Question

ABC is a triangle and PQ is a straight linemeeting AB in P and AC in. Q. If A P= 1cm, PB = 3 cm, AQ = 1.5 cm, QC =4.5 m, prove that area of $\Delta APQ$ is one- sixteenth of the area of $\Delta ABC$

Answer 1

AP = 1 cm , PB = 3cm , AQ= 1.5 cm , QC = 4.5cm ( given), we have to prove, (APQ) = 1/16 ar (ABC) proof: AP = 1 cm and PB = 3 cm then AB = 4 cm since the sides are in proportion then the line PQ is parallel to BC in triangle APQ and triangle ABC angle A = angle A (common) angle APQ = angle B (alt. int. angles) angle AQP = angle C (alt. int. angles) therefore, triangle APQ ~ triangle ABC ar(APQ)/ ar(ABC) = AP2 / AB2 (area of two similar triangles is equal to the square of their coressponding side) ar(APQ) / ar(ABC) = 12/ 42 ar (APQ) / ar (ABC) = 1/16 therefore ar(APQ) = 1/ 16 ar(ABC) hence proved.

Read