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NCERT Solutions For Class 9th Maths Chapter 10 : All Q&A Miscellaneous NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex Ex Class 9 Maths Question 1. In figure A,B and C are three points on a circle with centre 0 such that ?BOC = 30� and ? AOB = 60�. If D is a point on the circle other than the arc ABC, find ? ADC. Solution: We have a circle with centre O, such that ?AOB = 60� and. Mar 14, �� This video is prepared for your convenience. You can understand these problems and concepts at your own pace sitting at home without anyone's help. So enjoy. Exercise Class 9 Maths Solutions The questions in Exercise are based on all the concepts included in the chapter. It mainly covers the concept of the angle subtended by an arc of a circle and the cyclic quadrilaterals. The exercise consists of 12 questions, including a variety of short answers as well as comprehensive ones.
Abstract:

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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 Ch 6 Maths Class 10 Theorems Free 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 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. The longest Ch 6 Maths Class 10 Theorems Youtube chord in the circle is equal to the diameter.

Another related term to the circle is the sector. Students can refer to class 9 chapter 10 maths to get a better understanding of this section. Students are asked to draw a card in a circle. Then extend that line to another point which joins two line segments. It forms a triangle inside the circle.

This is known as an angle subtended by a chord at a point. Based on these two theorems were explained in the PDF. If the lengths of chords are the same, then their angles are the same and vice versa. In this section, students can learn about the perpendicular drawn from the centre of the chord by making an activity.

Here also two theorems were given. The perpendicular drawn from the centre can bisect the chord. The line is drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Chapter 10 Class 9 specifies the students to learn about the number of circles that can be drawn from a single point. They have already learned the basics of this concept in the 6th class. Here a slight difference to this concept is, if the circles can be drawn from multiple points, these points are known as collinear points centre. If we have more than two collinear points, then there is no chance to get more than a single circle.

The length of the perpendicular is between the point to a line and the distance of the line. 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. The students have to prove that it is cyclic.

They can do so by drawing two perpendiculars to the base.

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