Use the triangle at the right. Find the length of the missing side. Show your work Did i do the first one right? I don't want to do the second one wrong too. 1.a = 16, b = 63 2. b = 2.1, c = 2.9 16^2 =256, 63^2=3,969 256+3969 = 4225 now we need to find the square root by separating the 4225 into two separate numbers square root of 42 is 6 now find the largest root in 42. Which is 36. 6*6 = 36 and the square of 25 is 5 add the roots This give me the answer for c length. 6 and 5 is 65. a=256 b=3969 and c = 65
Do you have a picture or other description of the triangle you are dealing with in problem 1 and/or problem 2?
yup you need a picture
The picture probably wouldn't help it's just a right triangle but okay
@SolomonZelman
this picture will definitely help, because I would have thought that b is the hypotenuse
i mean in question 1 i would have thought so, but this picture gives me that c=hypotenuse in both cases, and I know it is a right triangle
Alright.
I want to correct you though. The set up is a little different, that is 1. Here is the rule. A right triangle with sides a, b, c (where the hypotenuse is side c), must satisfy the following statement: a²+b²=c²
this statement is known as the pythagorean theorem.
Now, you are given your two smaller legs a and b are 16 and 63 (respectively). And you are missing the hypotenuse, so this is what you would do. (16)² + (63)² = c²
then you simplify the left hand side, and solve for c (just by talking the square root of both sides)
So i had to square c?
Take the square root of a and b?
(16)² + (63)² = c² 4225 = c² then do this: \(\sqrt{4225}\) = \(\sqrt{{\rm c}^2}\)
this way you are able to solve for c. (normally the square root will give you the ±, as you know, but in this case since distance or sidelength can not be negative, you diregard any negative solutions)
Alright. is c 65? @SolomonZelman
yes
Did i do the problem right?
i wrote it all out and still doesn't take that much space and time..... ok, now question 2...
Kay.
oh, I deleted that.... ------------------ a²+b²=c² 16²+63²=c² 4225=c² √4225 = √c² c=65 ------------ reposted it.
ok, in question 2, you are given c=2.9 b=2.1
can you plug in this information into the a²+b²=c² ?
Sure a^2+ b^2=c^2
but, you are given the c and b, so you can g ahead and plug in 2.9 for c, and 2.1 for b.
2.9^2 + 2.1^2
no,
The pythagorean theorem is: a²+b²=c² where c is the hypotenuse and a & b are two legs. you are given that your c (which is hypotenuse as well) is 2.9 and you are given that your b is 2.1 (your missing side is a)
Alright
please take a shot to plug in your values (into a²+b²=c²)
2.9^+2.1=c^2
your c is given, but a is not
it is like this: a² + (2.1)² = (2.9)²
Oh sorry.
it's ok...
2.9^+2.1=c^2
a² + (2.1)² = (2.9)²
Wifi is bad okay a^2 +(2.1)^2 =( 2.9)^2
because you are given: c=2.9 b=2.1 so, the missing side is a. Our theorem is: a² + b² = c² so lets plug in everything plugging 2.9 for c plugging 2.1 for b you get: a² + (2.1)² = (2.9)²
ok now solve for a
Alright one moment
Do i add the c value or divide or do it the same thing i did in my last problem.
@SolomonZelman you there?
you first calculate the values of (2.1)² and (2.9)²
Alright.
4.41= 8.41. Do i subtract next?
a² + (2.1)² = (2.9)² without a calculator: 21•20=420 21•21=420+21=441 so 2.1² = 4.41 30•30=900 30•29=900-30=870 29•29=870-29=841 so 2.9²=8.41
just demonstrating another technique.
anyway a² + (2.1)² = (2.9)² a² + 4.41 = 8.41 yes you subtract 4.41 from both sides
Okay.
a² + 4.41 \(\small \color{red}{-4.41}\)= 8.41\(\small \color{red}{-4.41}\)
and from this u get?
4.00
yes, or just 4 :)
so, a²=4 correct?
Yeah it looks correct i don't think it can be factored anymore
no, there is no factoring here:) so we got a²=4 what do you think your next (and final) step is?
putting them together like 2.1^2+2.9^2= 4^2
u just take the square root of both sides \(a^2=4\) \(\color{red}{\sqrt{\color{black}{a^2}}}=\color{red}{\sqrt{\color{black}{4} }}\)
a = ?
4^2?
a square root of a 4 is?
2?
yes so a=2
Alright is there anything else?
no, you found the missing side in both of the problems.
Alright thanks for your help :)
yw
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