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Thus, D lies on BC. Case � I: If both the triangles are in the same semi-circle. Join BD. DC is a Byjus Class 9 Maths Circles Us chord. Case � II : If both the triangles are not in the same semi-circle. Prove that a cyclic parallelogram is a rectangle. Since, ABCD is a cyclic quadrilateral.
Thus, ABCD is a rectangle. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection. 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 centre.
If the distance between AB and CD is 6 cm, find the radius of the circle. Solution: We have a circle with centre O. Let r cm be the radius of the circle. 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 centre, what is the distance of the other chord from the centre? Parallel chords AB and CD are such that the smaller chord is 4 cm away from the centre. 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.
Proof: An exterior angle of a triangle is equal to the sum of interior opposite angles. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals. Taking AB as diameter, a circle is drawn. A circle drawn with Q as centre, will pass through A, B and O. ABCD is a parallelogram. ABCE is a cyclic quadrilateral. AC and BD are chords of a circle which bisect each other. Similarly, AC is a diameter.
Since, opposite angles of a parallelogram are equal. Two congruent circles intersect each other at points A and B. Solution: We have two congruent circles such that they intersect each other at A and B. A line segment passing through A, meets the circles at P and Q.
Let us draw the common chord AB. Since angles subtended by equal chords in the congruent circles are equal. The perpendicular bisector of BC passes through O. Join OB and OC. Suppose it cuts circumcirde at P.
Solution: i interior ii exterior iii diameter iv semicircle v the chord vi three Ex Solution: Let us draw different pairs of circles as shown below: We have Figure Maximum number of common points i nil ii one �i two Thus, two circles can have at the most two points in common. RD Sharma Class 12 Solutions. Watch Youtube Videos.
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