# 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 m,

Question:

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 m, prove that the area of ∆APQ is one-sixteenth of the area of ∆ABC.

Solution:

Given: In ΔABC, PQ is a line segment intersecting AB at P, and AC at Q. AP = 1cm, PB = 3cm, AQ = 1.5cm and QC = 4.5cm.

To find: $\operatorname{Ar}(\triangle \mathrm{APQ})=\frac{1}{16}(\triangle \mathrm{ABC})$

In ΔABC,

$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

$\frac{1}{3}=\frac{1.5}{4.5}$

$\frac{1}{3}=\frac{1}{3}$

According to converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Hence, $\mathrm{PQ} \| \mathrm{BC}$

In $\triangle \mathrm{APQ}$ and $\triangle \mathrm{ABC}$,

∠APQ=∠B              Corresponding angles

∠PAQ=∠BAC       Common

So, ∆APQ~∆ABC      (AA Similarity)

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence

$\frac{A r(\Delta \mathrm{APQ})}{A r(\Delta \mathrm{ABC})}=\frac{\mathrm{AP}^{2}}{\mathrm{AB}^{2}}$

$=\frac{\mathrm{AP}^{2}}{(\mathrm{AP}+\mathrm{BP})^{2}}$

$=\frac{1^{2}}{(1+3)^{2}}$ (given)

$\frac{\operatorname{Ar}(\Delta \mathrm{APQ})}{\operatorname{Ar}(\Delta \mathrm{ABC})}=\frac{1}{16}$

$A r(\Delta \mathrm{APQ})=\frac{1}{16} A r(\triangle \mathrm{ABC})$