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Figure shown below consists of two squares having
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Updated on: 20 Feb 2018, 11:39
Updated on: 20 Feb 2018, 11:39
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Question Stats:
79% (01:39) correct
20% (01:37) wrong based on 48 sessions
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Figure shown below consists of two squares having the same centers. Sides of the inner square, each of length 4 units, are parallel to the sides of the outer square each having length of 12 units.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The perimeter of the shaded region. |
Area of the shaded region. |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Attachments
IMG_20180220_142335.jpg [ 852.37 KiB | Viewed 8698 times ]
Originally posted by Haripriya517 on 20 Feb 2018, 01:31.
Last edited by Carcass on 20 Feb 2018, 11:39, edited 2 times in total.
Last edited by Carcass on 20 Feb 2018, 11:39, edited 2 times in total.
Edited by Carcass
Sherpa Prep Representative
Joined: 15 Jan 2018
Posts: 147
Given Kudos: 0
Re: Figure shown below consists of two squares having
[#permalink]
20 Feb 2018, 11:29
20 Feb 2018, 11:29
2
Expert Reply
For Quantity A, the vertical sides are obviously 4 and 12. Less obviously, the shaded region can be broken up into a square with a right isosceles triangle on the top and bottom. (The picture isn't drawn to scale at all.) The diagonals of the right triangles would thus be 4√2. Thus, the perimeter is 16 + 8√2.
For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.
Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.
Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.
(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)
For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.
Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.
Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.
(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)
Re: Figure shown below consists of two squares having
[#permalink]
24 Feb 2018, 10:50
24 Feb 2018, 10:50
SherpaPrep wrote:
For Quantity A, the vertical sides are obviously 4 and 12. Less obviously, the shaded region can be broken up into a square with a right isosceles triangle on the top and bottom. (The picture isn't drawn to scale at all.) The diagonals of the right triangles would thus be 4√2. Thus, the perimeter is 16 + 8√2.
For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.
Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.
Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.
(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)
excellent
For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.
Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.
Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.
(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)
