1. OAB is a sector with angle AOB=0.4 radians. C is the midpoint of OA and D lies on OB. If OC=3cm and the area of the shaded region is 4.5cm^2, calculate the length of DB.
2. The figure shows a circle, centre O of radius 10cm. The chord AB is the diameter of a semicircle AOB. Show that the length of AB=10√2 cm. Hence find the area of the shaded region.
Thank you very much! :)
1. It is known that the area of a sector of a circle is πr².
It is also known that there are 2π radians in a circle.
If the measure of the angle is x radians, the area is xr²/2.
Using this, given the angle at AOB is 0.4 radians, the area is .16r²/2 = 0.08r².
Now to look at the problem.
Since C lies in the midpoint of OA, and OC is 3 cm, this means the length of OA is 6 cm.
Since CD/3 = tan 0.4, it can be seen that CD = 3 tan 0.4.
This means the area of triangle OCD is (1/2)3(3 tan 0.4) = 4.5 tan 0.4.
The area of sector AB is (.4)(6²)/2 = .4*36/2 = 7.2.
This means that shaded area can be found as the difference between these two.
This is seen to be 7.2 - 4.5 tan 0.4 = 5.3.
2. Put point X in the center of AB. Since the point X is in the center of the diameter, the length of XA is the radius of the circle with center at x. It should also be noted that XO is the radius of the added circle.
That means that since XO and XA are the same, and they are perpendicular lines with a right angle at X, triangle AXO is a 45 degree triangle.
This means that both of the legs of this triangle are the same and have a length which is the hypotenuse over root 2. That is, the length of AX is 10/root(2).
Now since AX is half of AB, the length of AB is 10*2/root(2) = 10*root(2).
Now since the measure of AOB is 90°, we are trying to find the 3/4 of the area of the circle with center 0 plus the area of the triangle AOB.
That would be (3/4)100π + (1/2)(10*root(2))².
That is the same as 75π + 100 = 236 + 100 = 336.