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.
75/100 is less than 8/10.
8/10 can be equivalent to 80/100.
75/100 is less than 80/100.
Hope this helps.
Answer:
Perimeter would be 12 because you add 2a+2b
Step-by-step explanation:
Since perimeter is adding all the sides you have to add 4+4+2+2
Multiplication by a factor less than 1 will result in a smaller image.
Selections A, D, F.
