The area of a rectangle is 27 square meters. If the length is 6 meters less than 3 times the width, then find the dimensions of the rectangle. Round off your answers to the nearest hundredth.?

2 Answers

\color{blue}{6.487 m, 4.162m}

Explanation:

Let L & B be the length & width of rectangle then as per given conditions,

L=3B-6 \ .........(1)

LB=27 \ .........(2)

substituting the value of L from (1) into (2) as follows

(3B-6)B=27

B^2-2B-9=0

B=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-9)}}{2(1)}
=1\pm\sqrt{10}

since, B>0, hence we get

B=1+\sqrt{10} &
L=3(1+\sqrt{10})-6
L=3(\sqrt{10}-1)

Hence, length & width of given rectangle are

L=3(\sqrt{10}-1)\approx 6.486832980505138\ m
B=\sqrt{10}+1\approx 4.16227766016838\ m

Jun 24, 2018

length = m = 6.49
width = n = 4.16

Explanation:

Assume that length = m and width = n.
The rectangle's area will hence be mn.
The first statement states "The area of a rectangle is 27 square meters.
Hence mn=27.
The second statement states "If the length is 6 meters less than 3 times the width..."
Therefore m=3n-6
Now you can create a system of equations:
mn=27
m=3n-6

Replace m in the first equation with 3n-6:
(3n-6)*n=27
Expand the bracket:
3n^2-6*n=27
Make a quadratic equation:
3n^2-6*n-27=0
Simplify by dividing everything by 3:
n^2-2*n-9=0
Use (-b+-sqrt(b^2-4ac))/(2a), where a is 1, b is -2, and c is -9:
=(2+-sqrt(4+36))/(2)
=1+-sqrt10

Since dimensions must be positive, n will be 1+sqrt10, which to the nearest hundredths is 4.16.
Use mn=27 to find m:
m(1+sqrt10)=27
m=27/(1+sqrt10)
m=6.49