find the area of rhombus if each side measures 5 cm and the shorter diagonal is 6 cm.
so I see with the diagnol and sides, two triangles, of sides, 5,5,6
let s = 1/2 perimeter=16/2=8cm
total area=2*sqrt(s(s-5)(s-5)(s-6))
= 2(3)sqrt(8*2)=6*4=24cm^2
To find the area of a rhombus, you can use the formula:
Area = (d1 * d2) / 2
Where d1 and d2 are the lengths of the diagonals.
In this case, you are given the length of one diagonal (6 cm) and the length of each side (5 cm). To find the length of the other diagonal, you can use the properties of a rhombus.
In a rhombus, the diagonals are perpendicular bisectors of each other, meaning they cut each other at a 90-degree angle and bisect each other. This also means that the diagonals divide the rhombus into four congruent right triangles.
To find the length of the other diagonal, you can apply the Pythagorean theorem to one of these right triangles. Let's call the length of the other diagonal d2.
Using the given side length of 5 cm, one leg of the right triangle is half of that, which is 2.5 cm. The hypotenuse is the length of one diagonal, given as 6 cm.
Using the Pythagorean theorem:
(2.5)^2 + (d2/2)^2 = (6)^2
Simplifying the equation:
6.25 + (d2^2 / 4) = 36
Multiply the equation by 4 to eliminate the fraction:
25 + d2^2 = 144
Subtract 25 from both sides:
d2^2 = 119
Take the square root of both sides to find d2:
d2 = √119
Now that we have the values for both diagonals, we can calculate the area:
Area = (d1 * d2) / 2
Substituting the values:
Area = (6 * √119) / 2
Area = 3√119 square cm