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1)

In a Δ PQR , Let  PQR=300 and the side PQ and QR have lengths 10√3 and 10, respectively . Then , which of the following statement (s) is (are) TRUE?


A) QPR=450

B) The area of the PQR is 253 and QRP=1200

C) The radius of the incircle of the PQR is 10315

D) The area of the circumcircle of the PQR is 100π

Answer:

Option B,C,D

Explanation:

 In  Δ PQR 

          PQR=300

                  PQ = 103    , QR=10

             392019643_new.JPG

By cosine rule

   cos300=PQ2+QR2PR22PQ.QR

  32=300+100PR22003

       300=300+100-PR2

       PR=10

   Since , PR=QR= 10

       \angle QPR=30^{0}   and    \angle QRP=120^{0}

Area of ΔPQR = \frac{1}{2} PQ.QR \sin 30^{0}

              =  \frac{1}{2}\times10\sqrt{3}\times 10\times\frac{1}{2}=25\sqrt{3}

Radius of incircle of

            Δ PQR= \frac{Area of \triangle PQR }{Semi-perimetre of \triangle PQR}

 i.e   r=\frac{\triangle }{s}= \frac{25\sqrt{3}}{\frac{10\sqrt{3}+10+10}{2}}=\frac{25\sqrt{3}}{5(\sqrt{3}+2)}

  \Rightarrow       r= 5\sqrt{3}(2-\sqrt{3})=10\sqrt{3}-15

       and radius of the circumcircle 

                       (R)=\frac{abc}{4\triangle}= \frac{10\sqrt{3}\times10\times10}{4\times25\sqrt{3}}=10

\therefore   Area of circumcircle of Δ PQR=  \pi R^{2}=100\pi

         Hence, option (b), (c) and (d) are correct answer.