why area of triangle changes when measured as components of triangles?












0












$begingroup$


enter image description here



If we measure an area of triangle directly using a formula I got 90 square unit. But if we measure by components like 2 triangles and one rectangle and take some it counts 90.5



Why 0.5 square unit difference occur?



Any Help will be appreciated






geometry triangle







share|cite|improve this question











$endgroup$








  • 10




    $begingroup$
    Is this meant as a riddle? The union of the colored areas is not a triangle.
    $endgroup$
    – Martin R
    12 hours ago










  • $begingroup$
    yes.I wanted to know why such thing is happening
    $endgroup$
    – SRJ
    12 hours ago






  • 2




    $begingroup$
    Related: How come $32.5 = 31.5$? (The “Missing Square” puzzle.)
    $endgroup$
    – Andrew T.
    5 hours ago












  • $begingroup$
    See the addendum to THIS ANSWER
    $endgroup$
    – steven gregory
    4 mins ago


















0












$begingroup$


enter image description here



If we measure an area of triangle directly using a formula I got 90 square unit. But if we measure by components like 2 triangles and one rectangle and take some it counts 90.5



Why 0.5 square unit difference occur?



Any Help will be appreciated






geometry triangle







share|cite|improve this question











$endgroup$








  • 10




    $begingroup$
    Is this meant as a riddle? The union of the colored areas is not a triangle.
    $endgroup$
    – Martin R
    12 hours ago










  • $begingroup$
    yes.I wanted to know why such thing is happening
    $endgroup$
    – SRJ
    12 hours ago






  • 2




    $begingroup$
    Related: How come $32.5 = 31.5$? (The “Missing Square” puzzle.)
    $endgroup$
    – Andrew T.
    5 hours ago












  • $begingroup$
    See the addendum to THIS ANSWER
    $endgroup$
    – steven gregory
    4 mins ago
















0












0








0


1



$begingroup$


enter image description here



If we measure an area of triangle directly using a formula I got 90 square unit. But if we measure by components like 2 triangles and one rectangle and take some it counts 90.5



Why 0.5 square unit difference occur?



Any Help will be appreciated






geometry triangle







share|cite|improve this question











$endgroup$




enter image description here



If we measure an area of triangle directly using a formula I got 90 square unit. But if we measure by components like 2 triangles and one rectangle and take some it counts 90.5



Why 0.5 square unit difference occur?



Any Help will be appreciated






geometry triangle




geometry triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Todd Sewell

210314




210314










asked 12 hours ago









SRJSRJ

1,6061519




1,6061519








  • 10




    $begingroup$
    Is this meant as a riddle? The union of the colored areas is not a triangle.
    $endgroup$
    – Martin R
    12 hours ago










  • $begingroup$
    yes.I wanted to know why such thing is happening
    $endgroup$
    – SRJ
    12 hours ago






  • 2




    $begingroup$
    Related: How come $32.5 = 31.5$? (The “Missing Square” puzzle.)
    $endgroup$
    – Andrew T.
    5 hours ago












  • $begingroup$
    See the addendum to THIS ANSWER
    $endgroup$
    – steven gregory
    4 mins ago
















  • 10




    $begingroup$
    Is this meant as a riddle? The union of the colored areas is not a triangle.
    $endgroup$
    – Martin R
    12 hours ago










  • $begingroup$
    yes.I wanted to know why such thing is happening
    $endgroup$
    – SRJ
    12 hours ago






  • 2




    $begingroup$
    Related: How come $32.5 = 31.5$? (The “Missing Square” puzzle.)
    $endgroup$
    – Andrew T.
    5 hours ago












  • $begingroup$
    See the addendum to THIS ANSWER
    $endgroup$
    – steven gregory
    4 mins ago










10




10




$begingroup$
Is this meant as a riddle? The union of the colored areas is not a triangle.
$endgroup$
– Martin R
12 hours ago




$begingroup$
Is this meant as a riddle? The union of the colored areas is not a triangle.
$endgroup$
– Martin R
12 hours ago












$begingroup$
yes.I wanted to know why such thing is happening
$endgroup$
– SRJ
12 hours ago




$begingroup$
yes.I wanted to know why such thing is happening
$endgroup$
– SRJ
12 hours ago




2




2




$begingroup$
Related: How come $32.5 = 31.5$? (The “Missing Square” puzzle.)
$endgroup$
– Andrew T.
5 hours ago






$begingroup$
Related: How come $32.5 = 31.5$? (The “Missing Square” puzzle.)
$endgroup$
– Andrew T.
5 hours ago














$begingroup$
See the addendum to THIS ANSWER
$endgroup$
– steven gregory
4 mins ago






$begingroup$
See the addendum to THIS ANSWER
$endgroup$
– steven gregory
4 mins ago












3 Answers
3






active

oldest

votes


















12












$begingroup$

Because the joint point of the two segments is not on the point of $(11,5)$. As the equation of the line is $y = frac{9}{20} x$, hence the coordinate of the joint point is $(frac{20times 5}{9} = frac{100}{9}, 5)$ which is not exactly $(11,5)$.






share|cite|improve this answer









$endgroup$





















    8












    $begingroup$

    This picture is basically what happens here.



    The difference is that your picture try to make the slope of the orange and the blue triangles almost the same as the big one, hence giving the illusion of them being sub-triangles.



    fig 1






    share|cite|improve this answer









    $endgroup$





















      7












      $begingroup$

      The points A$(0,0)$ , P$(11,5)$ and B$(20,9)$ are not on the same straight line. It forms a very small triangle (APB). So, the half area of the rectangle is not the sum of the colored pieces.



      enter image description here



      $a=AP=sqrt{11^2+5^2}quad;quad b=PB=sqrt{11^2+4^2}quad;quad c=AB=sqrt{9^2+20^2}$



      The calculus of the area of triangle (APB) is easy thanks to the Heron's formula :



      $frac14sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=0.5$



      Similar fake problem with solution on page 23 in https://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        12












        $begingroup$

        Because the joint point of the two segments is not on the point of $(11,5)$. As the equation of the line is $y = frac{9}{20} x$, hence the coordinate of the joint point is $(frac{20times 5}{9} = frac{100}{9}, 5)$ which is not exactly $(11,5)$.






        share|cite|improve this answer









        $endgroup$


















          12












          $begingroup$

          Because the joint point of the two segments is not on the point of $(11,5)$. As the equation of the line is $y = frac{9}{20} x$, hence the coordinate of the joint point is $(frac{20times 5}{9} = frac{100}{9}, 5)$ which is not exactly $(11,5)$.






          share|cite|improve this answer









          $endgroup$
















            12












            12








            12





            $begingroup$

            Because the joint point of the two segments is not on the point of $(11,5)$. As the equation of the line is $y = frac{9}{20} x$, hence the coordinate of the joint point is $(frac{20times 5}{9} = frac{100}{9}, 5)$ which is not exactly $(11,5)$.






            share|cite|improve this answer









            $endgroup$



            Because the joint point of the two segments is not on the point of $(11,5)$. As the equation of the line is $y = frac{9}{20} x$, hence the coordinate of the joint point is $(frac{20times 5}{9} = frac{100}{9}, 5)$ which is not exactly $(11,5)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 12 hours ago









            OmGOmG

            2,322722




            2,322722























                8












                $begingroup$

                This picture is basically what happens here.



                The difference is that your picture try to make the slope of the orange and the blue triangles almost the same as the big one, hence giving the illusion of them being sub-triangles.



                fig 1






                share|cite|improve this answer









                $endgroup$


















                  8












                  $begingroup$

                  This picture is basically what happens here.



                  The difference is that your picture try to make the slope of the orange and the blue triangles almost the same as the big one, hence giving the illusion of them being sub-triangles.



                  fig 1






                  share|cite|improve this answer









                  $endgroup$
















                    8












                    8








                    8





                    $begingroup$

                    This picture is basically what happens here.



                    The difference is that your picture try to make the slope of the orange and the blue triangles almost the same as the big one, hence giving the illusion of them being sub-triangles.



                    fig 1






                    share|cite|improve this answer









                    $endgroup$



                    This picture is basically what happens here.



                    The difference is that your picture try to make the slope of the orange and the blue triangles almost the same as the big one, hence giving the illusion of them being sub-triangles.



                    fig 1







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 6 hours ago









                    BigbearZzzBigbearZzz

                    8,07321651




                    8,07321651























                        7












                        $begingroup$

                        The points A$(0,0)$ , P$(11,5)$ and B$(20,9)$ are not on the same straight line. It forms a very small triangle (APB). So, the half area of the rectangle is not the sum of the colored pieces.



                        enter image description here



                        $a=AP=sqrt{11^2+5^2}quad;quad b=PB=sqrt{11^2+4^2}quad;quad c=AB=sqrt{9^2+20^2}$



                        The calculus of the area of triangle (APB) is easy thanks to the Heron's formula :



                        $frac14sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=0.5$



                        Similar fake problem with solution on page 23 in https://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes






                        share|cite|improve this answer











                        $endgroup$


















                          7












                          $begingroup$

                          The points A$(0,0)$ , P$(11,5)$ and B$(20,9)$ are not on the same straight line. It forms a very small triangle (APB). So, the half area of the rectangle is not the sum of the colored pieces.



                          enter image description here



                          $a=AP=sqrt{11^2+5^2}quad;quad b=PB=sqrt{11^2+4^2}quad;quad c=AB=sqrt{9^2+20^2}$



                          The calculus of the area of triangle (APB) is easy thanks to the Heron's formula :



                          $frac14sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=0.5$



                          Similar fake problem with solution on page 23 in https://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes






                          share|cite|improve this answer











                          $endgroup$
















                            7












                            7








                            7





                            $begingroup$

                            The points A$(0,0)$ , P$(11,5)$ and B$(20,9)$ are not on the same straight line. It forms a very small triangle (APB). So, the half area of the rectangle is not the sum of the colored pieces.



                            enter image description here



                            $a=AP=sqrt{11^2+5^2}quad;quad b=PB=sqrt{11^2+4^2}quad;quad c=AB=sqrt{9^2+20^2}$



                            The calculus of the area of triangle (APB) is easy thanks to the Heron's formula :



                            $frac14sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=0.5$



                            Similar fake problem with solution on page 23 in https://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes






                            share|cite|improve this answer











                            $endgroup$



                            The points A$(0,0)$ , P$(11,5)$ and B$(20,9)$ are not on the same straight line. It forms a very small triangle (APB). So, the half area of the rectangle is not the sum of the colored pieces.



                            enter image description here



                            $a=AP=sqrt{11^2+5^2}quad;quad b=PB=sqrt{11^2+4^2}quad;quad c=AB=sqrt{9^2+20^2}$



                            The calculus of the area of triangle (APB) is easy thanks to the Heron's formula :



                            $frac14sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}=0.5$



                            Similar fake problem with solution on page 23 in https://fr.scribd.com/doc/15493868/Pastiches-Paradoxes-Sophismes







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 6 hours ago

























                            answered 12 hours ago









                            JJacquelinJJacquelin

                            43k21750




                            43k21750






























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