NCERT Solutions for Exercise 10.5 Class 9 Maths Chapter 10 - Circles
NCERT Solutions Class 9 Maths exercise 10.5 – This exercise includes some important points like When two circle chords are equal, their corresponding arcs are congruent, and when two arcs are congruent, their corresponding chords are equal.
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A circle's congruent arcs (or equal arcs) subtend equal angles at the centre.
The angle of an arc at its centre is double that of any point on the circle's remainder.
The above theorem establishes the link between the angles subtended by an object at its centre and at a point on the circle.
NCERT Solutions for Class 9 Maths exercise 10.5 uses the concepts-
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If all four vertices of a quadrilateral ABCD lie on a circle, it is called cyclic.
The sum of any pair of opposite angles of a cyclic quadrilateral is 180°.
Theorem 12: If a quadrilateral's sum of opposite angles is 180°, the quadrilateral is cyclic.
Along with NCERT book Class 9 Maths chapter, 10 exercise 10.5 the following exercises are also present.
Circles Exercise 10.1
Circles Exercise 10.2
Circles Exercise 10.3
Circles Exercise 10.4
Circles Exercise 10.6
Circles Class 9 Chapter 10 Exercise: 10.5
Q1 In Fig. , A,B and C are three points on a circle with centre O such that and . If D is a point on the circle other than the arc ABC, find .
AOC = AOB + BOC=
AOC = 2 ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)
Q2 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.
Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.
To find: ADB and ACB.
OA = AB (Given )
OA = OB (Radii of circle)
ABC is a equilateral triangle.
So, AOB =
AOB = 2 ADB
ACBD is a cyclic quadrilateral .
So, ACB+ ADB =
Q3 In Fig. , , where P, Q and R are points on a circle with centre O. Find .
Construction: Join PS and RS.
PQRS is a cyclic quadrilateral.
So, PSR + PQR =
Here, POR = 2 PSR
In OPR ,
OP=OR (Radii )
ORP = OPR (the angles opposite to equal sides)
In OPR ,
OPR+ ORP+ POR=
Q4 In Fig. , find
A+ ABC+ ACB=
A = BDC = (Angles in same segment)
Q5 In Fig. , A, B, C and D are four points on a circle. AC and BD intersect at a point E such that and . Find
DEC+ BEC = (linear pairs)
DEC+ = ( BEC = )
DEC = -
D+ DEC+ DCE =
D = BAC (angles in same segment are equal )
Q6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If , is , find . Further, if , find .
(angles in the same segment are equal )
If AB = BC ,then
Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
AC is the diameter of the circle.
Thus, and ............................1(Angle in a semi-circle is a right angle)
Similarly, BD is the diameter of the circle.
Thus, and ............................2(Angle in a semi-circle is a right angle)
From 1 and 2, we get
Hence, ABCD is a rectangle.
Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given: ABCD is a trapezium.
Construction: Draw AD || BE.
Proof: In quadrilateral ABED,
AB || DE (Given )
AD || BE ( By construction )
Thus, ABED is a parallelogram.
AD = BE (Opposite sides of parallelogram )
AD = BC (Given )
so, BE = BC
BE = BC (Proved above )
Thus, ...........1(angles opposite to equal sides )
...............2(Opposite angles of the parallelogram )
From 1 and 2, we get
Thus, ABED is a cyclic quadrilateral.
Q9 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 Fig. ). Prove that .
................1(vertically opposite angles)
..................2(Angles in the same segment are equal)
.................3(angles in the same segment are equal)
From 1,2,3 ,we get
Q10 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.
Given: circles are drawn taking two sides of a triangle as diameters.
Construction: Join AD.
Proof: AB is the diameter of the circle and ADB is formed in a semi-circle.
ADB = ........................1(angle in a semi-circle)
AC is the diameter of the circle and ADC is formed in a semi-circle.
ADC = ........................2(angle in a semi-circle)
From 1 and 2, we have
ADB+ ADC= + =
ADB and ADC are forming a linear pair. So, BDC is a straight line.
Hence, point D lies on this side.
Q11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
Given: ABC and ADC are two right triangles with common hypotenuse AC.
To prove :
Triangle ABC and ADC are on common base BC and BAC = BDC.
Thus, point A, B, C, D lie in the same circle.
(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)
CAD = CBD (Angles in the same segment are equal)
Q12 Prove that a cyclic parallelogram is a rectangle.
Given: ABCD is a cyclic quadrilateral.
To prove: ABCD is a rectangle.
In cyclic quadrilateral ABCD.
.......................1(sum of either pair of opposite angles of a cyclic quadrilateral)
........................................2(opposite angles of a parallelogram are equal )
From 1 and 2,
We know that a parallelogram with one angle right angle is a rectangle.
Hence, ABCD is a rectangle.
Also, Read| Circles Class 9 Notes
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Benefits of NCERT Solutions for Class 9 Maths Exercise 10
Equal Chords and Their Distances from the Center is the subject of exercise 10.5 Class 9 Maths.
Theorems relating to equal chords in circles are introduced in chapter 10 exercise 10.5 of Class 9 Maths.
Understanding the concepts from chapter 10 exercise 10.5 in Class 9 Math will help us understand the theorems of equal chords and their distance.
NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Solutions Class 9 Maths Chapter 10
NCERT Solutions of Class 10 Subject Wise
NCERT Solutions for Class 9 Maths
NCERT Solutions for Class 9 Science
Subject Wise NCERT Exemplar Solutions
- NCERT Exemplar Class 9 Maths
- NCERT Exemplar Class 9 Science