Drawing this square and then drawing in the four radii from the center of the cirble to each of the vertices of the square results in the construction of four triangular areas whose hypotenuse is 3 sqrt(2). Draw this to verify this statement. Note that the height of each such triangular area is (3 sqrt(2))/2.
So now we have the base and height of one of the triangular sections.
The area of a triangle is A = (1/2) (base) (height). Subst. the values discussed above, A = (1/2) (3 sqrt(2) ) (3/2) sqrt(2). Show that this boils down to A = 9/2.
You could also use the fact that the area of a square is (length of one side)^2, and then take (1/4) of this area to obtain the area of ONE triangular section. Doing the problem this way, we get (1/4) (3 sqrt(2) )^2. Thus,
A = (1/4) (9 * 2) = (9/2). Same answer as before.
Answer:
1 11/21
Step-by-step explanation:
6/7 + 2/3 = 32/21
32/21 simplified is 1 11/21
Answer:
<h3>B. m<3 = m<6 = 75degrees</h3>
Step-by-step explanation:
First you must understand that in geometry, the alternate interior angles are equal
Hence if ∠3 and ∠6 are alternate interior angles, then <3 = <6
Given that m<3 = 75 degrees
m<6 will also be 75 degrees since both angles are equal
Answer:
m 1 = 45 degrees because 1 and the angle measuring 135 degrees are supplementary angles.
m 2 = 95 degrees because 2 and the angle measuring 95 degrees are vertical angles.
m 3 = 40 degrees because 1, 2, and 3 form a triangle
Step-by-step explanation:
