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BigorU [14]
3 years ago
11

If a 4 x 16 riangle has the same area as a square, what is the length of a side of the square

Mathematics
1 answer:
m_a_m_a [10]3 years ago
7 0

Answer: \bold{4\sqrt2}

<u>Step-by-step explanation:</u>

A_{triangle}=\bigg(\dfrac{1}{2}\bigg)b\times h

A_{triangle}=\bigg(\dfrac{1}{2}\bigg)4\times 16

              =\bigg(\dfrac{1}{2}\bigg)64

              =32

A_{square}= s^2

32 = s^2

\sqrt{32}=s

\sqrt{\underline{2\cdot 2}\cdot \underline{2\cdot 2}\cdot 2}=s

2\cdot 2\sqrt{2}=s

4\sqrt{2}=s


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The two non-parallel sides of an isosceles trapezoid are each 7 feet long. The longer of the two bases measures 22 feet long.
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Let X represent the shorter base, and drop the heights from shorter to longer base.

 

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So the height is 7* sin 70 = 6.5778483455....

 

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Now the algebra gets ugly.

 

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(11-x/2)^2 + 49sin70^2 = 49

(11 - x/2)^2 = 49 - 49sin70^2

(11 - x/2)^2 = 49(1 - sin70^2)

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121 - 11x + x^2/4 = 49 cos70^2

484 - 44x + x^2 = 196 cos 70^2

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The two roots are:

26.78282 AND 17.2117331

The first one is disqualified and we must reject it

because if used will make the base of the

right triangles negative. Remember this is

supposed to be the smaller base. So it 

cannot be greater than 22.

 

SO the smaller base is 17.2117331

 

So 11 - x/2 = overhang = 2.39413345

 

Drawing the diagonal, it's length can

be found by pythagorean theorem.

square-root(17.2117331)^2 + (7sin70)^2)=18.4258472....<-- diagonal

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