Question

In $△ABC$ points $P$ and $Q$ trisect side $AB$ points $T$ and $U$ trisect side $AC$ and points $R$ and $S$ trisect side $BC$. Then perimeter of hexagon $PQRSTU$ is how many times of the perimeter of $△ABC$?

• A. $\frac{1}{3}$ times
• B. $\frac{2}{3}$ times
• C. $\frac{1}{6}$ times
• D. $\frac{1}{2}$ times

Answer 1

Answer: (B)

$\frac{2}{3}$ times

Let $AB$ be $x$
$\therefore$ $AQ=QP=BP=\frac{x}{3}$
Let $BC$ be $y$
$\therefore$ $BR=RS=SC=\frac{y}{3}$
Let $AC=z$
$AT=TU=UC=z/3$

Opposite sides of Hexagon are equal
$\therefore$ Perimeter of Hexagon $=PQ+QT+TU+US+RS+PR$
$=(\frac{x}{3}+\frac{y}{3}+\frac{z}{3})\times 2$
$\therefore \frac{PerimeterofHexagon}{Perimeterof△ABC}=\frac{2/3(x+y+z)}{(x+y+z)}=\frac{2}{3}$.

$\therefore$ Perimeter of hexagon is $2/3$ times the perimeter of $△ABC$.