The length of a rectangle is 5 m more than its width. If the area of the rectangle is 15 m2, what are the dimensions of the rectangle, to the nearest tenth of a metre?

Question

Isaiah Anthony · Accepted Answer

#"length " = 7.1 m" " # rounded to 1 decimal place
#"width" color(white)(..) = 2.1m" "# rounded to 1 decimal place

#color(blue)("Developing the equation")# Let length be #L# Let width be #w# Let area be #a# Then #a=Lxxw# ............................Equation(1) But in the question it states: "The length of a rectangle is 5m more than its width"#->L=w+5# So by substituting for #L# in equation(1) we have: #a=Lxxw" "->" "a=(w+5)xxw# Written as: #a=w(w+5)# We are told that #a=15m^2# #=>15=w(w+5)....................Equation(1_a)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Solving for value of width")# Divide the bracket by two. #15=w^2+5w# Take 15 off of both sides. #w^2+5w-15=0# Not that #3xx5=15# However, #3+-5!=5# So using the standardised formula: #y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)# #a=1; b=5; c=-15# #=>x=(-5+-sqrt((5)^2-4(1)(-15)))/(2(1))# #=>x=-5/2+-sqrt(85)/2# Since a negative value defies logic, we utilize #x=-5/2+sqrt(85)/2" "=" "2.109772..# #color(green)("The question instructs that are to use the nearest 10th")# #"width "= x=2.1" "# rounded to 1 decimal place #color(red)(" "uarr)# #color(red)(" This comment is very important")# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Solving for value of length")# #a=Lxxw" "-> 15=Lxx2.109772..# #=>L=15/2.109772.. = 7.1.9772..# length# = 7.1 # rounded to 1 decimal place

Nathan Axel · Answer

undefined Let's represent the width of the rectangle as x meters. Since the length is 5 meters more than the width, we can represent the length as (x + 5) meters.

The formula for the area of a rectangle is length multiplied by width. So, we have the equation:

Area = Length × Width

Substituting the given values, we get:

$15 = (x + 5) \times x$

Expanding and rearranging the equation:

$15 = x^2 + 5x$

Now, we can rewrite the equation in standard quadratic form:

$x^2 + 5x - 15 = 0$

Using the quadratic formula to solve for x:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$

where $a = 1$, $b = 5$, and $c = -15$

$x = \frac{{-5 \pm \sqrt{{5^2 - 4(1)(-15)}}}}{{2(1)}}$

$x = \frac{{-5 \pm \sqrt{{25 + 60}}}}{2}$

$x = \frac{{-5 \pm \sqrt{{85}}}}{2}$

$x \approx \frac{{-5 \pm 9.2}}{2}$

$x_1 \approx \frac{{-5 + 9.2}}{2} \approx 2.1$ meters
$x_2 \approx \frac{{-5 - 9.2}}{2} \approx -7.1$ meters

Since the width cannot be negative, we discard the negative solution. Therefore, the width of the rectangle is approximately 2.1 meters.

Now, we can find the length:

Length $= x + 5 = 2.1 + 5 = 7.1$ meters

Therefore, the dimensions of the rectangle are approximately 2.1 meters by 7.1 meters.