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3 years ago

7

How in the world do I do this can somebody help please.

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer:

Bl^2+4Bl*Fh

Step-by-step explanation:

I'm not quite certain what "draw a net" means here. But for part b, we are doing the formula. The bottom part is a square(assumingly so take this with a grain of salt), thus making the base equal to 3*3 cm or 9 cm^2. The triangular faces are each 3*2.24 cm or 6.72 cm^2. We then multiply this by 4 to get 26.88. Thus, the equation is Bl^2(Base length squared)+Bl*Fh(Face height, I forgot the official name sorry about that)*4 for part b.

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If W(-10, 4), X(-3, -1), and Y(-5, 11) classify ΔWXY by its sides. Show all work to justify your answer.

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Given:

The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).

To find:

Which type of triangle is ΔWXY by its sides.

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Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

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$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

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$WX=WY$

So, triangle is an isosceles triangles.

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$WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2$

$WX^2+WY^2=74+74$

$WX^2+WY^2=148$

$WX^2+WY^2=(2\sqrt{37})^2$

$WX^2+WY^2=WY^2$

So, triangle is right angled triangle.

Therefore, the ΔWXY is an isosceles right angle triangle.

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