Area of triangle

Dec 2009
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0
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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!
 

Grandad

MHF Hall of Honor
Dec 2008
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South Coast of England
Hello cakeboby
View attachment 16981
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

-------------------------------------------
(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