Question

ABC is a triangle. P,Q,R are the points of the sides AB,BC,CA such that the ratio of AP:PB=2:5, BQ:QC=2::5 , CR:RA=2:5 . find the ratio of the areas of the triangle ABC and triangle PQR.

Answer 1

Area of triangle $= 1/2$ b*c*sinA, therefore (Area PBQ/Area ABC) $= 5^{*}2/7^{*}7 = 10/49$ , because $1/2$ and $\sin \mathrm{PBQ}$ was reduced. We compute the areas of anothers triangles.

(Area CQR/Area ABC) $= 5^{*}2 / 7^{*}7 = 10 / 49$

(Area APR/Area ABC) $= 5^{*}2 / 7^{*}7 = 10 / 49$ , therefore, (Area PQR/Area ABC) $= 1$

$(10 / 49 + 10 / 49 + 10 / 49) = 19 / 49$ - ratio of the areas of the triangle ABC and triangle PQR.

Answer: 19/49