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3 years ago

Gabriel wants to paint the model. How

Mathematics

1 answer:

enyata [817]3 years ago

4 0

Answer:

<h2>The area of the base is 144 square inches.</h2><h2>The area of each triangular face is 66 square inches.</h2><h2>Grabiel needs 408 square inches of paint.</h2>

Step-by-step explanation:

The complete problem is attached.

Notice that the figure is a square pyramid, where its base dimensions are 12 inches by 12 inches, which represents an area of

$B=12 \times 12 = 144 \ in^{2}$

The slant height of the pyramid is 11 inches, which allow us to find the area of each triangle face

$A=\frac{1}{2}bh =\frac{1}{2}(12)(11)= 66 \ in^{2}$

But there are four triangle faces, so $A_{faces} =4(66)=264 \ in^{2}$ .

Therefore, the area of each triangular face is 66 square inches.

So, the total surface area would be the sum

$S=144+264=408 \ in^{2}$

Therefore, Gabriel needs 408 square inches to paint the whole model.

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3 years ago

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<u>Answer: </u>

sec squared 55 – tan squared 55 = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

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$\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}$

And,

$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$

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Substituting the values

$\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}$

Solving,

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}$

According to Pythagoras theorem,

$\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}$

Putting this in the equation;

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