Find the area of a right isosceles triangle with hypotenuse 10\sqrt{2}10

1/8.

Step-by-step explanation:

By Pythagoras:

[10/ sqrt(2)10)]^2 = x^2 + x^2     where x is length of the 2 equal legs.

(1/sqrt2)^2 = 2x^2

2x^2 = 1/2

x^2 = 1/4.

The area = 1/2 * x^2

= 1/2 * 1/4

= 1/8.

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure