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Isosceles right triangle ABC and Square EFGH have the same a
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29 Oct 2017, 02:01
29 Oct 2017, 02:01
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39% (01:46) wrong based on 46 sessions
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Isosceles right triangle ABC and Square EFGH have the same area.
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for correct solution.
Quantity A |
Quantity B |
Perimeter of ABC |
Perimeter of EFGH |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Kudos for correct solution.
Re: Isosceles right triangle ABC and Square EFGH have the same a
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Updated on: 08 Dec 2017, 02:41
Updated on: 08 Dec 2017, 02:41
2
Here we can set up the following equation with both Areas:
1) Triangle: (a^2)/2 = A
2) Sqaure: b^2 = A
A stands for Area, a for side of triangle and b for side of square.
If we solve these equations for the sides a and b respectively, we obtain the following
1) a^2 = 2A => a = sqRoot(2A)
2) b^2 = A => b =sqRoot (A)
Now we can set up perimeter equations with the comparable variable A
1) SqRoot2A * 3 => 3*SqRoot2A
2) SqRootA * 4 => 4*SQRootA
Hence the perimeter of the triangle is bigger than that of the square Therefore A>B
1) Triangle: (a^2)/2 = A
2) Sqaure: b^2 = A
A stands for Area, a for side of triangle and b for side of square.
If we solve these equations for the sides a and b respectively, we obtain the following
1) a^2 = 2A => a = sqRoot(2A)
2) b^2 = A => b =sqRoot (A)
Now we can set up perimeter equations with the comparable variable A
1) SqRoot2A * 3 => 3*SqRoot2A
2) SqRootA * 4 => 4*SQRootA
Hence the perimeter of the triangle is bigger than that of the square Therefore A>B
Re: Isosceles right triangle ABC and Square EFGH have the same a
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07 Dec 2017, 19:38
07 Dec 2017, 19:38
1
simon1994 wrote:
Here we can set up the following equation with both Areas:
1) Triangle: (a^2)/2 = A
2) Sqaure: b^2 = A
A stands for Area, a for side of triangle and b for side of square.
If we solve these equations for the sides a and b respectively, we obtain the following
1) a^2 = 2A
2) b^2 = A
Now we can set up perimeter equations with the comparable variable A
1) 2A * 3 => 6A
2) A*4 => 4A
Hence the perimeter of the triangle(6A) is bigger than that of the square(4A) Therefore A>B
1) Triangle: (a^2)/2 = A
2) Sqaure: b^2 = A
A stands for Area, a for side of triangle and b for side of square.
If we solve these equations for the sides a and b respectively, we obtain the following
1) a^2 = 2A
2) b^2 = A
Now we can set up perimeter equations with the comparable variable A
1) 2A * 3 => 6A
2) A*4 => 4A
Hence the perimeter of the triangle(6A) is bigger than that of the square(4A) Therefore A>B
Would you explain why you put "2A * 3" and "A*4"
you times their area, this is also okay to compare their sides?
thank you !
