Question

Find the area of a rhombus with sides of 8 in, and one diagonal is 8 in.

Answer 1

Note: The diagonals of a rhombus bisect each other at right angle

`\begin{gathered} \text{ The area of a rhombus = }(1)/(2)\text{ x d1 x d2} \\ \text{where d1 and d2 are the lengths of the two diagonals} \end{gathered}`

We need to calculate the length of the other diagonal, that is diagonal AC.

E is the midpoint of AC

We can use the Pythagoras theorem to find the value of m

`\begin{gathered} \text{ In }\Delta BEC,\text{ } \\ 8^2=4^2+m^2 \\ m^2=8^2-4^2 \\ m^2=64-16 \\ m^2=48 \\ m=\sqrt[]{48} \\ m=4\sqrt[]{3}\text{ in} \\ \text{But diagonal }AC\text{ = 2 x m = 2 x 4}\sqrt[]{3\text{ }}=8\sqrt[]{3}\text{ in} \end{gathered}`
`\begin{gathered} \text{ Area of the rhombus = }(1)/(2)\text{ x AC x BD} \\ =(1)/(2)\text{ x 8}\sqrt[]{3}\text{ x 8} \\ =32\sqrt[]{3}in^2\text{ } \\ or\text{ in decimal} \\ \text{Area = 55.43 in}^2 \end{gathered}`