A triangle can be formed by taking any two points of a straight line and third point on the second straight line
\(\rightarrow \)Number of ways =\((C\frac{10}{2}\times 11)+(C\frac{11}{2}\times 10)\)
\(\rightarrow 495+550=1045\)
The correct answer is (C) : 1045
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).