Exact circular-segment area and perimeter

Subtract the central triangle

Let chord AB subtend minor angle θ with 0<θ≤π. Its minor sector has area ½r²θ. Triangle OAB has area ½r²sin θ, so the minor segment area is ½r²(θ−sin θ). θ is in radians in the sector term; evaluate sin consistently

A circle has radius 6 cm and θ=π/2. Give the coefficient of π in its minor sector area

Original circle

\(\displaystyle r=6\,\mathrm{cm},\quad\theta=\frac{\pi}{2}\,\mathrm{rad}\)

Find the area of its central triangle in cm²

Assemble the exact segment

The minor segment is 9π−18 cm². The major segment is the whole disc minus this region: 36π−(9π−18)=27π+18 cm². The positive triangle term in the major expression follows from subtracting the entire minor-segment expression

For r=4 cm and θ=π/3, write the minor segment as Aπ+B√3. Find A.

Original circle

\(\displaystyle r=4\,\mathrm{cm},\quad\theta=\frac{\pi}{3}\,\mathrm{rad}\)

Original area form

\(\displaystyle \text{Minor segment area}=(A\pi+B\sqrt3)\,\mathrm{cm}^2\)

In the same expression, find the signed coefficient B.

A segment perimeter uses its chord

Bisect the isosceles triangle to get chord length c=2r sin(θ/2). The minor perimeter is rθ+c; the major perimeter is r(2π−θ)+c. The same chord belongs to both boundaries

For r=7 cm and θ=π/3, find the chord length in cm

Original circle

\(\displaystyle r=7\,\mathrm{cm},\quad\theta=\frac{\pi}{3}\,\mathrm{rad}\)

Give the coefficient of π in that major segment perimeter

Why does halving the central angle appear in the chord formula?

Derive both boundaries, ungraded

Draw a circle, chord, two radii and a perpendicular bisector. Derive the minor-segment area and chord length. Then obtain the major-segment area and perimeter and explain why its area and perimeter use different complements

Model: complement the correct quantity

Subtract the central triangle from the minor sector. For the major area subtract that whole segment from πr², giving πr²−½r²(θ−sin θ). For perimeter keep the same chord and replace only the arc angle by 2π−θ. The perpendicular bisector gives c/2=r sin(θ/2). At θ=π the central triangle is degenerate, the segment is a semicircle and the chord is a diameter