Finding percentage increase of area [closed]
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Find the percentage increase in the area of a triangle if its each side is doubled?since no information is given about the type of triangle in question, so should i take a equilateral or isosceles or a scalene triangle, also will all the situations yields the same result?
Any suggestion will be helpful.
triangle area percentages
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closed as off-topic by Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Find the percentage increase in the area of a triangle if its each side is doubled?since no information is given about the type of triangle in question, so should i take a equilateral or isosceles or a scalene triangle, also will all the situations yields the same result?
Any suggestion will be helpful.
triangle area percentages
$endgroup$
closed as off-topic by Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Find the percentage increase in the area of a triangle if its each side is doubled?since no information is given about the type of triangle in question, so should i take a equilateral or isosceles or a scalene triangle, also will all the situations yields the same result?
Any suggestion will be helpful.
triangle area percentages
$endgroup$
Find the percentage increase in the area of a triangle if its each side is doubled?since no information is given about the type of triangle in question, so should i take a equilateral or isosceles or a scalene triangle, also will all the situations yields the same result?
Any suggestion will be helpful.
triangle area percentages
triangle area percentages
edited Jan 6 at 16:48
asked Jan 6 at 16:35
user632216
closed as off-topic by Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho Jan 7 at 0:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, A. Pongrácz, Gibbs, Shailesh, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
We have $$A_1=frac{1}{2}absin(gamma)$$ then $$A_2=4frac{1}{2}absin(gamma)=4A_1$$
Can you finish?
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
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This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
add a comment |
$begingroup$
Set b to be the base of the triangle where the other point on the triangle is over b, Then set line h perpendicular to b. Finally set the two angles adjacent to b to π and θ. If you break b up into b1 and b2 (which add up to b) then sin^-1(π)*h=b1 and sin^-1(θ)*h=b2 so (sin^-1(π)*h+sin^-1(θ))*h=b.If the angles π and θ stay the same the h is proportional to b so if b is doubled then h will be doubled too so that means if bh/2 is the original triangle then 2b*2h/2 is the new triangle so we get the new triangle to be 400% the size of the original triangle.
answered Jan 6 at 17:50
J. DoeJ. Doe
1
$endgroup$
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
add a comment |
$begingroup$
Think about it geometrically. Here we have a triangle:
The type of triangle and its orientation don’t matter since we can perform the following steps starting with any triangle at all.
Now reflect it in the midpoint of $overline{AB}$, getting a paralellogram:
You will agree, I hope, that this figure has twice the area of the original triangle. Now, chop off and move a right-triangular piece to create a rectangle with the same area:
(All we’ve really done here is to recreate geometrically the formula $frac12bh$ for the area of a triangle, taking $overline{AC}$ as the base.)
Now, what happens to the area of this rectangle if you double the lengths of all of those lines? What does this mean for the area of the original triangle?
answered Jan 6 at 20:54
amdamd
31.1k21051
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $$A_1=frac{1}{2}absin(gamma)$$ then $$A_2=4frac{1}{2}absin(gamma)=4A_1$$
Can you finish?
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
$begingroup$
This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
add a comment |
$begingroup$
We have $$A_1=frac{1}{2}absin(gamma)$$ then $$A_2=4frac{1}{2}absin(gamma)=4A_1$$
Can you finish?
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
$begingroup$
This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
add a comment |
$begingroup$
We have $$A_1=frac{1}{2}absin(gamma)$$ then $$A_2=4frac{1}{2}absin(gamma)=4A_1$$
Can you finish?
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
$endgroup$
We have $$A_1=frac{1}{2}absin(gamma)$$ then $$A_2=4frac{1}{2}absin(gamma)=4A_1$$
Can you finish?
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
answered Jan 6 at 16:41
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78k42866
78k42866
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
$begingroup$
This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
add a comment |
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
$begingroup$
This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
$begingroup$
this formula is for which type of triangle, equilateral, scalene, or isosceles?
$endgroup$
– user632216
Jan 6 at 16:51
$begingroup$
This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
This formula holds for every Euclidian triangle.
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:56
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
$begingroup$
thanks, it helps.
$endgroup$
– user632216
Jan 6 at 17:07
add a comment |
$begingroup$
Set b to be the base of the triangle where the other point on the triangle is over b, Then set line h perpendicular to b. Finally set the two angles adjacent to b to π and θ. If you break b up into b1 and b2 (which add up to b) then sin^-1(π)*h=b1 and sin^-1(θ)*h=b2 so (sin^-1(π)*h+sin^-1(θ))*h=b.If the angles π and θ stay the same the h is proportional to b so if b is doubled then h will be doubled too so that means if bh/2 is the original triangle then 2b*2h/2 is the new triangle so we get the new triangle to be 400% the size of the original triangle.
answered Jan 6 at 17:50
J. DoeJ. Doe
1
$endgroup$
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
add a comment |
$begingroup$
Set b to be the base of the triangle where the other point on the triangle is over b, Then set line h perpendicular to b. Finally set the two angles adjacent to b to π and θ. If you break b up into b1 and b2 (which add up to b) then sin^-1(π)*h=b1 and sin^-1(θ)*h=b2 so (sin^-1(π)*h+sin^-1(θ))*h=b.If the angles π and θ stay the same the h is proportional to b so if b is doubled then h will be doubled too so that means if bh/2 is the original triangle then 2b*2h/2 is the new triangle so we get the new triangle to be 400% the size of the original triangle.
answered Jan 6 at 17:50
J. DoeJ. Doe
1
$endgroup$
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
add a comment |
$begingroup$
Set b to be the base of the triangle where the other point on the triangle is over b, Then set line h perpendicular to b. Finally set the two angles adjacent to b to π and θ. If you break b up into b1 and b2 (which add up to b) then sin^-1(π)*h=b1 and sin^-1(θ)*h=b2 so (sin^-1(π)*h+sin^-1(θ))*h=b.If the angles π and θ stay the same the h is proportional to b so if b is doubled then h will be doubled too so that means if bh/2 is the original triangle then 2b*2h/2 is the new triangle so we get the new triangle to be 400% the size of the original triangle.
answered Jan 6 at 17:50
J. DoeJ. Doe
1
$endgroup$
Set b to be the base of the triangle where the other point on the triangle is over b, Then set line h perpendicular to b. Finally set the two angles adjacent to b to π and θ. If you break b up into b1 and b2 (which add up to b) then sin^-1(π)*h=b1 and sin^-1(θ)*h=b2 so (sin^-1(π)*h+sin^-1(θ))*h=b.If the angles π and θ stay the same the h is proportional to b so if b is doubled then h will be doubled too so that means if bh/2 is the original triangle then 2b*2h/2 is the new triangle so we get the new triangle to be 400% the size of the original triangle.
answered Jan 6 at 17:50
J. DoeJ. Doe
1
answered Jan 6 at 17:50
J. DoeJ. Doe
1
answered Jan 6 at 17:50
J. DoeJ. Doe
1
answered Jan 6 at 17:50
J. DoeJ. Doe
1
1
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
add a comment |
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
$begingroup$
Hi ! Welcome to MSE. Just a quick thought here. Please review your answer as no angle in a triangle can be equal to $pi$ and also use Mathjax to write math equations. I'm attaching a link that can get you started math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sauhard Sharma
Jan 6 at 18:05
add a comment |
$begingroup$
Think about it geometrically. Here we have a triangle:
The type of triangle and its orientation don’t matter since we can perform the following steps starting with any triangle at all.
Now reflect it in the midpoint of $overline{AB}$, getting a paralellogram:
You will agree, I hope, that this figure has twice the area of the original triangle. Now, chop off and move a right-triangular piece to create a rectangle with the same area:
(All we’ve really done here is to recreate geometrically the formula $frac12bh$ for the area of a triangle, taking $overline{AC}$ as the base.)
Now, what happens to the area of this rectangle if you double the lengths of all of those lines? What does this mean for the area of the original triangle?
answered Jan 6 at 20:54
amdamd
31.1k21051
$endgroup$
add a comment |
$begingroup$
Think about it geometrically. Here we have a triangle:
The type of triangle and its orientation don’t matter since we can perform the following steps starting with any triangle at all.
Now reflect it in the midpoint of $overline{AB}$, getting a paralellogram:
You will agree, I hope, that this figure has twice the area of the original triangle. Now, chop off and move a right-triangular piece to create a rectangle with the same area:
(All we’ve really done here is to recreate geometrically the formula $frac12bh$ for the area of a triangle, taking $overline{AC}$ as the base.)
Now, what happens to the area of this rectangle if you double the lengths of all of those lines? What does this mean for the area of the original triangle?
answered Jan 6 at 20:54
amdamd
31.1k21051
$endgroup$
add a comment |
$begingroup$
Think about it geometrically. Here we have a triangle:
The type of triangle and its orientation don’t matter since we can perform the following steps starting with any triangle at all.
Now reflect it in the midpoint of $overline{AB}$, getting a paralellogram:
You will agree, I hope, that this figure has twice the area of the original triangle. Now, chop off and move a right-triangular piece to create a rectangle with the same area:
(All we’ve really done here is to recreate geometrically the formula $frac12bh$ for the area of a triangle, taking $overline{AC}$ as the base.)
Now, what happens to the area of this rectangle if you double the lengths of all of those lines? What does this mean for the area of the original triangle?
answered Jan 6 at 20:54
amdamd
31.1k21051
$endgroup$
Think about it geometrically. Here we have a triangle:
The type of triangle and its orientation don’t matter since we can perform the following steps starting with any triangle at all.
Now reflect it in the midpoint of $overline{AB}$, getting a paralellogram:
You will agree, I hope, that this figure has twice the area of the original triangle. Now, chop off and move a right-triangular piece to create a rectangle with the same area:
(All we’ve really done here is to recreate geometrically the formula $frac12bh$ for the area of a triangle, taking $overline{AC}$ as the base.)
Now, what happens to the area of this rectangle if you double the lengths of all of those lines? What does this mean for the area of the original triangle?
answered Jan 6 at 20:54
amdamd
31.1k21051
answered Jan 6 at 20:54
amdamd
31.1k21051
answered Jan 6 at 20:54
amdamd
31.1k21051
answered Jan 6 at 20:54
amdamd
31.1k21051
31.1k21051
add a comment |
add a comment |
