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The perimeter of a rectangular shop in the mall is $32$ meters. The area is $60$ square meters. What are the dimensions of the shop? $\newline$ $\_\_\_$ meters by $\_\_\_$ meters

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Area and perimeter: word problems

Full solution

Q. The perimeter of a rectangular shop in the mall is

32

meters. The area is

60

square meters. What are the dimensions of the shop?

\newline

\_\_\_

meters by

\_\_\_

meters

Perimeter Equation: Let's denote the length of the shop as $L$ meters and the width as $W$ meters. The perimeter $P$ of a rectangle is given by the formula $P = 2(L + W)$ . We are given that the perimeter is $32$ meters. $\newline$ So, we have the equation $2(L + W) = 32$ .
Simplify Perimeter Equation: We can simplify the equation by dividing both sides by $2$ to find $L + W$ . $\newline$ $L + W = \frac{32}{2}$ $\newline$ $L + W = 16$
Area Equation: The area $A$ of a rectangle is given by the formula $A = L \times W$ . We are given that the area is $60$ square meters. $\newline$ So, we have the equation $L \times W = 60$ .
Solve System of Equations: Now we have a system of two equations: $\newline$ $1$ ) $L + W = 16$ $\newline$ $2$ ) $L \times W = 60$ $\newline$ We can solve this system of equations to find the values of $L$ and $W$ .
Express $W$ in Terms of $L$ : Let's express $W$ in terms of $L$ from the first equation: $W = 16 - L$ . Now we can substitute $W$ in the second equation with $(16 - L)$ . $L \times (16 - L) = 60$
Expand and Rearrange Equation: Expanding the equation, we get: $\newline$ $16L - L^2 = 60$ $\newline$ Rearranging the terms, we get a quadratic equation: $\newline$ $L^2 - 16L + 60 = 0$
Solve Quadratic Equation: We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. The equation factors nicely as: $\newline$ $(L - 10)(L - 6) = 0$
Factor Quadratic Equation: Setting each factor equal to zero gives us the possible values for $L$ : $L - 10 = 0$ or $L - 6 = 0$ So, $L = 10$ or $L = 6$
Find Possible Values for $L$ : If $L = 10$ , then $W = 16 - L = 16 - 10 = 6$ . If $L = 6$ , then $W = 16 - L = 16 - 6 = 10$ . Both pairs ( $L = 10$ , $W = 6$ ) and ( $L = 6$ , $W = 10$ ) are valid solutions since they are interchangeable (length and width can be swapped).

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