Here's a good triangle problem for ya!?

Question

You have a right triangle
The height of the triangle from the hypotenuse to the right angle + the
shortest side = the hypotenuse.

If shortest side is 1 inch what are the lenghs of the other 2 sides?

Can you give me a set of equations to put in an equation solver to show how u got your answer?

Answer 1

That's impossible. If the height of the triangle from the hypotenuse to the right angle (let's call this "b") + the shortest side (let's call this "a") = the hypotenuse (let's call this "c"). Then the equation is (b + a = c). But if one side equals one and since all triangles follow the pythagorean theorem, (a^2 + b^2 = c^2) this creates a conflict.

For instance:
a = 1 and solve for "c"
a + b = c
1 + b = c

Solving for "c"
a^2 + b^2 = c^2
1^2 + b^2 = c^2
1 + b^2 = c^2
sqrt(1 + b^2) = c

Set the two equations together:
sqrt(1 + b^2) = 1+ b
1 + b^2 = (1+ b)^2
1 + b^2 = 1^2 + 2b + b^2
1 + b^2 = 1 + 2b + b^2
The "1" and "b^2" cancel each other out
0 = 2b
0 = b which would make c = 1

While this works for the equations, this does not make a triangle. A triangle cannot have sides: 1, 0, 1.

Unless the problem was miss stated, the answer is No Solution.

Answer 2

Let the other side be x and hypotenus = y
from pythagorus theorem
x^2+1 = y^2 ....1
area of the triange = x/2
so height of the triangle from hypotenuse = x/y
so x/y + 1= y
or x +y = y^2 .... 2
1 and 2 should be solved

now from 2 x = y^2-y
(y^2-y)^2 + 1 = y^2
y^4 +y^2-3y^3 + 1 = y^2
y^4-3y^3+1 = 0
I am unable to proceed further this should be used to solve for y

Answer 3

base^2 + height^2 = hypotenuse^2
If base (usually the shortest, but this is an assumption)
1+height^2 = hypo^2

sqrtRt (heigth^2+1) = hypo
You can solve this by limits where lower limit is >0 (as its a triangle cannot have a side of zero) and substitute height =1
Therefore the other two sides could be 1 and 1.41.

The answer is based on taking the lowest whole number on one side of the equation, you can also get different answers nased on what your upper limit is.

Answer 4

1&1/2 in.

Answer 5

Let the sides of the large triangle be a, b, & c.
a = a
(c-1)^2+1/c^2 = 1, 1 c^2(c-1)^2+ 1 = c^2
c^2(c^2-2c+1) + 1 = c^2
c^4-2c^3+1 = 0
(c-1)(c^3-c^2-c-1)=0 (c=1 is trivial)
1=-c-c^2+c^3
1/c=-1-c+c^2
1/c+1=-c+c^2
((1/c)+1)/c=-1+c
((1/c)+1)/c)+1=c
c≈1.8393
b≈1.5437
(There may be 2 other roots, but I'm too lazy to find them if there are.)

Answer 6

the ratio for right triangle 1: sqrt 3: 2

Answer 7

1 inch square plus the other side square is equal to the hypotenuse square. Now you work out the rest

Answer 8

take height AB= L shortest side AC = 1 hypot.comes out CD = L+1
solve
AD^2 + DC^2 = AC^2 comes out AD^2 = (L+1)^2 - 1..................1
solve
AB^2 + DB^2 = AD^2 comes AD^2 = L^2 +( L+1)^2..................2
solve 1 & 2 L= 1 hence base =1 hypo =2 height = 5

Answer 9

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Answer 10

square root of 3,over 3 and 2multiplied by square root of 3over 3

that is the answer for the other two lenght i hope you get what i mean