In $\triangle A B C, A C=8.4 \mathrm{~cm}$ and $B C=14 \mathrm{~cm}$. $\mathrm{P}$ is a point on $\mathrm{AB}$ such that $\mathrm{CP}=11.2 \mathrm{~cm}$ and $\angle A C P=\angle B$. What is the length (in $\mathrm{cm}$) of $\mathrm{BP}$ ?
Answer & explanation
Correct answer: option 3
It is given that ∠ACP = ∠B
In Triangle ACP and ABC ,
∠ACP = ∠B
∠A is common in both
So, △ACP ∼ △ABC
⇒ \(\frac{AC}{AB}\) = \(\frac{PC}{BC}\) = \(\frac{AP}{AC}\)
\(\frac{AC}{AB}\) = \(\frac{PC}{BC}\)
\(\frac{8.4}{AB}\) = \(\frac{11.2}{14}\)
AB = \(\frac{8.4 × 14 }{11.2}\)
AB = 10.5
Now, \(\frac{PC}{BC}\) = \(\frac{AP}{AC}\)
\(\frac{11.2}{14}\) = \(\frac{AP}{ 8.4}\)
AP = 6.72
Now, BP = AB - AP = 10.5 - 6.72
= 3.78