The diagonals of a rhombus measure 10 and 14. Find the length of one side of the rhombus.?

The diagonals of rhombus bisect each other at right angle. So the two halves of perpendicular diagonal and one side of rhombus, makes a rt angled triangle.

Hence, side^2 = (half of first diagonal)^2 + (half of second diagnoal)^2

S^2 = 5^2 + 7^2 = 25+49= 74

S = sqrt(74).

tricky devils, ain't dey?

First of all, draw a sketch. you can easily prove all the triangles are congruent, since all sides of a rhombus are equal. Prove it by SSS.

Consequently the central angles are all equal. Since the adjacent angles are supplementary (180) and equal, they must be 90 degrees.

If they are 90 degrees, then you have right triangles, and can use pythagoras's theroem.

You only have to go through the above if you do not recall that the diagonals of a rhombus are the perpendicular bisectors of each other.

If the diagonals measure 10 and 14, then the right triangle has legs of 7 and 5, so the hypotenuse is

sqrt (7^2 + 5^2) = sqrt (49 + 25) = sqrt (74)

always,

tony

5^2 + 7^2 = s^2

25 + 49 = s^2

64 = s^2

8 = s