Ex 5.3, 3

Transcript

Ex 5.3, 3 Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords. Let’s draw the figure Let our two chords be AB & CD with mid-points M & N respectively Since M is mid-point of AB AM = MB = 6/2 = 3 cm And, N is mid-point of CD CN = ND = 8/2 = 4 cm We need to find Distance between the midpoints of the chords i.e. we need to find MN Now, from Theorem 4 The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord Thus, OM ⊥ AB & ON ⊥ CD In ∆ OMA AM = 3 cm OA = Radius = 5 cm Since ∆ OMA is a right angled triangle By Pythagoras Theorem 𝐎𝐀^𝟐=𝐎𝐌^𝟐+𝐀𝐌^𝟐 5^2=OM^2+3^2 25=OM^2+9 25−9=OM^2 16=OM^2 〖𝑶𝑴〗^𝟐=𝟏𝟔 〖𝑂𝑀〗^2=4^2 𝐎𝐌=𝟒 cm In ∆ OCN CN = 5 cm OC = Radius = 5 cm Since ∆ OCN is a right angled triangle By Pythagoras Theorem 𝐎𝐂^𝟐=𝐎𝐍^𝟐+〖𝐂𝐍〗^𝟐 5^2=ON^2+4^2 25=ON^2+16 25−16=ON^2 9=ON^2 〖𝑶𝑵〗^𝟐=𝟗 〖𝑂𝑁〗^2=3^2 𝐎𝐍=𝟑 cm Thus, Distance between the midpoints of the chords = MN = OM + ON = 4 + 3 = 7 cm