Question

In a right triangle inscribed within a half-circle with a diameter of 16, one angle of the triangle is 30°. What is the ratio of the area of the triangle to the area of the semicircle? Reduce to the lowest terms.

a. √3/π
b. π√3
c. 2π
d. √2/π
e. π/√2

Answer 1

The ratio of the area of the right triangle to the area of the semicircle is A)√3/π after calculating the areas of both the triangle and the semicircle and simplifying the resultant fraction.

Step-by-step explanation:

To find the ratio of the area of the triangle to the area of the semicircle, we first need to calculate both areas separately.

The diameter of the semicircle is given as 16, which means the radius (r) is 8. The area of the semicircle is half the area of a full circle with radius 8, so the area (A) of the semicircle is A = ½ πr² = ½ π(8²) = 32π.

Since one angle of the triangle is 30°, this is a special right triangle. In a 30-60-90 triangle, the ratios of the sides are 1:√3:2. Given that the hypotenuse (the diameter of the semicircle) is 16, the length of the side opposite the 30° angle is half the hypotenuse, so it is 8. The other side, opposite the 60° angle, can be calculated as 8√3. Thus, the area (A) of the triangle is A = ½ × base × height = ½ × 16 × 8 = 64.

The ratio of the area of the triangle to the area of the semicircle is therefore 64 / 32π = 2/π. Simplifying this ratio we get √3/π which corresponds to option (a).