Break the problem into two parts: 1) the area of the this isosceles triangle whose hypotenuse is AB and 2) the area of the semicircle whose diameter is AB.
The triangle is isosceles because the lengths of the two shorter legs are the same (2 meters). Use the Pythagorean Theorem to find the length of the hypotenuse of this triangle. (AB)^2 = (2 m)^2 + (2 m)^2, or 8 m^2. Thus, AB = sqrt(8 m^2), or 2sqrt(2). This AB is also the base of the triangle. What is the area of the triangle?
Next, noting that the diameter AB of the semicircle is 2sqrt(2) and the radius is just sqrt(2), find the area of the semicircle. The area of a circle of radius r
is pi*r^2; here it's pi*(sqrt 2)^2, or pi*2, or 2pi.
Add the area of the triangle to this area of the semicircle (2pi) to find the total area of the figure.
Hint: the area of a triangle is (bh)/2, where h is the height, b is the base.
A, C, and E.
you can check them by putting into y=mx+b form. if m is negative then as domain increases range decreases.
Using the distance formula, when I plugged in the information, the answer I got was the square root of 85. (B)
