Given that the diagonals of a rhombus are always perpendicular bisectors of each other, what is the area of a rhombus with side length sqrt89 units and diagonals that differ by 6 units?

PlZ HELP

I DON'T UNDERSTAND

Aopshelp Apr 3, 2019

#1**+1 **

I wish I knew how to draw the picture, but I don't. I'll try to describe it.

Draw the rhombus. Draw the diagonals. This creates four right triangles inside.

Label each half of the short diagonal as X. Label each half of the long diagonal as X+3.

Label the sides of the rhombus as sqrt(89).

Each of the internal triangles will be figured the same, so I'll just describe one.

You have a right triangle with one side X, one side X+3, and the hypotenuse sqrt(89).

Pythagoras' Theorem (X)^{2} + (X+3)^{2} = sqrt(89)^{2}

X^{2} + X^{2} + 6X + 9 = 89

2X^{2} + 6X – 80 = 0

This will factor to (2X + 16)(X – 5) = 0

Disregard the negative solution. Half of the short diagonal is 5 units. Can you take it from there?

.

.

Guest Apr 3, 2019