Answer:
- perimeter: 16 +4/3π ≈ 20.19 units
- area: 16 +8/3π ≈ 24.38 units²
Explanation:
You are asked for the area and perimeter of a figure comprised of a square and two sectors.
Perimeter
Straight edges
The perimeter of the figure is the sum of the lengths of the outside edges. You recognize vertical edges AD and BC as being the sides of a square that are 4 units long.
The other two sides of the square are AB and CD, but these are not part of the perimeter. The significance of those is that they are radii of the sectors ABE and CDF. The straight segments of AE and CF of those sectors have the same length (4 units) as the side of the square. Those straight segments are part of the perimeter.
In effect, the four straight segments of the perimeter are all 4 units.
Curved edges
The curved edges of the two sectors have a length that is found using the formula ...
s = rθ
where r is the sector radius, and θ is the central angle in radians.
The angle is shown as 30°, which is 30°(π/180°) = π/6 radians. The radius is the square side length, 4, so each curved line has length ...
s = (4)(π/6) = 2/3·π
Full perimeter
The perimeter of the figure is the sum of the lengths of the straight segments and the curved arcs:
P = 4(4 units) +2(2/3π units) = 16 +4/3π units ≈ 20.19 units
Area
As with the perimeter, the area is composed of the area of a square and the areas of two sectors.
Square area
The area of the square is the square of its side length:
A = s²
A = (4 units)² = 16 units²
Sector area
The area of each sector is effectively the area of a triangle with base equal to the arc length (2/3π) and height equal to the radius of the arc (4 units). The sector area is ...
A = 1/2rs
A = 1/2(4 units)(2/3π units) = 4/3π units²
Total area
The area of the whole figure is the sum of the area of the square and the areas of the two sectors:
A = square area + 2×(sector area)
A = 16 units² + 2×(4/3π units²) = (16 +8/3π) units² ≈ 24.38 units²
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Additional comment
In general, you find the perimeter and/or area of a strange figure by decomposing it into parts whose perimeter and area you can compute. (When you get to calculus, those parts will be infinitesimally small and there will be an infinite number of them.) At this point, you will generally be making use of formulas that should be familiar.
The formula for the area of a sector is usually written ...
A = 1/2r²θ
Here, we have made use of our previous computation of s=rθ to write the area formula as A = 1/2rs. The similarity to the triangle area formula is not accidental.