

OPR = ORP [angles opposite to equal sides are equal] …..(ii)

Now in isosceles triangle OPR,

OPR + ORP + POR =

OPR + ORP + =

2OPR =[Using (i) & (ii)]

2OPR =

OPR =


4. In figure, ABC = ACB = find BDC.

Ans. In triangle ABC,

BAC + ABC + ACB =

BAC +

BAC =

BAC = …….(i)

Since, A and D are the points in the same segment of the circle.

BDC = BAC

[Angles subtended by the same arc at any points in the alternate segment of a circle are equal]

BDC = [Using (i)]


NCERT Solutions for Class 9 Maths Exercise 10.5

5. In figure, A, B, C, D are four points on a circle. AC and BD intersect at a point E such that BEC = and

ECD = Find BAC.

Ans. Given: BEC = and ECD =

DEC = BEC = [Linear pair]

Now in DEC,

DEC + DCE + EDC = [Angle sum property]

EDC =

EDC =

BAC = EDC = [Angles in same segment]


NCERT Solutions for Class 9 Maths Exercise 10.5

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. DBC = BAC is findBCD. Further if AB = BC, find ECD.

Ans. For chord CD

(Angles in same segment)

=
(Opposite angles of a cyclic quadrilateral)


AB = BC (given)
(Angles opposite to equal sides of a triangle)

We have


NCERT Solutions for Class 9 Maths Exercise 10.5

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

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


(Consider BD as a chord)(Cyclic quadrilateral)

(Considering AC as a chord)

(Cyclic quadrilateral)


Here, each interior angle of cyclic quadrilateral is of. Hence it is a rectangle.


NCERT Solutions for Class 9 Maths Exercise 10.5

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

Ans. Given: A trapezium ABCD in which ABCD and AD = BC.

To prove: The points A, B, C, D are concyclic.

Construction: Draw DECB.

Proof: Since DECB and EBDC.

EBCD is a parallelogram.

DE = CB and DEB = DCB

Now AD = BC and DA = DE

DAE = DEB

But DEA + DEB =

DAE + DCB =

[DEA = DAE and DEB = DCB]

DAB + DCB =

A + C =

Hence, ABCD is a cyclic trapezium.


NCERT Solutions for Class 9 Maths Exercise 10.5

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, P, Q respectively (see figure). Prove that ACP = QCD.

Ans. In triangles ACD and QCP,

A = P and Q = D [Angles in same segment]

ACD = QCP [Third angles] ……….(i)

Subtracting PCD from both the sides of eq. (i), we get,

ACD – PCD = QCP – PCD

ACPO = QCD

Hence proved.


NCERT Solutions for Class 9 Maths Exercise 10.5

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.

Ans. Given: Two circles intersect each other at points A and B. AP and AQ be their respective diameters.

To prove: Point B lies on the third side PQ.

Construction: Join A and B.

Proof: AP is a diameter.

1 =

[Angle in semicircle]

AlsoAQ is a diameter.

2 =

[Angle in semicircle]

1 + 2 =

PBQ =

PBQ is a line.

Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on PQ.


NCERT Solutions for Class 9 Maths Exercise 10.5

11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that CAD = ABD.

Ans. We have ABC and ADC two right triangles, right angled at B and D respectively.

ABC = ADC [Each ]

If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through of an arc AC, Because B and D are the points in the alternate segment of an arc AC.

Now we have subtending CBD and CAD in the same segment.

CAD = CBD

Hence proved.

NCERT Solutions for Class 9 Maths Exercise 10.5

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