SOLUTION: In the diagram, triangle ABC is inscribed with diameter AC. Semicircles are constructed having AB and BC as diameters. The total area represented by regions a and b, in square unit
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Question 1166972: In the diagram, triangle ABC is inscribed with diameter AC. Semicircles are constructed having AB and BC as diameters. The total area represented by regions a and b, in square units, is:
A) 105.125 π
B) 210
C) 420
D) 210.25 π
E) 105
https://imageshack.com/i/pmYzXW65j
The first things you need are the three simplest pieces of this puzzle, the radius of the large circle, and the area of the two smaller semi-circles.
The radius is just half of the diameter that you can find using Pythagoras on the two chord lengths which are the legs of a right triangle with the necessary diameter as the hypotenuse. The area of either of the semicircles is half of the associated chord length squared times and then divide by 2.
Now that you have the two semicircle areas, you need to find the area of the associated segments of the large circle. For each, you will need the radian measure of the subtended central angle. Use:
where is the measure of the chord and is the radius.
Once you have the central angle measure, you can calculate the area of the segment using:
Once you have calculated the two segment areas, you can subtract those areas from the corresponding semicircle areas to get the two desired areas which you can then sum to find your final answer.
John
My calculator said it, I believe it, that settles it
The description in the post is not clear enough to identify the areas of the interest by a unique way.
I will interpret the post as " to find the sum of the areas of the two lunes ".
For the meaning of the term " lune " in Geometry see this Wikipedia article
https://en.wikipedia.org/wiki/Lune_(geometry)
Solution
The diameter is d = (Pythagoras), or
d^2 = 20^2 + 21^2 = 841.
The area of the large upper-right semi-circle is therefore
= = .
From it, subtract the area of the right angled triangle with the legs 20 and 21; this area is = 210.
You will get the sum of areas of two segments of the large circle
segments_area = . (1)
The area of two smaller semi-circles is
= (2)
( ! same as the area of the large semi-circle !)
Finally, to find the area under the problem's question, you should subtract the area of two segments (1)
from the area of two small semicircles of (2).
You will get then the answer to the problem's question as 210 square units.
