## Front Porch Math

By Bobby Neal Winters

The following is a real problem that a carpenter with a great respect for trigonometry came to me with. Say you are going to add a porch of width

**to the side of a house and want to over it by adding to the roof as depicted in the picture below:**

*L*The question is how long do you want your rafter to be, i.e. what is the length of the segment

*. As it is there is not enough information in the problem. You have to know that the slope of the current roof is 7/12, (i.e. over a run of 12 feet the roof will rise 7) and they desire the porch roof to have a slope of 3/12 (i.e. over a run of 12 feet it will rise 3.) Yes, we could say that is a slope of 1/4, but these are how the numbers came to me, and it will all come out in the wash anyway.*

**D**I must confess that this problem took longer for me to solve than it might have because I was guided initially by the carpenters love of trigonometry. It can be solved that way, but this is a nice case study of how imposing a coordinate system can simplify the problem. Consider the following:

We've put on coordinates so that the outside point of the porch roof is a (0,0), the point where the old roof touches the house is (

*,0), and the point where the porch roof meets the old roof is (*

**L***).*

**x*, y***The equation for the line representing the porch roof is

**and for the old roof is**

*y=mx**. It is an exercise to work out that*

**y=nx-nL***and*

**x*= nL/(n-m)****. We may use the Pythagorean Theorem (or the distance formula) to calculate the value of D2=(x*)2+(y*)2 which after substitutions simplifies to:**

*y*=mnL/(n-m)*

*D2=(1+m2)n2L2/(n-m)***or**

*2*

*D=(1+m2)1/2***.**

*nL/(n-m)*For the given values of n and m, this works out to D=2.02598L.

Having done it this way, I also worked it by trigonometry, but I despair of reproducing those calculations without specialized mathematical software.