In the diagram below, \overline{ST} ST is parallel to \overline{PQ} PQ. If STST is 55 more than QTQT, RT=10RT=10, and PQ=15PQ=15, find the length of \overline{QT} QT. Figures are not necessarily drawn to scale. State your answer in simplest radical form, if necessary.PQRTS
Answers
The length of the [tex]\overline{QT}[/tex] can be determined using the given information and
the relationship between corresponding sides of similar triangles.
[tex]\underline{\overline{QT} = 5}[/tex]Reasons:
The given parameters are;
[tex]\overline{ST} \parallel \overline{PQ}[/tex]
ST = QT + 5
RT = 10
PQ = 15
Required: The length of [tex]\overline{QT}[/tex]
Solution:
Using the relationship between corresponding sides of similar triangles, ΔRST and ΔRPQ, we have;
[tex]\displaystyle \frac{\overline{ST}}{\overline{RT}} = \mathbf{\frac{\overline{PQ}}{\overline{RQ}}}[/tex]RQ = QT + RT
Therefore;
[tex]\displaystyle \frac{\overline{ST}}{\overline{RT}} = \mathbf{\frac{\overline{PQ}}{\overline{QT} + \overline{RT}}}[/tex]Which gives;
[tex]\displaystyle \frac{\overline{QT} + 5}{\overline{RT}} = \frac{\overline{PQ}}{\overline{QT} + \overline{RT}}[/tex]
Plugging in the known values gives;
[tex]\displaystyle \frac{\overline{QT} + 5}{10} = \mathbf{\frac{15}{\overline{QT} + 10}}[/tex]
[tex](\overline{QT} + 5) \times (\overline{QT} + 10) = 10 \times 15[/tex]
[tex]\overline{QT}^2 + 10\cdot \overline{QT} + 5\cdot \overline{QT} + 50 = 10 \times 15[/tex]
[tex]\overline{QT}^2 + 15\cdot \overline{QT} + 50 - 150= 0[/tex]
[tex]\mathbf{\overline{QT}^2 + 15\cdot \overline{QT} -100} = 0[/tex]
Factorizing with a graphing calculator gives;
[tex](\overline{QT} -5) \cdot (\overline{QT} + 20)= 0[/tex]
[tex]\mathbf{\overline{QT} = 5} \ or \ \overline{QT} = -20[/tex]
Therefore, given that [tex]\overline{QT}[/tex] is a natural number, we have;
[tex]\underline{\overline{QT} = 5}[/tex]
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