The diagonals of a rhombus have lenghts 16 and 30. Find the perimeter of the rhombus.
The sides, which are equal since it is a rhombus, have a length dimension of sqrt (8^2 + 15^2) = sqrt(289) = 17, because the diagonals are perpendicular. The perimeter is therefore 4x17 = 68
why would you add (8^2 + 15^2)
I am applying the Pythagorean theorem
a^2 + b^2 = c^2
c is the length of a side
a and b are half the lengths of the two diagonals. Draw yourself a picture with intersetinc diagonals and you will see the four congruent right triangles. Make sure you do it for a rhombus, which is a parallelogram with all sides equal.
To find the perimeter of a rhombus, we need to know the side length of the rhombus.
In a rhombus, the diagonals are perpendicular bisectors of each other. This means that when the diagonals intersect, they divide each other into two equal parts.
Let's label the length of one half of each diagonal as 'a' and 'b'. So we have 'a' and 'b' representing the lengths of the two halves of the diagonal that intersect at a right angle.
Therefore, in our case, we have:
a = 16/2 = 8
b = 30/2 = 15
Using the Pythagorean theorem, we can find the length of each side of the rhombus.
Using one half of each diagonal as the two legs of a right triangle, the hypotenuse (side length of the rhombus) can be found using the formula:
side length = √(a² + b²)
In our case:
side length = √(8² + 15²)
= √(64 + 225)
= √289
= 17
Now that we know the side length, we can find the perimeter of the rhombus.
Perimeter of a rhombus = 4 * side length
Perimeter = 4 * 17 = 68
Therefore, the perimeter of the rhombus is 68 units.