Question

Find the perimeter of the quadrilateral to the nearest tenth.

Answer 1

Since ABC is a 90-60-30 triangle, it is half of an equilateral triangle. This means that BC is half AC, i.e. 17.

So, we have

`AB = √(AC^2-BC^2)=\sqrt{AC^2-\left((AC)/(2)\right)^2}=(√(3))/(2)AC = √(3)\cdot BC =√(3)\cdot 17\approx 29.4`

So, the perimeter is

`AD+DC+BC+AB\approx34+34+17+29.4\approx 114.4`