Question

El lote de la familia lópez tiene un lote de 16m*10m y se desea construir una casa de base rectangular en el centro de el.¿que dimensiones debe tener el rectángulo del piso de la casa si tiene un área de 91m2?

Answer 1

The dimensions of the house should be 7 m by 13 m.

If the house is to be centered, we will take the same amount from the width as we do from the length of the lot for the dimensions.  This gives us (10-x) and (16-x) as the dimensions.

The area of a rectangle is found by multiplying the length and width:
(10-x)(16-x) = 91

Multiplying the binomials we have:
10*16 - x*10 - x*16 -x*(-x) = 91
160 - 10x - 16x --x² = 91
160-10x-16x+x² =91

Combine like terms:
160-26x+x²=91

Rewrite this in standard form:
x²-26x+160=91

Subtract 91 from both sides:
x²-26x+160-91 = 91-91
x²-26x+69 = 0

Factoring this, we look for factors of 69 that sum to -26.  -23*-3 = 69 and -23+-3 = -26, so:

(x-23)(x-3) = 0

Using the zero product property we know that either x-23=0 or x-3=0, so x=23 or x=3. 

x was the amount we take off of the width and length of the lot; if we took 23m off of it, 10-23 gives us a negative amount, which is not realistic. This means both the width and length are subtracted by 3.

10-3 = 7 and 16-3 = 13.

These are the dimensions of the house.