The diagonals of a rhombus (not rombhus) bisect one another at right angles. The sides of the rhombus form the hypotenuses of triangles whose other sides are half the diagonals. So use Pythagoras.
Find the midpoint of the two diagonals
Using the formula (x)(x-3)/2 = Diagonals ; simply replace the diagonals with the number of diagonals you're given. Then, you'll havev (x)(x-3)/2 = Diagonals. Simplify it, and you'll be given x(power of 2) - 3X = (2)(Diagonals). Subtract the amount of diagonals from both sides, and you'll have x(power of 2) - 3X - 2Diagonals = 0. From there, use the quadratic formula to find the number of sides the polygon has.
The answer depends on the measure of WHAT! Side length, angles, length of diagonals, area? And the answers to these depend on what information is given.
The diagonals bisect each other at right angles. So you could use Pythagoras on half the diagonals. So, if the diagonals are a and b units long, then half the diagonals are a/2 and b/2 units long. Then, by Pythagoras, the sides of the rhombus are sqrt[(a/2)2 + (b/2)2]
That depends on what you are given. The two diagonals will make four right triangles, and you can use the angles, along with other information given to find unknown values. See related link.
Find the area of a rhombs with diagonals that measure 8 and 10.
The diagonals are not equal in length but thet bisect each other at 90 degrees
1/2*(n2-3n) = diagonals when n equals number of sides.
nope.aviImproved Answer:-33 because 1/2*(332-99) = 495
It is the point where the diagonals cross.
They are of the same length (congruent).
Add the numbers.