Hello cakeboby Originally Posted by

**cakeboby**
Given PR = 20 , QS = 15 , PR = x , QS = y , angle PTQ = 60 degree

(a)

Find the area of triangle PQT

(b)

Find the area of triangle QRT

(c)

Using (a) and (b), find the area of PQRS

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(a)

triangle PQT =$\displaystyle \frac {\sqrt 3}{ 4 }xy $

(b)

triangle QRT = $\displaystyle \frac{\sqrt 3}{4}(20-x)y

$

(c)

don't know how to do (c)

Thanks in advance!

You've clearly made a couple of typos in the 'given', but judging by the answers to (a) and (b), I assume you mean

$\displaystyle PT = x$ and $\displaystyle QT = y$

and you've then used the $\displaystyle \tfrac12ab\sin C$ formula for the area of a triangle, where $\displaystyle \sin 60^o =\frac{\sqrt3}{2}$.

For (c), use the same method again to find the areas of $\displaystyle \triangle$ s $\displaystyle PTS$ and $\displaystyle STR$ (where $\displaystyle TS = 15-y$), and then add all four areas together.

Can you complete it now?

Grandad