Find Area of Circle

PLZ reply if you have any confusion

$OA=OB$

$\triangle EAO=\triangle EOB=\triangle CEB$

$OE=EC=\frac{1}{2}OC=\frac{1}{2}OA$

$\angle OEA=90^{\circ}$

$\triangle EAO$ is the special 30-60-90 triangle.

Since $AD$ is a chord, and $OC \bot AD$ , $AE=ED=\frac{1}{2}$

$OA=\frac{1}{\sqrt 3}$

The area is $\pi r^2=\frac{\pi}{3} \approx \boxed {1.047}$

Let the radius of the circle be r

Since AB is a diameter

Then angle ADB = 90

Then OE is parallel to BD (because angle AEO = 90)

Since O is midpoint of AB

Then E is midpoint of AD

Then AE = 1/2

Area of triangle AEO = area of triangle BEO ... (equal bases and common height)

Then

Area of triangle BEC = area of triangle BEO

Then

OE = CE = r/2

In the right triangle AEO

r^2 = (r/2)^2 + (1/2)^2

Then

r^2 = 1/3

So , the area of the circle = (Pi) r^2 = 1.047

Let $\angle AOE$ be $\theta$

Area of $\triangle AOE=$ Area of $\triangle BEC$

$\frac{1}{2}\times r\sin\theta \times r\cos\theta=\frac{1}{2}\times (r-r\cos\theta) \times r\sin\theta$

$\Rightarrow \cos\theta = 1 - \cos\theta$

$\cos\theta = \frac{1}{2}$

$\theta = 60^\circ$

$\Rightarrow\angle OAD=30^\circ$

$\triangle AOD$ is isosceles, so $\angle AOD=120^\circ$

Using cosine theorem on $\triangle AOD$

$1^2=r^2+r^2-2r^2cos120^\circ$ (AD=1)

$\Rightarrow r^2=\frac{1}{3} and$ $Area=\frac{\pi}{3}$

In my opinion the core of the problem is realizing that BE is the median of triangle BOC, since area of triangles AOE = OBE= BOC. As by hypothesis AD is perpendicular to OC, point E is in the intersection of the 2 auxiliaries circumferences as shown in drawing. AE equal half of AD and OE equals half of OA.

Therefore by Pythagoras OA = 1/sqrt{3}and the Area is Pi/3

$\large In ~\triangle \text{ AEO and BEO, base AO=BO, common hight.}\\\large \therefore~\ Area~same. ~\implies~ Areas~of~\triangle ~BEO~and~BEC~are\\\large same~ and~have~ the~ common~ hight. ~~~\therefore~bases~are~equal.\\ \large\implies E~is~ midpoint ~of~OC.~But ~OC=R,~radius~of~the\\ \large circle. ~~radius~OC\perp~chord~AD. ~~\therefore~ E~is~midpoint~of~AD.~\\\large CE = \dfrac{1}{2}*R.~~~~~~~~~~ AE= \dfrac{1}{2}. ~~~~~~~~~~~~~~So ~in ~the~ circle~~~\\\large \dfrac{1}{4}=AE^2=(2R-CE)*CE=(2R-\dfrac{1}{2}*R)*\dfrac{1}{2}*R.\\ \large \therefore~\dfrac{1}{4} =\dfrac{3}{4}*R^2.~~\therefore~AREA~of ~the~ circle~=\dfrac{\pi}{3}=~~~~\boxed{ \color{#D61F06}{ 1.047}}$

Angle OAE=30 Degree,AE=0.5 ,FROM RIGHT TRIANGLE (OAE) GET RADUIES OF CIRCLE (OA)=0.577 , HENCE: CIRCLE AREA=1.04719 SQUARE UNITE