Question

ABC is an equilateral triangle, P is a point on BC such that BP : PC = 2 : 1. Then $\frac{A{P}^2}{A{B}^2}=$

• A. $\frac{9}{7}$
• B. $\frac{7}{9}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

The correct option is B

$\frac{7}{9}$

Construction: Draw AD $\perp$ BC

Given: BP : PC = 2 : 1

$\Rightarrow BP=\frac{2}{3}BC=\frac{2}{3}AB$.

$AD\perp BC\Rightarrow BD=\frac{1}{2}BC=\frac{1}{2}AB$.

DP = BP - BD

$\Rightarrow DP=\frac{2}{3}AB-\frac{1}{2}AB=\frac{AB}{6}$ .... (1)

In right triangle ABD,

$A{D}^2+B{D}^2=A{B}^2$

$\Rightarrow A{B}^2=A{D}^2+{\left(\frac{AB}{2}\right)}^2$

$\Rightarrow A{D}^2=A{B}^2-\frac{A{B}^2}{4}=\frac{3A{B}^2}{4}$ .... (2)

In $\Delta ADP$

$A{P}^2=A{D}^2+D{P}^2$

$=\frac{3A{B}^2}{4}+{\left(\frac{AB}{6}\right)}^2$ [From 1 & 2]

$=\frac{3A{B}^2}{4}+\frac{A{B}^2}{36}$

$=\frac{27A{B}^2+A{B}^2}{36}$

$=\frac{28A{B}^2}{36}$

$\Rightarrow \frac{A{P}^2}{A{B}^2}=\frac{28}{36}=\frac{7}{9}$