More graphed triangles..

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July 20th 2006, 03:15 PM
zoso

More graphed triangles..

Summer school can really be a drag, ya know?

So..

"ABC's vertices A(-1, 1), B(5,1), and C(2,4)

Determine the perimeter of ABC. Express your answer as an exact value in simplest form.

What is the slope of AC?

Find the measure of $\angle$ A

What is the area of ABC?

So, pretty straightforward, I'd like to compare my work with somebody elses though.

By the way, thanks for all the help. You guys have no idea the amount of work I'm facing in quite a few subjects.
July 20th 2006, 03:32 PM
Quick

1 Attachment(s)

Perimeter

Quote:

Originally Posted by zoso

Summer school can really be a drag, ya know?

So..

"ABC's vertices A(-1, 1), B(5,1), and C(2,4)

Determine the perimeter of ABC. Express your answer as an exact value in simplest form.

the distance between two points is: $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

so let's find the distances...

AB:
$\overline{AB}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\overline{AB}=\sqrt{(5-\neg1)^2+(1-1)^2}$

$\overline{AB}=\sqrt{6^2+0^2}$

$\overline{AB}=\sqrt{36}$

$\overline{AB}=6$

BC:
$\overline{BC}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\overline{BC}=\sqrt{(5-2)^2+(1-4)^2}$

$\overline{BC}=\sqrt{3^2+(\neg3)^2}$

$\overline{BC}=\sqrt{9+9}$

$\overline{BC}=\sqrt{18}$

$\overline{BC}=\sqrt{9}\sqrt{2}$

$\overline{BC}=3\sqrt2$

CA:
$\overline{CA}=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$\overline{CA}=\sqrt{(2-\neg1)^2+(4-1)^2}$

$\overline{CA}=\sqrt{3^2+3^2}$

$\overline{CA}=\sqrt{9+9}$

$\overline{CA}=\sqrt{18}$

$\overline{CA}=\sqrt{9}\sqrt{2}$

$\overline{CA}=3\sqrt2$

now add those together: $\overline{AB}+\overline{BC}+\overline{CA}=$ $6+3\sqrt2+3\sqrt2=6+2(3\sqrt2)$ $=6+6\sqrt2$ $=\boxed{6(1+\sqrt2)}$
July 20th 2006, 03:43 PM
Quick

slope and angle

Quote:

Originally Posted by zoso

What is the slope of AC?

slope formula: $\frac{y_2-y_1}{x_2-x_1}$
therefore:
$\frac{y_2-y_1}{x_2-x_1}$

$\frac{4-1}{2-\neg1}$

$\frac{3}{3}$

$\boxed{1}$

Quote:

Find the measure of $\angle$ A

$\angle\text{A}=45^{\circ}$
July 20th 2006, 03:49 PM
Quick

Area

Quote:

Originally Posted by zoso

What is the area of ABC?

So, pretty straightforward, I'd like to compare my work with somebody elses though.

By the way, thanks for all the help. You guys have no idea the amount of work I'm facing in quite a few subjects.

The triangle happens to be a right triangle (angle C is 90 degrees) therefore you multiply the sides and divide by two...

$A=\frac{1}{2}\times\overline{BC}\times\overline{CA }$

substitute: $A=\frac{1}{2}\times3\sqrt2\times3\sqrt2$

multiply: $A=\frac{1}{2}\times9\times2$

multiply: $A=\frac{1}{2}\times18$

multiply: $A=9$

~ $Q^u_u\!u^i_i\!i^c_c\!c^k_k\!k$
July 20th 2006, 04:53 PM
zoso

K, Everything looks good, with a few discrepncies that were my fault.

What was the method you used for obtaining angle a?
July 20th 2006, 05:38 PM
Quick

Quote:

Originally Posted by zoso

K, Everything looks good, with a few discrepncies that were my fault.

What was the method you used for obtaining angle a?

To tell you the truth, I don't know the method for obtaining angles in graphs, I just know that angle A is 45 degrees because it's between two lines of slope 1 and 0
July 20th 2006, 06:07 PM
ThePerfectHacker

Nice responses and Good graph Quick.
---
Let me tell you how to find angles. Since AB is parallel to x-axis then the angle of A is same as AC creates with x-axis when extended because of parallel line. But the slope of a line is the same as the tangent of angle it forms with x-axis thus,
$\tan \theta =1$ thus $\theta=45^o$
July 20th 2006, 06:12 PM
Quick

Quote:

Originally Posted by ThePerfectHacker

Nice responses and Good graph Quick.

Thanx :D

Quote:

Let me tell you how to find angles. Since AB is parallel to x-axis then the angle of A is same as AC creates with x-axis when extended because of parallel line. But the slope of a line is the same as the tangent of angle it forms with x-axis thus,
$\tan \theta =1$ thus $\theta=45^o$

is there an equation to solve for theta? it seems to me like no mathematician (especially you) would be willing to do trial and error until they found the answer.
July 20th 2006, 06:27 PM
ThePerfectHacker

Quote:

Originally Posted by Quick

Thanx :D
is there an equation to solve for theta? it seems to me like no mathematician (especially you) would be willing to do trial and error until they found the answer.

It seems to me that many people do not accept trial and error as a solution, why? You guess a solution and then you know it is true, proved. Yes, you cannot solve it in finite number of steps. Which is why solutions to equations are used. But the basic angles are 0, 30, 45, 60, 90 and larger and all other can be build from their sum and diffrences, halves and thirds. This is one of those angles you need to have memorized.

You can use a calculator feature called "arc-tangent" basically it is opposite of tangent like square root is opposite of squaring. But then you gonna ask another question how can do it without calculator? There really is no way to do it, (in fact most math problems are like that). You can, however, approximate solutions as close as you want (as I am writing this I came up with a way I might show it to you).

What I said before, "In fact most math problems are like that". Let me explain, mathematicians are not mostly concerned about finding what satisfies a certain equation rather they are mostly concerned about prooving that a solution exists. For example, the fundamental theorem of algebra guarenntes the existence of a solution for any non-constant polynomial yet it does not provide a way to find it (in fact it is impossible to have such a method!) Yet, it never bothered mathematicians all they needed to know that such a solution exists.

Knowing you you will probably ask much about what I just wrote, but it is okay. I would love to have you as a student.