#### 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?

#### Solution

Given:

Length of the square field = 4 m

∴ A {rea of the square field = 4 x 4 = 16 m}

Given: Area of the rhombus = Area of the square field

Length of one diagonal of the rhombus = 2 m

∴ Side of the rhombus \[=\frac{1}{2}\sqrt{d_1^2 + d_2^2}\]

And, area of the rhombus \[=\frac{1}{2} \times ( d_1 \times d_2 )\]

∴ Area:

\[16 = \frac{1}{2}(2 \times d_2 )\]

\[ d_2 =16 m\]

Now, we need to find the length of the side of the rhombus.

∴ Side of the rhombus \[=\frac{1}{2}\sqrt{2^2 + {16}^2}=\frac{1}{2}\sqrt{260}=\frac{1}{2}\sqrt{4 \times 65}=\frac{1}{2}\times2\sqrt{65}=\sqrt{65}m\]

Also, we know: Area of the rhombus = Side X Altitude

\[ \therefore 16=\sqrt{65}\times \] Altitude

Altitude \[=\frac{16}{\sqrt{65}}m\]

Length of the square field = 4 m

∴ A {rea of the square field = 4 x 4 = 16 m}

^{2}Given: Area of the rhombus = Area of the square field

Length of one diagonal of the rhombus = 2 m

∴ Side of the rhombus \[=\frac{1}{2}\sqrt{d_1^2 + d_2^2}\]

And, area of the rhombus \[=\frac{1}{2} \times ( d_1 \times d_2 )\]

∴ Area:

\[16 = \frac{1}{2}(2 \times d_2 )\]

\[ d_2 =16 m\]

Now, we need to find the length of the side of the rhombus.

∴ Side of the rhombus \[=\frac{1}{2}\sqrt{2^2 + {16}^2}=\frac{1}{2}\sqrt{260}=\frac{1}{2}\sqrt{4 \times 65}=\frac{1}{2}\times2\sqrt{65}=\sqrt{65}m\]

Also, we know: Area of the rhombus = Side X Altitude

\[ \therefore 16=\sqrt{65}\times \] Altitude

Altitude \[=\frac{16}{\sqrt{65}}m\]

Is there an error in this question or solution?

Solution 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? Concept: Area of a Polygon.