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The figure ABCD is a rectangle with AD = 5 units and AE = EB. EF is perpendicular to DB and is half of DF. If the area of the triangle DEF is 9 sq units, what is the area of ABCD (in sq units)? (A) 405 (B) 505 (C) 205 (D) 305 (E)255
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The figure ABCD is a rectangle with AD = 5 units and AE = EB. EF is perpendicular to DB and is half of DF. If the area of the triangle DEF is 9 sq units, what is the area of ABCD (in sq units)? (A) 405 (B) 505 (C) 205 (D) 305 (E)255
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I am not able to draw the figure for given information. Can you please share the image for this question, if you have one.
The figure ABCD is a rectangle with AD = 5 units and AE = EB. EF is perpendicular to DB and is half of DF. If the area of the triangle DEF is 9 sq units, what is the area of ABCD (in sq units)? (A) 405 (B) 505 (C) 205 (D) 305 (E)255
I am not able to draw the figure for given information. Can you please share the image for this question, if you have one.
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I am thinking that it is a rectangle ABCD with a triangle EDF in it. I guess given the info you can solve the problem.
The figure ABCD is a rectangle with AD = 5 units and AE = EB. EF is perpendicular to DB and is half of DF. If the area of the triangle DEF is 9 sq units, what is the area of ABCD (in sq units)? (A) 405 (B) 505 (C) 205 (D) 305 (E)255
I am not able to draw the figure for given information. Can you please share the image for this question, if you have one.
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There was no figure given with this question and I am not able to upload the image which I have draw. Its not accepting ODG format.
Anyway, there can be 3 ways of placing E ( on AB, outside the rectangle the way you guys interpreted and inside Rectangle), I use the simpler approach to place E on AB but when I try to solve the equations I end up in a trap of some stupid equations
The figure ABCD is a rectangle with AD = 5 units and AE = EB. EF is perpendicular to DB and is half of DF. If the area of the triangle DEF is 9 sq units, what is the area of ABCD (in sq units)? (A) 405 (B) 505 (C) 205 (D) 305 (E)255
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Consider E is on line AB only. Draw diagonal DB, and a line EF perpendicular to DB, at point F on DB.
Given EF is half of DF and area of DEF = 9 sq units. From this, we can calculate DF = 6 and EF = 3. Now join DE, it'll be a common hypotenuse to AED and DEF. Thereby, we end up with AE = EB = 2*sqrt(5). AB will be 4*sqrt(5) and total area of the triangle should be 20*sqrt(5).
The options given are all wrong, each last digit 5 there should be inside a sqrt. C is the answer.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.