Question

A rhombus has a side length equal to one of the diagonals. The other diagonal is 4cm longer. Find the area of the rhombus.

Answer 1

Answer:

The area of the rhombus is 25.83 cm².

Step-by-step explanation:

The area of a rhombus is given by:

$A = \frac{d_{1} \times d_{2}}{2}$

Where:      

d₁: is one diagonal      

d₂: is the other diagonal = d₁ + 4 cm

We know that one side length of the rhombus is equal to d₁. We can imagine a right triangle inside the rhombus, with the following dimensions:

h: hypotenuse of the right triangle  

a: one side of the right triangle

b: is the other side of the right triangle  

From the above we know that:

h = d₁                                

$a = \frac{d_{2}}{2} = \frac{d_{1} + 4}{2}$

$b = \frac{d_{1}}{2}$            

We can find d₁ with Pitagoras:

$h^{2} = a^{2} + b^{2}$  

$d_{1}^{2} = (\frac{d_{1} + 4}{2})^{2} + (\frac{d_{1}}{2})^{2}$

$d_{1}^{2} = \frac{1}{4}(d_{1}^{2} + 8d_{1} + 16 + d_{1}^{2})$

By solving the above quadratic equation for d₁ and taking the positive solution we have:

$d_{1} = 5.46 cm$    

So, d₂ is:

$d_{2} = d_{1} + 4 = 5.46 cm + 4 cm = 9.46 cm$

Now, we can find the area:

$A = \frac{d_{1} \times d_{2}}{2} = \frac{5.46 cm \times 9.46 cm}{2} = 25.83 cm^{2}$

Therefore, the area of the rhombus is 25.83 cm².

I hope it helps you!