Question 63

In $$\triangle ABC, AC = 8.4 cm$$ and $$BC = 14 cm$$. P is a point on AB such that CP= 11.2 cm and $$\angle ACP = \angle B$$. What is the length (in cm) of BP?

Solution

$$\triangle ACP ~ \triangle ABC

($$\because \angle$$ A is common and $$\angle ACP = \angle B$$)

$$\frac{AB}{AC} = \frac{BC}{CP} = \frac{AC}{AP}$$

From first two,

$$\frac{AB}{AC} = \frac{BC}{CP}$$

$$\frac{AB}{8.4} = \frac{14}{11.2}$$

AB = 10.5

From last two,

$$\frac{BC}{CP} = \frac{AC}{AP}$$

$$\frac{14}{11.2} = \frac{8.4}{AP}$$

AP = 6.72 cm

BP = AB - AP = 10.5 - 6.72 = 3.78 cm

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