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:
The answer to your question is $ 8591.04
Step-by-step explanation:
Data
height = 11.4 m
diameter = 12 m
cost = $20/ft³
Process
1.- Calculate the volume of the cone
Formula
Volume = 1/3 πr²h
Substitution
Volume = 1/3 (3.14)(12/2)²(11.4)
Simplification
Volume = 1/3(1288.66)
Volume = 429.55 m³
2.- Calculate the cost
$20 ----------------- 1 m³
x ----------------- 429.55 m³
x = (429.55 x 20)/ 1
x = $ 8591.04
The answer is C Kay. Got it?
