Help.
How big is big triangle?
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You mean fucking field of that triangel?
1/2a*h or a2+b2=c2 I have no idea how to count that shit
>mfw this task overwhelmed me
bump for answer
Really fucking big.
You cant be so dumb
please give the fucking answer and how did you do that
>inb4 you have no idea either
>nobody remembers theorem of pythagoras in middle school
~ tree fiddy
The other side is 8 and the area is 30 now fuck off
it's 40,8. man you dumb.
Such a baka.
tell us how did you do that
or this is just fiction
Side BC is equal to sqrt(10^2 + 6^2) = sqrt(136) ~= 11.66
10 = AC
this fucking 10 is all AC line, not fucking part of it
this is why its tricky
AC² = BC²+AB²
>How big is big triangle?
About yay big.
Ah. then fuck if I know. I took middle school geometry like 18 yers ago
...
are you fucking retarded ? then it's fucking much easier.
it's 6 x 10 / 2 = 30. oh my fucking god
>mfw
>Sup Forums
>defeated
>by
>3th
>grade
>maths
>geometry
>question
KEKITY KEK LELITY LEL
MY SIDES
Lol, no way this triangle exist, If AC is 10, h can be a maximum of 5
just use, pythagorean theorem. jeez, it's easy
idiot , what is the question ?
>perimeter
>area
>which triangle
>he thought p+10 = AC and not AC=10
your eyes are retarded: "10" is in the middle of AC ergo AC=10
dude if |AC| is 10 the answer is 30. if not it's 40,8.
i'm not defeated. i'm a physics student and this is embrassing.
Which length is 10?
Question is philosophic mostly.
Use A^2+B^2=C^2 to find C. Then use sin or tan to find the angle at point C. Then recognize that 180-90-angle at C= Angle at A. Then use the law of sines or cosines (can't remember which) to find length AB. Finally, use A=1/2b*h to find the area. If you need perimeter, use A^2+B^2=C^2 to find length AC. Then kill yourself cause you're a stupid faggot.
The maximum possible altitude is half the hypotenuse, making it five and not six.
No such triangle can exist, therefore I can't tell you how big it is. Now gtfo or post tits
>if |AC| is 10...
Stopped reading right there
Your assumptions were wrong from the very beginning.
Three triangles
Each are 30-60-90 triangles(angle measures)
Side lengths are in proportion of 1:2:root3(EACH TRIANGLE IS LIKE THIS)
Set up an extended proportion
Solve
OP here
That triangle cant exist
hey faggots, look here
quora.com
a2 + b2 = c2
This.
It is impossible for the length inside to be 6, if the Hypotenuse is 10.
this
fucking troll.
solution: use pythagoras to find BC, then cosine theorem to find all angles and sine theorem to extract your missing AB.
I yawn at your general direction.
four times as big as a quarter of the triangle!
Uh... so what's that in the pic then dude?
Nice one, completely forgot about Thales.
SOLUTION:
In rectangular triangle:
h = sqr(c1*c2)
so:
c1 = x
c2 = 10
h = 6
6 = sqr(10*x);
36 = 10x
x = 36/10 = 18/5
c = 10+18/5 = 68/5
area = 68/5*6/2 = 204/5 = 40.8
Small triangle hypoth. = 7
Small triangle small side = ~ 3.602
theer u go user :-)
To clarify, |BD|^2 + |DC|^2 = |BC|^2 at first, and |AB|^2 +|AC|^2 = |BC|^2 for the second one.
Don't be an idiot. This isn't a trick question. |DC| = 10, not |AC| = 10.
|AC| = 10 is some Facebook tier shit like that fucking banana coconut math.
This
To start of I am assuming AC=10, the picture is a bit unclear about that part.
Let's name the point where AC is split D
The hypotenuse of BDC is the base for ABC
AC=10
DB=6
By using similarity we can get to this equation:
BC/BD=AC/BC
BC/6=10/BC
BC^2=6*10
BC=(6*10)^0,5
BC= 7,74597 (approximately)
Now we need to figure out what AB is so we can calculate the area of the big triangle, ABC.
AC^2=BC^2+AB^2
AB^2=AC^2-BC^2 (Pythagoras theorem)
AB=(AC^2-BC^2)^0,5
AB=(10^2-7,74597^2)^0,5
(Raising something to 0,5 is the same as taking the square root of it)
AB=6,325 (approximately)
Now we calculate the area with a simple formula
(h*b)/2= A
A = (AB*BC)/2
A = (6,325*7,74597)/2
A = 24,5 L.E
I wish I was high on potenuse.
I just realized that this kind of triangle is impossible to even form...
the longest distance the line which has value '6' in it can only be 5... So this is impossible to solve