Area

[jatinet]

A square ABCD, side length is 1. DE=1, AE=CE. What the area of the triangle ADE?

 

[arjmen]

A square ABCD, side length is 1. DE=1, AE=CE. What the area of the triangle ADE?

 

 

Is E inside the square or outside?

[dsaqwert]

Where is E?

 

it should be outside because if it is inside, the area is 0 and if it is outside, area is sqrt(2)/2

[rv_blitz]

Is it 0.5 sq units ?

[rv_blitz]

Assuming the following co-ordinates

1) A (0,1)

2) B (1,1)

3) C (1,0)

4) D (0,0)

 

then, E (-1,-1) to satisfy AE=CE.

 

Then we have the values of the three points A, D , E as (0,1), (0,0) and (-1,-1) resp.

[jatinet]

E is out side the square

[arjmen]

it should be outside because if it is inside the area is 0 and if it is outside area is sqrt(2)/2

 

Actually if E is inside the square, since AE=CE, E will lie on the diagonal DB and area of ADE = 1/(2*SQRT(2)) [since in a clock-wise or anti-clockwise naming system A and C will be diametrically opposite vertices]

[dsaqwert]

Actually if E is inside the square, since AE=CE, E will lie on the diagonal DB and area of ADE = 1/(2*SQRT(2)) [since in a clock-wise or anti-clockwise naming system A and C will be diametrically opposite vertices]

If outside we can still solve but...

 

yep my mistake, if it is inside it has area sqrt(2)/2, but E cannot lie on any diagonal

 

and for the actual question, the answer is (sqrt(3)/2+1)/2

[jatinet]

it should be outside because if it is inside, the area is 0 and if it is outside, area is sqrt(2)/2

 

Can you please elaborate how you arrived at the answer

[dsaqwert]

Can you please elaborate how you arrived at the answer

 

that was a wrong answer, check my previous post

[Paliwal800]

:hmm: If AE=CE then E has to be on a line which equidistant from both A and E. Given that DE=1 the only line possible is the diagonal BD.

If E is inside the square the angle ADE=45 and if it is outside the angle ADE is 135. But on this basis it is quite difficult to arrive on the answer.

 

It will be interesting to see how you solved this question. Please expedite.

 

Cheers,

Neel

[arjmen]

hmm, I get the answer to be SQRT(2)/4

 

Borrowing blitz's method:

A ---> (0,0)

B ---> (1,0)

C ---> (1,1)

D ---> (0,1)

 

E ---> (-1/Sqrt(2), 1+1/Sqrt(2)) = (-1/Sqrt(2), (SQRT(2)+1)/(Sqrt(2))

 

Using the method of drawing a rectangle of minimum area that encompasses A,D and E:

Area (ADE) = (SQRT(2)+1)/4 - 1/4 = SQRT(2)/4

 

Curiously enough this would also be the area if E were inside the square.

[rv_blitz]

Yeah, I get 1/[2*(sqrt)2]

[jatinet]

hmm, I get the answer to be SQRT(2)/4

 

Borrowing blitz's method:

A ---> (0,0)

B ---> (1,0)

C ---> (1,1)

D ---> (0,1)

 

E ---> (-1/Sqrt(2), 1+1/Sqrt(2)) = (-1/Sqrt(2), (SQRT(2)+1)/(Sqrt(2))

 

Using the method of drawing a rectangle of minimum area that encompasses A,D and E:

Area (ADE) = (SQRT(2)+1)/4 - 1/4 = SQRT(2)/4

 

Curiously enough this would also be the area if E were inside the square.

using the picture that is posted by blitz (without xy plane ) do we still need to work out the point E or is there any quicker way of gettting the solution. Also, in Ritz's figure please swap point B and D. Point E is in the same quadrant as D(quad I).

[arjmen]

using the picture that is posted by blitz (without xy plane ) do we still need to work out the point E or is there any quicker way of gettting the solution. Also, in Ritz's figure please swap point B and D. Point E is in the same quadrant as D(quad I).

 

Actually, I've just about stopped kicking myself.

 

So, we've a square ABCD of side of length 1 unit

Extend the diagonal DB to E outside the square such that DE = 1. Mark a point F on the Diagonal BD inside the square such that DF = 1. AE = CE and AF = CF.

 

Extend AD to a point G such that AEGF forms a rectangle with diagonals EF and AG where D is the point of intersection of the diagonals.

 

The diagonals of a rectangle divide the rectangle into 4 triangles of equal area.

 

So, just calculate the area of the triangle AFD (pretty straight-forward) and thats the area of ADE

[dsaqwert]

well, E should not necessarily be on a DB diagonal, it can be 75 degree up on the diagonal as well, hope you can imagine what i mean. so question is not clear

[rv_blitz]

Well , jatinet, do you think the figure posted will not fulfill the conditions ?

It satisfies DE=1, ABCD, is a square and AE = CE. I feel the diagram posted will also work.

Am i missing something ?

[jatinet]

Well , jatinet, do you think the figure posted will not fulfill the conditions ?

It satisfies DE=1, ABCD, is a square and AE = CE. I feel the diagram posted will also work.

Am i missing something ?

I have got a figure but I am not sure how I can past it or may be attach it. If you guys can help?

 

Also, my line of thinking is that we are dealing with a cube and we are asked to find the area of the half of one face. may be once I am able to post the picture things will be clearer

[rd_eastbay]

Insufficient Info. Is the triangle inside or outside the square?

If inside 1/(2 x 2^1/2)