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Softa [21]
3 years ago
11

the perimeter of a basketball court is 284 ft. the length is 40 ft longer than the width. find the dimensions of the court.

Mathematics
2 answers:
irakobra [83]3 years ago
8 0
We know that the perimeter is the distance around an object. This is also 2 times the length + 2 times the width. We know the width would be X and the length would be 40 feet longer so it is x+40. Now we have 2 (X) + 2 (x+40)=284
Doing the distributive property you would get 2x + 2x + 80=284
add like terms to get 4x+80=284
subtract 80 from both sides to get 4x=204
divide both sides by 4 to get x=51
Width is 51 and width is 40 more than that so length is 91
mrs_skeptik [129]3 years ago
7 0
284=2x+2(x+40)
284=4x+80
204=4x
51=x
dimensions 51 ft by 91ft
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Nady [450]
<h3>Answer: 10.1 cm approximately</h3>

=====================================================

Explanation:

The double tickmarks show that segments DE and EB are the same length.

The diagram shows that DB = 16 cm long

We'll use these facts to find DE

DE+EB = DB

DE+DE = DB

2*DE = DB

DE = DB/2

DE = 16/2

DE = 8

-------------

Now let's focus on triangle DEC. We just found the horizontal leg is 8 units long. The vertical leg is EC which is unknown for now. We'll call it x. The hypotenuse is CD = 9

Use the pythagorean theorem to find x

a^2+b^2 = c^2

8^2+x^2 = 9^2

64+x^2 = 81

x^2 = 81 - 64

x^2 = 17

x = sqrt(17)

That makes EC to be exactly sqrt(17) units long.

If you follow those same steps for triangle ADE, then you'll find the missing length is AE = 6

---------------

So,

AC = AE+EC

AC = 6 + sqrt(17)

AC = 10.1231056256177

AC = 10.1 cm approximately

4 0
2 years ago
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ICE Princess25 [194]

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5 0
3 years ago
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given: s is the midpoint of rt

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Yanka [14]
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3 years ago
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