The area of a playground is 168 yd2. The width of the playground is 2 yd longer than its length. Find the length and width of the playground. Then, enter the sum of the length and width in the provided grid.

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Length = 12yards

Width = 14yards

Sum = 26yards

Step-by-step explanation:

Let assume the play ground is rectangular in form.

Area of a rectangle = length × width

Given area of the playground = 168yd²

If the width of the playground is 2yd longer than its length, then:

W = L+2

Substituting W = L+2 into the formula above we have:

A = L × (L+2)

A = L²+2L

168 = L²+2L

L²+2L-168 = 0

L²+14L-12L-168 = 0

L(L+14)-12(L+14) = 0

(L-12)(L+14) = 0

L-12 = 0 and L+14 =0

L = 12 and -14

Taking the positive length, L = 12yards

If Length = 12yards

W = L+2

W = 12+2

Width = 14yards

Sum of the length and width will be:

L+W= 12+14

= 26yards

The length  of playground is 12 yd

The width of playground is 14 yd

The sum of the length and width of playground is 26 yd

Step-by-step explanation:

The area, A of the playground is given as 168 yd²

The formula for the area of the playground is given as follows;

Area, A of playground = Length, L × Width, W

The Width =  length  + 2 yd or W  = L + 2

Therefore, A = L × (L + 2)

A = L² + 2·L

Where we have A = 168 yd², we have the area equation presented as follows

168 = L² + 2·L or

L² + 2·L - 168 = 0

Factorizing the above equation gives

L = 12 or L = -14 we note that the appropriate solution is L = 12 yd

Therefore W = L + 2 = 12 + 2 = 14 yd

The length  of playground = 12 yd

The width of playground = 14 yd

The sum of the length and width of playground = 12 + 14 = 26 yd.

so the possibilities of the measurement of the side of the triangle are either it is longest side or not the longest side

when it is the longest side

s < 25 + 35s < 60

if it is not the longest side

35 < 25 + s10 < sso 10 < s < 60so the perimeter70 < p < 130

could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?