SOLUTION: A circle is centered at O and has an area of 48pi. Let Q and R be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO, and triang

Algebra ->  Triangles -> SOLUTION: A circle is centered at O and has an area of 48pi. Let Q and R be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO, and triang      Log On


   

Question 1202262: A circle is centered at O and has an area of 48pi. Let Q and R be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO, and triangle PQR is equilateral, then find the area of triangle PQR.
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20063) About Me  (Show Source):
You can put this solution on YOUR website!


OQ and OR are radii of the large circle.
PO, PQ, and PR are radii of the smaller circle.
QR is also equal to the radius of the smaller circle, because triangle PQR is
equilateral. Let x = the lengths of those 4 line segments.

Area of large circle = 
pi%2Ar%5E2=48
r%5E2=48%2Fpi
r=sqrt%2848%2Fpi%29=OQ=OR

Extend OP to meet QR at point S (in green), which bisects QR at S.



Use the Pythagorean theorem on right triangle PQS



Since PS is the height of equilateral triangle PQR, and the base is x,

the area of equilateral triangle PQR is 

A=expr%281%2F2%29base%2Aheight

A=expr%281%2F2%29QR%2ASP

A=expr%281%2F2%29x%2Aexpr%28sqrt%283%29%2F2%29%2Ax

A=expr%28sqrt%283%29%2F4%29x%5E2

So we need to find x%5E2 so we can substitute it there and have the answer
we are looking for.



We apply the Pythagorean theorem on right triangle OQS:



Now that we have x2, we can substitute in

A=expr%28sqrt%283%29%2F4%29x%5E2

A=expr%28sqrt%283%29%2F4%29%28%2848%282-sqrt%283%29%29%29%2Fpi%29

The 4 divides into the 48 to give 12

A=sqrt%283%29%28%2812%282-sqrt%283%29%29%29%2Fpi%29

A=%2812sqrt%283%29%282-sqrt%283%29%29%29%2Fpi%29

A=%2812%282sqrt%283%29-3%29%29%2Fpi

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Edwin McCravy has made a slight mistake.

The area of the larger circle is 48pi, so it should lead to a radius of sqrt(48) and not sqrt(48/pi)

This means that x^2 = 48*(2-sqrt(3)) when following a similar outline of steps he wrote

Then,
area of equilateral triangle PQR = (sqrt(3)/4)*x^2
area of equilateral triangle PQR = (sqrt(3)/4)*48*(2-sqrt(3))
area of equilateral triangle PQR = 12*sqrt(3)*(2-sqrt(3))
area of equilateral triangle PQR = 24*sqrt(3) - 36 which is the final answer.

You can use a tool like GeoGebra to confirm the answer is correct.

That expression is approximately equal to 5.56921938165306