There is in fact a quite easier way to solve this problem. As AB is equal to AD, and as AE is equal to EC, then the triangle ABE can be moved to the top of the quadrilateral, creating a square, which has an area of 24. Then its side is sqrt(24)=2×sqrt(6)

@erikzdrumpi

2 жыл бұрын

Wooooooooow!!!

Why not just rotate ∆ABE by 90° to get square AE'CE, then the area of AE'CE = 24 which means the length of AE = √24 = 2√6

good explanation and even better diagram 👍👍👍

Really smart

my method. extend CD vertically. draw a line horizontally(parallel to BC) mark the intersection X. AECX is a rectangle since angles at E, C, X are right. Hence angle EAX is right. since BAD is also right, it follows that angles BAE=XAD. and using the right angles and the equality of AB and AD, we have that triangles BAE=XAD. hance AE=AX So, we have that AECX is a square. Its area is 24 so its sidelength AE is sqrt(24) or 2sqrt(6)

Let F be a point on AE such that AE is perpendicular to FD at F. Let AB = AD = y, AE = h, FD = n, BE = p and FE = x implies AF = h - x. triangle(ABE) ~ triangle(AFD) implies y/h = y/n implies n = h and p/y = h - x/y implies p = h - x implies A(EFDC) = xh, A(triangle(ABE)) = 1/2(h - x)h = A(triangle(AFD)) implies 24 = A(ABCD) = xh + h^2 - xh = h^2 implies h = 2 * sqrt(6).

draw horizontal line from D to AE. AE=x, BE=a. 24= a*x/2+a*x/2+(x-a)*x.

There is a simpler and faster way : After completing the square with the triangle (the two triangles are similar, so ,they have the same area), the area of ​​the figure (24) is equal to the area of ​​the square , Hence, the required length is equal to the square root of 24 which is 2√6 .

There is an extremely easy way to solve it which I call triangulation which often works for quads and triangles we know that AB=AD and there are many quad we can make satisfying this property. but since the answer is a constant we know that in all suck cases AE has to remain the same length now we can make assumptions about the figure which will make the problem easier to solve.1 such case will be when angle bae becomes shorter and shorter until it is 0 degrees since we've not been given any info about the angle or the length BE so we will assume it to be 0 and make get ae equal to AD now we can sole but an even easier way is now to assume that it is a square since we've been given that the angles are 90 degrees we won't be breaking any rules(note that this does not always work as there are some cases when the figure of the square will never have the angle to be 0 but we see that assuming this is a square doesnt change anything) now we know s2= √24 = 2√6

Easy method to solve this cons.

Mam , after taking AE common why again AE in the bracket

Answer is 2✓6 I watched the thumbnail and just came out to comment my answer and now i will watch how they did it........

A cleaver way to complicate the solution to the problem. A simpler method is given by Rocco Dalto "Let F be a point on AE such that AE is perpendicular to FD at F. Let AB = AD = y, AE = h, FD = n, BE = p and FE = x implies AF = h - x. triangle(ABE) ~ triangle(AFD) implies y/h = y/n implies n = h and p/y = h - x/y implies p = h - x implies A(EFDC) = xh, A(triangle(ABE)) = 1/2(h - x)h = A(triangle(AFD)) implies 24 = A(ABCD) = xh + h^2 - xh = h^2 implies h = 2 * sqrt(6)."

What?