Geometry, Circumscribe

[anishprabhu]

What is the radius of the circle that circumscribes the triangle, measure of whose sides is 9, 40 and 41?

A. 20

B. 20.5

C. 4

D. 8

E. 45

[krisraam]

41,40,9 forms a right triangle.

So the diameter is 41.

radius is 20.5

Thanks
raama

[franciskyle]

41,40,9 forms a right triangle.

So the diameter is 41.

Is the hypotenuse of an inscribed (circle) right triangle always the diameter?

[cramya]

9,40,41 forms what is know as a pythogorean triplet i.e. right traingle with a hypotenuse of 41 and the otehr 2 sides being 9 and 40. If u draw the circle 41 the hypotenuse would be joining the circle to the midpoint of circle to either ends of circle forming the diameter.

Radius = 1/2 diameter = 20.5


Some useful pythagorean triplets to know for the GMAT:

3,4,5
5,12,13
7,24,25
8,15,17
9,40,41
11,60,61

The multiples or sub multiples of these triplets are also pythagorean triplets.

Regards,
Cramya

[Ian Stewart]

41,40,9 forms a right triangle.

So the diameter is 41.

Is the hypotenuse of an inscribed (circle) right triangle always the diameter?

It is. There are a few ways to see this, though it's easier to demonstrate with diagrams, of course. We could do as follows:

-Draw a right triangle RAB in quadrant 1 of the coordinate plane, with the right angle R at the origin, A on the x-axis, and B on the y-axis. So our vertices are R = (0,0), A = (a, 0) and B = (0,b). The midpoint M of the hypotenuse AB is at (a/2, b/2). If r is equal to the distance AM (which must be equal to BM), then r^2 = (a/2)^2 + (b/2)^2. But, if d = the length of MR, then d^2 = (a/2)^2 + (b/2)^2, so d = r, and R, A and B are all the same distance from M. Thus, you could draw a circle centered at M passing through A, R and B.

Alternatively, you could draw your triangle RAB, with a right angle at R, with RA a horizontal line and RB a vertical line. Now, draw a second right triangle ABS, by reflecting triangle RAB through the hypotenuse. This should make a rectangle RASB, where the hypotenuse AB is a diagonal of the rectangle. Draw the other diagonal RS, and since the diagonals cut each other in half, we can see that the distance from the centre of the rectangle to any of our points A, B, R or S is the same, so they could all be on the circumference of a circle drawn using the centre of the rectangle.