3. Find the area of parallelogram ABCD, given that AB = BE = ED = 1 and the measure of angle ABE is 90 degrees

Question

here E is a one point in between AD

Answer 1

ok. first find the height. let F be the foot on the perpendicular from A to CD extended. ie. AF = height

since AB = BE and ABE is 90. therefore AE is root(2)

so AD is 1 + root(2)

using the sine rule

AF = root(2) x AD = [2 + root(2)]/2

base will be DC or AB. ie. 1

so area of ABCD is [2 + root(2)]/2 = 1.707 units squared

Answer 2

The base, AD, = √2 + 1
The height drawn from E to BC is √2/2. Here's why:


Because triangle ABE is an isosceles right triangle, angle A = 45. AE = √2 since it's the hypotenuse of isosceles right triangle ABE with legs equal to 1. AE + ED = AD = √2 + 1

The height drawn from E to BC also forms an isosceles right triangle because angle EBC = 45 due to congruent alternate interior angles in the parallelogram. The hypotenuse of this triangle, BE = 1, so the leg (which is the height) = 1/√2 =√2/2.

Thus, Area = √2/2(√2+1) = 1 + √2/2 sq. units

Answer 3

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Answer 4

You must mean that E lies between C and D;otherwise the problem is impossible.

So area = bh =1*1 = 1 unit squared

Answer 5

Its physically impossible to have E between A and D, and have angle ABE = 90 degrees. Unless the naming of the vertices doesn't follow and clockwise or anti-clock wise direction.

Answer 6

We need a clearer description of the location of A, B, C, and D.
AB
CD
or
AC
BD
or
AB
DC
etc.

Answer 7

base times height friend

Answer 8

Do your own damn homework!