Question

In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of $\Delta APQ$ is $\frac{1}{16}$ of the area of $\Delta ABC$.

Answer 1

Given :
AP = 1 cm , PB = 3cm , AQ= 1.5 cm , QC = 4.5cm

To prove: ar(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)