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?

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?

Answer: Option C

Solution

Answer: Option C

Solution:

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