The triangle ABC shown is a right triangle. The semicircles have the sides of the triangle as diameters. The areas of two of the semicircles are shown. What is the area of the third semicircle?
The diameter of APC is 2sqrt(7), and the diameter of CQB is 2sqrt(13). You can then use the Pythagorean Theorem to find the third side:
2sqrt(7)^2+2sqrt(13)^2 = x^2
The diameter of AB is 2sqrt(5). The area of the third semicircle is then sqrt(5)^2*pi=5pi.
Note that, area of semicircle = πr2/2, where
r - radius of semicircle.
The area of semicircle APC = 7
Let r1 be radius of semicircle APC.
By above note,
Area of semicircle APC = πr12/2
=> πr12/2 = 7
=> r12 = 14/π
r1 = √14/π.
length AC = 2r1 = 2√14/√π
Again, the area of semicircle CQB = 13
Let r2 be radius of semicircle CQB.
By above rule,
Area of semicircle CQB = πr22/2
=> πr22/2 = 13
=> r22 = 26/π
r2 = √26/π.
length CB = 2r2 = 2√26/√π
Now, in △ACB, MLC = 90°
By Pythagoras theorem,
AB2 = AC2 + CB2
= 4(14)/π + 4(26)/π
=> AB2 = 160/π
AB = √160/π
Therefore, radius of third circle = r3 = 1/2 √160/π.
Area of third circle = πr32/2
= π(1/4 * 160/π)/2
= 40/2 = 20
Therefore area of third circle is 20 square unit.