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Install NowNCERT 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 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 = 2
ADC
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image007.png)
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.
= 2
BDA
BOA = 2
BDA
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
= 2
AEB
AOB = 2
AEB
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
= 2
PQR
ROP = 2
PQR
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 +
=
2
OPR =
[Using (i) & (ii)]
2
OPR =
OPR =
4. In figure,
ABC = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image042.png)
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
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 = ![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image055.png)
BAC is
find
BCD. 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 EB
DC.
EBCD is a parallelogram.
DE = CB and
DEB =
DCB
Now AD = BC and DA = DE
DAE =
DEB
But DEA +
DEB =
DAE +
DCB =
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image007.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image001.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image001.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image001.png)
![](https://media-mycbseguide.s3.amazonaws.com/images/static/ncert/09/maths/10_5/image001.png)
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 =
AlsoAQ is a diameter.
2 =
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
NCERT Solutions for Class 9 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 9 Maths have total 15 chapters. 9 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 9 solutions PDF and Maths ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.
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Very well explained answers
Very well explained answers
It’s very useful
Very easily explained answers through valuable tips
Please keep Q12 also!!
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Usefull ?Thanks ??
I can understand it easily ?.Thanks for your help. ?
I can understand it easily ?.Thanks for your help. ?
It’s very useful and I understood easily. Thanks..
Nice
Good job
Not given the diagram properly
Your solution are brilliant. I can easily understand
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Very nice put the question 12 also