Question

The length of a side of a square field is 4 m. what will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{Length of the square field = 4 m} \\ \therefore \mathrm{A}{\text{rea of the square field = 4×4 = 16 m}}^2 \\ \mathrm{Given}\text{: Area of the rhombus = Area of the square field} \\ \text{Length of one diagonal of the rhombus = 2 m} \\ \therefore \mathrm{S}\text{ide of the rhombus=}\frac{1}{2}\sqrt{d_1^2+d_2^2} \\ \text{And, area of the rhombus=}\frac{1}{2}\times (d_1\times d_2) \\ \therefore A\text{rea:} \\ 16=\frac{1}{2}(2\times d_2) \\ {d}_2\text{=16 m} \\ \text{Now, we need to find the length of the side of the rhombus.} \\ \therefore \text{Side of the rhombus =}\frac{1}{2}\sqrt{2^2+{16}^2}\text{=}\frac{1}{2}\sqrt{260}\text{=}\frac{1}{2}\sqrt{4\times 65}\text{=}\frac{1}{2}\text{×2}\sqrt{65}\text{=}\sqrt{65}\text{m} \\ \text{Also, we know: Area of the rhombus = Side × Altitude} \\ \therefore \text{16=}\sqrt{65}\text{×Altitude} \\ \text{Altitude=}\frac{16}{\sqrt{65}}\text{m} \end{gathered}$$