# NCERT Solutions for Class 9 Maths Exercise 10.5

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NCERT solutions for Class 9 Maths Circles

## 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 =

[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]

###### 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.

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)

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.

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.

Hence proved.

## NCERT Solutions for Class 9 Maths Exercise 10.5

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### 16 thoughts on “NCERT Solutions for Class 9 Maths Exercise 10.5”

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