Systems of equations (but there might be trig in this?)

Question

A path across a diagonal of a rectangular lot is 65 m long. If the area of the lot is 1500 m^2, what are the dimensions of the lot?

Answer 1

Math problems often have “magic numbers” (numbers that make the calculations easy so you can concentrate on solving the problem and the grading is easier).  This problem has:

    a diagonal = 65  =  5*13

    an area of rectangle = 1500 =  2*2*3*5*5*5           [the prime factors]

          the area also is:              5*(5) * 5*(2*2*3)

                             or:                   5*(5)  * 5*(12)

Now, there some “Pythagorean Triples” often used for triangles that use integers: 

       3, 4, 5

     5, 12, 13           (or   5*5,  5*12,  5*13)

Sides of triangle:           25,     60,      65

                               

Answer 2

Hi Haley.

 

Let's let L be the length of the lot and W be the width of the lot.

 

Then LW = the area of the lot which is 1500 m²

But we also know that the diagonal across the lot is 65m.

 

Recall the relationship between the two legs of a right triangle (a and b)  and the hypotenuse (c).

a² + b² = c²

or in our case L² + W² = 65²

From the first equation LW = 1500, 

we can say that L = 1500/W

and can substitute this into our equation above,

 

(1500/W)2 + W² = 65²

 

We can use algebra to solve this.

2250000/W² + W² = 4225

2250000 + W⁴ = 4225W²

Continuing

W⁴ - 4225W² + 2250000 = 0

(W2 - 625)(W2 - 3600) = 0

W = ±25, ±60

 

Since we can't have a negative distance, discard the negative possible answers.  Therefore 

W = 25 or 60

 

Back to the area of the lot

LW = 1500

if W = 25, L = 1500/25 = 60

if W = 60, L = 1500/60 = 25

 

So the dimensions of the lot are 25m x 60m