In the given figure ‘O’ is the centre of the circle.
Arc AB = Arc BC= Arc CD
If ∠OAB = 48 °,
Find:i)∠AOB ii) ∠BOD iii) ∠OBD
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Let AB be the chord of the given circle with centre O and a radius of 5 cm.
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cm
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2⇒ 25 = 9 + MB2
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2⇒ 25 = 9 + MB2⇒ MB2 = (25 − 9) = 16
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2⇒ 25 = 9 + MB2⇒ MB2 = (25 − 9) = 16⇒ MB=16 −−−√cm=4 cm
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2⇒ 25 = 9 + MB2⇒ MB2 = (25 − 9) = 16⇒ MB=16 −−−√cm=4 cmSince the perpendicular from the centre of a circle to a chord bisects the chord, we have:
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2⇒ 25 = 9 + MB2⇒ MB2 = (25 − 9) = 16⇒ MB=16 −−−√cm=4 cmSince the perpendicular from the centre of a circle to a chord bisects the chord, we have:AB = 2 × MB = (2 × 4) cm = 8 cm
Let AB be the chord of the given circle with centre O and a radius of 5 cm.From O, draw OM perpendicular to AB.Then OM = 3 cm and OB = 5 cmFrom the right ΔOMB, we have:OB2 = OM2 + MB2 (Pythagoras theorem)⇒ 52 = 32 + MB2⇒ 25 = 9 + MB2⇒ MB2 = (25 − 9) = 16⇒ MB=16 −−−√cm=4 cmSince the perpendicular from the centre of a circle to a chord bisects the chord, we have:AB = 2 × MB = (2 × 4) cm = 8 cmHence, the required length of the chord is 8 cm.
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