If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap?Area of...
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If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap?
If a 12 by 16 sheet of paper is folded on its diagonal, what is the area of the region of the overlap?
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$begingroup$
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
edited 4 hours ago
Blue
48.6k870154
asked 4 hours ago
NJCNJC
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.
$endgroup$
add a comment |
$begingroup$
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
edited 4 hours ago
Blue
48.6k870154
asked 4 hours ago
NJCNJC
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited 4 hours ago
Blue
48.6k870154
asked 4 hours ago
NJCNJC
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.
$endgroup$
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
edited 4 hours ago
Blue
48.6k870154
asked 4 hours ago
NJCNJC
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.
edited 4 hours ago
Blue
48.6k870154
asked 4 hours ago
NJCNJC
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.
edited 4 hours ago
Blue
48.6k870154
edited 4 hours ago
Blue
48.6k870154
edited 4 hours ago
Blue
48.6k870154
48.6k870154
asked 4 hours ago
NJCNJC
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.
asked 4 hours ago
NJCNJC
161
asked 4 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
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
add a comment |
$begingroup$
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
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
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 $EH=15/2$.
answered 4 hours ago
AretinoAretino
24k21443
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered 4 hours ago
bitesizebobitesizebo
1,50618
$endgroup$
add a comment |
$begingroup$
The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
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.)
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 $frac 12*20*7.5 = 75$.
answered 4 hours ago
fleabloodfleablood
71.4k22686
$endgroup$
add a comment |
$begingroup$
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.
answered 3 hours ago
D.R.D.R.
1,467721
$endgroup$
add a comment |
$begingroup$
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]
$$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$
answered 1 hour ago
CRawsonCRawson
11
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $EH=15/2$.
answered 4 hours ago
AretinoAretino
24k21443
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $EH=15/2$.
answered 4 hours ago
AretinoAretino
24k21443
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $EH=15/2$.
answered 4 hours ago
AretinoAretino
24k21443
$endgroup$
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 $EH=15/2$.
answered 4 hours ago
AretinoAretino
24k21443
answered 4 hours ago
AretinoAretino
24k21443
answered 4 hours ago
AretinoAretino
24k21443
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered 4 hours ago
bitesizebobitesizebo
1,50618
$endgroup$
add a comment |
$begingroup$
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.
answered 4 hours ago
bitesizebobitesizebo
1,50618
$endgroup$
add a comment |
$begingroup$
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.
answered 4 hours ago
bitesizebobitesizebo
1,50618
$endgroup$
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.
answered 4 hours ago
bitesizebobitesizebo
1,50618
answered 4 hours ago
bitesizebobitesizebo
1,50618
answered 4 hours ago
bitesizebobitesizebo
1,50618
answered 4 hours ago
bitesizebobitesizebo
1,50618
1,50618
add a comment |
add a comment |
$begingroup$
The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
The area is (1/2)(20)(120/16) = 75 which is half the base times the hight. The hight is found by similar triangles.
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered 4 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
$begingroup$
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.)
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 $frac 12*20*7.5 = 75$.
answered 4 hours ago
fleabloodfleablood
71.4k22686
$endgroup$
add a comment |
$begingroup$
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.)
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 $frac 12*20*7.5 = 75$.
answered 4 hours ago
fleabloodfleablood
71.4k22686
$endgroup$
add a comment |
$begingroup$
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.)
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 $frac 12*20*7.5 = 75$.
answered 4 hours ago
fleabloodfleablood
71.4k22686
$endgroup$
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.)
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 $frac 12*20*7.5 = 75$.
answered 4 hours ago
fleabloodfleablood
71.4k22686
edited 3 hours ago
answered 4 hours ago
fleabloodfleablood
71.4k22686
answered 4 hours ago
fleabloodfleablood
71.4k22686
answered 4 hours ago
fleabloodfleablood
71.4k22686
71.4k22686
add a comment |
add a comment |
$begingroup$
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.
answered 3 hours ago
D.R.D.R.
1,467721
$endgroup$
add a comment |
$begingroup$
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.
answered 3 hours ago
D.R.D.R.
1,467721
$endgroup$
add a comment |
$begingroup$
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.
answered 3 hours ago
D.R.D.R.
1,467721
$endgroup$
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.
answered 3 hours ago
D.R.D.R.
1,467721
answered 3 hours ago
D.R.D.R.
1,467721
answered 3 hours ago
D.R.D.R.
1,467721
answered 3 hours ago
D.R.D.R.
1,467721
1,467721
add a comment |
add a comment |
$begingroup$
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]
$$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$
answered 1 hour ago
CRawsonCRawson
11
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$endgroup$
add a comment |
$begingroup$
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]
$$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$
answered 1 hour ago
CRawsonCRawson
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.
$endgroup$
add a comment |
$begingroup$
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]
$$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$
answered 1 hour ago
CRawsonCRawson
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.
$endgroup$
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]
$$Area=frac 1 2 cdot Base cdot Height Rightarrow frac 1 2 cdot AC cdot EH Rightarrow frac 12 cdot 20 cdot 7.5 = 75 $$
answered 1 hour ago
CRawsonCRawson
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.
answered 1 hour ago
CRawsonCRawson
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.
answered 1 hour ago
CRawsonCRawson
11
answered 1 hour 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.
add a comment |
add a comment |
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.
NJC is a new contributor. Be nice, and check out our Code of Conduct.
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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