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Mathematics 18 Online
Nicole:

Classify the triangle based on side lengths 5, 12 and 13. right acute obtuse no triangle can be formed with given side lengths

Nicole:

@mhchen

Nicole:

@Hero

Nicole:

and why is that

Nicole:

@Gdeinward @Vocaloid any ideas>

Nicole:

@dude

justjm:

5, 12, and 13 is a Pythagorean triplet. hopefully that identifies your answer.

Nicole:

so right triangle?

justjm:

yup! However in the future, this will not always work. As a result, you must youse the Pythagorean Inequality Theorem, if you know what that is

Nicole:

http://prntscr.com/q5atsz

Nicole:

@justjm

justjm:

Just plug 280 as area and base as 70 for a=1/2 bh

Nicole:

8?

justjm:

yeah

Nicole:

http://prntscr.com/q5avbj true?

justjm:

So they're saying a side is congruent, and since they are parallel then they are alternate interior angles that are cut by a transversal. Then LK is shared. So it looks like SSA. is SSA ever true?

Nicole:

no

justjm:

so that should direct you to your answer

justjm:

By the way, I did geometry a while back so I might not be of best knowledge when it comes to euclidean triangle axioms and theorems.

Nicole:

so you're not 100% sure of this?

justjm:

Sorry it's SAS

justjm:

It's SAS guaranteed. Sorry I confused you with the SSA

Nicole:

its okay but its still false correct?

darkknight:

I did geometry last year

justjm:

no it's true because it's congruent by SAS

darkknight:

it is true

Nicole:

okay http://prntscr.com/q5axhz and this is also true ?

darkknight:

yes, SSS

darkknight:

are you in pre-calc.

Nicole:

Okay I see

justjm:

Took that last year buddy

Nicole:

oh so that is true?

justjm:

Yeah it's true due to SSS

Nicole:

http://prntscr.com/q5ay27 SSS or SAS?

darkknight:

what did I say?

Nicole:

Yes sorry I was confused for a second

darkknight:

SAS

darkknight:

it is SAS

darkknight:

no what am I saying. It is SSS

darkknight:

They share the same side and the other 2 sides are congruent

darkknight:

SSS

Nicole:

Yes okay thats why I paused for a second lmao

Nicole:

http://prntscr.com/q5ayde

justjm:

Oh this is CPCTC isn't that what it's called?

Nicole:

I thought it is the last option

darkknight:

FDE is similar to CAB

Nicole:

not CBA?

darkknight:

nope

darkknight:

Take the second triangle and flip it so that B is on top and C is the furthest left. Now the triangles are lined up. It makes it easier

Nicole:

Oh okay got it http://prntscr.com/q5az5m

darkknight:

The first one, obviously

darkknight:

it is given that ABC is congruent to DEF and option 1 has everything in the right order

Nicole:

Wow im totally out of it today lol my bad http://prntscr.com/q5b021

darkknight:

use the special right triangle rule

darkknight:

do you know it?

Nicole:

multiplying this length by the square root of 2?

Nicole:

to find the hypotenuse

darkknight:

yes

Nicole:

10 square root 2

darkknight:

I'm not sure you did it right, did you do 5 times sqrt 2 times sqrt 2?

Nicole:

I did \[5\sqrt{2}*2\]

darkknight:

yes it is 10, you said 10 times sqrt 2 before

Nicole:

yes because it is \[10\sqrt{2}\]

darkknight:

its not

Nicole:

http://prntscr.com/q5b1mn

darkknight:

plug 5√2(√2) into a calculator

darkknight:

(√2)^2=2, multiply by 5 you get 10!

Nicole:

omg see I told you im out of it today I forgot to put \[\sqrt{2}\]

Nicole:

sorry again

darkknight:

lol

Nicole:

http://prntscr.com/q5b27u

darkknight:

try to do this one on your own

Nicole:

12.5?

Nicole:

since 25/2=12.5

darkknight:

yes

Nicole:

One face on a dice is an example of a regular polygon True False im not sure tbh

darkknight:

well yes it would be, when you look at a dice (cube or not) you see all sides that are regular. A regular polygon that has the same length sides and same angles

Nicole:

oh okay well that makes sense now http://prntscr.com/q5b415

Nicole:

False?

darkknight:

check your theorms/postulates. Is there one that can prove this true?

Nicole:

well we automatically know because opposite sides are parallel, which does not show on the shape, which means....its true

darkknight:

it is true, not in that way. We have 2 angles that are congruent and the triangles share a side so what theorem/postulate shows this is true?

darkknight:

ASA

Nicole:

sorry my internet cut off but yes ASA

Nicole:

5 questions left please truly appreciated http://prntscr.com/q5b8ch

darkknight:

What grade r u in?

Nicole:

10th but taking geometry since im homeschooled and thats the only math im missing

darkknight:

I'm in 9th

Nicole:

basically im ahead but yeah

Nicole:

you're a smart 9th grader Mars

darkknight:

thx

Nicole:

wow I said a "curse" word and it turned into "mars" ight

Nicole:

and np

darkknight:

This should be easy. You know that the perimeter is 136. Set 2L+2W (length and width) equal to 136

Nicole:

isnt it 2(17)+2(x+3)=136 ?

darkknight:

yes, the equation is right

Nicole:

x=48

darkknight:

x=48

darkknight:

yes

darkknight:

I did that wrong

darkknight:

the area is 136. So (x+3)(17)=136. sorry about that

Nicole:

x=5

Nicole:

50?

Nicole:

is the answer? since when you plug 5 in it turns out that the answer is 50

darkknight:

yes

Nicole:

oky http://prntscr.com/q5bb1d

darkknight:

do the same thing as above

Nicole:

x=11

Nicole:

since (x+3)(17)=238 makes x equal to 11

Nicole:

correct?

Nicole:

http://prntscr.com/q5bbr3

mhchen:

Area = length x width Perimeter = length + length + width + width = 38 length = 10 10 + 10 + width + width = 38 width = (?)

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