NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.
- Circles Class 9 Ex 10.1
- Circles Class 9 Ex 10.2
- Circles Class 9 Ex 10.3
- Circles Class 9 Ex 10.4
- Circles Class 9 Ex 10.5
- Circles Class 9 Ex 10.6
|Number of Questions Solved||10|
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6
Prove that the line of centers of two intersecting circles subtends equal angles at the two points of intersection.
Given Two circles with centers O and O’ which intersect each other at C and D.
To prove: ∠OCO’ = ∠ODO’
Construction: Join OC, OD, O’C and O’ D.
Proof: In ∠OCO’ and ∠ODO’, we have
OC = OD [radii of the same circle]
O’C = O’D [radii of the same circle]
OO’ = OO’ [common sides]
∴ Δ OCO’ = Δ ODO’ [by SSS congruence rule]
Then, ∠OCO’ = ∠ODO’ [by CPCT]
Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, then find the radius of the circle.
Let 0 be the center of the given circle and its radius be b cm.
AB = 5 cm and CD = 11 cm are two chords on opposite sides of its center.
Draw ON ⊥ AB and OM ⊥ CD.
Since, ON ⊥ AB, OM ⊥ CD, and AB||CD, therefore the points N, O and M are collinear.
Given, distance between AB and CD = MN = 6 cm
Let ON=a cm
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the center?
Let PQ and RS be two parallel chords of a circle with center O such that PQ = 6 cm and RS = 8 cm.
Let a be the radius of the circle. Draw ON ⊥ RS and OM ⊥ PQ.
Since, PQ || RS, and ON ⊥ RS and OM ⊥ PQ.
Therefore, points O, N, and M are collinear. v OM = 4 cm and M and N are the mid-points of PQ and RS, respectively.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half of the difference of the angles subtended by the chords AC and DE at the center.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Given: Let PQRS be a rhombus and PR and SQ are its two diagonals which bisect each other at right angles at 0.
To prove: A circle drawn on PQ as diameter will pass through O.
Construction: Through O, draw MN || PS and EF || PQ.
ABCD is a parallelogram. The circle through A, B, and C intersect CD (produced, if necessary) at E. Prove that AE = AD.
AC and BD are chords of a circle which bisects each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
(i) Let BD and AC be two chords of a circle bisect at P.
In Δ ASP and Δ CPD, we have PA = PC
Bisectors of angles A, B and C of the triangle ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the Δ DEF are 90° – A, 90° – B and 90° – C.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn, so that P and Q lie on the two circles. Prove that BP = BQ.
In any Δ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the Δ ABC.
(i) Let bisector of ∠A meet the circumcircle of Δ ABC at M.
Join BM and CM.
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