Question

1. the window of a building is in the shape of a rhombus. the area covered by a window is 1.8 m². one of the diagonals of the window is 3.0 m long. calculate the a. length of the window. b. height of the window.

Explanation:

Step1: Recall area formula of rhombus

The area formula of a rhombus is $A=\frac{1}{2}d_1d_2$, where $A$ is the area, $d_1$ and $d_2$ are the lengths of the diagonals.

Step2: Substitute known values

We know that $A = 1.8\ m^{2}$ and $d_1=3.0\ m$. Substitute into the formula $1.8=\frac{1}{2}\times3.0\times d_2$.

Step3: Solve for $d_2$

First, simplify the right - hand side of the equation: $\frac{1}{2}\times3.0\times d_2 = 1.5d_2$. Then, solve the equation $1.8 = 1.5d_2$ for $d_2$. Divide both sides by 1.5: $d_2=\frac{1.8}{1.5}=1.2\ m$.

However, the problem asks for length and height of the window. For a rhombus, if we assume the diagonals are $d_1$ and $d_2$, and the side length $s$ can be found using the fact that the diagonals of a rhombus bisect each other at right - angles. Let the half - lengths of the diagonals be $a=\frac{d_1}{2}$ and $b = \frac{d_2}{2}$. Here $a=\frac{3.0}{2}=1.5\ m$ and $b=\frac{1.2}{2}=0.6\ m$. Using the Pythagorean theorem $s=\sqrt{a^{2}+b^{2}}=\sqrt{1.5^{2}+0.6^{2}}=\sqrt{2.25 + 0.36}=\sqrt{2.61}\approx1.62\ m$.

If we consider the base as the diagonal $d_1 = 3.0\ m$, then using the area formula $A = base\times height$, and since $A = 1.8\ m^{2}$ and base $=3.0\ m$, the height $h=\frac{A}{base}=\frac{1.8}{3.0}=0.6\ m$.

Answer:

Length (side - length) of the window is approximately $1.62\ m$ and height of the window is $0.6\ m$.