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@Will.H
There's a glitch M supposed to be (3x, 2y) yet they already submitted the x value as 4x. Not that it matters but yeah.
The length of the base in triangle KMN is NM we can find the length using distance formula. N(3x,0) M(3x, 2y) Using distance formula NM= SQRT(4Y^2) NM= 2Y
The height can be found as we mentioned before. 1st find the midpoint of the base then you'll manage to connect the vertex by that point to eventually have the height. Midpoint of Nm = (3x, y) Thus height is = (0,0)(3x,y) Find the length of that Height = sqrt(9x^2 + y^2) Height = 3x+y Not sure if your allowed to have these values it's always confusing when it comes to variables
Did you ask me to solve it? lol √(2y-0)^2+(3x-3x)^2= √2^2+√3^2=√4+9=2+3=5
Lol I don't even know what you wrote above but no
Anyways the height is 3x
I just randomly solved it I didnt know if you wanted me to solve or not I know it was wrong but like i was just like uuuhhhh lmao
Okay
what is the base?
So the area of KMN = 0.5 * 2y * 3x That's the expression
We already did the base. It is 2y
okay 2y and 3x gotcha
They are dealing with variables not numbers, constants
And the expression is above in case you've missed it tho. So now we are done with triangle KMN. Make sense so far?
Yes
KMN = 0.5 * 2y * 3x Do I solve this to get 3 or no?
Wait I'd dispute here. In the expression we were told to find. Were you able to write it down entirely or were u asked to simplify.
You can't solve anything tho
We're dealing with variables meaning that the expression we're getting are already the answer
oh okay
Alright so the expression of the area of triangle KMN is 0.5 * 2y *3x. However I doubt that you'd be able to write it down that way so instead we can have it like B*h)/2 Thus (2y * 3x)/2 That can be simplified into Y*3x/2 So you pick which one you can write down
okay thank you
Moving on to triangle Klm The expression would be Area =( b* h)/2 Area = 3x* 2y)/2 Area = 3x/2 *y Notice that it is the same exact area of triangle KMN So that means our work is entirely correct and we just proved that
You're welcome. Let me know if you have other questions
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