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Mathematics 6 Online
OpenStudy (anonymous):

What is the perimeter of triangle JKL? J =-1, 2 k=3, 2 L=3, 4 Round your answer to the nearest tenth. A. 8.6 units B. 9.6 units C. 10.5 units D. 11.5 units

OpenStudy (anonymous):

Could you help? @jim_thompson5910

jimthompson5910 (jim_thompson5910):

the perimeter is simply the sum of all of the sides (basically you add up the side lengths)

jimthompson5910 (jim_thompson5910):

this triangle has the sides: JK, KL, JL so you need to use the distance formula to find the lengths of the sides listed above

jimthompson5910 (jim_thompson5910):

does that make sense @vicyoria15 ?

OpenStudy (anonymous):

Yes it does

OpenStudy (anonymous):

so i basically just add up all the points??

jimthompson5910 (jim_thompson5910):

ok great, let me know what you get for the side lengths

jimthompson5910 (jim_thompson5910):

no not the points

jimthompson5910 (jim_thompson5910):

what is the distance from J to K?

OpenStudy (anonymous):

4?....

jimthompson5910 (jim_thompson5910):

correct

jimthompson5910 (jim_thompson5910):

so JK is 4 units long

OpenStudy (anonymous):

okay so i added all the sides and i got 10

jimthompson5910 (jim_thompson5910):

how long is KL ?

OpenStudy (anonymous):

kl is 2

jimthompson5910 (jim_thompson5910):

JK = 4 KL = 2 how about JL ?

OpenStudy (anonymous):

4 also

jimthompson5910 (jim_thompson5910):

no

jimthompson5910 (jim_thompson5910):

J = (-1,2) L = (3,4)

OpenStudy (anonymous):

5?

jimthompson5910 (jim_thompson5910):

subtract the coordinates, square the result, then add the squares and take the square root

OpenStudy (anonymous):

it looks like 4 but its hard to telll

jimthompson5910 (jim_thompson5910):

notice we go from x = -1 to x = 3 that's a difference of ???

OpenStudy (anonymous):

1?

jimthompson5910 (jim_thompson5910):

-1 minus 3 = ??

OpenStudy (anonymous):

oh its -4

OpenStudy (anonymous):

sorry

jimthompson5910 (jim_thompson5910):

square that to get ??

OpenStudy (anonymous):

16

jimthompson5910 (jim_thompson5910):

now do the same for the y coords

jimthompson5910 (jim_thompson5910):

and you get ??

OpenStudy (anonymous):

...You lost me...

jimthompson5910 (jim_thompson5910):

J = (-1,2) L = (3,4) what are the y coordinates?

OpenStudy (anonymous):

2 and 4

jimthompson5910 (jim_thompson5910):

subtract them, then square that result

OpenStudy (anonymous):

i got 4

jimthompson5910 (jim_thompson5910):

yes

jimthompson5910 (jim_thompson5910):

for the x coordinates, we subtracted and squared the result to get 16 for the y coordinates, we subtracted and squared the result to get 4 add the squares: 16+4 = 20

jimthompson5910 (jim_thompson5910):

then you take the square root \[\Large \sqrt{20}\]

jimthompson5910 (jim_thompson5910):

that's the exact distance from J to L

OpenStudy (anonymous):

4.4

jimthompson5910 (jim_thompson5910):

what comes after the second 4?

OpenStudy (anonymous):

7

jimthompson5910 (jim_thompson5910):

so roughly 4.47, yes

jimthompson5910 (jim_thompson5910):

JK = 4 KL = 2 JL = 4.47 (approximate)

jimthompson5910 (jim_thompson5910):

add up: JK, KL and JL to get the perimeter

OpenStudy (anonymous):

i got 10.47

jimthompson5910 (jim_thompson5910):

that rounds to 10.5

OpenStudy (anonymous):

so rounded to the nearest tenth its 10.5 right?

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

sorry i typed without seeing what you said haha Thanks so much, you teach me better than my teachers!

jimthompson5910 (jim_thompson5910):

I'm glad I could help out

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