Question

The area of quadrilateral PRQS, is

• A. $\frac{3\sqrt{15}}{2}$
• B. $\frac{15\sqrt{3}}{2}$
• C. $\frac{5\sqrt{3}}{2}$
• D. $\frac{5\sqrt{15}}{2}$

Answer 1

No option matches the calculated area

Answer 2

The problem asks for the area of the quadrilateral PRQS.

We are given a circle with center O and radius $r = 5$.

P is a point on the circle, so OP = 5.

Q is an external point. QR and QS are tangents from Q to the circle at R and S respectively.

We are given the length of the line segment PQ = 13.

From the diagram, it is evident that points P, O, and Q are collinear, with O lying between P and Q.

Therefore, the distance PQ is the sum of PO and OQ.

$PQ = PO + OQ$

$13 = 5 + OQ$

$OQ = 13 - 5 = 8$.

Now, consider the right-angled triangle ORQ.

OR is the radius, so OR = 5.

OQ is the hypotenuse, OQ = 8.

QR is the tangent length.

Using the Pythagorean theorem in $\triangle ORQ$:

$QR^2 = OQ^2 - OR^2$

$QR^2 = 8^2 - 5^2$

$QR^2 = 64 - 25 = 39$

$QR = \sqrt{39}$.

Since QR and QS are tangents from the same external point Q, their lengths are equal: $QS = QR = \sqrt{39}$.

To find the area of quadrilateral PRQS, we can divide it into two triangles: $\triangle PQR$ and $\triangle PQS$.

The vertices are P, R, Q, S. This implies the quadrilateral is PRQS, with diagonals PS and RQ.

Let's set up a coordinate system to calculate the area easily.

Let the center O be at the origin (0,0).

Since P, O, Q are collinear and O is between P and Q, and P is on the circle, P must be at $(-5,0)$.

Since OQ = 8 and Q is on the positive x-axis (from the diagram's implied orientation), Q must be at $(8,0)$.

Now, we need the coordinates of R and S.

R and S are the points of tangency from Q(8,0) to the circle $x^2+y^2=5^2=25$.

The radius OR is perpendicular to the tangent QR.

Let R be $(x_R, y_R)$.

The slope of OR is $m_{OR} = \frac{y_R - 0}{x_R - 0} = \frac{y_R}{x_R}$.

The slope of QR is $m_{QR} = \frac{y_R - 0}{x_R - 8} = \frac{y_R}{x_R - 8}$.

Since OR $\perp$ QR, $m_{OR} \cdot m_{QR} = -1$.

$\left(\frac{y_R}{x_R}\right) \left(\frac{y_R}{x_R - 8}\right) = -1$

$y_R^2 = -x_R(x_R - 8)$

$y_R^2 = -x_R^2 + 8x_R$.

Since R is on the circle, $x_R^2 + y_R^2 = 25$.

Substitute $y_R^2$:

$x_R^2 + (-x_R^2 + 8x_R) = 25$

$8x_R = 25$

$x_R = \frac{25}{8}$.

Now find $y_R$:

$y_R^2 = 25 - x_R^2 = 25 - \left(\frac{25}{8}\right)^2 = 25 - \frac{625}{64} = \frac{25 \times 64 - 625}{64} = \frac{1600 - 625}{64} = \frac{975}{64}$.

$y_R = \pm \sqrt{\frac{975}{64}} = \pm \frac{\sqrt{25 \times 39}}{8} = \pm \frac{5\sqrt{39}}{8}$.

So, the coordinates of R are $\left(\frac{25}{8}, \frac{5\sqrt{39}}{8}\right)$ and S are $\left(\frac{25}{8}, -\frac{5\sqrt{39}}{8}\right)$.

Now we can calculate the area of quadrilateral PRQS by splitting it into $\triangle PQS$ and $\triangle RQS$.

The line segment RS is a vertical line segment with x-coordinate $25/8$.

The length of the base RS is the difference in y-coordinates:

$RS = \frac{5\sqrt{39}}{8} - \left(-\frac{5\sqrt{39}}{8}\right) = \frac{10\sqrt{39}}{8} = \frac{5\sqrt{39}}{4}$.

Area of $\triangle RQS$:

Base = RS = $\frac{5\sqrt{39}}{4}$.

The height of $\triangle RQS$ with respect to base RS is the perpendicular distance from Q to the line containing RS (which is $x=25/8$).

The x-coordinate of Q is 8.

Height $h_Q = |x_Q - x_R| = \left|8 - \frac{25}{8}\right| = \left|\frac{64 - 25}{8}\right| = \left|\frac{39}{8}\right| = \frac{39}{8}$.

Area($\triangle RQS$) = $\frac{1}{2} \times RS \times h_Q = \frac{1}{2} \times \frac{5\sqrt{39}}{4} \times \frac{39}{8} = \frac{5 \times 39\sqrt{39}}{64} = \frac{195\sqrt{39}}{64}$.

Area of $\triangle PRS$:

Base = RS = $\frac{5\sqrt{39}}{4}$.

The height of $\triangle PRS$ with respect to base RS is the perpendicular distance from P to the line containing RS (which is $x=25/8$).

The x-coordinate of P is -5.

Height $h_P = |x_P - x_R| = \left|-5 - \frac{25}{8}\right| = \left|\frac{-40 - 25}{8}\right| = \left|\frac{-65}{8}\right| = \frac{65}{8}$.

Area($\triangle PRS$) = $\frac{1}{2} \times RS \times h_P = \frac{1}{2} \times \frac{5\sqrt{39}}{4} \times \frac{65}{8} = \frac{5 \times 65\sqrt{39}}{64} = \frac{325\sqrt{39}}{64}$.

Total Area of quadrilateral PRQS = Area($\triangle RQS$) + Area($\triangle PRS$)

Area(PRQS) = $\frac{195\sqrt{39}}{64} + \frac{325\sqrt{39}}{64} = \frac{(195+325)\sqrt{39}}{64} = \frac{520\sqrt{39}}{64}$.

Simplify the fraction $\frac{520}{64}$ by dividing by 8:

$\frac{520 \div 8}{64 \div 8} = \frac{65}{8}$.

So, Area(PRQS) = $\frac{65\sqrt{39}}{8}$.

This result does not match any of the given options. Let's re-check the problem statement and options.

The options contain $\sqrt{15}$ or $\sqrt{3}$. My result contains $\sqrt{39}$.

This implies there might be a different interpretation of the problem or a typo in the question/options.

However, based on the standard interpretation of the diagram and given values, the calculation is robust.

Given that this is a multiple choice question and my calculated answer is not among the options, there might be an error in the question itself or the provided options. However, since I must provide an answer from the options, I will state that the calculated answer does not match.