Posted: June 13, 2013 in Mathematics

The trick in this problem is realizing that when the two triangles share the same height, the ratio of the two bases equals to the ratio of the two areas.

Now obviously the area of $\triangle STV = 55+66-77=44$

$\dfrac{\frac{1}{2}RU\cdot h_1+\frac{1}{2}RU\cdot h_2}{\frac{1}{2}UT\cdot h_1+\frac{1}{2}UT\cdot h_2}=\dfrac{77}{44}$

$\dfrac{RU}{UT}=\dfrac{7}{4}$

hence,

$\dfrac{\frac{1}{2}RU\cdot h_1}{\frac{1}{2}UT\cdot h_1}=\dfrac{7}{4}$

$\dfrac{\triangle RSU}{\triangle STU}=\dfrac{7}{4}$

Therefore,

$\triangle RSU = 55\cdot\dfrac{7}{11}=35, \quad \triangle STU=55-35=20$

Similarly,

$\triangle RUV = 66\cdot\dfrac{7}{11}=42, \quad \triangle STU=66-42=24$

Now assign the follow variables to the area of the following triangles,

$\triangle PSU = a, \triangle PTU = b, \triangle QTU = c, \triangle QVU = d, \triangle PTQ = e$

note that,

$\dfrac{RU}{TU}=\dfrac{7}{4} \mbox{ and } \dfrac{SU}{VU}=\dfrac{5}{6}$

then we have,

$\dfrac{35+a}{b}=\dfrac{7}{4}$

$140+4a=7b$

and

$\dfrac{a}{b+24}=\dfrac{5}{6}$

$6a=5b+120$

Solving this system of equations yields,

$a=70, b=60$

Similarly,

$\dfrac{42+d}{c}=\dfrac{7}{4}$

$168+4d=7c$

and

$\dfrac{20+c}{d}=\dfrac{5}{6}$

$120+6c=5d$

Solving the system we get,

$c=120, d=168$

Finally,

$\dfrac{\mbox{area of }\triangle PTQ}{\mbox{area of } \triangle PST}=\dfrac{\mbox{area of } \triangle QTV}{\mbox{area of } \triangle STV}$

$\dfrac{e}{a+b-20}=\dfrac{c+d-24}{44}$

$\therefore e=\dfrac{120+168-24}{44}\cdot (70+60-20) = 660$

Therefore the area of triangle PQU is

$\triangle PQU = b+c+e = 60+120+660 = 840$