1
Get the whole circle first — the slice is only a fraction of it.
Two formulas, and the SAT expects both from memory.
circumference
C = 2πr = πd
the distance around — a length
area
A = πr²
the space inside — square units
r = 8 → C = 2π(8) = 16π
≈ 50.3
r = 8 → A = π(8)² = 64π
≈ 201.1
Only the radius gets squared
π(8)² is 64π, not (8π)² = 64π². And a stem that hands you a
diameter of 8 means r = 4, so the area is 16π — a quarter of the size.
Read which one you were given before you write anything down.
2
A central angle cuts an arc, and its share of 360° is its share of the circle.
Central angle AOB is 135° in a circle of radius 8. The arc is the piece of rim between A and B; the sector is the wedge inside.
135360
=
38
so this slice — arc and sector alike — is three eighths of the circle
The drawn wedge really is 135°: it opens from the horizontal radius OA up and over
past the vertical, three eighths of the way round. A sector that looks like a
quarter but is labelled 135° is a figure you should not trust.
3
Apply that fraction twice — once to C for the arc, once to A for the sector.
arc AB = 135360 × 2π(8)
a fraction of the circumference
= 38 × 16π = 6π
≈ 18.85, a length
sector AOB = 38 × π(8)²
the same fraction of the area
= 38 × 64π = 24π
≈ 75.4, square units
arc = 6π · sector = 24π
| You want | Degrees | Radians |
|---|
| arc length | (θ360) × 2πr | s = rθ |
|---|
| sector area | (θ360) × πr² | A = ½r²θ |
|---|
| perimeter of a sector | arc + 2r — you walk back down both radii |
|---|
Never leave a π behind
The SAT usually wants the exact answer, so 6π stays 6π — do not round it to 18.85
unless the stem says "closest to". And the units tell you if you used the wrong
whole: an arc can never come out in square units, and a sector can never come out
as a plain length.
The opening problem is this run backwards. C = 2π(15) = 30π, and the arc 5π is
5π30π = 16 of it, so the angle is (16)(360°) = 60°. Whenever the stem gives you
the slice and wants the angle, form the fraction first and multiply by 360 last.
4
In radians the whole turn is 2π, and the formulas get shorter.
Same fraction, different denominator: θ out of 2π instead of θ out of 360.
135° × π180 = 3π4
the same angle, in radians
s = rθ = 8 × 3π4 = 6π
the arc — same answer as before
A = ½r²θ = ½ × 64 × 3π4 = 24π
the sector — same answer as before
s = rθ and A = ½r²θ are only the θ360 versions with the 360 replaced by 2π and the
2π cancelled. They work only in radians — feeding 135 into rθ gives 1080,
which is longer than the whole circle.
Watch the calculator's mode
Desmos reads sin, cos and tan in radians by default, and it will happily evaluate
an angle you meant as degrees. Type an angle as 135° with the degree symbol, or
convert it yourself first, and the graph and the arithmetic will agree.
5
A chord plus the perpendicular from the centre makes a right triangle.
That perpendicular always lands on the midpoint of the chord — and that is what turns a circle question into Pythagoras.
AM = MB = 16 ÷ 2 = 8
the perpendicular bisects the chord
OM² + MB² = OB²
right angle at M, and OB is a radius
OM² + 8² = 10²
substitute
OM² = 100 − 64 = 36, OM = 6
a 6-8-10 triangle
OM = 6
Half the chord, not the whole chord
The leg of the right triangle is 8, not 16. Using the full chord gives
OM² = 100 − 256, a negative number, which is the answer telling you the
bisection step was skipped. The hypotenuse is always the radius, so it is always
the longest of the three.
The rule runs in every direction: the perpendicular from the centre bisects the
chord, a line from the centre to the midpoint of a chord is perpendicular to it,
and equal chords sit equally far from the centre. Whichever two the stem gives you,
the third is free.
6
A tangent meets the radius at a right angle, always.
A tangent touches the circle at exactly one point, and at that point it is perpendicular to the radius drawn there.
∠OPQ = 90°
tangent ⊥ radius at the point of contact
OP² + PQ² = OQ²
so triangle OPQ is right-angled at P
r = 9, PQ = 12 → OQ² = 81 + 144 = 225
a 9-12-15 triangle
OQ = 15
the distance from the centre to Q
The right angle is at P, where the tangent touches — not at O and not at Q. So the
hypotenuse is OQ, the segment from the centre to the outside point, and it is the
longest side every time.
tangent ⊥ radiusone point of contact
two tangents from a point are equal
✓
Check: do the eight slices rebuild the circle?
eight arcs
6π ÷ 3 × 8 = 16π
the full circumference ✓
eight sectors
24π ÷ 3 × 8 = 64π
the full area ✓
chord
6² + 8² = 100 = 10²
the radius is the hypotenuse ✓
sensible?
135° is more than a quarter turn
and 38 > 14 ✓
Remember the flow
Whole circle first · θ360 (or θ2π) · Apply it twice
Arc from C, sector from A — and leave the π where it is.
