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Elden [556K]
3 years ago
12

What is the surface area of the right rectangular prism?

Mathematics
1 answer:
scoundrel [369]3 years ago
3 0
<h3>Answer:  Choice A) 252 in^2</h3>

=======================================================

Explanation:

Refer to the diagram below.

The red rectangle is 6 inches by 12 inches, so its area is 6*12 = 72 square inches. There are two of these rectangles (one on top, one on the bottom). That accounts for 2*72 = 144 square inches so far.

The purple rectangles are 3 inches by 12 inches, yielding an area of 3*12 = 36 square inches each. That's another 2*36 = 72 square inches when we account for the front and back purple rectangles.

The green rectangles are 6 inches by 3 inches. Each green rectangle is 6*3 = 18 square inches. Having two of them means we'll add on 2*18 = 36 square inches.

Overall, the entire surface area is the sum of all the areas we calculated: 144+72+36 = 252 square inches, which is choice A

-------------------------

Alternative Method:

  • L = 12 = length
  • W = 6 = width
  • H = 3 = height

SA = surface area of the rectangular prism

SA = 2*(L*W + L*H + W*H)

SA = 2*(12*6 + 12*3 + 6*3)

SA = 2*(72 + 36 + 18)

SA = 2*(126)

SA = 252 in^2

So we get the same answer either way

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Answer:

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Step-by-step explanation:

we have

The coordinates of triangle ABC are

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we know that

An isosceles triangle has two equal sides and two equal internal angles

The formula to calculate the distance between two points is equal to

d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}

step 1

Find the distance AB

substitute in the formula

d=\sqrt{(5-2)^{2}+(2-0)^{2}}

d=\sqrt{(3)^{2}+(2)^{2}}

dAB=\sqrt{13}\ units

step 2

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substitute in the formula

d=\sqrt{(7-5)^{2}+(-1-2)^{2}}

d=\sqrt{(2)^{2}+(-3)^{2}}

dBC=\sqrt{13}\ units

step 3

Find the distance AC

substitute in the formula

d=\sqrt{(7-2)^{2}+(-1-0)^{2}}

d=\sqrt{(5)^{2}+(-1)^{2}}

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step 4

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dBC=\sqrt{13}\ units

dAC=\sqrt{26}\ units

dAB=dBC

therefore

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Applying the Pythagoras Theorem

(AC)^{2} =(AB)^{2}+(BC)^{2}

substitute

(\sqrt{26})^{2} =(\sqrt{13})^{2}+(\sqrt{13})^{2}

26=13+13

26=26 -----> is true

therefore

Is an isosceles right triangle

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\qquad\qquad\huge\underline{{\sf Answer}}♨

Let's solve ~

\qquad \sf  \dashrightarrow \:  - \frac{  3}{5} m + 8 = 20

\qquad \sf  \dashrightarrow \:  - \frac{  3}{5} m  = 20 - 8

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\qquad \sf  \dashrightarrow \: - m = 4 \times 5

\qquad \sf  \dashrightarrow \:m =  - 20

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