Rhombus ABCD has diagonals AC = 6 and BD = 8 that intersect at point E. Find the perimeter of rhombus ABCD.
To find the perimeter of rhombus ABCD, we need to know the length of at least one side of the rhombus.
In a rhombus, the diagonals bisect each other at right angles. This means that segment AE is equal to segment CE, and segment BE is equal to segment DE.
Let's denote the length of one of the sides of the rhombus as 's'.
Since the diagonals bisect each other and are perpendicular, we can form four right-angled triangles: ABE, BCE, CDE, and DAE.
Using the Pythagorean theorem, we can find the length of each side of the rhombus in terms of the diagonals:
In triangle ABE:
AB² + AE² = BE² (applying the Pythagorean theorem)
s² + s² = 4² (substituting AB = s and BE = 8)
2s² = 16
s² = 8
s = √8 = 2√2
Similarly, in triangles BCE, CDE, and DAE, we will get the same length for each side of the rhombus: 2√2.
Now, since a rhombus has four equal sides, the perimeter will be:
Perimeter = 4 * s = 4 * 2√2 = 8√2
Therefore, the perimeter of rhombus ABCD is 8√2.
The diagonals of a rhombus intersect each other at right angles.
So look at your diagram, each of the 4 triangles is a right-angled triangle
with short sides 3 and 4. Do yo recognize the 3-4-5 triangle??
then the sides of the rhombus are 5 units each
and the perimeter is 20 units