A farmer wishes to start a 100 , m^2 rectangu | vedclass

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(B) Let the length of the side parallel to the house wall be $l$ and the width of the other two sides be $b$.The area of the rectangular garden is giv

Vedclass · Accepted Answer

$20 \, m 	imes 5 \, m$

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(B) Let the length of the side parallel to the house wall be $l$ and the width of the other two sides be $b$.The area of the rectangular garden is given by $A = l 	imes b = 100 \, m^2$.The total length of the barbed wire used for three sides is $l + 2b = 30 \, m$.From the area equation,$l = 100/b$.Substituting this into the perimeter equation: $(100/b) + 2b = 30$.Multiplying by $b$: $100 + 2b^2 = 30b$,which simplifies to $2b^2 - 30b + 100 = 0$.Dividing by $2$: $b^2 - 15b + 50 = 0$.Factoring the quadratic equation: $(b - 10)(b - 5) = 0$.Thus,$b = 10$ or $b = 5$.If $b = 5 \, m$,then $l = 100/5 = 20 \, m$. Checking the perimeter: $20 + 2(5) = 30 \, m$ (Matches).If $b = 10 \, m$,then $l = 100/10 = 10 \, m$. Checking the perimeter: $10 + 2(10) = 30 \, m$ (Matches).Comparing with the given options,the dimensions $20 \, m 	imes 5 \, m$ are provided.

$A$ farmer wishes to start a $100 \, m^2$ rectangular vegetable garden. Since he has only $30 \, m$ of barbed wire,he fences three sides of the garden,using his house compound wall as the fourth side. The dimensions of the garden are:

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