Scott Woodbury-Stewart

GMAT OFFICIAL GUIDE DS – If arc PQR above is a semicircle…

If arc PQR above is a semicircle…

Solution:

We are given a triangle inscribed inside a semicircle. When a triangle is inscribed inside a semicircle, the triangle must be a right triangle. Therefore, triangle PQR is a right triangle with a right angle at Q. With this knowledge, let’s start by sketching the diagram.

semicircle

Notice that in sketching the diagram we added point S to create line segment QS. Thus, we see that a represents the length of side PS and b represents the length of side SR. We also added in that angle PQR is a 90-degree angle because it’s a right angle.
We see now that we have 3 right triangles.

1) Right triangle PQR

2) Right triangle SQR

3) Right triangle PQS

We can let side PQ = x and side QR = y.

semicircle

Using the Pythagorean theorem, we can create a few equations.

Equation 1

a^2 + 2^2 = x^2

a^2 + 4 = x^2

Equation 2

b^2 + 2^2 = y^2

b^2 + 4 = y^2

Equation 3

x^2 + y^2  = (a + b)^2

x^2 + y^2  = a^2 + b^2 + 2ab

We can combine equations one and two by adding them together. We obtain:

a^2 + 4 + b^2 + 4  = x^2 + y^2

a^2 + b^2 + 8  =  x^2 + y^2

Since a^2 + b^2 + 8 = x^2 + y^2, we can substitute a^2 + b^2 + 8 for x^2 + y^2 in the equation x^2 + y^2  = a^2 + b^2 + 2ab.

a^2 + b^2 + 8 = a^2 + b^2 + 2ab

8 = 2ab

4 = ab

We are asked to determine the value of diameter PR, which is the sum of a and b. If we can determine a value for a or b, we will have enough information to determine the values for a and b, and thus determine a value for the sum of a and b.

Statement One Alone: 

a = 4

Since a = 4 and ab = 4, we know that b = 1. Thus, the sum of a and b is 5. Statement one alone is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone: 

b = 1

Since b = 1 and ab = 4, we know that a = 4. Thus, the sum of a and b is 5. Statement two alone is also sufficient to answer the question.

Answer: D 

Alternate solution: 

semicircle

When the altitude (or height) is drawn from the right angle to the hypotenuse of a right triangle, it not only creates two more right triangles (like in our problem), but also, all three right triangles are similar to one another. We know that the sides of similar triangles are proportional. Therefore, the ratio of the longer leg to the shorter leg of one triangle must be equal to the ratio of the longer leg to the shorter leg of another triangle. Using triangles QPS and QRS, this will translate to a/2 = 2/b. Cross multiplying, we have ab = 4. Once again, statement one alone is sufficient and statement two alone is sufficient.

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