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







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
    5 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
    5 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
    5 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
    5 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






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.









asked 5 hours ago









NJCNJC

161




161




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.





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.






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












  • $begingroup$
    I either don't understand the question, or the answer is 96.
    $endgroup$
    – Floris Claassens
    5 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
    5 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
    5 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
    5 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
5 hours ago




$begingroup$
I either don't understand the question, or the answer is 96.
$endgroup$
– Floris Claassens
5 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
5 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
5 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

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
    5 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.






          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });






            NJC is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3124273%2fif-a-12-by-16-sheet-of-paper-is-folded-on-its-diagonal-what-is-the-area-of-the%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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
              5 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
              5 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 5 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
              5 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
              5 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
            5 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
            5 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 5 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 5 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 5 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]
                                              $$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$






                                              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












                                                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.
                                                Check out our Code of Conduct.






                                                $endgroup$



                                                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








                                                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.









                                                share|cite|improve this answer



                                                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.









                                                answered 2 hours ago









                                                CRawsonCRawson

                                                11




                                                11




                                                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.





                                                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.






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






















                                                    NJC is a new contributor. Be nice, and check out our Code of Conduct.










                                                    draft saved

                                                    draft discarded


















                                                    NJC is a new contributor. Be nice, and check out our Code of Conduct.













                                                    NJC is a new contributor. Be nice, and check out our Code of Conduct.












                                                    NJC is a new contributor. Be nice, and check out our Code of Conduct.
















                                                    Thanks for contributing an answer to Mathematics Stack Exchange!


                                                    • Please be sure to answer the question. Provide details and share your research!

                                                    But avoid …



                                                    • Asking for help, clarification, or responding to other answers.

                                                    • Making statements based on opinion; back them up with references or personal experience.


                                                    Use MathJax to format equations. MathJax reference.


                                                    To learn more, see our tips on writing great answers.




                                                    draft saved


                                                    draft discarded














                                                    StackExchange.ready(
                                                    function () {
                                                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3124273%2fif-a-12-by-16-sheet-of-paper-is-folded-on-its-diagonal-what-is-the-area-of-the%23new-answer', 'question_page');
                                                    }
                                                    );

                                                    Post as a guest















                                                    Required, but never shown





















































                                                    Required, but never shown














                                                    Required, but never shown












                                                    Required, but never shown







                                                    Required, but never shown

































                                                    Required, but never shown














                                                    Required, but never shown












                                                    Required, but never shown







                                                    Required, but never shown







                                                    -

                                                    Popular posts from this blog

                                                    Loup dans la culture

                                                    Image
                                                    Loups sur une enluminure du Bestiaire d'Aberdeen Le loup est l'animal le plus emblématique de l'histoire de l'Eurasie, il était à l'honneur durant l'Antiquité chez la totalité des anciens peuples européens [ 1 ] . Sommaire 1 Le loup dans la mythologie 2 Les loups dans le folklore 2.1 L'évolution des mentalités 2.2 Le loup dans la psychanalyse 3 Le loup dans les œuvres culturelles 3.1 Le loup dans la littérature 3.1.1 Fables 3.1.2 Poésie 3.1.3 Romans 3.1.4 Littérature jeunesse 3.1.4.1 Quelques titres d'albums contemporains 3.1.5 Bande dessinée 3.2 Le loup à l'écran 3.2.1 Le cinéma et la télévision 3.2.2 Documentaires 3.2.3 Dessins animés 3.2.4 Jeux vidéo 3.3 Le loup en peinture et en sculpture 3.4 Le loup dans les œuvres musicales 4 Le loup et les noms propres 5 Bibliographie 6 Notes et références 7 Voir aussi 7.1 Articles connexes ...
                                                    Read more

                                                    How to solve the problem of ntp “Unable to contact time server” from KDE?

                                                    Image
                                                    1 0 I want to set date and time automatically but this error appears when I try to .. I'm using openSUSE, I tried different servers but it didn't work either, help please! It doesn't seem to be a network problem: $ ping pool.ntp.org 64 bytes from 41.78.128.17: icmp_seq=1 ttl=41 time=260 ms To the question of whether ntpd is installed and running: $ ps -C ntpd PID TTY TIME CMD i.e. there's no ntpd process running. $ rpm -qa | grep ntp yast2-ntp-client-3.1.12-1.7.noarch ntp-4.2.6p5-25.2.1.i586 Further information from the comments: $ sudo ntpdate pool.ntp.org 25 Sep 19:41:51 ntpdate[3162]: no server suitable for synchronization found networking kde opensuse ntp ntpd sh...
                                                    Read more

                                                    ASUS Zenbook UX433/UX333 — Configure Touchpad-embedded numpad on Linux

                                                    Image
                                                    0 1 I have an ASUS Zenbook UX433,on which ASUS decided to innovate by embedding a Numpad inside the touchpad (see picture 1). Toggling between classic touchpad mode and numpad (also turns the touchpad LEDs on) can be done through the golden button in the top-right corner of the touchpad. However, on my Manjaro arch Linux config, this behavior does not work by default and I can only use the touchpad, no numpad. Has anyone been able to configure the numpad to work on Linux, or have any idea how this could be achieved? EDIT1: I have had a look at the official Windows drivers for the UX433FA TouchPad, but was not able to make anything of it. I tried installing them on Linux using tools like NDISWrapper, but they do not detect the TouchPad/Numpad hardware, and my understanding of drivers is too limited. ...
                                                    Read more