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You can put this solution on YOUR website! the perimeter of a rectangle is 60 meters and the area is 221 square meters. Find its dimensions *** let x=length of rectangle let y=width of rectangle 2x+2y=perimeter 2x+2y=60 x+y=30 y=30-x length*width=area xy=221 x(30-x)=221 30x-x^2=221 x^2-30x+221=0 (x-13)(x-17) x=13 or x=17 dimensions: length=17 m width=13 m Andres D. 1 Expert Answer
William W. answered • 08/19/20 Math and science made easy - learn from a retired engineer
Let the rectangle have a length (L) and a width (W) like this: Then the perimeter is L + W + L + W or 2L + 2W And the area is L•W So, since perimeter is given as 60 and perimeter is 2L + 2W, then 2L + 2W = 60 And, since the area is given as 200, then L•W = 200 If L•W = 200 then L = 200/W. That means, in the first equation, we can plug in "200/W" wherever there we see "L" so: 2L + 2W = 60 2(200/W) + 2W = 60 400/W + 2W = 60 [multiply both sides by W to get: 400 + 2W2 = 60W [This is a quadratic equation. To solve a quadratic, we need to set it equal to zero. So subtract 60W from both sides to get: 2W2 - 60W + 400 = 0 {divide both sides by 2 to get: W2 - 30W + 200 = 0 [factor it to get: (W - 10)(W - 20) = 0 [now, set each piece equal to zero W - 10 = 0 or W = 10 and W - 20 = 0 or W = 20 So, when W = 10, we can plug 10 in for W in the equation L = 200/W to get L = 200/10 or L = 20 And, when W = 20, we can plug 10 in for W in the equation L = 200/W to get L = 200/20 or L = 10 So the rectangle is either 10 meters by 20 meters, or it is 20 meters by 10 meters |