Mathematics, Part 12

Question 1: In the figure, find the area of the entire region formed by triangle ABC and the semicircle having AB as its diameter.a) 2 + ? b) 2 + 2? c) 1 + ? d) 4 + 4 ? e) 2 + 4?

Answer 1: a) 2 + ?For the area of the triangle, A = ½bh, where b = base and h = height. The base and height of the triangle are both 2 units. A = 1/2 (2)(2)A = ½ (4)A = 2 Pythagorean’s Theorem can be used to determine the length of AB, which is the diameter of the circle. AB is the triangle’s hypotenuse.A2 + b2 = c222 + 22 = c24 + 4 = c28 = c2v8 = c, the hypotenuse length and the diameter length. Therefore, the radius = (v8)/2Area of a semicircle = (?r2)/2A = [(?)(v8/2)2]/2A = ?Therefore, the triangle area = 2 and the area of the semicircle = ?

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Question 2: In the figure below, RS is tangent to circle O. If PO is 5 and QR is 8, what is the value of RS?a) 5 b) v39 c) 8 d) 12 e) 144

Answer 2: d) 12 When you first look at this problem, it appears that there are lots of things not given; however, they can be deduced. The length of PO, OQ, and OS are all the same, which is 5. OS and SR form a 90 degree angle. To get the length of the hypotenuse, add the length of OQ and QR, which is 5 + 8 = 13. You need to know the length of RS. SOHCAHTOA means that the sine of angle ORS is given by 5/13. The inverse sine of 5/13 is 22.6, which means the angle of ORS is 22.6 degrees. Now you know the cosine ORS is equal to length SR/13. The cosine of 22.6 degrees is equal to SR/13. Cosine (22.6) = .923 Multiplying .923*13 = 12, to solve for length SR, which is choice d.

Question 3: In the figure, if the small circle has a radius R, and the larger circle has a diameter 6R, what is the area of the region inside the large circle and outside the small circle?a) ?/2 b) 3?R/4 c) ?R/2 d) 8?R² e) 15?

Answer 3: d) 8?R²In order to answer the question, you have to figure out the area of the big circle and the area of the smaller circle. The big circle has a diameter of 6R, which means it would have a radius of 3R. Area = ?r2; therefore, the big circle would have an area of ?(3R)2, which is equal to 9?R2. The smaller circle has a radius R, which means it has an area of ?R2. 9?R2 - ?R2 = 8?R2, which is choice d.

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