Geometry Help?

Question

The diagonals of rectangle ABCD intersect at point E. If measure of angle BDC is 30 degrees and
AE =8 sqrt 3 units, find the area of the rectangle.

The area of a rhombus is 7.5 sq. in. Find the length of the two diagonals if one of them is one more than twice the other.

Thank so much!

Answer 1

Angle BDC is the same as angle CAB, which is also therefore 30 degrees.

A to E to an altitude from E to AB is a 30-60-90 triangle with a hypotenuse AE, which measures 8*sqrt(3). One leg of that triangle is half the rectangle's height and the other leg is half the rectangle's width.

Half of AB = 8 * sqrt(3) * cos(30)
Half of AB = 8 * sqrt(3) * sqrt(3)/2
Half of AB = 12

Half of AD = 8 * sqrt(3) * sin(30)
Half of AD = 8 * sqrt(3) * 1/2
half of AD = 4 * sqrt(3)

Area of rectangle = AB * AD
= (12 * 2) * (4 sqrt(3) * 2)
= 192 sqrt(3) units
= about 332.55 square units

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The area of a rhombus is half of the product of the diagonals:

a = 1/2 * d1 * d2

Since we know one diagonal is "one more than twice the other" we can use "x" for the shorter diagonal and rewrite that as:

a = 1/2 * x * (2x + 1)
7.5 = 0.5 * (x) * (2x + 1)
15 = 2x^2 + x

2x^2 + x - 15 = 0
(2x - 5)(x + 3) = 0
x = 5/2 or x = -3.

Discounting the negative solution, the shorter diagonal is 5/2, or 2.5. The longer diagonal is twice as big, plus one, or 2*2.5 + 1 = 5 + 1 = 6.

The diagonals are 2.5 and 6 inches.

Answer 2

If you can let me know the positions of A, B, C, and D on the first one I can assist you.