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The surface of a mirror is composed of a rectangular piece..

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The surface of a mirror is composed of a rectangular piece..

by swerve » Sun Dec 10, 2017 1:24 pm
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The surface area of a mirror is composed of a rectangular piece that is 9 feet long and two semicircular pieces whose diameters are equal to the width of the recatngular piece, as shown in the figure. If the ratio of the area of the rectangular piece to the total area of the two semicircular pieces is 9/pi, what is the width of the rectangular piece, in feet?

A. 1
B. 2
C. 3
D. 4
E. 5

The OA is D.

I solved it this way:
Area of rectangular piece = 9b ( Let b = width of rectangle = diameter of the 2 semicircles)
Area of 1 semicircular piece = pi.(D^2)/4
Area of 2 semicircular pieces = pi.(D^2)/2

So, putting the ratio = 9/pi, I get D = b =2 -> which is not the OA.

Experts, can you help me with this PS question please? Thanks!
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by OWN » Sun Dec 10, 2017 1:52 pm
Hi swerve,

You seem to have doubled the the area of the two semicircular sections.

Here is how I did it:

Given:
Length = 9
Assume w = width;
Diameter = w, therefore, radius = w/2
Area of rectanlge/area of two semicircles = 9/pi

Calculations
Area of the rectangular section = 9*w = 9w
Combined area of two semicircles with radius w/2 = Area of one full circle = Pi* r^2 = Pi * [(w/2)^2] (note that the two is included in the squaring operation), so Area of circle = Pi*w^2/4
Since we know that dividing the area of the rectangle with the combined area of the semi-circles gives us 9/Pi, we can say:
$$\frac{9w}{\left[\frac{\pi\times w^2}{4}\right]}=\frac{9}{\pi}$$
As you can see, the numerator of both expressions is multiplied by 9 and the denominator of both expressions is multiplied by Pi, so we can get rid of them:
$$\frac{1\times w}{\left[\frac{w^2}{4}\right]}=\frac{1}{1}$$
We can now multiply both the numerator and denominator by 4 for the left-hand expression, which gives:
$$\frac{4w}{\left[w^2\right]}=1$$
Then multiply both sides of the equation by w^2:
$$4w\ =\ w^2\ ;\ w^2-4w\ =\ 0\ ;\ w\left(w-4\right)\ =\ 0;\ so\ w\ =\ \left\{0,4\right\}$$
Since we know that w is nonzero (otherwise the shape would not exist), w must equal 4, which is OA D.
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