Question 43

In ΔPQR, PS and PT are bisectors of ∠QPR and ∠QPS respectively. If ∠QPT = 30°, PT = 9 cm and TR = 15 cm, then what is the area $$(in cm^2)$$ of ΔPTR ?

Solution

PS and PT are angle bisectors, => ∠QPT = ∠PTS = 30° and ∠SPR = 60°

Thus, ∠TPR = 30° + 60° = 90° and TPR is a right angled triangle.

=> $$(PR)^2=(TR)^2-(PT)^2$$

=> $$(PR)^2=(15)^2-(9)^2$$

=> $$(PR)^2=225-81=144$$

=> $$PR=\sqrt{144}=12$$ cm

$$\therefore$$ Area of ΔPTR = $$\frac{1}{2}\times(PT)\times(PR)$$

= $$\frac{1}{2}\times9\times12=54$$ $$cm^2$$

=> Ans - (B)

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