Question

solve.
there is a grassy plain $2x^2 + 20$ft by $3x^2 + 12$ft. there is a walkway with a width of $x$ all the way around the grass.

1. find the area of the grass plus the walkway.
2. find the area of the walkway.

Explanation:

Sub - Question 1:

Step 1: Determine the length and width including the walkway

The grassy plain has dimensions \(2x^{2}+20\) ft (let's say length \(l = 2x^{2}+20\)) and \(3x^{2}+12\) ft (width \(w=3x^{2}+12\)). The walkway has a width of \(x\) all around, so we need to add \(2x\) to both the length and the width ( \(x\) on each side).
New length \(L=(2x^{2}+20)+2x=2x^{2}+2x + 20\)
New width \(W=(3x^{2}+12)+2x=3x^{2}+2x + 12\)

Step 2: Calculate the area of the grass plus walkway

The area \(A\) of a rectangle is \(A = L\times W\). So we multiply the new length and new width:

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Step 1: Calculate the area of the grassy plain

The area of the grassy plain \(A_{grass}\) is given by the product of its original length and width. So \(A_{grass}=(2x^{2}+20)(3x^{2}+12)\)

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Step 2: Calculate the area of the walkway

The area of the walkway \(A_{walkway}\) is the area of the grass plus walkway minus the area of the grass. From part 1, we know the area of grass plus walkway is \(6x^{4}+10x^{3}+88x^{2}+64x + 240\) and the area of grass is \(6x^{4}+84x^{2}+240\)

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Answer:

The area of the grass plus the walkway is \(6x^{4}+10x^{3}+88x^{2}+64x + 240\) square feet.

Sub - Question 2: