In trapezoid PQRS, PQ ∥ SR.

What is the area of PQRS in simplest radical form?

Image: www(dot)connexus(dot)com/content/media/941080-7222013-111302-AM-598825301(dot)gif

a. 50 + 50√3
b. 110 + 50√3
c. 110√3 - 50
d. 210√3 = 150

1 answer

We can find the area of trapezoid PQRS using the formula:

A = 1/2 (base1 + base2) x height

Since PQ and SR are parallel, we can see that base1 = PQ and base2 = SR. To find the height, we can draw a perpendicular line segment from P to SR, and call the point where it intersects SR "T". Then, we can see that triangle PQT is a 30-60-90 triangle (because PQ = QR), and the height of the trapezoid from P to SR is equal to QT.

Using the properties of 30-60-90 triangles, we can see that QT = PQ/2 times the square root of 3. Therefore:

A = 1/2 (PQ + SR) x QT
A = 1/2 (PQ + SR) x (PQ/2 times the square root of 3)
A = 1/4 (PQ^2) times the square root of 3 + (PQ x SR) / 2

To find the area of PQRS, we need to find PQ and SR. We can see from the diagram that PQ = 10 and SR = 20.

Plugging these values into our formula:

A = 1/4 (10^2) times the square root of 3 + (10 x 20) / 2
A = 25 times the square root of 3 + 100

Simplifying:

A = 100 + 25 times the square root of 3

Therefore, the answer is (a) 50 + 50 times the square root of 3.
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