O is the centre of the circle and CAB is the secant , CO=41 cm ,CA=28 cm and OB=15 cm find AE
Answers
Answer:
ANSWER
In MQNP,
m∠CAO=90
0
(Radius is perpendicular to the tangent)
m∠CBO=90
0
(Radius is perpendicular to the tangent)
∴m∠AOB+m∠ACB=360
0
−m∠CAO−m∠CBO=360
0
−90
0
−90
0
=180
0
Hence, angle AOB and angle ACB are supplementary.
Answer:
12 cm
Step-by-step explanation:
OA = OB ( radius of the circle)
In Triangle OAC
Semi perimeter s = (41+28+15)/ 2 = 42
ar(OAC) = √[42 × (42-41) × (42-28) × (42-15) ]
(By Heron's formula)
= 126 cm² ....(1)
Now area OAC is also equal to 1/2 × CA × OE =
1/2 × 28 × OE
=> 14 ×OE = 126 ( From 1)
=> OE = 9 cm
Now OA² - OE² = AE² ( Pythagoras theorem)
=> 225 - 9 = AE²
=> 144 = AE²
=> AE = 12
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