If FG || BC, which equation is necessarily true?

Mathematics

1 answer:

vazorg [7]2 years ago

7 0

First, let's prove that triangle ABC and AFG are similar.

Line segment FG and BC are parallel to each other, that's a given.
With this information, we can state that these sides are similar to each other since you can transform BC on to FG using only rigid transformations and dilations.

Another thing is that both of these triangles share angle A, that's a given. Therefore, using the SA similar theorem, we can conclude that these triangles are similar.

Now that we know that triangle ABC and AFG are similar we may start going to the answer box.

We can eliminate the first three answer choices for having sides we cannot make full conclusions due to limited knowledge.

The only answer remaining is the fourth choice.

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A closed box has a square base with side length feet and height feet given that the volume of the box is 34 ft.³ express the sur

Deffense [45]

Answer:

$\dfrac{2s^3+136}{s}$

Step-by-step explanation:

Let the side length of the square base =s feet

Let the height of the box = h

Given that the volume of the box = $34$ ft^3$

Volume of the box = $s^2h$

Then:

$s^2h=34$ ft^3\\$Divide both sides by s^2\\h=\dfrac{34}{s^2}$

Surface Area of a Rectangular Prism =2(lb+bh+lh)

Since we have a square base, l=b=s feet

Therefore:

Surface Area of our closed box $= 2(s^2+sh+sh)$

$S$urface Area= 2s^2+4sh\\Recall: h=\dfrac{34}{s^2}\\$Surface Area= 2s^2+4s\left(\dfrac{34}{s^2}\right)\\=2s^2+\dfrac{136}{s}\\$Surface Area in terms of length only=\dfrac{2s^3+136}{s}$