Question

A rectangular pool 12 meters by 8 meters is surrounded by a walkway of width x meters. At

Answer 1

Answer:

x=3 meters

Step-by-step explanation:

step 1

Find the area of the rectangular pool

$A=LW$

we have

$L=18\ m\\W=12\ m$

substitute

$A=18(12)=216\ m^2$

step 2

Find the area of rectangular pool including the area of the walkway

Let

x ----> the width of the walkway

we have

$L=(18+2x)\ m\\W=(12+2x)\ m$

substitute

$A=(18+2x)(12+2x)$

step 3

Find the area of the walkway

To find out the area of the walkway subtract the area of the pool from the area of rectangular pool including the area of the walkway

so

$A=(18+2x)(12+2x)-216$

step 4

Find the value of x if the area of the walkway equal the area of the pool

so

$(18+2x)(12+2x)-216=216$

Solve for x

$(18+2x)(12+2x)=432\\216+36x+24x+4x^{2}=432\\4x^{2} +60x-216=0$

Solve the quadratic equation by graphing

The solution is x=3 meters

see the attached figure

Answer 2

Answer:

$x=-4$

Step-by-step explanation:

To calculate the area of the walkway, we have to subtract the total area (pool plus walkway) with the pool area. So first we calculate these two:

Pool area:

$A_{pool}=b.h=(12m)(8m)=96m^{2}$

Total area (pool plus walkway):

$A_{total}=(x+12+x)(x+8+x)=(12+2x)(8+2x)\\A_{total}=96+24x+16x+4x^{2} \\A_{total}=4x^{2}+40x+96$

Walkway area:

The walkway area is obtained by subtracting the pool are from the total area:

$A_{total}-A_{pool}=A_{walkway}$

$A_{walkway}=4x^{2}+40x+96-96=4x^{2}+40x$

Now, the problem is asking for x value that makes equal the pool area and the walkway area.

$A_{walkway}=4x^{2}+40x=96\\4x^{2}+40x-96=0\\\frac{4x^{2}+40x-96}{4} =0\\x^{2}+10x+24=0$

Now, we factorize this quadratic expression:

$x^{2}+10x+24=0\\(x+6)(x+4)=0$

Therefore,

$x=-6$ and $x=-4$

The correct answer is -4, we demonstrate this by replacing this value inside the equation for the area of the walkway:

$A_{walkway}=4x^{2}+40x=4(-4)^{2}+40(-4)=4(16)-160=64-160=-94m^{2}$

(We ignore the negative sign)