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3 years ago

5

Gabriel wants to paint the model. How

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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Given: ABCD is a trapezoid, AB=CD, MN is a midsegment, MN=30, BC=17, AB=26 Find: m∠A, m∠B, m∠C, and m∠D

enyata [817]

Answer:

The measures of angle ∠A, ∠B, ∠C, and ∠D are 60, 120, 120 and 60 degree respectively.

Step-by-step explanation:

Given information:AB=CD, MN is a midsegment, MN=30, BC=17, AB=26

Since two opposites sides are equal, therefore we can say that two no parallel sides are equal.

The length of midsegment is average of length of parallel lines.

$MN=\frac{AD+BC}{2}$

$30=\frac{AD+17}{2}$

$60=AD+17$

$43=AD$

Draw perpendiculars on AD from B and C. Let angle A be θ. D

$AD=AE+EF+FD$

Since ABCD is an isosceles trapezoid, therefore AE=FD and EF=BC

$AD=AE+EF+AE$

$43=2(AE)+17$

$26=2(AE)$

$13=AE$

$\cos\theta=\frac{base}{hypotenuse}$

$\cos\theta=\frac{AE}{AB}$

$\cos\theta=\frac{13}{26}$

$\cos\theta=\frac{1}{2}$

$\theta=\cos^{-1}\frac{1}{2}$

$\theta=60$

Since ABCD is an isosceles trapezoid, therefore angles A and D are same. Angle B and C are same.

$\angle A=\angle D=60^{\circ}$

The sum of two consecutive angles of a trapezoid is 180 degree by consecutive interior angle theorem.

$\angle A+\angle B=180^{\circ}$

$60^{\circ}+\angle B=180^{\circ}$

$\angle B=120^{\circ}$

$\angle B=\angle C=120^{\circ}$

Therefore measures of angle ∠A, ∠B, ∠C, and ∠D are 60, 120, 120 and 60 degree respectively.

3 0

3 years ago

Solve:#1 :8 1/5+(4 2/5+ 3 3/10). #2: 2 1/10+ (1 2/7+4 9/10)

nadya68 [22]

#1 <span><span><span>815</span>+<span>425</span></span>+<span>3310</span></span><span>=<span>27910</span></span><span>(Decimal: 27.9)

#2</span><span><span><span>2110</span>+<span>127</span></span>+<span>4910</span></span><span>=<span>617</span></span><span>(Decimal: 8.714286)</span>

3 0

3 years ago

Help!??!!???????????????????

vagabundo [1.1K]

I honestly really do not know so sorry if it is wrong..i would say -0.25 sry if wrong

6 0

3 years ago

What is the standered form of the number shown in the calculator display? <br> 3.82e+7

ella [17]

the answer is 38200000

6 0

3 years ago

If <br> B is the midpoint of AC, AB=x+6 and AC=5x-6, then what is BC?

nikklg [1K]

If B is the midpoint of AC, then AB=BC. It is also BC=AC-AB

AB = x + 6

AC = 5x - 6

BC = (5x - 6) - (x + 6)

We can kinda use distributive property, like this:

BC = 5x - 6 -x - 6

Combine like terms:

BC = 4x - 12

To verify, we can add AB and BC to see if their sum is equal to AC.

(x + 6) + (4x - 12) = 5x - 6

If we are correct, this equation will be true.

x + 6 + 4x - 12 = 5x - 6

Combine like terms:

6 + 5x - 12 = 5x - 6

5x - 6 = 5x - 6

It's correct!

BC = 4x - 12. Final Answer.

Let me know if you have any questions about this! :)

7 0

2 years ago