# Find Dimensions of a Log

• February 6th 2013, 12:40 AM
BobBali
Find Dimensions of a Log
Good day to All,

The question asked what are the dimensions of a square that can be cut from a cross-section of a tree log. The diameter of the log is given as 40cm. In my pic below it may not look like a square, but it is; sorry for the error produced in drawing it.

So, what i first deduced is that if x cm of equal length is cut to make a square, we are left with a square of dimensions (40-2x).
Then i found the area of the log's cross-section using:

$A= \pi (r)^2. This is = 1257cm^2$

Then I used length x width = Area to make a quadratic equation to solve for x:

$(-2x+40)(-2x+40) = 1257$
$4x^2 - 160x +1600=1257$
$4x^2 -160x +343= 0$

Solving the Quad equ gives: x = 37.4 or 2.3 (37.4 exceeds)
The logical ans is 2.3, So> 40 - 2(2.3)= 36.4 cm
However, the ans is 33.9cm...
• February 6th 2013, 04:37 PM
chiro
Re: Find Dimensions of a Log
Hey BobBali.

The thing where you slipped up was when you equated the area of the circle with the area of the square: you need to account for the left over area.

I am going to use a technique involving integration to get the value of the square length.

Let x^2 + y^2 = r^2 denote the relationship of a circle at the origin with radius r. We will now solve a boundary value problem where

2*Integral (-r,r) SQRT(y^2 - x^2)dx = Area_Of_Circle = x^2 + LeftOverArea = 400*pi.

The "LeftOverArea" is the area that is not part of the square, but part of the circle.

We have 2*[Integral(-r,-r+y0)SQRT(r^2 - x^2)dx + Integral(r-y0,r)SQRT(r^2 - x^2)dx] = LeftOverArea = 400*pi - x^2

By solving LeftOverArea, we solve x, and do to this, we solve for y0 where x = r - y0.

I will complete this a little later on, but if you can pick up where I am going with this then feel free to complete it yourself.
• February 7th 2013, 08:32 PM
bjhopper
Re: Find Dimensions of a Log
using simple geometry I get an answer of 28.3 cm