Find the area & perimeter of the figure

- anonymous

Find the area & perimeter of the figure

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- anonymous

##### 1 Attachment

- mathstudent55

You were given a triangle like this:
|dw:1437006808180:dw|

- mathstudent55

What do the circled marks mean?
|dw:1437006848302:dw|

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## More answers

- anonymous

@11.seventeen what do the circled marks mean

- anonymous

doesn't it mean its equal

- anonymous

@mathstudent55

- mathstudent55

Yes. All sides are congruent.
That means every side of this triangle measures 2a mm.
That makes the perimeter easy to find.
What is the perimeter of a triangle?

- mathstudent55

\(\Large P_{triangle} = a + b + c\)
where a, b, and c are the lengths of the sides of the triangle.

- mathstudent55

\(P = 2a + 2a + 2a = 6a\)
The perimeter of the triangle is 6a mm

- mathstudent55

Ok so far?

- anonymous

Yes. It's the same as what i got @mathstudent55

- mathstudent55

Now you have to find the area.

- anonymous

And I would do that how ? Because I don't have the heigh . I have the base is 2

- mathstudent55

To find the area, you need to find the height of the triangle since the formula for the area involves the height.

- mathstudent55

Have you heard about the ratios of the lengths of the sides of a 30-60-90 triangle?

- anonymous

Yes but I never understood it

- mathstudent55

Ok. I'll explain it to you.
Here is a triangle with a right angle and a 30-deg angle and a 60-deg angle.
|dw:1437011007751:dw|

- mathstudent55

Since the right angle is 90 deg, this triangles' three angles have measures 30, 60, and 90 degrees. This is what is called a 30-60-90 triangle.

- mathstudent55

In a 30-60-90 triangle, the length of the hypotenuse is twice the length of the short leg.
If you call the length of the short leg 1, then the hypotenuse has length 2.
|dw:1437011223749:dw|

- mathstudent55

The length of the long leg is \(\sqrt 3\) times the length of the short leg.

- mathstudent55

|dw:1437011291789:dw|

- mathstudent55

So far all we have is that the lengths of the sides of a 30-60-90 triangle are in the ratio of
1 : \(\sqrt 3 : 2\)

- mathstudent55

Do you understand so far?

- anonymous

sort of

- mathstudent55

It is simpler than you think.
Think of \(\sqrt 3\) as being approximately 1.7
All this means is that:
In a 30-60-90 triangle, the long leg is 1.7 times the length of the short leg.
The hypotenuse is 2 times the length of the short leg.

- mathstudent55

If a 30-60-90 triangle has a
short leg of 1, then
the long leg is 1.7 * 1 = 1.7
the hypotenuse is 2 * 1 = 2

- anonymous

how do I get the height out of this?

- mathstudent55

Another example:
If a 30-60-90 triangle has
a short leg of length 4, then
the long leg is 1.7 * 4 = 6.8
and the hypotenuse is 2 * 4 = 8

- mathstudent55

Ok. Let's get back to our problem.

- mathstudent55

|dw:1437011694076:dw|

- mathstudent55

You see our triangle above.
Since all sides are congruent, all angles are also congruent.
That means all angles measure 60 degrees.
Ok?

- mathstudent55

|dw:1437011786387:dw|

- mathstudent55

We are looking for the length of the height.
By drawing the height, we created two new triangles.
Notice each small triangle is a 30-60-90 triangle.

- mathstudent55

|dw:1437011853170:dw|

- mathstudent55

Look at the figure above.
Each of the two small triangles is a 30-60-90 triangle. The 30 degree angle is above.
Since the side of the large triangle has length 2a, half of that is just a, and that is the length of the short side of the 30-60-90 triangle.

- mathstudent55

Look at the small triangle in the circle and ignore the rest.
|dw:1437012005959:dw|

- mathstudent55

The short leg has length a.
The long leg has length \(\sqrt 3\) times a, which is simply: \(\sqrt 3 a\)
|dw:1437012059638:dw|

- mathstudent55

|dw:1437012157126:dw|

- mathstudent55

The long leg of the small triangle is the height of the large triangle. That was the height we needed for the area.

- mathstudent55

|dw:1437012211126:dw|

- anonymous

so is the height 3a?

- mathstudent55

Now we have all the info we need to find the area of the large triangle.

- mathstudent55

The height is \(\sqrt 3 a\), not 3a.
\(\Large A_{triangle} = \dfrac{bh}{2} \)

- mathstudent55

\(\Large A = \dfrac{(a)(\sqrt 3 a)}{2} = \dfrac{\sqrt 3}{2}a^2 ~mm^2\)

- anonymous

thanks!!!

- mathstudent55

You're welcome.

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