Question

Question 5
In $\Delta PQRand\Delta MST,\angle P={55}^{\circ },\angle Q={25}^{\circ },\angle M={100}^{\circ }and\angle S={25}^{\circ }.Is\Delta QPR\sim \Delta TSM$? Why?

Answer 1

False, $\Delta QPR$ is not similar to $\Delta TSM$.
According to the angle sum property of a triangle, the sum of three angles of a triangle is ${180}^{\circ }$.
In $\Delta PQR,$
$\angle P+\angle Q+\angle R={180}^{\circ }$
$\Rightarrow {55}^{\circ }+{25}^{\circ }+\angle R={180}^{\circ }$
$\Rightarrow \angle R={180}^{\circ }-({55}^{\circ }+{25}^{\circ })={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

In $\Delta TSM,$
$\angle T+\angle S+\angle M={180}^{\circ }$
$\Rightarrow \angle T+\angle {25}^{\circ }+{100}^{\circ }={180}^{\circ }$
$\Rightarrow \angle T={180}^{\circ }-({25}^{\circ }+{100}^{\circ })$
$\angle T={180}^{\circ }-{125}^{\circ }={55}^{\circ }$



In $\Delta PQR,and\Delta TSM$,
$\angle P=\angle T,\angle Q=\angle S$
and $\angle R=\angle M$
$\therefore$ $\Delta PQR\sim \Delta TSM$, [since, all corresponding angles are equal ]
Hence, $\Delta QPR$ is not similar to $\Delta TSM$, since the correct correspondence is $P\leftrightarrow T,Q\leftrightarrow SandR\leftrightarrow M$.