ab is achord of circle with radius 8 cm distance from the circle is 4 cm find the length of ab
Answers
Answer:
hey buddy here's your answer hope you Will understand plz Mark me as brainliest
Step-by-step explanation:
Solution :
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we have
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we haveOA2=OL2+AL2
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we haveOA2=OL2+AL2⇒AL2=OA2−OL2=[(17)2−(8)2]cm2
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we haveOA2=OL2+AL2⇒AL2=OA2−OL2=[(17)2−(8)2]cm2=(17+8)(17−8)cm2=225cm2
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we haveOA2=OL2+AL2⇒AL2=OA2−OL2=[(17)2−(8)2]cm2=(17+8)(17−8)cm2=225cm2⇒AL=225−−−√=15cm.
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we haveOA2=OL2+AL2⇒AL2=OA2−OL2=[(17)2−(8)2]cm2=(17+8)(17−8)cm2=225cm2⇒AL=225−−−√=15cm.Since the perpendicular fromthe centre of a circle to a chord bisects the chord, we have
Solution : Let AB be a chord of a circle with centre O and radius 17 cm.Draw OL ⊥ AB. Join OA.Then , OL=8cm and OA=17cm.From the right ΔOLA , we haveOA2=OL2+AL2⇒AL2=OA2−OL2=[(17)2−(8)2]cm2=(17+8)(17−8)cm2=225cm2⇒AL=225−−−√=15cm.Since the perpendicular fromthe centre of a circle to a chord bisects the chord, we haveAB=2×AL=(2×15)cm=30cm
Answer:
can you frame the question properly. I think its not properly put. Thanks!
Step-by-step explanation: