Find area of the fourth triangle given the area of three triangles.












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This is the question that I got in TCS Ninja under the Quantitative section.



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    0












    $begingroup$


    This is the question that I got in TCS Ninja under the Quantitative section.



    Help



    How shall I do this? Help !!










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      This is the question that I got in TCS Ninja under the Quantitative section.



      Help



      How shall I do this? Help !!










      share|cite|improve this question









      $endgroup$




      This is the question that I got in TCS Ninja under the Quantitative section.



      Help



      How shall I do this? Help !!







      area






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 20 at 11:03









      jayjay

      1034




      1034






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Denote by $[...]$ the area of the polygon $...$




          Lema 1



          Let $ABCD$ be a rectangle a $P$ a point inside as shown. It follows that $$[APB]+[DPC]=[APD]+[CPB]$$ enter image description here




          Proof



          Draw the parallels to $AD$ and $AB$ through $P$. Then
          $$[APB]+[DPC]=frac{AB·FP}{2}+frac{DC·EP}{2}=frac{AB·FP+AB·DC}{2}=frac{AB·EF}{2}=frac{[ABCD]}{2}$$



          Now back to your problem $$A1+A3=2040=A2+A4=1057+A4iff A4=2040-1057=983$$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Hint:



            Prove that $A_1+A_3=A_2+A_4=frac{1}{2}(A_1+A_2+A_3+A_4)$






            share|cite|improve this answer









            $endgroup$













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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Denote by $[...]$ the area of the polygon $...$




              Lema 1



              Let $ABCD$ be a rectangle a $P$ a point inside as shown. It follows that $$[APB]+[DPC]=[APD]+[CPB]$$ enter image description here




              Proof



              Draw the parallels to $AD$ and $AB$ through $P$. Then
              $$[APB]+[DPC]=frac{AB·FP}{2}+frac{DC·EP}{2}=frac{AB·FP+AB·DC}{2}=frac{AB·EF}{2}=frac{[ABCD]}{2}$$



              Now back to your problem $$A1+A3=2040=A2+A4=1057+A4iff A4=2040-1057=983$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Denote by $[...]$ the area of the polygon $...$




                Lema 1



                Let $ABCD$ be a rectangle a $P$ a point inside as shown. It follows that $$[APB]+[DPC]=[APD]+[CPB]$$ enter image description here




                Proof



                Draw the parallels to $AD$ and $AB$ through $P$. Then
                $$[APB]+[DPC]=frac{AB·FP}{2}+frac{DC·EP}{2}=frac{AB·FP+AB·DC}{2}=frac{AB·EF}{2}=frac{[ABCD]}{2}$$



                Now back to your problem $$A1+A3=2040=A2+A4=1057+A4iff A4=2040-1057=983$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Denote by $[...]$ the area of the polygon $...$




                  Lema 1



                  Let $ABCD$ be a rectangle a $P$ a point inside as shown. It follows that $$[APB]+[DPC]=[APD]+[CPB]$$ enter image description here




                  Proof



                  Draw the parallels to $AD$ and $AB$ through $P$. Then
                  $$[APB]+[DPC]=frac{AB·FP}{2}+frac{DC·EP}{2}=frac{AB·FP+AB·DC}{2}=frac{AB·EF}{2}=frac{[ABCD]}{2}$$



                  Now back to your problem $$A1+A3=2040=A2+A4=1057+A4iff A4=2040-1057=983$$






                  share|cite|improve this answer









                  $endgroup$



                  Denote by $[...]$ the area of the polygon $...$




                  Lema 1



                  Let $ABCD$ be a rectangle a $P$ a point inside as shown. It follows that $$[APB]+[DPC]=[APD]+[CPB]$$ enter image description here




                  Proof



                  Draw the parallels to $AD$ and $AB$ through $P$. Then
                  $$[APB]+[DPC]=frac{AB·FP}{2}+frac{DC·EP}{2}=frac{AB·FP+AB·DC}{2}=frac{AB·EF}{2}=frac{[ABCD]}{2}$$



                  Now back to your problem $$A1+A3=2040=A2+A4=1057+A4iff A4=2040-1057=983$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 20 at 12:21









                  Dr. MathvaDr. Mathva

                  2,094324




                  2,094324























                      0












                      $begingroup$

                      Hint:



                      Prove that $A_1+A_3=A_2+A_4=frac{1}{2}(A_1+A_2+A_3+A_4)$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Hint:



                        Prove that $A_1+A_3=A_2+A_4=frac{1}{2}(A_1+A_2+A_3+A_4)$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Hint:



                          Prove that $A_1+A_3=A_2+A_4=frac{1}{2}(A_1+A_2+A_3+A_4)$






                          share|cite|improve this answer









                          $endgroup$



                          Hint:



                          Prove that $A_1+A_3=A_2+A_4=frac{1}{2}(A_1+A_2+A_3+A_4)$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 20 at 11:39









                          LarryLarry

                          2,41331129




                          2,41331129






























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