Question

In the given figure, ABC is a triangle. P, R, T and Q, S, U are points on side AB and AC respectively such $AP=PR=TR=TB$ and $AQ=QS=SU=UC$. If $BC=24$ cm, then $(PQ+RS+TU)$ is

• A. 24 cm
• B. 36 cm
• C. 42 cm
• D. 48 cm

Answer 1

The correct option is B 36 cm


Given that $AP=PR=TR=TB$ and $AQ=QS=SU=UC$.
In $\Delta ABC$, R and S are mid-points of AB and AC respectively.
$\Rightarrow RS=\frac{1}{2}BC$ (Mid-point theorem)
$\Rightarrow RS=\frac{24}{2}=12cm$ ..........(i)
Similarly, in $\Delta ARS$, P and Q are mid-points of AR and AQ respectively.
$\therefore PQ=\frac{1}{2}RS$ (Mid-point theorem)
$\therefore PQ=6$ cm [From (i)] …..(ii)
In $\Delta ATU$ and $\Delta ARS$,
$AT=AR+RT=\frac{3}{2}AR$
and, $AU=AS+SU=\frac{3}{2}AS$
Thus, the line RS divide two sides of ΔATU in the same ratio.
$\therefore TU=\frac{3}{2}RS$
$\Rightarrow TU=\frac{3}{2}\times 12$ [From (i)]
$\Rightarrow TU=18$ cm
From (i), (ii) and (iii), we get
$PQ+RS+TU=6+12+18$
$=36$ cm
Hence, the correct answer is option (b).