Problem Solving

BTGmoderatorDC wrote:

The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?


A. √2

B. 2

C. 3

D. 4√2

E. 6

Let's add the radius of 6 feet to the diagram to get:


From here, we can see the right triangle hiding within the diagram, which means we can apply the Pythagorean Theorem.
Se can write: h² + 2² = 6²
Simplify: h² + 4 = 36
So, we get: h² = 32
This means h = √32
Check the answer choices. . . √32 is not among them.
Looks like we need to simplify √32

We'll use the fact that √(xy) = (√x)(√y)
So, √32 = √[(16)(2)] = (√16)(√2) = (4)(√2) = 4√2
Check the answer choices. . . D

Answer: D

Cheers,
Brent

Brent provided a nice solution, but there's also an easy way to GUESS on this problem:

Since h is not the radius but is perpendicular to the diameter, we can infer that it's shorter than the radius:


Therefore, we can eliminate E.

We can also infer that h must be greater than 4, since the angle that's across from must be greater than 45 degrees. Compare these two:





We don't have to calculate the angle of x to see that it's opening wider than 45*. Thus angle y must be less than 45 degrees. Since larger sides are across from larger angles, h must be larger than 4.

Eliminate any answer choices less than 4: A, B, and C.

The only thing left must be the answer: D.

For another exact approach, you could use SIMILAR TRIANGLES:

We can create 2 similar right triangles inside the semicircle using h as a side length for each:



Since side lengths in similar triangles are proportional, we can infer:
(4/h) = (h/8)
h² = 32
h = √32
h = 4√2

BTGmoderatorDC wrote:

The figure above represents a semicircular archway over a flat street. The semicircle has a center at O and a radius of 6 feet. What is the height h, in feet, of the archway 2 feet from its center?


A. √2

B. 2

C. 3

D. 4√2

E. 6

Since the radius is 6 feet, we can create a right triangle with hypotenuse of 6, leg of 2, and leg of h; thus, we have:

2^2 + h^2 = 6^2

4 + h^2 = 36

h^2 = 32

h = √16 x √2 = 4√2

Answer: D