How do you prove the rhombus area theorem?

Question

Given: WXYZ is a rhombus with diagonal WY and XY

Prove: Area ABCD= 1/2 h(b1+b2)

please help!

Answer 1

You have written the formula for a trapezoid, but asked about a rhombus.
I will assume you meant

Area ABCD = 1/2 d1 d2 (the product of the diagonals)

Just draw the figure and its diagonals. Since the diagonals of a rhombus are perpendicular and bisect each other you get 4 right triangles with base 1/2 of d1 and height 1/2 of d2. Each triangle's area is 1/2 (1/2 d1)(1/2 d2) = 1/8 d1 d2. But there are 4 of them so the combiled area is 4 (1/8 d1 d2) or 1/2 d1 d2

Hopw that helps

Answer 2

The area of the square has TWO FORMULAS! A = s² and A = 1/2 d²; try the second formula! 50 = 1/2 d² 100 = d² d = 10m.

Answer 3

remember that the diagonals of any rhombus intersect at right angles.

That gives you 4 right triangles.
The area of a triangle is 1/2 b x h

the area of all 4 triangles is....you can do it!