SOLUTION: A circle has an inscribed right triangle haveing and altitude of 4 and an area of 28. What are the circumference and area of the circle?

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Question 38582: A circle has an inscribed right triangle haveing and altitude of 4 and an area of 28. What are the circumference and area of the circle?
Found 2 solutions by rapaljer, AnlytcPhil:
Answer by rapaljer(4671) About Me  (Show Source):
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If a right triangle is inscribed in a circle, then the legs of that right triangle will be the base and altitude of the triangle, and the hypotenuse of the right triangle will be the diameter of the circle. To find the area and circumference of the circle, you need to find the diameter (and radius!) of the circle.
If altitude = 4 and area = 28, then
A=+%281%2F2%29%2Ab%2Ah
28+=+%281%2F2%29%2A+4%2Ah
28=2h
h=+14

Next, Let d= diameter of the circle, which is also the hypotenuse of the right triangle. According to the Theorem of Pythagoras: a%5E2+%2B+b%5E2+=+c%5E2
4%5E2+%2B+14%5E2+=+d%5E2
16+%2B+196+=+d%5E2
212+=+d%5E2
d+=+sqrt%28212%29+=+2%2Asqrt%2853%29+

Remember that the radius is HALF of the diameter, so if r= the radius, then
r=+sqrt%2853%29+ and r%5E2+=+53

Area = A=+pi%2Ar%5E2=++pi%2A53+=+53%2Api square units.

Circumference = C+=+pi%2Ad=+pi%2A+2%2Asqrt%2853%29+=2%2Api%2Asqrt%2853%29+ units.

R^2 at SCC

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
A circle has an inscribed right triangle having an altitude of 4 and an area of
28.  What are the circumference and area of the circle?

=============================================================================

Note.  A triangle has three altitudes, so the problem is not clear.  It should
have stated which altitude is 4.

In the case of a right triangle, each of the two legs are altitudes, and the
third altitude is drawn from the vertex of the right angle to the hypotenuse.

Regardless of whether the altitude is one of the legs or the altitude drawn
from the vertex to the hypotenuse, we will use this fact:

The hypotenuse of any right triangle inscribed in any circle is a diameter.

Case 1.  The altitude given is a leg.  then h = 4. Since A = 28

The area of a triangle is A = bh/2

                         28 = b(4)/2
                         28 = b(2)
                         14 = 2b
                          7 = b
                      _______    _______    __
So the hypotenuse is Ö4² + 7² = Ö16 + 49 = Ö65 
                                                __
The hypotenuse is a diameter, so the radius is Ö65/2
                                               __        __ 
The circumference of a circle is C = 2pr = 2p(Ö65/2) = pÖ65
                                 __
The area of a circle is pr² = p(Ö65)² = p(65) = 65p


Case 2.  The altitude given is the one from the vertex of the right angle to
the hypotenuse.  Then h = 4.

The area of a triangle is A = bh/2

                         28 = b(4)/2
                         28 = b(2)
                         14 = 2b
                          7 = b

So the hypotenuse is the base b = 7        

The hypotenuse is a diameter, so the radius is 7/2
                                                      
The circumference of a circle is C = 2pr = 2p(7/2) = 7p
                                 
The area of a circle is pr² = p(7/2)² = p(49/4) = 49p/4

Edwin