Question

Please solve q 109 ?

Answer 1

`text{area} = tan 15^circ =2-sqrt{3}`

Explanation:

More from Rahul's book!

Closer to B than A means on B's side of the perpendicular bisector between the two. Similarly closer to B than C means B's side of the perpendicular bisector of BC.

Let's call E the midpoint of AB and F the midpoint of BC. D is far away so we can skip it.

We'll call G the meet of the perpendicular bisectors or AB and BC. So BEG and BFG are right triangles, right angles E and F.

`angle EBG=angle FBG={angle ABC}/2=30^circ/2= 15^circ`

`EB=1/2 AB=1`

So the base of the triangle is `EG= EB tan(angle EBG)=tan 15^circ`

So the area of EBG+FBG is `2( 1/2 (1) tan 15^circ ) = tan 15^circ`

One of my favorite half angle formulas is

`tan(theta/2) = { 1 - cos theta}/ sin theta`

`text{area} = tan 15^circ = {1- cos 30^circ}/{sin 30^circ} = {1-sqrt{3}/2}/{1/2}=2-sqrt{3}`

Choice (1)