Find the area of the quadrilateral OCBAO

[chwala]

Homework Statement

This is my own question. I just made some extension to original question( see diagram)

Relevant Equations

understanding of circle properties.

My challenge was on trying to determine the angles: My approach;

came up with a number of equations: ie

$m+n=70^0$
$r=p+40^0$
$q-2r=100^0, ⇒ r=50^0 + \dfrac{1}{2} q$

then it follows that,
$2q+100^0=180^0$
$⇒q=40^0, r=70^0, p=m=30^0, n=40^0$

$m+40^0+t=180^0, ⇒t=110^0$

and
$q+p+s=180^0$
$40+30+s=180^0, s=110^0$

problem here...i will need to check on this...

I need to have $t+s=180^0$.

I know once i am certain on the angles then finding area is as easy as abc...

i see my own mistake...i will go through this again...I may need to use cosine rule...coming back in a moment.

 

[SammyS]

Homework Statement: This is my own question. I just made some extension to original question( see diagram)
Relevant Equations: understanding of circle properties.

View attachment 329308

My challenge was on trying to determine the angles: My approach;
. . .

i see my own mistake...i will go through this again...I may need to use cosine rule...coming back in a moment.

It looks like you failed to provide enough information.

 

[Frabjous]

Nothing in the problem description limits C to a single location.

 

[Steve4Physics]

As already noted by @SammyS and @Frabjous, the position of point C on the circle needs to be defined (unless maybe AC and OB are meant to be perpendicular?).

Your hand-drawn Post #1 diagram is wrong. ∠ACB is not 40º.

∠ACB is the angle subtended by AB at a point C on the circumference.
∠AOB is the angle subtended by AB at the centre.
There is a simple relation between these two angles but they are not equal.

 

[chwala]

...I realised that the question has many unknowns...been looking at it for last 30 minutes...it cannot be solved...and true $C$ can lie at any point on the circle circumference.

 

[chwala]

...supposing we are told that the length AD = DC... so that we have the point C fixed at a point. Are we going to have some breakthrough? i need to analyse this later...

 

[Frabjous]

While you added a condition, you also added a new unknown. While for a given D, C is determined, D is not limited to a single point.

You are too eager to jump into the analysis, instead of sitting back and developing a strategy.

 

[SammyS]

...supposing we are told that the length AD = DC... so that we have the point C fixed at a point. Are we going to have some breakthrough? i need to analyse this later...

View attachment 329311

The equations that i had last were:
$2r+q=180^0$
$4x-q=40^0$

As pointed out by @Steve4Physics "∠ACB is not 40º".

 

[chwala]

As pointed out by @Steve4Physics "∠ACB is not 40º".

It is 40^0 check on the circle properties... angles subtended by the same chord....

aaaargh its $20^0$ ...you are correct.