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Sagot :
SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Represent the sides of the rectangle
Let the length be represented by l
Let the width be represented by w
STEP 2: Interpret the statements in the question.
[tex]\begin{gathered} length\text{ is 1m more than the double of the width:} \\ double\text{ of the width}\Rightarrow2w \\ 1m\text{ more than the double}\Rightarrow2w+1 \\ \therefore l=2w+1 \end{gathered}[/tex]STEP 3: Equate the area of the rectangle to given measure
[tex]\begin{gathered} Area=length\times width \\ length=2w+1,width=w \\ Area=(2w+1)\cdot w=28 \\ By\text{ simplification,} \\ w(2w+1)=28 \end{gathered}[/tex]STEP 4: Solve for the width
[tex]\begin{gathered} w(2w+1)=28 \\ By\text{ expansion,} \\ 2w^2+w=28 \\ Subtract\text{ 28 from both sides} \\ 2w^2+w-28=28-28 \\ 2w^2+w-8=0 \end{gathered}[/tex]STEP 5: Solve the equation using quadratic formula
[tex]quadratic\text{ formula}\Rightarrow\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]From the equation,
[tex]a=2,b=1,c=-28[/tex]By substitution,
[tex]\begin{gathered} w_{1,\:2}=\frac{-1\pm\sqrt{1^2-4\cdot\:2\left(-28\right)}}{2\cdot\:2} \\ \sqrt{1^2-4\times2(-28)}=15 \\ By\text{ substitution,} \\ w_{1,\:2}=\frac{-1\pm \:15}{2\cdot \:2} \\ \mathrm{Separate\:the\:solutions} \\ w_1=\frac{-1+15}{2\cdot \:2},\:w_2=\frac{-1-15}{2\cdot \:2} \\ w=\frac{-1+15}{2\times2}=\frac{14}{4}=\frac{7}{2}=3.5 \\ w=\frac{-1-15}{2\cdot\:2}=\frac{-16}{4}=-4 \end{gathered}[/tex]Since the width cannot be negative, this means that the value of the width is 3.5m
STEP 6: Solve for the length
By substitution into the formula in step 2, we have:
[tex]\begin{gathered} l=2w+1 \\ l=2(3.5)+1=8 \\ l=8 \end{gathered}[/tex]Hence,
length = 8m
width = 3.5m
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