/sci/ - Science & Math

[Anonymous No.7435969]

science idol thread?

science idol thread.

[Anonymous No.7435981]

Pic related.

[Anonymous No.7435987]

>>7435981
32*16 = 20*x

[Anonymous No.7436000]

>>7435981
I don't get it. Is this supposed to be tricky?

[Anonymous No.7436133]

mine idol

[Anonymous No.7436148]

I think Gibbs was a great man.

[Anonymous No.7436150]

>>7436148
I grew up in Connecticut all my life and I JUST figured out who Gibbs was and where he did what he did. I feel like a total pleb, but Gibbs is my new found hero.

[Anonymous No.7436160]

>>7436150
Nice, what do you do anon? I do Chem and a little bit of Math on the side, gotta learn some of it so I can at least be a fraction of what Gibbs was.

[Anonymous No.7436752]

>>7435981

Here's what I got. Let a_1 = 16, a_2 = 20, a_3 =32 be the known areas given by the figure. Let x be the unknown area. Let the side length of the square be given as s. Then the area of the square is s^2. We also have x = s^2 - a_1 - a_2- a_3. In order to find x, it suffices to find s^2.

Subdivide the square as follows: from each corner of the square, draw a line to the point of intersection within the square. This should subdivide each region into two triangles, and will subdivide the entire square into eight triangles.

Using the triangle area formula A = (1/2)bh, we obtain the following constraints:

a_1 = (1/2)(1/2s{h_1}) + (1/2)(1/2s{h_2})

= (1/4)s(h_1 + h_2)

similarly, for the other areas we have

a_2 = (1/4)s(h_2 + h_3)

and

a_3 = (1/4)s(h_3 + h_4)

Now examine the sum of these three numbers:

a_1 + a_2 + a_3 = (1/4)s[(h_1 + h_3) + (h_2 + h_4) + (h_2 + h_3)]

But notice that h_1 + h_3 = s, and h_2 + h_4 = s, and from the above constraint we see that h_2 + h_3 = (4{a_2}/s)

Then the sum is equal to (1/4)s(2s + 4{a_2}/s), continuing:

= (1/2)s^2 + a_2

remembering where this started, this is equal to a_1 + a_2 + a_3, so:

(1/2)s^2 = a_1 + a_3 and s^2 = 2a_1+2a_3

Then x = s^2 - a_1 - a_2 - a_3 = a_1 - a_2 + a_3

= 28 cm^2

Can anyone else confirm using a different method?

[Anonymous No.7436972]

My nigga right here

[Anonymous No.7436984]

>>7436000
That's not a good way to solve it.

[Anonymous No.7436985]

>>7435987
Incorrect, but close.

[Anonymous No.7437209]

>>7436000

Nice trips, but unless you're asspulling some complicated theorem I'm not sure how this isn't just an unnecessary convolution of a method like >>7436752