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Mathematics 10 Online
OpenStudy (ibrafacts):

I need help

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?

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

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

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

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

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

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

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

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 \[\Large d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\]

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

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)

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

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

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

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

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

OpenStudy (ibrafacts):

oh

OpenStudy (ibrafacts):

thank you

jimthompson5910 (jim_thompson5910):

no problem

OpenStudy (ibrafacts):

i think i understand how to do it

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