Question

A rhombus has sides \[8{\text{ }}cm\] in length and its shortest diagonal is \[{\text{10 }}cm\] in length. What is the area of the rhombus to the nearest \[c{m^2}\]?

Answer 1

Hint: To solve this question, we will first find the length of another diagonal using the given length of the side and length of the given diagonal, by applying Pythagoras theorem. Then we know that the area of a rhombus is equal to \[\left( {\dfrac{1}{2} \times {d_1} \times {d_2}} \right)\], where \[{d_1}\] and \[{d_2}\] are the diagonals of the rhombus. Using this and by further calculation, we will calculate the area.

Complete step by step answer:
Given, a rhombus has sides \[8{\text{ }}cm\] in length and its shortest diagonal is \[{\text{10 }}cm\] in length.

As we know that area of a rhombus is equal to \[\left( {\dfrac{1}{2} \times {d_1} \times {d_2}} \right)\], where \[{d_1}\] and \[{d_2}\] are the diagonals of the rhombus.The diagonals of a rhombus bisect each other at \[{90^ \circ }\]. By applying the Pythagoras theorem in \[\vartriangle AOB\], we get
\[ \Rightarrow A{B^2} = A{O^2} + B{O^2}\]

Putting the values, we get
\[ \Rightarrow {\left( 8 \right)^2} = A{O^2} + {\left( 5 \right)^2}\]
\[ \Rightarrow 64 = A{O^2} + 25\]
On simplification, we get
\[ \Rightarrow A{O^2} = 64 - 25\]
\[ \Rightarrow A{O^2} = 39\]
Taking square root both sides, we get
\[ \Rightarrow AO = \sqrt {39} \]
\[\therefore BD = 2 \times AO\]
On putting the value, we get
\[ \Rightarrow BD = 2 \times \sqrt {39} \]
\[ \Rightarrow BD = 2\sqrt {39} \]

Therefore, we get the two diagonals \[{d_1}\] and \[{d_2}\] as \[{\text{10 }}cm\] and \[2\sqrt {39} {\text{ }}cm\].
\[ \Rightarrow \] Area of the rhombus \[ = \dfrac{1}{2} \times {d_1} \times {d_2}\]
Putting the obtained values, we get
\[ \Rightarrow \] Area of the rhombus \[ = \dfrac{1}{2} \times 10 \times 2\sqrt {39} \]
On simplifying, we get
\[ \Rightarrow \] Area of the rhombus \[ = 10\sqrt {39} \]
On calculating, we get
\[ \therefore \] Area of the rhombus \[ = 62.45\,c{m^2}\]

Therefore, the area of the rhombus to the nearest \[c{m^2}\] is \[62.45{\text{ }}c{m^2}\].

Note: In a rhombus, all sides are of equal length. The diagonal of a rhombus divides it into two congruent triangles and the diagonals of a rhombus bisect each other at \[{90^ \circ }\]. We can calculate the height and the base of one of these triangles and multiply the result by two to get the area of the rhombus.