Question

In a circle with centre O and radius 5 cm , AB is a chord of length 5√3 cm .find the area of sector AOB.

Answer 1

Solution:

In the attached figure

OA = OB = 5cm [radius of circle]

AB = 5√3 cm [ chord length]

Construction: Draw a perpendicular line OE on chord AB

As we know that perpendicular OE bisect the chord AB,so


$AE = EB = \frac{5 \sqrt{3} }{2}$
Now in Right triangle ∆OEA

Angle OAE =>
$\cos( \alpha ) = \frac{base}{hypotenuse} \\ \\ \cos( \alpha ) = \frac{ \frac{5 \sqrt{3} }{2} }{5} \\ \\ = \frac{ \sqrt{3} }{2}$
we know that √3/2 is value of cos 30°

so,angle OAE = 30°

Thus angle AOE = 60°[ angle sum property of triangle]

By the same way in another triangle ∆OBE

Angle OBE = 30°

So,center angle of the chord is 120°

Now area of the sector OAB
$= \frac{\theta}{360} \times \pi \times {r}^{2} \\ \\ = \frac{120}{360} \times \frac{22}{7} \times 25 \\ \\ = \frac{22 \times 25}{21} \\ \\ = 26.190 \: {cm}^{2}$
Hope it helps you.

Answer 2

hope it is help for you

answer in this photo