Question: Circle Area with Triangle

Comment on Circle Area with Triangle

Question can be easily solved by using Isosceles Triangle property only, why you have used Similar Triangle property?
Please explain
gmat-admin's picture

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 really useful, but in this example, realizing that the triangle is an isocele right one is enough since pythagorean theorem may be a faster aproach :
Now, since we do not really need more than that because the area is (Pi r²) = 18pi
gmat-admin's picture


The center of a circle is (5, -2). (5, 7) is outside the circle, and (1, -2) is inside the circle. If the radius, r, is an
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 ?
gmat-admin's picture

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 inscribed angles.May i get the additional information about the inscribed angles.Kindly get me in the pdf format if possible.
gmat-admin's picture

Here's our video on the topic:

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

Applet #1:

Applet #2:

Applet #3:

Applet #4:

I hope that helps.


If the figure is not drawn to scale and you're only given one angle how can you assume this is a 45-45-90?
gmat-admin's picture

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?


Is this a 600 level question? or less?
gmat-admin's picture

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?

gmat-admin's picture

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:

Does that help?

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