16

firstly show your working

how did you confirm it to be 7? my friend is also saying it's 7 but I'm not sure why

nope

wolframalpha.com/input/?i=xy = 4, xz = 8, (x-y)(x-z) = 6

it's the only solution when x > 0, x > y, x > z

...

>xy = 4
>xz = 8
Red and green don't share side with the same length, try again

>mfw non-engineers actually struggle with this problem

so it's actually 9, not 7

fucking nice!
7 it is.
/thread

you're a fucking moron, dude

meant for

Are we supposed to assume it's a square?

nice

no, you

4 and 3 = 7 but there still is that slither of blue leftover

so the blue area is 9, since 7 and 2 is 9

It's 7 lol. The green is height 4, base 2. And the red is height height 4 base 1. This all works out to allow for the yellow to be height 3 base 2. All the areas work for the values given. Square is 42=16. 16-(4+3+2)=7

The 2 is red wtf

Ya I'm assuming square, since no other way it would be possible

lol you still cant solve this without more info about the problem.
are right. its an impossible problem meant to bait morons like you

it's not 9 can't you understand?

it's possible if you assume it's a square

yeah but whats the area of the square if its moving at .95*c

16-9 = 7

how are you so fucking dumb, seriously??

you dont count the red part!
the red bit was moved to fill up that blue part that's the same size, so it all fits as right angled triangles

You don't have to assume it's a square

I'm assuming it's a square unless it says it's not drawn to scale.

(x*y)/2=2
(x*z)/2=4
((x-y)*(x-z))/2=3
(x^2)-9=?

Also I'm too lazy for LaTeX

Maybe 3, maybe 9.

>I'm assuming it's a square unless it says it's not drawn to scale.
Brainlet

7, if square

It's 7 and it's not necessarily square

Well, how would I know those are even right angles at the corners? The whole shape could actually be a rhombus, a trapezoid or a triangle, or even a poor approximation of a circle! It could just be not drawn to scale.

This thing is impossible without assuming information. It could be asking for the area of the outlines of the shapes for all I care!.. They're blue.

>This thing is impossible without assuming information.
Occams Razor, and people don't usually make puzzles just for someone else to "solve" it with a ruler

what guarantee do you have that the red triangle intersects the black line at exactly the halfway point?

not convinced. proof by picture is often flawed.

whats the area of it if its in 5 dimensional space and we rotate it 30° about the x axis and accelerate it to infinatly close to the speed of light?

this is the correct answer but the wrong way to solve it.

7

xy=8
xz=4
(x-y)(x-z)=6

x-6/(x-z) = y
z=4/x

x[x-6/(x-4/x)]=8

x=4

x*x=16
?=16-4-3-2=7

>Occams Razor
that was the point

The drawing and those three numbers are all the information given. The image shows a square, I assume it's a square. Those areas look like triangles, I assume they are triangles. If it says it's not drawn to scale, or if you disregard the properties of the shapes in the drawing, it's impossible.

oh, that makes sense then.
>upset that I didn't realize this

Likely the most intelligent person in this thread.
/thread

The real answer. Good job

1

Because the shapes are numbered 1, 2, 3, 4 and only the #1 is missing.

...

looks more like 5 or 5.5 desu

*I meant 6.5

The blue area is blue.

Nope, assuming it is a square when the only reason to think it is one is "because it looks like it"

Yes. You could stretch it out sideways, but the area of all 16 cells would stay proportionate.

Since some anons don't trust that the shape is a square: because I guess they don't trust the drawing(which is ironic in this case). I'm perfectly justified to not trust that the shape is even rectangular; yielding even more countless solutions.
Why assume those three numbers are the areas of the red/yellow/green triangles?

They could be the side lengths of the blue area. In which case:

[math]area = \sqrt{s(s-4)(s-3)(s-2)}[/math]

[math]s = \frac{4+3+2}{2} = 4.5[/math]

[math]area = \sqrt{4.5(0.5)(1.5)(2.5)}[/math]

[math]area = \sqrt{8.4375}[/math]

[math]area \approx 2.9047[/math]

Because the height of the triangle is the square's height.

You cant assume that the right side is evenly bisected like that.

assuming it's a square with side length x:

green triangle has legs of length:
>x
>8/x

red triangle has legs of length:
>x
>4/x

since the product of the length of the legs has to be double the area ( A = l1 * l2 / 2 )

therefore the yellow triangle has legs of length:
>x-8/x
>x-4/x
(x-8/x) * (x-4/x) = 6

x ϵ {-4. -sqrt2, sqrt2, 4}
but all the triangle legs must have positive length, so x=4
from there it's simple. total area of square is 16, so subtract the areas of the known triangles and you're left with 7

not that hard if you can do high school level math

There's no way both the green triangle and blue triangle on top could have the same area that's bullshit

holy fuck did you even read it? its one part 4 and the other would be 3. 3+4=???? 4????

Yeh I get it now

baka all these brainlets

it literally tells you to think outside the box and you start doin shit inside the box

>believing the values you are told when they might not be true in the first place

approximate pixel counts for each area

get on my level, tools.

I'm not re-approaching it assuming the numbers on the figures represent surface areas. (I only realized this interpretation as question when I finished my drawing)

they just can't understand the concept it seems. Veeky Forums is lost

short red = 1

long red = 4

long yellow (bottom) = 3

short yellow = 2

long green = 4

short green = 2

sides are 4 each, total area 16

sum of all non-blue triangles = 9

blue triangle = 7

debate me

How do you know the 4 and 3 triangles meet the right side at the middle?

5 duh

>2
>3
>4
Gee i wonder what comes next

It's not 5 if that's what you're implying.

Assume the lengths may be different, then,
[eqn]\begin{align}\frac{2}{x+y}+\frac{3}{y}&=\frac{4}{x}\\ \frac{3x+5y}{(x+y)y}&=\frac{4}{x}\\ 3x^2+5xy&=4xy+4y^2\\ 3x^2+xy-4y^2&=0\\ (x-y)(3x+4y)&=0\\ \implies x=y &\text{ or } x=\frac{-4y}{3}\end{align}[/eqn]

The negative result makes no sense in the context, so they must be equal.

There should be an extra factor of 2 on those fractions, but it cancels straight out though, so the result holds.

Answer is 7.

Sides of box = 4.
Total area = 16.

16 - 4 - 3 - 2 = 7

the blue area is a triangle

The 4 triangle is a 30, 60, 90 triangle. The other 3 are irregular.

1/2b*h=4
b*h=8

My trig is pretty rusty from here but I'm pretty sure you can always use, like a √3 for the base in relationship to the hypotenuse. And then if its possible to solve for the 4 triangle, you can use the law of exterior and interior angles to get the exact measurements of the rest of the sides.

let top/bottom side be a and left/right sides be b.
Blue area B = ab - 9
ax = 8
by = 4
(a-y)(b-x) = 6
==> ab - ax - by + xy = 6
ab - 9 = -3 + ax + by - xy = B
abxy = 32
B = 9 - xy = 9 - 32/ab = ab - 9
(ab)^2 - 18ab + 32 = 0
(ab - 16)(ab-2) = 0
ab = 16 or 2
Hence B = 16 - 9 = 7.

Like the other guy said, the extra factor of two, since A=(1/2)BH is important, but luckily cancels out; Otherwise, This is a great proof.

...

math schemes usually have codification and aren't supposed to mislead you, ergo:
-all these are triangles, no way the ends of yellow and green don't match
-straight edges are straight edges, so the thingy's a quadrilateral and not a curvy shit in disguise or a hexagon with angles really really close to 180°
-absence of codification of right angles and same lengths should imply angles ain't necessarily right, and (opposite) lengths ain't necessarily equal.
-i'll give you that, those actually look right and equal, which is misleading, but absence of codification vote for the opposite, so you can say "problem is badly specified"

still assuming rectangle tho :D
good job factorizing btw

how bout pic related?

I knew that not going to compsci was the right choice, I would have never gone past calc1

I DONT GET IT

pls explain

If you assume the figure is a rectangle, then it works out that it must be a square, and the answer as many pointed out is 7. I haven't yet figured out if theres a unique answer if the angles aren't 90 degrees like in your pic, but I will keep working at it.

I didn't assume it, that's the result of the calculation solved for a single variable. The only I assumptions I made to get that far is that it's a rectangle.

If you don't believe it, try coming up with a counterexample or calculate it yourself

>If you assume the figure is a rectangle, then it works out that it must be a square
No. See
You can set a to 8 or to 1 and neither of them is a square, but both solve the problem

you're fucking retarded it says "green" area

7

there are often many ways to solve a problem, this one is just fine.

Ha! Detected the Mensa applicant.

Maybe there is an answer related to the colors as well.

could be any area, no restriction (as long as the quad doesn't need to be convex)

see pic related

i'll check what the conditions are if we want the quad convex

i'm a dumbass!!

replace 4/(2a), 3/(2b) and 2/(2c)
with 2*4/a, 2*3/b and 2*2/c