Area of a Rhombus: Using Pythagoras' theorem

Remember there are two options to calculate the area of a rhombus:

1: The diagonal multiplied by the diagonal divided by 2.

2: The base multiplied by the height.

In the question, we are only given the data for one of the diagonals and one of the sides, which means we cannot use either of the above formulas.

We need to find more data. Let's begin by finding the second diagonal:

Remember that the diagonals of a rhombus are perpendicular to one another, which means that they form a 90-degree angle.

Therefore, all the triangles in a rhombus are right-angled.

Now we can focus on the triangle where the side and the height are given, and we will calculate the third side using the Pythagorean theorem:

$a²+b²=c²$Insert the given data:

$3^2+x^2=5^2$$9+x^2=25$$x^2=25-9=16$$x=\sqrt{16}=4$

Now that we have found the second half of the diagonal, we can calculate the area of the rhombus by multiplying the two diagonals together.

Since the diagonals in a rhombus are perpendicular and cross each other, they are equal. Hence, our diagonals are equal:

$3+3=6$$4+4=8$Therefore, the area of the rhombus is:

$\frac{6\times8}{2}=\frac{48}{2}=24$