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OpenStudy (ibrafacts):
Question
OpenStudy (fortytherapper):
Number 32 does seem to have 4 triangles
OpenStudy (ibrafacts):
i drew the lines inside to count
OpenStudy (ibrafacts):
@jim_thompson5910 help me plz
OpenStudy (fortytherapper):
Do you know the area of a triangle?
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OpenStudy (ibrafacts):
i try to do them but i kept getting 6 for the area for number 32
jimthompson5910 (jim_thompson5910):
notice you have 2 triangles
the first triangle is in quadrant 1
the second triangle is in quadrant 4
OpenStudy (ibrafacts):
yes
jimthompson5910 (jim_thompson5910):
both triangles have a base of 6, and a height of 3
OpenStudy (ibrafacts):
how u get base of 6 and height of 3 it doesnt tell u
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jimthompson5910 (jim_thompson5910):
count out the number of square units along the base
ie count the number of spaces between Q and S
jimthompson5910 (jim_thompson5910):
each tick on the grid represents 1
OpenStudy (ibrafacts):
okay i see it
OpenStudy (ibrafacts):
wat next?
jimthompson5910 (jim_thompson5910):
use the formula
A = (b*h)/2
A = area of triangle
b = base
h = height
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OpenStudy (ibrafacts):
9
jimthompson5910 (jim_thompson5910):
that's the area of ONE triangle, but there are 2 of them
both triangles are congruent (one is a mirror image of the other)
OpenStudy (ibrafacts):
oh oh 18
jimthompson5910 (jim_thompson5910):
yep the quadrilateral (in this specific case, a kite) has area 18 square units
jimthompson5910 (jim_thompson5910):
for #32
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OpenStudy (ibrafacts):
hb number 33?
OpenStudy (ibrafacts):
it slooks harder since its on a angle
jimthompson5910 (jim_thompson5910):
what's the distance from T to S?
OpenStudy (ibrafacts):
3
jimthompson5910 (jim_thompson5910):
so that's the base
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OpenStudy (ibrafacts):
height is 3 right?
OpenStudy (ibrafacts):
too
jimthompson5910 (jim_thompson5910):
if you started point R, and you went straight down, how many units do you count?
keep going til you hit the x axis
OpenStudy (ibrafacts):
5
jimthompson5910 (jim_thompson5910):
that's the height
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jimthompson5910 (jim_thompson5910):
b = 3
h = 5
jimthompson5910 (jim_thompson5910):
use the same formula to find the area of the triange
jimthompson5910 (jim_thompson5910):
triangle*
OpenStudy (ibrafacts):
7.5
jimthompson5910 (jim_thompson5910):
yes, there are 2 of these triangles so the total area is 15
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OpenStudy (ibrafacts):
oh
OpenStudy (ibrafacts):
number 34?
jimthompson5910 (jim_thompson5910):
are you able to determine the coordinates of Q, R, S, or T?
OpenStudy (ibrafacts):
yes
jimthompson5910 (jim_thompson5910):
tell me what they are
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OpenStudy (ibrafacts):
q=-3,-2
r=-2,2
s=2,3
t=1,-1
jimthompson5910 (jim_thompson5910):
don't forget parenthesis around each ordered pair
but you have the correct coordinates
OpenStudy (ibrafacts):
okay
OpenStudy (ibrafacts):
what do i do with coor?
jimthompson5910 (jim_thompson5910):
Q is (-3,-2) and the opposite point S is (2,3)
what is the distance from Q to S?
to find out, use the distance formula
\[\Large d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]
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OpenStudy (ibrafacts):
2
OpenStudy (ibrafacts):
the distance is base right?
jimthompson5910 (jim_thompson5910):
Q = (-3,-2)
\[\Large Q = ({\color{red}{-3}},{\color{blue}{-2}}) = ({\color{red}{x_1}},{\color{blue}{y_1}})\]
S = (2,3)
\[\Large S = ({\color{green}{2}},{\color{purple}{3}}) = ({\color{green}{x_2}},{\color{purple}{y_2}})\]
\[\Large d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]
\[\Large d = \sqrt{({\color{red}{x_1}}-{\color{green}{x_2}})^2+({\color{blue}{y_1}}-{\color{purple}{y_2}})^2}\]
\[\Large d = \sqrt{({\color{red}{-3}}-{\color{green}{2}})^2+({\color{blue}{-2}}-{\color{purple}{3}})^2}\]
I'll let you finish up
OpenStudy (ibrafacts):
20
jimthompson5910 (jim_thompson5910):
no
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jimthompson5910 (jim_thompson5910):
what is `-3-2` equal to?
OpenStudy (ibrafacts):
-5
OpenStudy (ibrafacts):
sqrt2 equal 10?
OpenStudy (ibrafacts):
no no my bad25
jimthompson5910 (jim_thompson5910):
So we have...
\[\Large d = \sqrt{({\color{black}{-3}}-{\color{black}{2}})^2+({\color{black}{-2}}-{\color{black}{3}})^2}\]
\[\Large d = \sqrt{(-5)^2+(-5)^2}\]
\[\Large d = \sqrt{25+25}\]
\[\Large d = \sqrt{50}\]
the exact distance from Q to S is \(\Large \sqrt{50}\) units. The approximate distance is \(\Large \sqrt{50} \approx 7.0710678\) (use a calculator with a square root function)
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OpenStudy (ibrafacts):
that is base right?
jimthompson5910 (jim_thompson5910):
no, but it will help us find the area
jimthompson5910 (jim_thompson5910):
what is the distance from R to T ?
OpenStudy (ibrafacts):
can i use the diatance formula for the 32 and 33 too?
jimthompson5910 (jim_thompson5910):
you could but it's faster to do what we did beforehand
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OpenStudy (ibrafacts):
why cant we do 34 like the other problems?
jimthompson5910 (jim_thompson5910):
we could use the distance formula to find out the distance from Q to S in #32
but since Q and S lie along the same horizontal line, it's much faster to count out the spaces between the two
OR you could subtract the x coordinates
jimthompson5910 (jim_thompson5910):
#34 is different because none of the sides are completely vertical or horizontal
jimthompson5910 (jim_thompson5910):
and it's not easy to split up the figure into triangles that have one side that is completely vertical or horizontal
OpenStudy (ibrafacts):
oh
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OpenStudy (ibrafacts):
18
jimthompson5910 (jim_thompson5910):
I think you meant \(\Large \sqrt{18}\) right?
OpenStudy (ibrafacts):
yes
OpenStudy (ibrafacts):
4.24
jimthompson5910 (jim_thompson5910):
The exact distance from R to T is \(\Large \sqrt{18}\). Correct
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OpenStudy (ibrafacts):
yes
jimthompson5910 (jim_thompson5910):
why did we bother with all this? because we can use these diagonal lengths to find the area
formula
area of kite = (p*q)/2
p = length of one diagonal
q = length of other diagonal
jimthompson5910 (jim_thompson5910):
In this case, the length of one diagonal is sqrt(50) while the other length is sqrt(18)
OpenStudy (ibrafacts):
oh
OpenStudy (ibrafacts):
450
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jimthompson5910 (jim_thompson5910):
\[\Large A = \frac{p*q}{2}\]
\[\Large A = \frac{\sqrt{50}*\sqrt{18}}{2}\]
\[\Large A = \frac{\sqrt{50*18}}{2}\]
\[\Large A = \frac{\sqrt{900}}{2}\]
\[\Large A = \frac{30}{2}\]
\[\Large A = 15\]
the area of the entire figure is 15 square units