ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm. and BP = 3 cm., AQ = 1.5 cm., CQ = 4.5 cm. Prove that (area of ∆APQ) = \frac{1}{16} (area of ∆ABC)

Step 1: According to the given measurements construct an imaginary figure.
We have,
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 m
In ΔAPQ and ΔABC
Step 2: Make sure that the triangles (APQ and ABC) are similar.
\angle A = \angle A (common angle)
\frac{AP}{AB} = \frac{1}{3+1} = \frac{1}{4}
\frac{AQ}{AC} = \frac{1.5}{1.5 + 4.5} = \frac{1}{4}
So, \frac{AP}{AB} = \frac{AQ}{AC}
since the sides are in proportion then the line PQ is parallel to BC
Then \triangle APQ,\ \triangle ABC are similar (By SAS similarity)
Step 3: To prove the required equation, apply the Area of similar triangle theorem
\frac{Area \triangle APQ}{Area \triangle ABC} = (\frac{AP}{AB})^2
\frac{Area \triangle APQ}{Area \triangle ABC} = (\frac{1}{4})^2
Area \triangle APQ = (\frac{1}{16}) * Area \triangle ABC
Hence proved