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Comment on Circle Area with Triangle
Question can be easily solved
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
Now, since we do not really need more than that because the area is (Pi r²) = 18pi
The center of a circle is (5,
integer, how many possible values are there for r?
(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)
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.
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?
Is this a 600 level question?
I'd say the difficulty level
I'd say the difficulty level is between 550 and 600
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?
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?