Solution

From the figure we do know that those are right triangles. So we can use the Pythagorean theorem to find the area of the two and then seeing which quantity is greater

The triangle \(PQR\) we do have that the area is \(\frac{1}{2}*2\sqrt{5}*\sqrt{5} = 5\)

For the second triangle we have the side RS = 3 and PR is unknown but PR is also the hypotenuse of the triangle PQR. As such, using the theorem we do have that

\((PR)^2=(PQ)^2 + (QR)^2\)

\((2\sqrt{5})^2 + (\sqrt{5})^2 = 20+5 = \sqrt{25}=5\)

Using again the theorem for the triangle \(PSR\)

\(3^2+ (PS)^2=5^2\)

\((PS)^2=16\)

\(PS=4\)

The area of \(PSR\) is \(\frac{1}{2}*4*3=6\)

As such, \(5<6\)

The answer is \(B\)

Quantity A is pretty straightforward since we are given the lengths of the two legs.
So one leg can be the base, and the other leg can be the height.

Area of triangle = (base)(height)/2
= (2√5)(√5)/2
= 10/2
= 5
So the area of triangle PQR = 5

In order to find the area of triangle PSR (Quantity B), we must first find the lengths of the two missing sides: side PR and side PS
Let's first find the length of side PR
Let's let x = the length of side PR
Since the red triangle (highlighted above) is a RIGHT triangle, we can apply the Pythagorean theorem to write: (2√5)² + (√5)² = x²
Simplify to get: 20 + 5 = x²
Simplify again: 25 = x²
Solve: x = 5 or x = -5
Since lengths cannot be negative, we can be certain that x = 5
In other words, the length of side PR = 5, which we'll add to the diagram below



At this point we need to find the length of side PS.
Since we already know that two sides have lengths 3 and 5, we might recognize that those lengths are part of the 3-4-5 Pythagorean Triplet, which means the missing side must have length 4
We get:


Now that we have all of the necessary lengths, we can find the area of a triangle PSR (Quantity B)
Area of triangle = (base)(height)/2
= (4)(3)/2
= 6
So the area of triangle PSR = 6

We get:
QUANTITY A: 5
QUANTITY B: 6

Answer: B

Cheers,
Brent
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