In rectangle ABCD, AD=BC=5units. Diagnals AC and BD, each of length 10 units, intersect at E. What is the degree measure of angle AEB?

THANKSS

look at the diagram. Drop a line from E to AB. E is in the center of the rectangle, so if angle AEB=θ,

tan(θ/2) = (10/2) / (5/2) = 2
θ/2 = 63.4°
θ = 126.8°

To find the degree measure of angle AEB, we can use the properties of a rectangle.

In a rectangle, opposite sides are equal in length, and the diagonals are equal in length and bisect each other.

Given that AD = BC = 5 units and the diagonals AC and BD are each 10 units, we can determine that each diagonal bisects the rectangle into two congruent right triangles.

Let's label the point of intersection of the diagonals as E.

Now, we can see that ∆AEB and ∆CEB are congruent right triangles. This is because they share a side (BE) and have another side (AE and CE) with the same length (half the length of the diagonal, which is 5 units).

Since the opposite sides in a rectangle are parallel, we know that the sum of the angles AEB and CEB is 180 degrees.

Thus, the angle AEB and CEB each have a degree measure of 90 degrees (180 degrees ÷ 2).

Therefore, the degree measure of angle AEB is 90 degrees.