False
-3 is not greater than positive 3
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.
<h3>
Answer: 1260</h3>
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Work Shown:
n = number of sides = 9
S = sum of the interior angles of any polygon with n sides
S = 180(n-2)
S = 180(9-2)
S = 180(7)
S = 1260
Answer:
∠4 = 78°, assuming out measurements have been taken in degrees.
Step-by-step explanation:
If ∠2 = 8x + 10 and
∠4 = 42 + 6x
We will assume that x in both cases is represented by the same number, therefore, we will first need to solve for x. We will do so by equating both angle measurement expressions.
8x + 10 = 42 + 6x Take away 6x from both sides
2x + 10 = 42 Take 10 away from both sides to combine like terms
2x = 32 Divide both sides by 2 to isolate x
x = 16
Knowing x, we can solve for the measure of ∠4 by plugging in 16 for x
∠4 = 42 + 6x
∠4 = 42 + 6(6)
∠4 = 42 + 36
∠4 = 78°, assuming out measurements have been taken in degrees.
7 1/2 years ago to be precise
If correct brainley please
Thank you
