If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap?












3












$begingroup$


I have tried this problem and keep on getting 96 as my answer, where the correct answer is 75.



Problem:




If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap (the region where paper is on top of paper)?







geometry







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  • $begingroup$
    I either don't understand the question, or the answer is 96.
    $endgroup$
    – Floris Claassens
    4 hours ago










  • $begingroup$
    @FlorisClaassens I get 75. How do you two get $96$? If we compare work we can maybe see where one (or both) of us are making the error.
    $endgroup$
    – fleablood
    4 hours ago










  • $begingroup$
    The answer can't be $96$. An oblong rectangle won't fold evenly onto itself so the overlapping area must be less than one half the area of the rectangle. $96$ is exactly equal to half the area so $96$ is an impossible answer.
    $endgroup$
    – fleablood
    4 hours ago
















3












$begingroup$


I have tried this problem and keep on getting 96 as my answer, where the correct answer is 75.



Problem:




If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap (the region where paper is on top of paper)?







geometry







share|cite|improve this question









New contributor




NJC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I either don't understand the question, or the answer is 96.
    $endgroup$
    – Floris Claassens
    4 hours ago










  • $begingroup$
    @FlorisClaassens I get 75. How do you two get $96$? If we compare work we can maybe see where one (or both) of us are making the error.
    $endgroup$
    – fleablood
    4 hours ago










  • $begingroup$
    The answer can't be $96$. An oblong rectangle won't fold evenly onto itself so the overlapping area must be less than one half the area of the rectangle. $96$ is exactly equal to half the area so $96$ is an impossible answer.
    $endgroup$
    – fleablood
    4 hours ago














3












3








3





$begingroup$


I have tried this problem and keep on getting 96 as my answer, where the correct answer is 75.



Problem:




If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap (the region where paper is on top of paper)?







geometry







share|cite|improve this question









New contributor




NJC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have tried this problem and keep on getting 96 as my answer, where the correct answer is 75.



Problem:




If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap (the region where paper is on top of paper)?







geometry




geometry






share|cite|improve this question









New contributor




NJC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




NJC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









Blue

48.6k870154




48.6k870154






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asked 5 hours ago









NJCNJC

161




161




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NJC is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    I either don't understand the question, or the answer is 96.
    $endgroup$
    – Floris Claassens
    4 hours ago










  • $begingroup$
    @FlorisClaassens I get 75. How do you two get $96$? If we compare work we can maybe see where one (or both) of us are making the error.
    $endgroup$
    – fleablood
    4 hours ago










  • $begingroup$
    The answer can't be $96$. An oblong rectangle won't fold evenly onto itself so the overlapping area must be less than one half the area of the rectangle. $96$ is exactly equal to half the area so $96$ is an impossible answer.
    $endgroup$
    – fleablood
    4 hours ago


















  • $begingroup$
    I either don't understand the question, or the answer is 96.
    $endgroup$
    – Floris Claassens
    4 hours ago










  • $begingroup$
    @FlorisClaassens I get 75. How do you two get $96$? If we compare work we can maybe see where one (or both) of us are making the error.
    $endgroup$
    – fleablood
    4 hours ago










  • $begingroup$
    The answer can't be $96$. An oblong rectangle won't fold evenly onto itself so the overlapping area must be less than one half the area of the rectangle. $96$ is exactly equal to half the area so $96$ is an impossible answer.
    $endgroup$
    – fleablood
    4 hours ago
















$begingroup$
I either don't understand the question, or the answer is 96.
$endgroup$
– Floris Claassens
4 hours ago




$begingroup$
I either don't understand the question, or the answer is 96.
$endgroup$
– Floris Claassens
4 hours ago












$begingroup$
@FlorisClaassens I get 75. How do you two get $96$? If we compare work we can maybe see where one (or both) of us are making the error.
$endgroup$
– fleablood
4 hours ago




$begingroup$
@FlorisClaassens I get 75. How do you two get $96$? If we compare work we can maybe see where one (or both) of us are making the error.
$endgroup$
– fleablood
4 hours ago












$begingroup$
The answer can't be $96$. An oblong rectangle won't fold evenly onto itself so the overlapping area must be less than one half the area of the rectangle. $96$ is exactly equal to half the area so $96$ is an impossible answer.
$endgroup$
– fleablood
4 hours ago




$begingroup$
The answer can't be $96$. An oblong rectangle won't fold evenly onto itself so the overlapping area must be less than one half the area of the rectangle. $96$ is exactly equal to half the area so $96$ is an impossible answer.
$endgroup$
– fleablood
4 hours ago










6 Answers
6






active

oldest

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1












$begingroup$

The overlapping region is triangle $AEC$, with base $AC=20$ and altitude $EH$. To find $EH$, observe that $GH=BF=48/5$ and $DB'=AC-2CF=28/5$. By similitude one then gets $EH=15/2$.



enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
    $endgroup$
    – Jacky Chong
    4 hours ago



















0












$begingroup$

The overlap is not exactly half the piece of paper, try folding a piece of paper along the diagonal and you'll see why. I can confirm the answer is indeed 75.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      The diagonal is $20$ inches long. (Pythagorean Theorem).



      The angles well be $A= arcsin frac {12}{20}=arccos frac {16}{20}$ and $B= arcsin frac {16}{20}=arccos frac {12}{20}$ with $A < B$ and $A + B = 90^circ$. (Basic trig definitions. Draw a picture to verify.)



      ![enter image description here



      The resulting shape will be an isosceles triangle with a base of $20$ and two base angles of $A$. This cuts in half into two congruent right triangles.



      The proportions of one of these right triangles is Hypotenuse = $h$. Base = $h*cos A= h*cosarccos frac {16}{20} = h*frac 45 = 10$. And height will = $h*sin A=hsin arcsin frac {12}{20} = h*frac 35$.



      $h* frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*frac 35 = 7.5$.



      And the area of the triangle is $frac 12*20*7.5 = 75$.






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        Folding the piece of paper, you get an isosceles triangle with 2 congruent right triangles on its sides. Each of these right triangles has side lengths $a, 12, h$, where $a$ is the shortest side and $h$ is the hypotenuse. Using the pythagorean theorem, and the fact that $a+h=16$, you get two equations in two variables, the other one being $h^2-a^2=(h-a)(h+a)=144$. This says that $h-a=9$, and so $h=12.5$ and $a=3.5$. The area of the overlap is the area of a 12x16 triangle minus a 3.5x12 triangle, which is in fact 75.






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          Referring to the Diagram from Aretino



          By the Pythagorean theorem the diagonal [AC] is 20 inches.

          Therefore 1/2 the diagonal [AH] is 10 inches.



          By similar triangles: [AEH] is similar to [ACB].
          $$frac{EH}{AH} = frac{BC}{AB} Rightarrow frac {h} {10} = frac {12}{16} $$
          Solve for height
          $$h=frac{120}{16}=7.5$$
          Solve for area of triangle [AEC]
          $$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$






          share|cite|improve this answer








          New contributor




          CRawson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The overlapping region is triangle $AEC$, with base $AC=20$ and altitude $EH$. To find $EH$, observe that $GH=BF=48/5$ and $DB'=AC-2CF=28/5$. By similitude one then gets $EH=15/2$.



            enter image description here






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
              $endgroup$
              – Jacky Chong
              4 hours ago
















            1












            $begingroup$

            The overlapping region is triangle $AEC$, with base $AC=20$ and altitude $EH$. To find $EH$, observe that $GH=BF=48/5$ and $DB'=AC-2CF=28/5$. By similitude one then gets $EH=15/2$.



            enter image description here






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
              $endgroup$
              – Jacky Chong
              4 hours ago














            1












            1








            1





            $begingroup$

            The overlapping region is triangle $AEC$, with base $AC=20$ and altitude $EH$. To find $EH$, observe that $GH=BF=48/5$ and $DB'=AC-2CF=28/5$. By similitude one then gets $EH=15/2$.



            enter image description here






            share|cite|improve this answer









            $endgroup$



            The overlapping region is triangle $AEC$, with base $AC=20$ and altitude $EH$. To find $EH$, observe that $GH=BF=48/5$ and $DB'=AC-2CF=28/5$. By similitude one then gets $EH=15/2$.



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            AretinoAretino

            24k21443




            24k21443












            • $begingroup$
              What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
              $endgroup$
              – Jacky Chong
              4 hours ago


















            • $begingroup$
              What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
              $endgroup$
              – Jacky Chong
              4 hours ago
















            $begingroup$
            What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
            $endgroup$
            – Jacky Chong
            4 hours ago




            $begingroup$
            What tools do you use to sketch the picture? I have been trying to find a fast sketching tool to generate good pictures to answer geometric questions.
            $endgroup$
            – Jacky Chong
            4 hours ago











            0












            $begingroup$

            The overlap is not exactly half the piece of paper, try folding a piece of paper along the diagonal and you'll see why. I can confirm the answer is indeed 75.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The overlap is not exactly half the piece of paper, try folding a piece of paper along the diagonal and you'll see why. I can confirm the answer is indeed 75.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The overlap is not exactly half the piece of paper, try folding a piece of paper along the diagonal and you'll see why. I can confirm the answer is indeed 75.






                share|cite|improve this answer









                $endgroup$



                The overlap is not exactly half the piece of paper, try folding a piece of paper along the diagonal and you'll see why. I can confirm the answer is indeed 75.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                bitesizebobitesizebo

                1,50618




                1,50618























                    0












                    $begingroup$

                    The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.






                    share|cite|improve this answer









                    $endgroup$


















                      0












                      $begingroup$

                      The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.






                      share|cite|improve this answer









                      $endgroup$
















                        0












                        0








                        0





                        $begingroup$

                        The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.






                        share|cite|improve this answer









                        $endgroup$



                        The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 4 hours ago









                        Mohammad Riazi-KermaniMohammad Riazi-Kermani

                        41.6k42061




                        41.6k42061























                            0












                            $begingroup$

                            The diagonal is $20$ inches long. (Pythagorean Theorem).



                            The angles well be $A= arcsin frac {12}{20}=arccos frac {16}{20}$ and $B= arcsin frac {16}{20}=arccos frac {12}{20}$ with $A < B$ and $A + B = 90^circ$. (Basic trig definitions. Draw a picture to verify.)



                            ![enter image description here



                            The resulting shape will be an isosceles triangle with a base of $20$ and two base angles of $A$. This cuts in half into two congruent right triangles.



                            The proportions of one of these right triangles is Hypotenuse = $h$. Base = $h*cos A= h*cosarccos frac {16}{20} = h*frac 45 = 10$. And height will = $h*sin A=hsin arcsin frac {12}{20} = h*frac 35$.



                            $h* frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*frac 35 = 7.5$.



                            And the area of the triangle is $frac 12*20*7.5 = 75$.






                            share|cite|improve this answer











                            $endgroup$


















                              0












                              $begingroup$

                              The diagonal is $20$ inches long. (Pythagorean Theorem).



                              The angles well be $A= arcsin frac {12}{20}=arccos frac {16}{20}$ and $B= arcsin frac {16}{20}=arccos frac {12}{20}$ with $A < B$ and $A + B = 90^circ$. (Basic trig definitions. Draw a picture to verify.)



                              ![enter image description here



                              The resulting shape will be an isosceles triangle with a base of $20$ and two base angles of $A$. This cuts in half into two congruent right triangles.



                              The proportions of one of these right triangles is Hypotenuse = $h$. Base = $h*cos A= h*cosarccos frac {16}{20} = h*frac 45 = 10$. And height will = $h*sin A=hsin arcsin frac {12}{20} = h*frac 35$.



                              $h* frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*frac 35 = 7.5$.



                              And the area of the triangle is $frac 12*20*7.5 = 75$.






                              share|cite|improve this answer











                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                The diagonal is $20$ inches long. (Pythagorean Theorem).



                                The angles well be $A= arcsin frac {12}{20}=arccos frac {16}{20}$ and $B= arcsin frac {16}{20}=arccos frac {12}{20}$ with $A < B$ and $A + B = 90^circ$. (Basic trig definitions. Draw a picture to verify.)



                                ![enter image description here



                                The resulting shape will be an isosceles triangle with a base of $20$ and two base angles of $A$. This cuts in half into two congruent right triangles.



                                The proportions of one of these right triangles is Hypotenuse = $h$. Base = $h*cos A= h*cosarccos frac {16}{20} = h*frac 45 = 10$. And height will = $h*sin A=hsin arcsin frac {12}{20} = h*frac 35$.



                                $h* frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*frac 35 = 7.5$.



                                And the area of the triangle is $frac 12*20*7.5 = 75$.






                                share|cite|improve this answer











                                $endgroup$



                                The diagonal is $20$ inches long. (Pythagorean Theorem).



                                The angles well be $A= arcsin frac {12}{20}=arccos frac {16}{20}$ and $B= arcsin frac {16}{20}=arccos frac {12}{20}$ with $A < B$ and $A + B = 90^circ$. (Basic trig definitions. Draw a picture to verify.)



                                ![enter image description here



                                The resulting shape will be an isosceles triangle with a base of $20$ and two base angles of $A$. This cuts in half into two congruent right triangles.



                                The proportions of one of these right triangles is Hypotenuse = $h$. Base = $h*cos A= h*cosarccos frac {16}{20} = h*frac 45 = 10$. And height will = $h*sin A=hsin arcsin frac {12}{20} = h*frac 35$.



                                $h* frac 45 = 10$ so $h = 12.5$ and so the height is $12.5*frac 35 = 7.5$.



                                And the area of the triangle is $frac 12*20*7.5 = 75$.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 4 hours ago

























                                answered 4 hours ago









                                fleabloodfleablood

                                71.4k22686




                                71.4k22686























                                    0












                                    $begingroup$

                                    Folding the piece of paper, you get an isosceles triangle with 2 congruent right triangles on its sides. Each of these right triangles has side lengths $a, 12, h$, where $a$ is the shortest side and $h$ is the hypotenuse. Using the pythagorean theorem, and the fact that $a+h=16$, you get two equations in two variables, the other one being $h^2-a^2=(h-a)(h+a)=144$. This says that $h-a=9$, and so $h=12.5$ and $a=3.5$. The area of the overlap is the area of a 12x16 triangle minus a 3.5x12 triangle, which is in fact 75.






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      Folding the piece of paper, you get an isosceles triangle with 2 congruent right triangles on its sides. Each of these right triangles has side lengths $a, 12, h$, where $a$ is the shortest side and $h$ is the hypotenuse. Using the pythagorean theorem, and the fact that $a+h=16$, you get two equations in two variables, the other one being $h^2-a^2=(h-a)(h+a)=144$. This says that $h-a=9$, and so $h=12.5$ and $a=3.5$. The area of the overlap is the area of a 12x16 triangle minus a 3.5x12 triangle, which is in fact 75.






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        Folding the piece of paper, you get an isosceles triangle with 2 congruent right triangles on its sides. Each of these right triangles has side lengths $a, 12, h$, where $a$ is the shortest side and $h$ is the hypotenuse. Using the pythagorean theorem, and the fact that $a+h=16$, you get two equations in two variables, the other one being $h^2-a^2=(h-a)(h+a)=144$. This says that $h-a=9$, and so $h=12.5$ and $a=3.5$. The area of the overlap is the area of a 12x16 triangle minus a 3.5x12 triangle, which is in fact 75.






                                        share|cite|improve this answer









                                        $endgroup$



                                        Folding the piece of paper, you get an isosceles triangle with 2 congruent right triangles on its sides. Each of these right triangles has side lengths $a, 12, h$, where $a$ is the shortest side and $h$ is the hypotenuse. Using the pythagorean theorem, and the fact that $a+h=16$, you get two equations in two variables, the other one being $h^2-a^2=(h-a)(h+a)=144$. This says that $h-a=9$, and so $h=12.5$ and $a=3.5$. The area of the overlap is the area of a 12x16 triangle minus a 3.5x12 triangle, which is in fact 75.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 4 hours ago









                                        D.R.D.R.

                                        1,467721




                                        1,467721























                                            0












                                            $begingroup$

                                            Referring to the Diagram from Aretino



                                            By the Pythagorean theorem the diagonal [AC] is 20 inches.

                                            Therefore 1/2 the diagonal [AH] is 10 inches.



                                            By similar triangles: [AEH] is similar to [ACB].
                                            $$frac{EH}{AH} = frac{BC}{AB} Rightarrow frac {h} {10} = frac {12}{16} $$
                                            Solve for height
                                            $$h=frac{120}{16}=7.5$$
                                            Solve for area of triangle [AEC]
                                            $$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$






                                            share|cite|improve this answer








                                            New contributor




                                            CRawson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                            Check out our Code of Conduct.






                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Referring to the Diagram from Aretino



                                              By the Pythagorean theorem the diagonal [AC] is 20 inches.

                                              Therefore 1/2 the diagonal [AH] is 10 inches.



                                              By similar triangles: [AEH] is similar to [ACB].
                                              $$frac{EH}{AH} = frac{BC}{AB} Rightarrow frac {h} {10} = frac {12}{16} $$
                                              Solve for height
                                              $$h=frac{120}{16}=7.5$$
                                              Solve for area of triangle [AEC]
                                              $$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$






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                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Referring to the Diagram from Aretino



                                                By the Pythagorean theorem the diagonal [AC] is 20 inches.

                                                Therefore 1/2 the diagonal [AH] is 10 inches.



                                                By similar triangles: [AEH] is similar to [ACB].
                                                $$frac{EH}{AH} = frac{BC}{AB} Rightarrow frac {h} {10} = frac {12}{16} $$
                                                Solve for height
                                                $$h=frac{120}{16}=7.5$$
                                                Solve for area of triangle [AEC]
                                                $$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$






                                                share|cite|improve this answer








                                                New contributor




                                                CRawson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                                $endgroup$



                                                Referring to the Diagram from Aretino



                                                By the Pythagorean theorem the diagonal [AC] is 20 inches.

                                                Therefore 1/2 the diagonal [AH] is 10 inches.



                                                By similar triangles: [AEH] is similar to [ACB].
                                                $$frac{EH}{AH} = frac{BC}{AB} Rightarrow frac {h} {10} = frac {12}{16} $$
                                                Solve for height
                                                $$h=frac{120}{16}=7.5$$
                                                Solve for area of triangle [AEC]
                                                $$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$







                                                share|cite|improve this answer








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                                                share|cite|improve this answer






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                                                answered 1 hour ago









                                                CRawsonCRawson

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