# Land Parcel - maths question

So there’s this piece of land (scale has gone out of the window):

AB is 1m, AC is 30m, CD = 5m.
The total area is 90sq m, that’s the rectangle ABXC = 30 & triangle BXD = 60.

How do I divide that piece of land into 6 equal slices, each 15sq m, without trial and error?

The dividing lines need to run perpendicularly from line AC to join BD.
[strike]For Part I (yes, there will be a Part II),[/strike] what are the positions of the dividing lines along line AC?

Are you allowed to use calculus in solving the problem?

(If it needs calculus, then I was never going to get there…)

And I’ve realised there’ll be no Part II.

Calculus isn’t necessary. All that’s needed is to find a formula that describes the areas involved, which is achievable though understanding how areas of rectangles and triangles are calculated.

Consider Point A to be the origin on a graph, and the line AC to be on the x-axis of the graph. Then work out the formula for the area from 0 to a point x on the x-axis. (x is not to be confused with the vertex named with the capital letter X!) Then solve for each x when the area = 15, 30, 45, 60, and 75.

Just pointing out the total area is 105 sq m. the triangle BXD = 75.

Or calculate the area of the trapezoid from the formula. area = 1/2(base1 + base2).height …1/2(1+6).30 … 3.5x30 = 105

So your calculating 6 areas, or 6 trapezoids of 17.5 sqm each.

No Mick the triangle is 60 meters squared. b=4. h=30.

First divide the rectangle into two equal plots. Then take the triangle and divide it into 4 plots. Do this by drawing a line from pt B. To a point at x+1 up line CD. Then another line from B to a point x+2. Then a final line from B to a point x+3 up line CD.

That should do it. Please check this I’m typing this on a phone and visualizing the lines in my head.

Edit: Sorry I just reread the question and missed the last requirement regarding the perpendicular lines. (like I said tiny screen sorry)

You are correct, scrub what I just said. Dont know why I was looking DX and thinking that was 5, then adding 1 for XC. Stupid.

Think I’m there, but that was quite painful:

So trying to find the first x, where it’s a point on AC, so that the area is 15.
Area = (x) [rectangle] + x(y/2) [triangle]
15 = x + xy/2
15 - x = xy/2
2(15 - x) = xy
(30 - 2x) / x = y

But then we’ve got 2 unknowns, x and y (which is the base of the triangle part).

So to find y in terms of x, we need the equation of the line BD. That’s y = 2x/15, so substitute it in:
(30 - 2x) / x = 2x/15, which simplifies to a quadratic 2x^2 + 30x - 450 = 0.
Solutions are -24.27 and 9.27.

So check if 9.27 works.
That’s (9.27 x 1) + (9.27 x y/2)
9.27 + (9.27 x (1.236/2))
9.27 + 5.73
15 :yay:

The other 4 points ramp up from the 1st equation:
2x^2 + 30x - 450 = 0.

got a bit confused again, but it’s only the 450 that changes:
2x^2 + 30x - 450 = 0.
2x^2 + 30x - 900 = 0. etc.

Which gives points along the line as:
9.27
15
19.54
23.42
26.86

Rather than attempt this algebraically, I think I’d probably go for trial and error. It avoids thinking.

Start your favourite spreadsheet program and calculate the cumulative area starting at the left and working over to the right, in steps of (say) 0.1m. You’ll end up with a list of 300 numbers.

Then pick out the answers closest to 15,30,45 … etc square meters. Will take you about 3 minutes.

EDIT: got nothing better to do, so:
9.3
15
~19.55
23.4
26.9

Which is pretty much what Nuit said.

Great!

If you use integral calculus, you express the line BD as a function of slope 4/30 and y-intercept 1. In other words,

f(x)=(2/15)x + 1

To find the area under the line defined by this function, take the integral of this function. We get:

F(x)=(1/15)x^2 + x. (Remember, the integral of x^n = (1/(n+1))x^(n+1).)

Note that F(0) = 0, which simplifies things a bit.

To figure out the point x where the area from 0 to x is 15, set (1/15)x^2 + x = 15. (Note that this is equivalent to Nuit’s 2x^2 + 30x - 450 = 0.) Then solve for x (through the Quadratic Formula, or factoring, or whichever means is easiest). Do the same for the areas 30, 45, etc.

The results should be the same as Nuit’s.

Nice work Nuit!! Also nice application of intergal calculus Chris.

Ah, of course. That phrase takes me back to extra double maths.

How could I forget!?

Nicely worked .