Problem Solving

I cant seem to figure out how to get the area of the chords.

thank you

Calculate the area of the sector OAB and deduct the the area of the equilateral triangle OAB.

Arc AB makes an angle 60 (= 2*the interior angle (ie 30)) at the center O.
Sector OAB area = 60/360 * 36 * pi


OAB is an equilateral triangle with side = radius
Area = (root 3)/4 * 36

Ans is 6pi-9(root 3).

Not sure where this is from, but this is a very difficult circle/triangle question.

To begin to solve, we need to know a semi-obscure rule about circles: Any triangle formed with the diameter and another point on the circle will form a right angle triangle, with the right angle at that other point.

The reverse will also be true: if we have a right angle triangle made up of 3 points on a circle, the line opposite the right angle will always be the diameter.

So, the first thing we need to know is that AC is the diameter of the circle.

If we know that AC is 12 (radius = 6, so diameter = 12), we can use our 30/60/90 triangle ratio (x : xroot3 : 2x) to determine that side AB is 6.

Let's call the centre of the circle O. If we draw a line from B to O we'll create equilateral triangle ABO.

(We know that it's equilateral because AO and BO are both radii, so we have at least an isosceles triangle; we also know that angle A is 60 degrees, which means that angle B is 60 degrees, and by default angle O is also 60 degrees).

Next we need to recognize that the shaded region is the difference between a 60 degree wedge of the circle (since angle O is 60 degrees) and our equilateral triangle.

To find the area of the wedge, we take 1/6 of the area of the circle (since our degree arc is 60 and 60/360 = 1/6).

r=6 (from above), so:

(1/6)pi(r)^2 = (1/6)pi(36) = 6pi

At this point we know the answer is 6pi - area of the triangle, so eliminate (a), (d) and (e). Worst case, we have a 50/50 guess.

Next, we calculate the area of the triangle.

We know that Area = (1/2)(base)(height) = (1/2)(6)(h) = 3h. Therefore, now we need the height.

To find the height of an equliateral triangle, we need to split it down the middle and create 2 30/60/90 triangles (our old friend). The height will always be opposite the 60 degree angle, so it's the "root3" side. Since our base is 6, half the base is 3, which means that 3 is our "multiplier" for this triangle (in a simple 30/60/90, we have a base of 1). Consequently, the height is 3(root3).

Plugging that in, we get:

area = 3h = 3(3(root3)) = 9(root3)

So, the area of the shaded region is:

6pi - 9(root3): choose (c).

Again, in case it wasn't abundantly clear from that thesis-like explanation, this is a VERY hard question (if for no other reason than the time it takes to work through all the steps).

There's another method we could have used to solve involving the area of the semicircle and the ratios of the two bits not part of the triangle, but it's probably more confusing and/or time consuming.

i feel a tad better knowing its supposed to difficult.

castorpollux wrote:i feel a tad better knowing its supposed to difficult.

In all honesty, although difficult, this seems representative of what's actually on the GMAT.

Can anyone just copy it tothe forum?

Another rule we can use is that the angle made by an arc at the center is double of the angle made by the same arc at the perimeter of the circle. Since we know from Stuart's post that AC is the diameter and Angle ACB = 30.

Give this, we can calculate Angle AOB=60. This tells us that triangle AOB is a equilateral triangle.

Since we know AOB is equilateral with sides = r = 6 and we know that angle AOB = 60

we can find the area of sector AOB and triangle AOB and subtract one from another to find the answer.

Actually this is very simple question and can be solved in seconds, if you know the formula:-

Area of segment = (square of radius X ((arc angle X Pi/180) - Sin(angle)))/2

radius = 6
Square of radius = 36

Now from the above you must know that how to calculate the arc AB angle

arc angle = 60 degree

Sin (60) = root(3)/2
therefore.

Area of segment = (36 X ((60 X Pi/180) - (root(3)/2)))/2

= 6Pi - 9 root(3)