Question

Lines $PS,QT$ and $RU$ intersect at a common point $O$ as shown. $P$ is joined to $Q,R$ to $S$ and $T$ to $U$ to form triangles. The value of $\angle P+\angle Q+\angle R+\angle S+\angle T+\angle U$ is:

• A. $270{}^{\circ }$
• B. $360{}^{\circ }$
• C. $450{}^{\circ }$
• D. $540{}^{\circ }$

Answer 1

The correct option is B $360{}^{\circ }$

In $△POQ$
$\angle P+\angle Q+\angle POQ={180}^o$
$\angle POQ={180}^{\circ }-\angle P-\angle Q$
Now In $△UOT$
$\angle UOT={180}^{\circ }-\angle U-\angle T$
In $△ROS$
$\angle ROS={180}^{\circ }-\angle S-\angle R$
Now, $\angle ROS=\angle POU$ ...... (Vertically opposite angles)
$\angle POQ=\angle SOT$ ...... (Vertically opposite angles)
$\angle UOT=\angle QOR$ ...... (Vertically opposite angles)
Now, $\angle O=2(\angle ROS+\angle POQ+\angle UOT)$
${360}^o=2({180}^o-\angle R-\angle S+{180}^o-\angle P-\angle Q+{180}^o-\angle U-\angle T)$
$2(\angle P+\angle Q+\angle R+\angle S+\angle T+\angle U)={720}^o$
$\angle P+\angle Q+\angle R+\angle S+\angle T+\angle U={360}^o$