Step-by-step explanation:
it filled up half the circle (up to the center point) - if we had a full circle. but a little bit is cut off (below AB).
what we see is that the shaded area is the sum of the area of the triangle AOB and 2 equally sized circle segment areas left and right of AOB.
since we are dealing with a half-circle, we have 180° in total. 120° are taken by AOB, so, that leaves us with 180-120 = 60° for both circle segments (so, one has an angle of 30°).
and 2×30° = 1×60°, so we can calculate the area of one 60° segment instead of two 30° segments.
AOB is an isoceles triangle (the legs are equally long, and therefore also the 2 side angles are equal).
the area of this triangle AOB is
1/2 × a × b × sin(C) = 1/2 × 3 × 3 × sin(120) =
= 3.897114317... m²
a circle segment area of 60° is 60/360 = 1/6 of the full circle area (as a full circle = 360°).
so, it's area is
pi×r² × 1/6 = pi×3²/6 = pi×3/2 = 4.71238898... m²
so, the total area of the shaded area is
3.897114317... m² + 4.71238898... m² =
= 8.609503297... m²
Answer:
No they are equal
Step-by-step explanation:
Find the slope:
y₂ - y₁ / x₂ - x₁
-1 - 2 / 0 - 4
-3 / -4
3/4
y = mx + b
y = 3/4x + b
Substitute any of the point's coordinate in the equation.
I'll pick (0,-1)
y = 3/4x + b
-1 = 3/4(0) + b
-1 = 0 + b
-1 = b
y-intercept = -1
y-intercept Equation:
y = 3/4x - 1
Point-slope form:
y - 2 = 3/4(x - 4)
Standard form:
-3/4x + y = -1
Volume = 4/3×π×143<span> = </span><span>11494.040321934 centimeters3</span><span>
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