Answer: Part a: x=sqrt(104) in; Part b: x=sqrt(5) in
Step-by-step explanation:
Part a:
Because the figure has a right angle (signified by the square), it is a right triangle. Therefore, the Pythagorean Theorem, a²+b²=c² where c is the hypotenuse (longest side) and a and b are the legs (shorter sides), applies. Thus:
a²+b²=c²
The 15 in side is opposite the right angle, so it is the hypotenuse.
The 11 in side and x in side are not opposite the right angle, so they are legs.
11²+x²=15²
121+x²=225
(121+x²)-121=225-121
x²=104
sqrt(x²)=sqrt(104)
x=±sqrt(104)
Reject x=-sqrt(104) because a negative side length in a triangle is impossible. Therefore,
x=sqrt(104) in
Part b:
The figure is a right triangle. Therefore, the Pythagorean Theorem (a²+b²=c²) applies. Hence,
5²+(2x)²=(3x)²
25+4x²=9x²
(25+4x²)-4x²=9x²-4x²
25=5x²
5=x²
sqrt(5)=sqrt(x²)
x=±sqrt(5)
Reject x=-sqrt(5) because a negative side length is impossible.
The answer must be x=sqrt(5) in.
67.5 degrees
3pi/8 * 180/pi = 67.5
seperate it into a square and right triangle
area of square: 2*2=4m^2
area of traignle: (1*2)/2= 1m^2
add them
5m^2
Answer:

Step-by-step explanation:
We can use the Law of Sines to find segment AD, which happens to be a leg of
and the hypotenuse of
.
The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is
. The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

Now use this value in the Law of Sines to find AD:

Recall that
and
:

Now that we have the length of AD, we can find the length of AB. The right triangle
is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio
, where
is the side opposite to the 30 degree angle and
is the length of the hypotenuse.
Since AD is the hypotenuse, it must represent
in this ratio and since AB is the side opposite to the 30 degree angle, it must represent
in this ratio (Derive from basic trig for a right triangle and
).
Therefore, AB must be exactly half of AD:
