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Solve for x. The vertex on the right is labelled as R.

Solve for x. The vertex on the right is labelled as R.-example-1
User Kampau Ocu
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1 Answer

11 votes
11 votes

Notice that the marks over the segments QT and TR tell us that those segments are congruent.

Since QT+TR=QR and QT=TR, then QT+TR=QT+QT=2QT.

Therefore:


QR=2QT

For a similar argument, we can conclude that:


QP=2\cdot QS

Notice that the segments ST and PR are parallel, because the triangles QST and QPR are similar (this can be demonstrated by the SAS criterion, since QR/QT=QP/QS and both triangles share tha angle Q).

Since ST and PR are parallel, then the angles QST and QPR are corresponding angles and they have the same measure. Furthermore, the angles QST and PST are supplementary angles. Therefore:


m\angle QST+m\angle PST=180\~

Since the measure of QST is the same as the measure of QPR, which is 5x-9, and the measure of the angle PST is 11x-3, then susbtitute those expressions into the equation:


(5x-9)+(11x-3)=180

Solve for x:


\begin{gathered} \Rightarrow5x-9+11x-3=180 \\ \Rightarrow16x-12=180 \\ \Rightarrow16x=180+12=192 \\ \Rightarrow x=(192)/(16) \\ \Rightarrow x=12 \end{gathered}

Therefore, x=12.

User Amanjot
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