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by gmattesttaker2 » Sat Aug 11, 2012 4:55 pm
Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri
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by theCEO » Sat Aug 11, 2012 5:37 pm
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
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by gmattesttaker2 » Sat Aug 11, 2012 10:13 pm
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
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by theCEO » Sun Aug 12, 2012 11:52 am
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
How to get area of ABCD = 2 x area of EFGH

Area of ABCD = area of frame + area of EFGH
The question says the area of EFGH is equal to the area of the picture frame
therefore area of frame = area of EFGH
therefore we can substitute the area of the frame with area of EFGH since they are the same

Area of ABCD = area of frame + area of EFGH
Area of ABCD = area of EFGH + area of EFGH = 2 x area of EFGH

Let me know if this helps!
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by gmattesttaker2 » Sun Aug 12, 2012 5:39 pm
theCEO wrote:
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
How to get area of ABCD = 2 x area of EFGH

Area of ABCD = area of frame + area of EFGH
The question says the area of EFGH is equal to the area of the picture frame
therefore area of frame = area of EFGH
therefore we can substitute the area of the frame with area of EFGH since they are the same

Area of ABCD = area of frame + area of EFGH
Area of ABCD = area of EFGH + area of EFGH = 2 x area of EFGH

Let me know if this helps!
Hello theCEO,

Thanks for the explanation again. It is clear now.

Best Regards,
Sri
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by theCEO » Sun Aug 12, 2012 6:56 pm
deleted!
Last edited by theCEO on Sun Aug 12, 2012 6:59 pm, edited 1 time in total.
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by theCEO » Sun Aug 12, 2012 6:58 pm
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:
theCEO wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this problem:

ABCD is a square picture frame (see figure). EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH (for the picture) is equal to the area of the picture frame (the area of ABCD minus the area of EFGH). If AB = 6, what is the length of EF?

Thanks a lot.

Best Regards,
Sri

Area of ABCD = area of frame + area of EFGH
area of ABCD = 2 x area of EFGH
6X6 = 2 X area of EFGH
18 = area of EFGH
18 = (EF)^2
EF = sqrt (18) = sqrt (2x9) = 3sqrt(2)
Hello theCEO,

Thanks for your reply. I was just wondering how you got :

area of ABCD = 2 x area of EFGH

Thanks.

Best Regards,
Sri
How to get area of ABCD = 2 x area of EFGH

Area of ABCD = area of frame + area of EFGH
The question says the area of EFGH is equal to the area of the picture frame
therefore area of frame = area of EFGH
therefore we can substitute the area of the frame with area of EFGH since they are the same

Area of ABCD = area of frame + area of EFGH
Area of ABCD = area of EFGH + area of EFGH = 2 x area of EFGH

Let me know if this helps!
Hello theCEO,

Thanks for the explanation again. It is clear now.

Best Regards,
Sri
Your welcome Sri!
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