   SEARCH HOME Math Central Quandaries & Queries  Question from Todd: Good Day. I have to figure out the square footage of a quonset style roof that's not playing by the rules The building dimensions (rectangular) are 63'x135' the height of the roof is 9.25'. It not an entire Quonset, It's that style of roof,(curved). There are concrete block walls 10' up to the metal roof. Thank you! Todd Hi Todd,

I drew a sketch of the end of the building, not to scale, and added some lines and labels. The area of the roof is the length of the arc ADB times the length of the building which is 135 feet. I am assuming that the arc ADB is an arc of a circle. C is the center of the circle and M is the midpoint of the chord AB. The length of the chord ADB is the radius of the circle times the measure of the angle BCA measured in radians.

First let's find the radius $r=|CA|$ of the circle. The triangle AMC is a right triangle with sides of length $|AM| = \large \frac{63}{2} \normalsize = 31.5$ feet, $|MC| = |DC| - |DM| = r - 9.25$ feet and a hypotenuse of length $r$ feet. Hence from Pythagoras Theorem

$r^2 = 31.5^2 + (r-9.25)^2 = 31.5^2 + r^2 - 18.5 r + 85.56.$

Solving for $r$ gives $r = 104.06$ feet.

The sine of the angle MCA is $\large \frac{|AM|}{|CA|} = \frac{31.5}{104.06} \normalsize = 0.3027$ and using the inverse sine function on my calculator I get that the measure of the angle MCA is 0.3075 radians. Hence the measure of the angle BCA is $2 \times 0.3075 = 0.6150$ radians.

Finally the area of the roof is the length of the arc ADB times the length of the building which is

$104.06 \times 0.6150 \times 135 = 8640 \mbox{ square feet.}$

I hope this helps,
Harley      Math Central is supported by the University of Regina and the Imperial Oil Foundation.