In the diagram, each of the circles with centres $X, Y$ and $Z$ is tangent to the two other circles. Also, the circle with centre $X$ touches three sides of rectangle $P Q R S$ and the circle with centre $Z$ touches two sides of rectangle $P Q R S$, as shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-028.jpg?height=431&width=656&top_left_y=1422&top_left_x=691)

If $X Y=30, Y Z=20$ and $X Z=40$, the area of rectangle $P Q R S$ is closest to
(A) 3900
(B) 4100
(C) 4050
(D) 4000
(E) 3950

## Part C: Each correct answer is worth 8.
Let the radii of the circles with centres $X, Y$ and $Z$ be $x, y$ and $z$, respectively.

The distance between the centres of two touching circles equals the sum of the radii of these circles.

Therefore, $X Y=x+y$ which means that $x+y=30$.

Also, $X Z=x+z$ which gives $x+z=40$ and $Y Z=y+z$ which gives $y+z=20$.

Adding these three equations, we obtain $(x+y)+(x+z)+(y+z)=30+40+20$ and so $2 x+2 y+2 z=90$ or $x+y+z=45$.

Since $x+y=30$ and $x+y+z=45$, then $30+z=45$ and so $z=15$.

Since $y+z=20$, then $y=20-z=5$.

Since $x+z=40$, then $x=40-z=25$.

Knowing the radii of the circles will allow us to calculate the dimensions of the rectangle.

The height of rectangle $P Q R S$ equals the "height" of the circle with centre $X$, which is length of the diameter of the circle, or $2 x$.

Thus, the height of rectangle $P Q R S$ equals 50 .

To calculate the width of rectangle $P Q R S$, we join $X$ to the points of tangency (that is, the points where the circle touches the rectangle) $T$ and $U$ on $P S$ and $S R$, respectively, $Z$ to the points of tangency $V$ and $W$ on $S R$ and $Q R$, respectively, and draw a perpendicular from $Z$ to $H$ on $X U$.

Since radii are perpendicular to tangents at points of tangency, then $X T, X U, Z V$, and $Z W$ are perpendicular to the sides of the rectangle.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-057.jpg?height=434&width=672&top_left_y=1255&top_left_x=781)

Each of $X T S U$ and $Z W R V$ has three right angles, and so must have four right angles and so are rectangles.

Thus, $S U=T X=25$ (the radius of the circle with centre $X$ ) and $V R=Z W=15$ (the radius of the circle with centre $Z$ ).

By a similar argument, $H U V Z$ is also a rectangle.

Thus, $U V=H Z$ and $H U=Z V=15$.

Since $X H=X U-H U$, then $X H=10$.

By the Pythagorean Theorem, $H Z=\sqrt{X Z^{2}-X H^{2}}=\sqrt{40^{2}-10^{2}}=\sqrt{1500}=10 \sqrt{15}$ and so $U V=10 \sqrt{15}$.

This means that $S R=S U+U V+V R=25+10 \sqrt{15}+15=40+10 \sqrt{15}$.

Therefore, the area of rectangle $P Q R S$ is $50 \times(40+10 \sqrt{15})=2000+500 \sqrt{15} \approx 3936.5$.

Of the given choices, this answer is closest to (E) 3950.

ANSWER: (E)

#