Question

The diagram shows the entrance to a tunnel.

Answer 1

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