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schepotkina [342]
3 years ago
7

Solve for the given length. The length of a rectangle is 3m longer than its width. The area of the rectangle is 154^2. What is t

he width of the rectangle?
Show all work
Mathematics
2 answers:
Andre45 [30]3 years ago
7 0
If the length of a rectangle is 3m longer than its width, then:
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m
VikaD [51]3 years ago
4 0
By definition the area of a rectangle is given by:
 A = (w) * (l)
 Where,
 w: width
 l: long
 We rewrite the expression based on:
 "The length of a rectangle is 3m longer than its width"
 A = (w) * (w + 3)
 We substitute the value of the area:
 154 ^ 2 = (w) * (w + 3)
 Rewrite:
 23716 = w ^ 2 + 3w
 w ^ 2 + 3w-23716 = 0
 We solve the polynomial:
 w = 152.5
 Note: the other root is negative.
 Answer: 
 the width of the rectangle is: 
 w = 152.5 m
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Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

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In this problem

Let

a=7\ cm\\c=13\ cm

substitute

13^{2}=7^{2}+b^{2}

Solve for b

169=49+b^{2}

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we have that

<em>Jayden's Work</em>

a^{2}+b^{2}=c^{2}

a=7\ cm\\b=13\ cm

substitute and solve for c

7^{2}+13^{2}=c^{2}

49+169=c^{2}

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Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

<em>Sheridan's Work</em>

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7^{2}+b^{2}=13^{2}

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b^{2}=169-49

b^{2}=120

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