The two legs (labeled x) of the right triangle below have equal length. if the hypotenuse has length 5√2 , solve for x.

x=5

Step-by-step explanation:

Other than using the plain special aspect of a 45-45-90 triangle where the legs are x, x, and x√2, you can solve for this.

Since the two legs have equal length, they are both x. Using the pythagorean theorem:

(x^2)+(x^2)=50 (Because 5 squared is 25 and √2 squared is 2, multiplying them gives you 50).

You can add (x^2) and (x^2) because they are the same terms (x squared).

Simplifying like so gives you:

2x^2=50

Dividing by two on both sides:

x^2=25

Taking the square root of both sides:

x=5

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!