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## Comment on

Circle Area with Triangle## Question can be easily solved

Please explain

## We're comparing the 45-45-90

We're comparing the 45-45-90 triangle in the diagram with the base 45-45-90 triangle (with sides 1-1-√2). So, we just need to use a similar triangle property to do so.

Can you show me your solution?

## Special right triangle can be

2x²=6²

2x²=36

x²=18=r²

Now, since we do not really need more than that because the area is (Pi r²) = 18pi

## Perfect!

Perfect!

## The center of a circle is (5,

integer, how many possible values are there for r?

(A) 4

(B) 5

(C) 6

(D) 7

(E) 8

(5,-2 ). Etc how will we use this to solve ?

## If we sketch the circle's

If we sketch the circle's center, (5, -2), and the point (5, 7), we can see that (5, 7) is 9 UNITS away from the center.

Since (5, 7) is OUTSIDE the circle, the radius of the circle must be LESS THAN 9 units

Using the same logic, we can see that (1, -2) is 4 UNITS away from the center. Since (1, -2) is INSIDE the circle, the radius of the circle must be GREATER THAN 4 units

Combining these results, we get: 4 < RADIUS < 9

Since the radius in an INTEGER, the radius can be 5, 6, 7, or 8 (4 values)

Answer: A

## Sir i am quite confused about

## Here's our video on the topic

Here's our video on the topic: https://www.gmatprepnow.com/module/gmat-geometry/video/880

Also, the links below have some nice applets to better understand inscribed angles:

Applet #1: http://www.mathopenref.com/circleinscribed.html

Applet #2: http://www.mathopenref.com/thalestheorem.html

Applet #3: http://www.mathopenref.com/arccentralangletheorem.html

Applet #4: http://www.mathopenref.com/tangentline.html

I hope that helps.

Cheers,

Brent

## If the figure is not drawn to

## Since O is the center of the

Since O is the center of the circle, we know that OB and OC are both radii of the circle.

Since OB and OC are both radii, we know that OB = OC

If OB = OC, then ∆OBC is an isosceles triangle, which means ∠OCB also equals 45°

Once we know two of the angles (45° and 45°), we know that the third angle must be 90°

Does that help?

Cheers,

Brent

## Is this a 600 level question?

## I'd say the difficulty level

I'd say the difficulty level is between 550 and 600

## Hi Brent,

Long time.

Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.

Is this a rule that we need to remember? Could you help me figure out how this has been derived?

Thanks,

Erik

## That's not quite right (but

That's not quite right (but it's close!).

You're referring to a property of ISOSCELES triangles (not RIGHT triangles). That said, the property DOES apply to ISOSCELES right triangles

I cover that property starting at 7:05 in the following video: https://www.gmatprepnow.com/module/gmat-geometry/video/865

Does that help?