3.2 Radian Measure & Arc Length

In the following diagram, O is the centre of the circle and (AT) is the tangent to the circle at T.

If OA = 12 cm, and the circle has a radius of 6 cm, find the area of the shaded region.

`\hat {OTA} = 90^@`

`AT=\sqrt{12^2-6^2} = 6sqrt3`

`hat{TOA}=60^@=pi/3`

Area `=` area of triangle `-` area of sector

`=1/2 times 6 times 6sqrt3 - 1/2 times 6 times 6 times pi/3`

`=12.3` cm² (or `18sqrt3 - 6pi`)

The diagram below shows a sector AOB of a circle of radius 15 cm and centre O. The angle `theta` at the centre of the circle is 2 radians.

(a) Calculate the area of the sector AOB.

(b) Calculate the area of the shaded region.

(a) Area `=1/2 r^2 theta = 1/2 (15^2)(2)=225 \ (cm^2)`

(b) Area `Delta OAB = 1/2 (15^2) sin 2= 102.3`

Area `=225 - 102.3 = 122.7 \ (cm^2)`

`=123 \ (3 \ s.f.)`

The following diagram shows a circle of centre O, and radius 15 cm. The arc ACB subtends an angle of 2 radians at the centre O.

Find

(a) the length of the arc ACB;

(b) the area of the shaded region.

(a) `ACB = 2 times OA =30 \ cm`

(b) `hat{AOB}` (obtuse) `=2pi-2`

Area `=1/2 theta r^2 = 1/2 (2pi-2)(15)^2 = 482 \ cm^2 \ (3 \ s.f.)`

The diagram below shows a triangle and two arcs of circles.

The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the midpoint of [BC].

The arc BDC is part of a circle with centre A.

The arc BEC is part of a circle with centre P.

(a) Calculate the area of the segment BDCP. 

(b) Calculate the area of the shaded region BECD. 

(a) area of sector ΑΒDC `=1/4 pi (2)^2=pi`

area of segment BDCP `=pi-` area of `Delta ABC= pi-2`

(b) BP `= sqrt 2` 

area of semicircle of radius BP `=1/2 pi (sqrt 2)^2 = pi`

area of shaded region `=pi - (pi-2)=2`

The diagram below shows two circles which have the same centre O and radii 16 cm and 10 cm respectively. The two arcs AB and CD have the same sector angle `theta = 1.5` radians.

Find the area of the shaded region.

Area of large sector `1/2 r^2 theta= 1/2 16^2 times 1.5=192`

Area of small sector `1/2 r^2 theta= 1/2 10^2 times 1.5=75`

Shaded area `=` large area `-` small area `=192-75=117`

The following diagram shows a circle with centre A and radius 6 cm.

The points B, C, and D lie on the circle, and `hat{BAC} = 2` radians. 

(a) Find the area of the shaded sector. 

(b) Find the perimeter of the non-shaded sector ABDC. 

(a) correct substitution 

e.g. `1/2 (2)(6^2)`

area `=36 \ (cm^2)`

(b) valid approach to find major arc length 

e.g. angle `=2pi-2`, circumference `-` arc BC

correct working for major arc length

e.g. `6(2pi-2), (2 times 6 times pi)-(6 times 2), 12pi-12`

valid approach to find perimeter of a sector (seen anywhere)

e.g. arc `+ 2` (radius), `12pi-12 + 2(6)`

perimeter `=12pi`

The following diagram shows a circle with centre O and radius `r` cm.

The points A and B lie on the circumference of the circle, and `hat{AOB}=theta`. The area of the shaded sector AOB is 12 cm² and the length of arc AB is 6 cm.

Find the value of `r`.

evidence of correctly substituting into circle formula (may be seen later)

e.g. `1/2 theta r^2=12, r theta =6`

attempt to eliminate one variable

e.g. `r=6/theta, theta=l/r, {1/2 theta r^2}/{r theta}=12/6`

correct elimination

e.g. `1/2 times 6/r times r^2=12, 1/2 theta times (6/theta)^2=12, A=1/2 times r^2 times l/r, r^2/{2r}=2`

correct equation

e.g. `1/2 times 6r=12, 1/2 times 36/theta=12, 12=1/2 times r^2 times 6/r`

correct working

e.g. `3r=12, 18/theta = 12, r/2=2, 24=6r`

`r=4` (cm)

The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and radius 6cm. The points A, B and D are on the same line.

(a) Find `hat{ABC}`.

(b) Find the exact area of the sector BDC. 

(a) (using height of triangle ABC by drawing perpendicular segment from C to AD)

correct substitution into formula for area of triangle

e.g. `1/2 (2sqrt3)(h)= 3sqrt3, \ hsqrt3`

correct working

e.g. `hsqrt3=3sqrt3`

height of triangle is 3

`hat {CBD}=pi/6 \ (30^@)`

`hat {ABC}={5pi}/6 \ (150^@)`

(b) recognizing supplementary angle 

e.g. `hat{CBD}=pi/6`, sector `=1/2 (180-hat{ABC})(6^2)`

correct substitution into formula for area of sector

e.g. `1/2 times pi/6 times 6^2, pi(6^2)(30/360)`

area `=3pi \ (cm^2)`

Points A and B lie on the circumference of a circle of radius `r` cm with centre at O.

The sector OAB is shown on the following diagram. The angle `hat {AOB}` is denoted as `theta` and is measured in radians. 

The perimeter of the sector is 10 cm and the area of the sector is 6.25 cm².

(a) Show that `4r^2-20r+25=0`.

(b) Hence, or otherwise, find the value of `r` and the value of `theta`.

(a) `2r + r theta = 10`

`1/2 r^2 theta = 6.25`

The following diagram shows a circle with centre O and radius 4 cm.

The points P, Q and R lie on the circumference of the circle and `hat{POR}=theta`, where `theta` is measured in radians.

The length of arc PQR is 10 cm.

(a) Find the perimeter of the shaded sector. 

(b) Find `theta`.

(c) Find the area of the shaded sector. 

(a) attempts to find perimeter 

arc `+ 2 times` radius `=18` (cm)

(b) `10=4theta`

`theta = 10/4 \ (=5/2, 2.5)`

(c) area `=1/2 (10/4)(4^2)= 1.25 times 16 = 20 \ (cm^2)`