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Mathematics 18 Online
OpenStudy (safia21):

− Calculate the length of LM in the isosceles right triangle ∆ KLM

OpenStudy (safia21):

OpenStudy (heisenberg):

You can use the pythagorean theorem since it has a right angle. Are you familiar with it?

OpenStudy (safia21):

no im kinda confused

OpenStudy (heisenberg):

do you know what a hypotenuse is?

OpenStudy (safia21):

yes

OpenStudy (heisenberg):

so the pythagorean theorem is: \[ a^2 + b^2 = c^2 \] where c is the hypotenuse. since this is an isosceles triangle, a = b, don't you think?

OpenStudy (safia21):

yes

OpenStudy (heisenberg):

so can you take it form here? since a = b, \( a^2 + a^2 = c^2 \)

OpenStudy (safia21):

what is a and what is like what do you plug in

OpenStudy (heisenberg):

well let me ask, what is a hypotenuse?

OpenStudy (safia21):

the longset side of a right triangle

OpenStudy (heisenberg):

very good! which is the side *directly* across from the right angle. so the other two sides would be 'a' and 'b', but this triangle is isosceles so a = b. therefore a is an unknown that we want to solve for and 'c' is the length of the hypotenuse

OpenStudy (heisenberg):

so if we have \(a^2 + a^2 = c^2\) where \( c = 36 \) we only have 1 unknown so we should be able to solve this like a regular algebra problem.

OpenStudy (safia21):

ok

OpenStudy (heisenberg):

are you still confused? all that's left is to simplify this equation and solve.

OpenStudy (safia21):

okay i got 36?

OpenStudy (heisenberg):

show me your steps. i don't that's right.

OpenStudy (heisenberg):

think*

OpenStudy (heisenberg):

what is \( a^2 + a^2 \) ?

OpenStudy (heisenberg):

(@victor, please consider that the best way for a person to learn something is to have the person discover it on their own rather than just giving that person an answer.)

OpenStudy (safia21):

yes i need the steps!

OpenStudy (heisenberg):

there's no reason you can't do it yourself, though. you have an equation. where are you getting stuck? show me your work and i can help you.

OpenStudy (safia21):

can we start from the beginning im soo sorry!

OpenStudy (heisenberg):

the pythagorean theorem states that (for a right triangle): \[ a^2 + b^2 = c^2\]where c is the hypotenuse. we have the hypotenuse in this case, as you pointed out, and it equals 36, agree?

OpenStudy (anonymous):

It's an isosceles triangle so it has two equal sides and therefor two equal angles. If you add the inner angles of any triangle the result will be 180º. You already know 1 angle = 90º. 180-90=90 And since the two other angles are equal you have 90/2=45. Now that you have all the angles you can use trigonometry and achieve the value of the sides. I recommend you use "sin" or "tan" (as expressed on a calculator).

OpenStudy (safia21):

yes

OpenStudy (heisenberg):

but this is an isosceles triangle, so the two remaining sides are equal in length, would you agree?

OpenStudy (safia21):

yes

OpenStudy (anonymous):

Of course, you only need the value of one side.

OpenStudy (heisenberg):

so since the two remaining sides are \(a\) and \(b\), we know that \(a =b \) therefore we can simplify the equation to this: \( a^2 + a^2 = c^2 \)

OpenStudy (safia21):

okay and c2 is 36 right so a2+a2= 36^2

OpenStudy (heisenberg):

exactly! so what does \( a^2 + a^2\) equal? we can reduce this to one term.

OpenStudy (safia21):

a^3

OpenStudy (heisenberg):

not quite. how about this, what does \( x + x \) equal?

OpenStudy (heisenberg):

what if i wrote it this way: \( 1x+ 1x \)

OpenStudy (safia21):

2x^2?

OpenStudy (heisenberg):

that wouldn't work. think about if x = 2, 1x + 1x = 1(2) + 1(2) = 4, but 2x^2 = 2(2)^2 = 8 so those two expressions are not equal. when you add terms of the same variable, you just add their "coefficients," the number in front of the variable.

OpenStudy (safia21):

o okay

OpenStudy (heisenberg):

so what is x + x?

OpenStudy (safia21):

4

OpenStudy (heisenberg):

no x is a variable, not a number. x can be *any* number

OpenStudy (safia21):

okay x^2

OpenStudy (heisenberg):

no just add the numbers *in front* of the variable. x = 1x

OpenStudy (safia21):

2x

OpenStudy (heisenberg):

right! so now let's look at the original equation, what is \( a^2 + a^2 \)

OpenStudy (heisenberg):

keep in mind that \(a^2 \) is the same thing as \(1a^2\)

OpenStudy (safia21):

2x^4

OpenStudy (heisenberg):

there are no x's in this equation. that was just as an aside example. the question here is to simplify \(a^2 + a^2\) you know that \( x + x = 2x\), so use that same *idea* and apply it to the other equation.

OpenStudy (safia21):

2a^4

OpenStudy (heisenberg):

you don't add the exponents, just the coefficients. \( x = 1x = x^1 = 1x^1 \) are all the same same

OpenStudy (safia21):

2a^2

OpenStudy (safia21):

isnt it a2=B2

OpenStudy (heisenberg):

very good! so let's look at our original equation: \(a^2 + a^2 = 36^2 \) = \(2a^2 = 36^2\)

OpenStudy (heisenberg):

yes it is, that's how we eliminated the b^2 and replaced it with another a^2

OpenStudy (safia21):

ok

OpenStudy (heisenberg):

so now our equation is \(2a^2 = 36^2\) do you know how to solve algebra problems? that's all this is.

OpenStudy (safia21):

so i multiply 36 times 36 and divide it by 2

OpenStudy (heisenberg):

absolutely!

OpenStudy (heisenberg):

that would be represented like so: \(a^2 = \frac{36^2}{2} \) then you just take the square root of both sides to get 'a' = something

OpenStudy (heisenberg):

a^2 = 36^2 / 2

OpenStudy (safia21):

36 right

OpenStudy (heisenberg):

not quite, remember, it's: 36 * 36 / 2

OpenStudy (safia21):

ok

OpenStudy (heisenberg):

first do 36 * 36, then divide that answer by 2

OpenStudy (safia21):

648

OpenStudy (heisenberg):

very good! so we're left with: \( a^2 = 648\) just take the square root of both sides and you have your answer!

OpenStudy (safia21):

i got 26

OpenStudy (safia21):

but thats not one of the anwsers

OpenStudy (heisenberg):

what are the answers?

OpenStudy (safia21):

a) 18 b) 18√2 c) 36 d) 36√2

OpenStudy (heisenberg):

well one of those answers roughly equals the number you got.

OpenStudy (safia21):

b

OpenStudy (heisenberg):

that's right! congrats :)

OpenStudy (safia21):

thanks soo much!

OpenStudy (heisenberg):

no problem :) i'm glad you wanted to learn rather than just want the answer. it'll always work out better that way. trust me ;)

OpenStudy (safia21):

haha thanks! :)

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