Question 53

In $$\triangle PQR, \angle Q = 85^\circ$$ and $$\angle R = 65^\circ$$. Points S and T are on the sides PQ and PR, respectively such that $$\angle STR = 95^\circ$$, and the ratio of the QR and ST is 9 : 5. If PQ = 21.6 cm, then the length of PT is:

Solution

$$\angle PTS + \angle STR = 180^\circ$$

$$\angle PTS = 180^\circ - 95^\circ = 85^\circ$$

$$\triangle$$ PTS is similar to $$\triangle$$ PQR So,

($$\angle$$ P is common and $$\angle PTS = \angle SQR$$)

$$\frac{ST}{QR} = \frac{PT}{PQ}$$

(QR : ST = 9 : 5 and PQ = 21.6 cm)

$$\frac{5}{9} = \frac{PT}{21.6}$$

$$21.6 \times 5 = PT \times 9$$

PT = 108/9 = 12 cm

$$\therefore$$ The length of PT is 12 cm.

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