Question 1171867

To greet the 2020 SHS Graduates, a tarpaulin is to be set up along the national highway.If the area of the tarp is to be 35/4 m^2 and its perimeter is 27/2 meters, what should be the dimensions of the tarpaulin? Please help me answer this question. Arigathanks in advance.
<pre>Let the width and length be W, and L, respectively
Then we get: {{{matrix(4,3, 2L + 2W, "=", 27/2, (2L + 2W)/2, "=", (27/2)/2, L + W, "=", 27/4, L, "=", 27/4 - W)}}}
{{{matrix(1,3, LW, "=", 35/4)}}}
{{{matrix(1,3, (27/4 - W)W, "=", 35/4)}}} ------ Substituting {{{27/4 - W}}} for L
{{{matrix(1,3, 27W/4 - W^2, "=", 35/4)}}}
{{{matrix(1,3, 27W - 4W^2, "=", 35)}}} ------- Multiplying by LCD, 4
{{{matrix(2,3, 4W^2 - 27W + 35, "=", 0, 4W^2 - 20W - 7W + 35, "=", 0)}}}
4W(W - 5) - 7(W - 5) = 0
(4W - 7)(W - 5) = 0  
4W - 7 = 0                  OR                W - 5 = 0
4W = 7                      OR                W = 5
Width or {{{highlight_green(matrix(1,6, W, "=", 7/4, or, 1.75, m))}}}
If width = {{{matrix(1,2, 7/4, m)}}}, length, or {{{highlight_green(matrix(1,8, L, "=", 27/4 - 7/4, "=", 20/4, "=", 5, m))}}}
This can be VICE-VERSA as well, so {{{highlight_green(matrix(1,7, DIMENSIONS, "are:", 1.75, m, x, 5, m))}}}