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Re: If ABCD is a rectangle, what is the area of the shaded region? [#permalink]

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22 Jan 2015, 13:46

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Area of Triangle BCD =1/2*20*15=150

BD= sqrt (20^2+15^2)=25 also area of BCD can be determined as 1/2* BD* height from C to BD lets height from C meets BD at X Area of triangle BCX will be 1/2* BD* CX =150 =>1/2*25* CX=150 =>CX=300/25=12

Now, BX= sqrt (20^2-12^2)=16 So, area of shaded region will be 1/2* BX*CX=1/2*16*12=96

Answer is C _________________

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a way different from the one mentioned ....ans C we can use similar triangles property between triangles abd and bcp and get the sides of triangle bcp and so area comes as 96..
_________________

When the GMAT puts a right triangle in front of you, it is probably NOT a random right triangle. While there are some exceptions, in most cases, you're looking at a 3/4/5, 5/12/13, 30/60/90 or 45/45/90 right triangle. If you see ANY clues that point to a specific right triangle, then you should think about how you can prove that the triangle IS what you think it is.

Here, we have 3 right triangles and they all appear to be multiples of a 3/4/5 (notice how each has a diagonal that is a MULTIPLE of 5).

Assuming they're all 3/4/5s, the sides SHOULD be....

Big triangle: 15/20/25 Middle triangle: 12/16/20 Small triangle: 9/12/15

From the picture, there are a couple of other things that need to "line up" to prove that we're correct:

1) One side of the "small" and "middle" triangles MUST be the same. 2) The "other" sides of each of those triangles MUST add up to 25 (the diagonal of the big triangle)

The common "shared" side would be 12, which means the "other" sides (16 and 9) add up to 25. So we have the EXACT situation that we thought we did. With knowledge of ALL of the side lengths, we can now answer any question that's asked.

the topic is locked there hence opening new thread experts please tell what wrong with this let x be the point from c to digonal BD now required area = area of rectangle - area of trangle BAD - area of trangle cxd angle CDx and XCD are 45 digree , hence area of trangle CDE = 1/2 x 15/root2 x 15/root2 = approx 46 required area = 300 - 150 - 46 = 104

The problem with your solution is that angle CXD and XCB are not 45 degrees - if they were, CX would bisect BD, but it does not (unless ABCD were a rhombus).

The correct solution: We are looking for \(\frac{CX*BX}{2}\) We know: BD is 25 (3-4-5 right triangle * 5), Area of BCD = 150, and therefore CX = 12 (because \(\frac{CX*BD}{2} = 150\)) So therefore: \(BX^2 + 144 = 400\)-> \(BX^2 = 256\) -> \(BX = 16\) -> Area is \(\frac{16*12}{2} = 96\)

All these triangles inscribed are proportionate, since BD is a diagonal of this rectangle Since BC = 20 and BD = 25, BC = = 4/5*BD So BX = 4/5 * 20 = 16 Solve for CX = A^2 + B^2 = C^2 256 + B^2 = 400 B^2 = 400-256 B^2 = 144 Sqrt(144) = 12

Re: If ABCD is a rectangle, what is the area of the shaded region? [#permalink]

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08 Jun 2015, 19:04

bluesquare wrote:

The problem with your solution is that angle CXD and XCB are not 45 degrees - if they were, CX would bisect BD, but it does not (unless ABCD were a rhombus).

The correct solution: We are looking for \(\frac{CX*BX}{2}\) We know: BD is 25 (3-4-5 right triangle * 5), Area of BCD = 150, and therefore CX = 12 (because \(\frac{CX*BD}{2} = 150\)) So therefore: \(BX^2 + 144 = 400\)-> \(BX^2 = 256\) -> \(BX = 16\) -> Area is \(\frac{16*12}{2} = 96\)

Hi, Thanks for prompt response, I an't saying angle CXD and angle XCB are 45 digrees, what i am saying angle CXD is 90 and angle CDX are 45 digrees, since DB is diognal of restangle hence in trangle CXD , angle CXD is 90 , CDX is 45 then XCD must be 180-90-45=45, by that CX and DX must be 15/root2, which is not inline with correct explanation

The problem with your solution is that angle CXD and XCB are not 45 degrees - if they were, CX would bisect BD, but it does not (unless ABCD were a rhombus).

The correct solution: We are looking for \(\frac{CX*BX}{2}\) We know: BD is 25 (3-4-5 right triangle * 5), Area of BCD = 150, and therefore CX = 12 (because \(\frac{CX*BD}{2} = 150\)) So therefore: \(BX^2 + 144 = 400\)-> \(BX^2 = 256\) -> \(BX = 16\) -> Area is \(\frac{16*12}{2} = 96\)

Hi, Thanks for prompt response, I an't saying angle CXD and angle XCB are 45 digrees, what i am saying angle CXD is 90 and angle CDX are 45 digrees, since DB is diognal of restangle hence in trangle CXD , angle CXD is 90 , CDX is 45 then XCD must be 180-90-45=45, by that CX and DX must be 15/root2, which is not inline with correct explanation

Re: If ABCD is a rectangle, what is the area of the shaded region? [#permalink]

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20 Aug 2017, 10:25

SMEs please let me know what's wrong with my approach as I am getting 100 which is not an answer option but close to it.

I saw that its a rectangle. Therefore angle CBD has to be 45 since segment BD is a diagonal. Therefore, the shaded region is a 45-45-90 triangle (one angle 90, other 45, so remaining can't be anything but 45).

Given that BC is 20 and side opposite to 90 angle is root2*x, x=20/root2=base=height.

Solving for 1/2 * b * h = 1/2 * 20 / root2 * 20 / root2=100.

SMEs please let me know what's wrong with my approach as I am getting 100 which is not an answer option but close to it.

I saw that its a rectangle. Therefore angle CBD has to be 45 since segment BD is a diagonal. Therefore, the shaded region is a 45-45-90 triangle (one angle 90, other 45, so remaining can't be anything but 45).

Hi aashishagarwal2,

The diagonal of a SQUARE will form 45 degree angles; since this is a rectangle with different side lengths, we know that those angles CANNOT be 45 degrees (since that shape is NOT a square). That incorrect assumption is why the rest of your calculation is incorrect.

Re: If ABCD is a rectangle, what is the area of the shaded region? [#permalink]

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20 Aug 2017, 12:53

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Answer is C

ABCD is rectangle therefore diogonal can be calculated by pythogoras theorm and will be equal to 25 and applying simple trigonometry we can calculate required sides. and finally area of shaded region

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Re: If ABCD is a rectangle, what is the area of the shaded region? [#permalink]

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20 Aug 2017, 13:00

chetan2u wrote:

a way different from the one mentioned ....ans C we can use similar triangles property between triangles abd and bcp and get the sides of triangle bcp and so area comes as 96..

i guess using trigonometry and catching angles 37 and 53 is fastest way, as in my solution.. although trigonometry is not included in gmat , but if one knows , can help him a lot.

Thanks.
_________________

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