1. 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 ?





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  • By: anil on 05 May 2019 01.46 am
    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 $$ herefore$$ Area of ΔPTR = $$frac{1}{2} imes(PT) imes(PR)$$ = $$frac{1}{2} imes9 imes12=54$$ $$cm^2$$ => Ans - (B)
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