Question

The diagram shows four straight lines AD, BC, AC, and BD. lines AC BD intersect at O such that angle AOB is `pi/6radians`. AB is an arc of the circle, centre O and radius 10cm, and CD is an arc of the circle, centre O and radius 20cm?

Find the perimeter of ABCD

Answer 1

`73.89color(white)(88)` 2.d.p.

Explanation:

From the diagram:

`/_COB=pi-pi/6=(5pi)/6`

These are angles on the straight line `bb(AC)`

`/_DOA=pi-pi/6=(5pi)/6`

These are angles on the straight line `bb(DB)`

`/_DOC=pi-(5pi)/6=pi/6`

These are angles on the straight line `bb(DB)`

Solving `Delta color(white)(88)AOD`

Using cosine rule:

`AD^2=OD^2+OA^2-2(OD)(OA)cos(O)`

`AD^2=(20)^2+(10)^2-2(20)(10)*cos((5pi)/6)`

`AD^2=400+100-400*(-sqrt(3)/2)`

`AD^2=500-400*(-sqrt(3)/2)=500+200*sqrt(3)`

`AD=sqrt(500+200*sqrt(3))=29.09color(white)(88)` 2 .d.p.

`CB=AD` Same dimensions

Arc length is: `color(white)(88)rtheta`

Length of arc `CD`

`20*pi/6=(10pi)/3`

Length of arc `AB`

`10*pi/6=(5pi)/3`

Perimeter is:

`AD +CB +(10pi)/3+(5pi)/3`

`2(sqrt(500+200*sqrt(3))) +5pi=73.89color(white)(88)` 2.d.p.