SOLUTION: a rectangle has a perimeter of 46cm and an area of 120cm squared. find its dimensions by writing an equation and using the quadratic formula to solve it. Thanks!

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Question 459849: a rectangle has a perimeter of 46cm and an area of 120cm squared. find its dimensions by writing an equation and using the quadratic formula to solve it.
Thanks!
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Find the dimensions of the rectangle using the quadratic formula.
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First, let's assign some variables.
W = the width of the rectangle
L = the length of the rectangle
A = the area of the rectangle (cm squared) = 120
P = perimeter of the rectangle (cm) = 46
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Now, use the area formula for a rectangle [L * W = A] to express L in terms of W.
L%2AW=A
L%2AW=120
L=120%2FW
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Use the perimeter formula for a rectangle [2(W + L) = P] to create an equation using only W. (Remember that perimeter is the distance around the rectangle.)
2%28W%2BL%29=P
2%28W%2BL%29=46
W%2BL=46%2F2
W%2BL=23
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Now we use the power of algebra! We know from our area equation, that L = 120/W. Substitute (120/W) for L in the perimeter equation.
W%2BL=23
W%2B%28120%2FW%29+=+23
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We'd like to have an equation without the W in the the denominator of the 2nd term. Multiply both sides of the equation by W. We get
W%5E2%2B120=23W
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We see that we have a quadratic equation. In order to apply the quadratic formula, we need our equation in the form
ax%5E2%2Bbx%2Bc=0
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We'll subtract 23W from both sides, and rearrange terms to get our equation into this form.
W%5E2-23W%2B120=0
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We see that a = 1, b = -23, c = 120. We are ready to apply the quadratic formula.
W=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
W=%28-%28-23%29%2B-sqrt%28%28-23%29%5E2-4%281%29%28120%29%29%29%2F2%281%29
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Simplify, and find the value of W.
W=%2823%2B-sqrt%28529-480%29%29%2F2
W=%2823%2B-sqrt%2849%29%29%2F2
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The "+ or -" sign gives us two values for W. We expect this because we are working with a quadratic equation.
W=%2823%2B7%29%2F2 OR W=%2823-30%29%2F2
W=15 OR W=-7.5
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The rectangle cannot have a width of -7.5, so we discard this value; W = 15. Now we need to find the length of the rectangle. We can use the area equation or the perimeter equation for this, but the area equation looks like less work. Recall that
L=120%2FW
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We substitute 15 for W, and solve for L.
L=120%2F15=8
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The length of the rectangle is 8 cm.
The width of the rectangle is 15 cm.
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I always like to check problems like these to make sure I didn't make an error along the way. We go back to the original problem and make sure these values give us an area of 120 cm squared and a perimeter of 46 cm. Substitute 15 for W and 8 for L in our perimeter and area equations.
L%2AW=A
8%2A15=120 (Yes, it does!)
.
2%28L%2BW%29=P
2%288%2B15%29=46
2%2823%29=46 (Yes, it does!)
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There you go! I hope this helps.
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Send me a message if you still have questions.
Ms.Figgy
math-vortex