Question

In a 45-45-90 triangle, what is the length of the hypotenuse when the length of one of the legs is 11 in.?

Answer 1

Part A)In a 45-45-90 triangle, what is the length of the hypotenuse when the length of one of the legs is 11 in.?

we know that
cos 45°=√2/2
[he length of the hypotenuse]=11/cos 45-----------> 11/(√2/2)----> (11*2)/√2
=22/√2-------> 11√2 in

the answer Part A) is 11√2 in

Part B) What is the exact value of sin 45° ?

we know that
sin 45°=11/(11√2)-------> 1/√2---------> (1/√2)*(√2/√2)-----> √2/2
the answer part b) is √2/2

Part C)
What is the area of a regular hexagon with a side length of 4 m?

we know that

In case of a regular hexagon  each of the six triangles that are formed by connecting its center with all six vertices is an equilateral triangle with a side equaled to 4 m.
The area of this hexagon is six times greater than the area of such a triangle

In an equilateral triangle with a side d 
the altitude h can be calculate from the Pythagorean Theorem as
h²=d²−(d/2)²=(3/4)d²
Therefore, 
h=d√3/2

Area of such a triangle is
A=d*h/2------------> d²*√3/4
From this the area of the regular hexagon with a side d is
S=6*A----------> d²3√3/2
for d=4 m
S=4²3√3/2------> 24√3 m²------------> 41.57 m²

the answer Part C) is 41.57 m²

Part D) In a 30-60-90 triangle, what is the length of the hypotenuse when the shorter leg is 5 cm?
[he length of the hypotenuse]=5/sin 30--------> 5/(1/2)---------> 10 cm

the answer part D) is 10 cm