The triangle ABD is inscribed in a circle, such that AD is a diameter of the circle and angle BAD is 45(degrees). If the area of triangle ABD is 72 square units, how much larger is the area of the circle than the area of the triangle ABD?
I follow the problem completely until the last step.
If AD is the diameter, then it is a 45-45-90, and thus base and height are equal. Therefore (x^2)/2=72... x = 12.. therefore the hypotenuse (AD) is 12sqrt(2) and thus the diameter is 6sqrt(2)..
So, area of a circle is pi(r^2) --> pi(6sqrt(2)^2)
6 squared is 36.. but the book says 6 squared is 72.. I have a feeling I am missing something..
Could someone please help out.. this is the 3rd edition book.
Solution is on page 110.
Thanks.