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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 Ncert Book Of Class 10th Maths University 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. Hence, a circle has infinite number of equal chords.
Now we may observe that for minor arc BDC. CAB is major arc. In this situation our chord will be passing through centre of circle. So it will be the diameter of circle.
A circle is a collection of points which are equidistant from a fix point. This fix point is called as the centre of circle and this equal distance is called as radius of circle. And thus shape of a circle depends on the radius of the circle. So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.
So, two circles are congruent if they have equal radius. Let us consider two congruent circles circles of same radius with centres as O and O'. Consider the following pair of circles.
So the circles have 1 point in common. So the circles have two points in common. We may observe that there can be maximum 2 points in common. We can have a situation in which two congruent circles are superimposed on each other, this situation can be referred as if we are drawing circle two times. Following are the steps of construction: Step1. Take the given circle centered at point O. Take any two different chords AB and CD of this circle and draw perpendicular bisectors of these.
Let these perpendicular bisectors meet at point O. Now, O is the centre of given circle. Consider two circles centered at point O and O' intersect each other at point A and B respectively. Join AB. Clearly centres of these circles lie on the perpendicular bisector of common chord.
Let radius of circle centered at O and O' be 5 cm and 3 cm respectively. Let PQ and RS are two equal chords Class 10 Maths Ch 7 Ex 7.3 Youtube of a given circle and there are intersecting each other at point T.
Draw perpendiculars OV and OU on these chords. Hence, the line joining the point of intersection to the centre makes equal angles with the chords. In this question, the students need to find out the angle OPR in the circle. This can be done implementing the theorem - the angle subtended by an arc at the centre is twice the angle on remaining parts of the circle, followed by the angle sum property of a triangle.
Ths students need to follow the property that the angles in the same segment of a circle are equal. Following this, they can use the angle sum property of a triangle to find the angles. This question follows the same approach as the previous one - that the angles in the same segment are equal. Following this, the students can solve it by applying the exterior angle property of a triangle.
In this question, a cyclic quadrilateral is given. So, the students can solve it using the opposite angle property of a cyclic quadrilateral. So, the angles opposite these sides will also be equal. In this question, a cyclic quadrilateral is given in which the diagonals are equal to the diameter of the circle. Also, according to the property of semi-circle, angles in a semi-circle are equal.
Therefore, all the internal angles will be right angles. In the given question, the non-parallel sides of the trapezium are equal.
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