![]() In the same way we are going to discuss about the distance from the centre, when the equal chords are given. If the angles subtended by two chords at the centre of a circle are equal, then the chords are equal. Now let us write the converse result as follows: ![]() That is, ∠ AOB = ∠ COD and the two sides which include these angles of the Δ AOB and Δ COD are radii and are equal.īy SAS rule, Δ AOB ≡ Δ COD. Now we are going to find out the length of the chords AB and CD, given the angles subtended by two chords at the centre of the circle are equal. Measure the distance from O to the chords. Construct perpendicular line to the chords AB and CD passing through the centre O. What do you observe? Is Δ OAB ≡ Δ OCD ? Answer: Yes Δ OAB ≡ Δ OCDĢ. Place these triangles ΔOAB and Δ OCD one on the other.ġ. Using trace paper, take the replicas of triangle ΔOAB and Δ OCD. Join the radius to get the ΔOAB and Δ OCD. We get one more crease line on the another part of semi circle, name it as CD ( observe AB = CD)ĥ. Make crease along AB in the semi circles and open it.Ĥ. Draw a circle with centre O and with suitable radius.Ģ. Theorem 8 Equal chords of a circle subtend equal angles at the centre.ġ. By SSS rule, the triangles are congruent, that is ΔOAB ≡ OCD. The other sides are radii, therefore OA= OC and OB= OD. Let us consider two equal chords in the circle with centre O. Join the end points of the chords with the centre to get the triangles Δ AOB and Δ OCD, chord AB = chord CD (because the given chords are equal). Now we are going to discuss another property. Instead of a single chord we consider two equal chords. Draw three circles passing through the points P and Q, where PQ = 4cm.Ģ. Find the distance of the chord from the centre.Ģ. The radius of the circle is 25 cm and the length of one of its chord is 40cm. (1) (Perpendicular drawn from centre to chord bisect it)ġ. Given : Chord AB of the outer circle cuts the inner circle at C and D. In the concentric circles, chord AB of the outer circle cuts the inner circle at C and D as shown in the diagram. “In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides”.Īpplication of this theorem is most useful in this unit. One of the most important and well known results in geometry is Pythagoras Theorem. Let AB be the chord and C be the mid point of AB Theorem 7 The perpendicular from the centre of a circle to a chord bisects the chord.Ĭonverse of Theorem 7 The line joining the centre of the circle and the midpoint of a chord is perpendicular to the chord.įind the length of a chord which is at a distance of 2√11 cm from the centre of a circle of radius 12cm. This argument leads to the result as follows. RHS criterion tells us that Δ AOC and Δ BOC are congruent. ∠ OCA = ∠ OCB = 90 ° ( OC ⊥ AB ) and OA = OB is the radius of the circle. Here, easily we get two triangles Δ AOC and Δ BOC (Fig.4.56).Ĭan we prove these triangles are congruent? Now we try to prove this using the congruence of triangle rule which we have already learnt. Perpendicular from the Centre to a ChordĬonsider a chord AB of the circle with centre O. Now, we are going to discuss some properties based on chords of the circle.Ĭonsidering a chord and a perpendicular line from the centre to a chord, we are going to see an interesting property.ġ. Using all the properties of these, we get some standard results one by one. ![]() Recently we have seen a new member circle. In this chapter, already we come across lines, angles, triangles and quadrilaterals.
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