In the given figure, in triangle STU, ST = 8 cm, TU = 9 cm and SU = 12 cm. QU = 24 cm, SR = 32 cm and PT = 27 cm. What is the ratio of the area of triangle PQU and area of triangle PTR?

Solution (By Examveda Team)

$$\eqalign{ & {\text{Let }}\angle PUQ = \theta \cr & {\text{Then }}\angle SUT = 180 - \theta \cr & {\text{And}}\,{\text{let }}\angle RTP = \alpha \cr & {\text{Then }}\angle STU = 180 - \alpha \cr & \therefore {\text{Ratio of area of }}\frac{{\Delta PUQ}}{{\Delta SUT}} \cr & = \frac{{\frac{1}{2} \times 18 \times 24 \times \sin \theta }}{{\frac{1}{2} \times 12 \times 9 \times \sin \left( {180 - \theta } \right)}} = \frac{{4\sin \theta }}{{\sin \theta }} \cr & \frac{{\Delta PUQ}}{{\Delta SUT}} = \frac{4}{1} \cr & {\text{Now, ratio of area of }}\frac{{\Delta PTR}}{{\Delta SUT}} \cr & = \frac{{\frac{1}{2} \times 27 \times 24 \times \sin \alpha }}{{\frac{1}{2} \times 9 \times 8 \times \sin \left( {180 - \alpha } \right)}} = \frac{{9\sin \alpha }}{{\sin \alpha }} \cr & \frac{{\Delta PTR}}{{\Delta SUT}} = \frac{9}{1} \cr & \therefore {\text{Ratio of area of }}\frac{{\Delta PQU}}{{\Delta PTR}} = \frac{4}{9} \cr} $$