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/sci/ - Science & Math

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1936861 No.1936861 [Reply] [Original]

curious on how to solve this, help?

>> No.1936888

x+6, x+4, 10 make up a right triangle

6,8,10 triangle
so one side is 8 the other is 6
x = 2

>> No.1936907

>>1936888
nobody said anything about right triangles

>> No.1936908

>>1936888
I don't think so because if x = 2, then that triangle would be 6,8,10 and not a Pythagorean triad.

>> No.1936926

try equating the areas of the 3 smaller triangles to the larger one by using heron's formula where you only need to use the side lengths, might work.

>> No.1936936

>>1936908
36+64=100, son

/thread

>> No.1937387

>>1936888
There's nothing there to indicate that angle will be 90º

Use the cosine rule

>> No.1937405

>>1936936
>Formula for calculating directly from KNOWN values is
>No known right angle

>> No.1937407

I think this is solved with area. All 3 smaller triangles add up to the area of the larger triangle. use the Heron's formula A=sqrt(s*(s-a)*(s-b)*(s-c) where s = (a+b+c)/2 to get equations for the area of all 3 triangles. Sum these 3 and equate it to the total area, giving you a 1 variable 1 equation problem.

Good luck op

>> No.1937412

>>1937387
can't, don't know enough angles

>> No.1937416

Law of Cosines

>> No.1937424

>>1937412
System of equations, bitch!

>> No.1937450

>>1937424
nope can't do. 3 equations, 4 unknowns.

>> No.1937455

>>1937450
I didn't look at the problem. I was just joking.

>> No.1938029
File: 51 KB, 485x386, 1287807278908.jpg [View same] [iqdb] [saucenao] [google]
1938029

U= 10+10+10 / 2 = 15

A(ABC) = in root square U.(U-A).(U-B).(U-C)

this is your A(ABC) formula if you try this to other 3 triangles and create logaritmetic f(x) it will be like that.

>> No.1938058
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1938058

Just draw rectangles, use the pythagorean theorem several times, and then solve the resulting equations.

After several pages of algebra I got the following ugly result.

Let <span class="math">D=a+b+c[/spoiler], <span class="math">E=a^2+b^2+c^2[/spoiler], <span class="math">F=a^3+b^3+c^3[/spoiler], and
<span class="math">G=a^4+b^4+c^4[/spoiler]

Then x is the solution of the following quadratic equation:

<div class="math">(36 E - 12 D^2 - 18 L^2)x^2 + (28 F - 4 D^3 - 12 L^2 D + 24 a b c)x + 9 G - 3 E^2 - 6 L^2 E + 6 L^4 = 0</div>

>> No.1938066

>>1938058
or in short, no you can't solve it.

>> No.1938069

>>1938058
or in terms of a, b, and c:
<div class="math">(4a^2 + 4b^2 + 4c^2 - 4ab -4ac -4bc -3L^2)x^2 + (4a^3 + 4b^3 + 4c^3 -2a^2 b -2a^2 c -2b^2 a -2b^2 c - 2c^2 a -2c^2 b -2aL^2 - 2bL^2 - 2cL^2)x + a^4 + b^4 + c^4 - a^2 b^2 - a^2 c^2 - b^2 c^2 - a^2 L^2 - b^2 L^2 - c^2 L^2 +L^4 = 0</div>

>> No.1938070
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1938070

law of cosines

stay in school /sci/

/thread

>> No.1938082

>>1938069
I'm fairly sure this answer is correct, because even after all that awful algebra the answer is invariant if you switch a <-> b and a <-> c and b <-> c

>> No.1938083

lol this is shit

>> No.1938179
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1938179

hey.

look at the right side of the triangle, from points A to C we move 10, yes? and thats the closest distance between A and C. another route, A-O-C we move X+X+4, so 2X+4> 10, so 2X>6, so X>3

furthermore X<4 because otherwise the line BO would leave the triangle, as it would be 10 units long

so we know 3<X<4

hope this helps.. ive no idea how to work out exactly what X is tho..