Question: Circle Area with Triangle

Comment on Circle Area with Triangle

is there another approach that can be used to solve this ?
greenlight-admin's picture

You bet.

Once we recognize that ∠BOC = 90°, we know that ∆BOC is a RIGHT TRIANGLE. We're also told that the hypotenuse of ∆BOC has length 6.

Also notice that sides OB and OC are the RADII of the circle. So, these two legs have the SAME LENGTH.
So, let's say that sides OB and OC have length x.

Apply the Pythagorean Theorem to get: x² + x² = 6²

Simplify: 2x² = 36

Divide by 2 to get: x² = 18

Now we COULD solve this equation to get x = √18, however this step is unnecessary. Here's why:

We're asked to find the AREA of the circle.

Area of circle = π(radius²)

Since we let x = the radius, we can write:
Area of circle = π(x²)
= π(18)
= 18π
= D

Cheers,
Brent

Hi,
As per the ques:

https://gre.myprepclub.com/forum/in-the-figure-above-a-is-the-center-of-the-circle-12255.html

Is there any other way instead of using Pythagoras theorem to solve this. Can we take the help of triplets to solve this as the triangle formed inside a circle and have 2 sides turn out to be radius
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/in-the-figure-above-a-is-the-center-of-...

Sorry, but I'm not sure what you mean.
Can you rephrase your question?

Cheers,
Brent

I have approached in another way

In the triangle EDA, since it is a right angled triangle and the sides are arranged in triplets (as AE = AD and AD =AF +5)

as one side is 25 so the other two sides are arranged in 15 : 20 : 25

AE and AD is the radius of the circle and

therefore AE = 20 and AF = 15

Now for the triangle EBD , this is also a right angled triangle with ∠C=90∠C=90

Since AE = 20 therefore EB = 40

Again as this is also a right angled triangle so the sides are

24 : 32 : 40

EC cannot be 24, since EF = 25

Therefore EC = 32 and BC = 24

Hence FC = EC - 25 = 32 -25 = 7
greenlight-admin's picture

That's great reasoning!
The tricky part is recognizing that FD = 5 is a hint.

ASIDE: Another way to determine the lengths of the sides of ∆AEF is as follow:
Let x = the length of side AE (aka the radius)
So x - 5 = the length of side AF

Applying the Pythagorean Theorem, we get: x² + (x - 5)² = 25²
Expand: x² + x² - 10x + 25 = 625
Simplify: 2x² - 10x - 600 = 0
Divide both sides by 2 to get: x² - 5x - 300 = 0
Factor: (x - 20)(x + 15) = 0
Since x must be POSITIVE, it must be the case that x = 20....etc

Cheers,
Brent

Thanks very much

Hi Brent, I understand the concept of a 45-45-90 triangle, but why can we assume this is one? Is it because we know O is the center?
greenlight-admin's picture

Since O is the circle's center, we know that OB and OC are both radii, which means OB = OC, which means ∆OBC is an ISOSCELES triangle.

Since ∆ is ISOSCELES, we know that ∠OCB is also 45°

Does that help?

Cheers,
Brent

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