Vtyjkyuk

I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem.

Here is the problem:

The diagram shows a circle of radius 8 metres. The points ABCD lie on the circumference of the circle.

BC = 14 m, CD = 11.5 m, AD = 8 m, $angle ADC$ = 104°, and $angle BCD$ = 73°.

a. Find AC.

AC is the diameter, right? So you would have double the radius which is 8 metres. So AC = 16?

b. (i) Find $angle ACD$

(ii) Hence, find $angle ACB$

If angle C is 73°, would ACD be half of that?

And then ACB is the other half?

c. Find the area of triangle ADC.

I believe the area of the triangle is 1/2 x base x height. So, 1/2 x 8 x 11.5?

d. Hence or otherwise, find the total area of the shaded regions.

This is confusing and I do not know how to do this part.

IMPORTANT: The angle subtended by the diameter on the circumference of a circle is always $90^circ$

Since $angle ACD$ subtend $104^circ$, AC is NOT a diameter.

Use Cosine rule to find AC
$$AC^2=AD^2+DC^2-2ADcdot DCcdotcos104^circ$$
$$AC^2=8^2+11.5^2-2times8times11.5times(-0.241)$$
$$ AC=15.51m$$

Then use Sine rule to find $angle ACD$
$$fracsinangle ACD8=fracsinangle ADCAC$$
$$fracsinangle ACD8=fracsinangle 10415.51$$
$$sinangle ACD=0.5$$
$$angle ACD=30^circ$$
So $angle ACB=73^circ-30^circ=43^circ$

Best way to find $triangle ADC $ area is to choose $AC$ as the base then find height using $DCcdotsinangle ACD$ (i.e perpendicular distance from D to side AC)

Area of $triangle ADC=frac12times ACtimes11.5times sin30 =44.59m^2$

Similarly, find Area of $triangle ABC$,

$$triangle ABC= frac12times ACtimes BCtimes sin angle ACB$$
$$= frac12times 15.51times 14times sin43^circ$$
$$= 74.18m^2$$

Shaded Area=$ pitimes8^2-(triangle ABC +triangle ADC)$