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Solve the following correct to 3 decimal place:

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(a)

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Find the width, length and area of a rectangular room which has a perimeter of $\\var{per2} m$, if its length is $\\var{ratio2}$ times its width.

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Let the width be $w$ and the length be $l$. Then from the information in the question we obtain the following:

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$l=\\var{ratio2}w$                    (Equation 1)

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$2l+2w=\\var{per2}$                (Equation 2)          (by considering the perimeter of a rectangle)

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Replacing $l$ in Equation 2 by $\\var{ratio2}w$ from Equation 1, we obtain:

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$2(\\var{ratio2}w)+2w=\\var{per2}$

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$\\var{2*ratio2+2}w=\\var{per2}$

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$w=\\var{precround(per2/(2*ratio2+2),3)}$ m

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Substituting into Equation 1 gives:

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$l=\\var{ratio2}\\times \\var{precround(per2/(2*ratio2+2),3)} =\\var{precround(ratio2*per2/(2*ratio2+2),3)}$ m

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And area = $l \\times w$ = $\\var{precround(ratio2*per2/(2*ratio2+2),3)}\\times\\var{precround(per2/(2*ratio2+2),3)}=\\var{precround((ratio2*per2/(2*ratio2+2))*(per2/(2*ratio2+2)),3)}$ m$^2$

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(b)

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The length of a rectangle is $\\var{lent3}$% greater than its width. If the perimeter of the rectangle is $\\var{per3} m$, find the length and width.

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Let the width be $w$ and the length be $l$. Then from the information in the question we obtain the following:

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$l=\\var{1+lent3/100}w$                    (Equation 1)         

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$2l+2w=\\var{per3}$                (Equation 2)          (by considering the perimeter of a rectangle)

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Replacing $l$ in Equation 2 by $\\var{1+lent3/100}w$ from Equation 1, we obtain:

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$2(\\var{1+lent3/100}w)+2w=\\var{per3}$

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$\\var{2*(1+lent3/100)+2}w=\\var{per3}$

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$w=\\var{precround(per3/(2*(1+lent3/100)+2),3)}$ m

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Substituting into Equation 1 gives:

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$l=\\var{(1+lent3/100)}\\times \\var{precround(per3/(2*(1+lent3/100)+2),3)} =\\var{precround((1+lent3/100)*per3/(2*(1+lent3/100)+2),3)}$ m

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Let the width be $w$ and the length be $l$. Now use the information in the question to set up 2 simultaneous equations involving $w$ and $l$.

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Find the width, length and area of a rectangular room which has a perimeter of $\\var{per2} m$, if its length is $\\var{ratio2}$ times its width.

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Width:   [[0]]$m$

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Length: [[1]]$m$

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Area:     [[2]]$m^2$

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Convert the precentage to decimals. E.g. 30% bigger would mean the length was 1.3x.

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The length of a rectangle is $\\var{lent3}$% greater than its width. If the perimeter of the rectangle is $\\var{per3} m$, find the length and breadth.

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Length:   [[0]]$m$

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Width:  [[1]]$m$

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Algebra word problems using area and perimeter.

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rebelmaths

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