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 1/12
Step-by-step explanation:
3/4 + 1/3
Get a common denominator of 12
3/4 * 3/3 = 9/12
1/3 * 4/4 =4/12
9/12 + 4/12
13/12
12/12 + 1/12
1 1/12
Answer:
<h2>-4/3</h2>
Step-by-step explanation:
3x-12=12x
-12=12x-3x
-12=9x
-12/9=x
-4/3=x
x=-4/3
Answer:
rhombus = a quadrilateral with four congruent sides
rectangle = a quadrilateral with four right angles
parallelogram = a quadrilateral with two pairs of parallel sides
trapezoid = a quadrilateral with one set of parallel sides
square = a quadrilateral with four right angles and four congruent sides
Answer:
2
Step-by-step explanation:
