let E be the intersection of the diagonals AC andBD
In a rhombus the diagonals right-bisect each other
so triangle ABE has
AB=12 , AE = 6 , and by Pythagoras BE = 8
Area of rhomus = 1/2 of the product of its diagonals
= (1/2)(12)(16) = 96
In the rhombus ABCD, AB=10 and AC=12. Find the area of the rhombus.
I am confused about how to solve this problem.
4 answers
How do I know the value of the diagonals?
You told me one was 12 , AC = 12
and I found half of the other for you
I said BE = 8, so BD = 16
and I found half of the other for you
I said BE = 8, so BD = 16
Ok, I understand that. But I am confused as to how you found the second diagonal (BD)?