Question

P,Q,R are the three points of a triangle ABC such that AP:PB =2:5 and BQ:QC=2:5 and CR:RA =2:5 . find the ratio of the area of the triangle ABC and triangle PQR .

Answer 1

P, Q, R are the three points of a $\Delta ABC$ such that AP:PB =2:5 and BQ:QC=2:5 and CR:RA =2:5. Find the ratio of the area of the $\Delta ABC$ and $\Delta PQR$ .

Solution.

$$\frac {A P}{P B} = \frac {2}{5}, P B = \frac {5 A P}{2}, A B = A P + P B = A P + \frac {5 A P}{2} = \frac {7 A P}{2}, A P = \frac {2 A B}{7}, P B = \frac {5 A B}{7}.$$ $$\frac {B Q}{Q C} = \frac {2}{5}, Q C = \frac {5 B Q}{2}, B C = B Q + Q C = \frac {7 B Q}{2}, B Q = \frac {2 B C}{7}, Q C = \frac {5 B C}{7}.$$ $$\frac {C R}{R A} = \frac {2}{5}, \text { so } C R = \frac {2 A C}{7}, R A = \frac {5 A C}{7}.$$

Let's consider $\Delta$ APR:

$$S _ {\Delta A P R} = \frac {1}{2} A P \cdot R A \sin \angle A, S _ {\Delta A B C} = \frac {1}{2} A B \cdot A C \sin \angle A$$ $$\frac {S _ {\Delta A P R}}{S _ {\Delta A B C}} = \frac {\frac {1}{2} A P \cdot R A \sin \angle A}{\frac {1}{2} A B \cdot A C \sin \angle A} = \frac {\frac {1}{2} \cdot \frac {2}{7} A B \cdot \frac {5}{7} A C \sin \angle A}{\frac {1}{2} A B \cdot A C \sin \angle A} = \frac {1 0}{4 9}$$

Let consider $\Delta$ PBQ:

$$S _ {\Delta P B Q} = \frac {1}{2} B Q \cdot P B \sin \angle B, S _ {\Delta A B C} = \frac {1}{2} A B \cdot B C \sin \angle B$$ $$\frac {S _ {\Delta P B Q}}{S _ {\Delta A B C}} = \frac {\frac {1}{2} B Q \cdot P B \sin \angle B}{\frac {1}{2} A B \cdot B C \sin \angle B} = \frac {1 0}{4 9}$$

Let consider $\Delta$ QCR:

$$S _ {\Delta Q C R} = \frac {1}{2} Q C \cdot C R \sin \angle C, S _ {\Delta A B C} = \frac {1}{2} B C \cdot A C \sin \angle C$$ $$\frac {S _ {\Delta Q C R}}{S _ {\Delta A B C}} = \frac {\frac {1}{2} Q C \cdot C R \sin \angle C}{\frac {1}{2} B C \cdot A C \sin \angle C} = \frac {1 0}{4 9}$$

Let's consider $\Delta$ PQR:

$$S _ {\Delta P Q R} = S _ {\Delta A B C} - S _ {\Delta A P R} - S _ {\Delta P B Q} - S _ {\Delta Q C R} = S _ {\Delta A B C} - \frac {1 0}{4 9} S _ {\Delta A B C} - \frac {1 0}{4 9} S _ {\Delta A B C} - \frac {1 0}{4 9} S _ {\Delta A B C} = \frac {1 9}{4 9} S _ {\Delta A B C}.$$ $$\frac {S _ {\Delta A B C}}{S _ {\Delta P Q R}} = \frac {S _ {\Delta A B C}}{\frac {1 9}{4 9} S _ {\Delta A B C}} = \frac {4 9}{1 9}$$

Answer.

$$\frac {S _ {\Delta ABC}}{S _ {\Delta PQR}} = \frac {49}{19}$$