The lateral area of the right regular triangular prism is 18cm^2. Find the total surface area.Give answer in decimal form rounded of to 2 decimal places.

- anonymous

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- anonymous

##### 1 Attachment

- anonymous

im looking for the area of the rectangle face and triangle base

- terenzreignz

This shouldn't be too hard :)
Let's consider just one rectangular face...
|dw:1377800484158:dw|

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## More answers

- terenzreignz

You with me?

- anonymous

yes

- terenzreignz

Now, it'd be much nicer if we knew how long this is:|dw:1377800558650:dw|

- terenzreignz

Which we do, actually... we know that the area is 18...

- terenzreignz

And the area is equal to the product of the two side-lengths...3 and an unknown number...
So... three times a number is 18, what is that number?

- anonymous

6

- terenzreignz

That's right :)|dw:1377800697260:dw|

- anonymous

how come looking at the picture 3cm looks alot longer then the 6cm side.

- terenzreignz

These things are rarely drawn to scale.

- anonymous

?

- terenzreignz

If it bothers you that much, I'll draw a slightly more accurate figure :P|dw:1377800780026:dw|

- anonymous

ohhh i see

- terenzreignz

Okay, that was the easy part, though... ready for the actual 'finding the surface area'?

- anonymous

yes

- terenzreignz

As you can see, there are three-rectangles around the sides, each with area 18.. so that makes how many total?

- anonymous

??? how many what?

- terenzreignz

total area around the three rectangles...

- anonymous

54 cm^2

- terenzreignz

That's right, and that's the total area around the three rectangles, now find the area of one triangular base.

- anonymous

okay one sec

- anonymous

gonna try it right now

- terenzreignz

Area of an equilateral triangle given a side-length s is
\[\Large A_\Delta=\frac{s^2\sqrt3}{4}\]

- anonymous

idk i got about 7.8

- anonymous

oh ill use the formula u gave me

- terenzreignz

Yes, please do :P

- anonymous

wait can u draw that formula on the triangle so i can see it? i dont really get it

- terenzreignz

Well, when you have an equilateral triangle, you know all sides measure the same, right?
So just square the side-length and mutltiply to \(\LARGE \frac{\sqrt3}{4}\)

- anonymous

Howd u get|dw:1377801669518:dw|

- terenzreignz

You want to derive formulas? :/ Okay.... but it can't be done without trigonometry... a fair warning XD
Are you sure you want to derive it?

- anonymous

oh so thats the formula to find the area of an equilateral triangle?

- anonymous

I was trying to find the area of the right triangle first.

- anonymous

by splitting the triangle in half

- terenzreignz

yes... you know 30-60-90 triangles?

- anonymous

yea

- terenzreignz

Okay, let's have an arbitrary equilateral triangle...|dw:1377802040102:dw|

- terenzreignz

with side-length equal to s.

- terenzreignz

Now, the area of a triangle is always equal to \[\Large \frac{bh}2\]

- terenzreignz

Its base is here...|dw:1377802092050:dw|
Which is equal to s, since this is just one side of the triangle... right?

- anonymous

yes

- anonymous

but isnt the base 6 since its an equilateral triangle?

- terenzreignz

We're just deriving the formula. This is the general case :)

- terenzreignz

So now, we need the height:|dw:1377802208651:dw|

- terenzreignz

We need the height h in terms of s.
So, what we do, we remember that this angle measures 60 degrees (due to it being an equilateral triangle)|dw:1377802273303:dw|

- anonymous

ok

- anonymous

and the other angle is 30 degrees?

- terenzreignz

Of course :)
So this side measures s/2, by the properties of the 30-60-90 triangle...|dw:1377802388850:dw|

- terenzreignz

And furthermore, by the properties of the 30-60-90 triangle, this side (h) measures|dw:1377802425279:dw|

- terenzreignz

So it turns out,
\[\Large b= s\]\[\Large h = \frac{s\sqrt3}{2}\]
Therefore, the area
\[\Large A = \frac{bh}2= \frac{s^2\sqrt3}4\]

- terenzreignz

Understood?

- anonymous

but isnt the height just \[\sqrt{3}\]

- terenzreignz

That's if the side-length of the triangle is 2.

- anonymous

im sorry|dw:1377802649031:dw|

- terenzreignz

|dw:1377802675764:dw|

- anonymous

why over 2?

- terenzreignz

Because, in a 30-60-90 triangle, the longer leg is equal to the shorter leg times \(\large \sqrt3\) Now, in our case, the shorter leg measures \(\Large \frac{s}2\) (not s!)|dw:1377802763415:dw|

- terenzreignz

Understood?

- anonymous

|dw:1377802835960:dw| can u use this also.

- terenzreignz

No... you got it backwards...|dw:1377802980320:dw|

- anonymous

thats what i meant ^ is it the same as the other way of looking at it?

- terenzreignz

No, it's very different... |dw:1377803116229:dw| in fact, that's wrong^

- anonymous

not the way i wrote it im

- anonymous

|dw:1377803208015:dw|

- anonymous

is that drawing^ technically the same as this one|dw:1377803274662:dw|

- terenzreignz

Yes... just multiplied everything by 2.

- anonymous

so how do we use this formula in my problem?

- terenzreignz

To find the area of one of the triangle faces... one of the sides is 6, just apply it to the formula
\[\Large A - \frac{s^2\sqrt3}4\]

- terenzreignz

\[\Large A \color{red}=\frac{s^2\sqrt3}4\] sorry

- anonymous

so about 15.588?

- anonymous

15.588 cm^2

- terenzreignz

Yeah, but there are two triangle faces, so double that.

- anonymous

about 31.176 cm^2 is the area od the two triangles.

- anonymous

*of

- terenzreignz

Okay, and add that to the area of the three rectangles, and that's the total surface area.

- anonymous

So the surface area is about 85.18 cm^2

- terenzreignz

Seems about right.

- anonymous

ALRIGHT! Thanks!

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