5.3. The lengths of the diagonals $A C$ and $B D$ of a rhombus are 15 and 8 cm. The first diagonal is directed along the $O x$ axis, and the second along the $O y$ axis. Form the equations of the sides of the rhombus and find the distance from the origin to a side of the rhombus.
5.3. According to the condition, $A C=15 \text{~cm}, B D=8 \text{~cm}$, from which we find the coordinates of the vertices of the rhombus: $A(-7.5 ; 0), B(0 ; 4), C(7.5 ; 0), D(0 ;-4)$ (Fig. 5.3) The slope of the line $B C$ is $k_{1}=\operatorname{tg} \angle B C N=$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-599.jpg?height=338&width=464&top_left_y=106&top_left_x=45)

Fig. 5.3

$=-\operatorname{tg} \angle B C O=-\frac{4}{7.5}=-\frac{8}{15}$. Then, knowing the coordinates of point $B$, we form the equation of side $B C$: $y-4=-\frac{8}{15}(x-0)$, or $8 x+15 y-60=0$. The equation of side $A D$ will be found using its known slope $k_{1}=-\frac{8}{15}$ (since $\left.A D \| B C\right)$

and the coordinates of point $D: y+4=-\frac{8}{15}(x-0)$, or $8 x+15 y+60=0$. Next, the slope of the line $A B$ is $k_{2}=\operatorname{tg} \angle B A O=\frac{4}{7.5}=\frac{8}{15}$ and the equation of side $A B$ is written as $8 x-15 y+60=0$, and the equation of side $D C$ - as $8 x-15 y-60=0$. Now, we draw $O K \perp B C$ and to find $O K$ we use the fact that $S_{\triangle O B C}=\frac{1}{2} O C \cdot O B=\frac{1}{2} O K \cdot B C$, from which $O K=\frac{O C \cdot O B}{B C}$. Since $B C=\sqrt{7.5^{2}+4^{2}}=8.5$, we find $O K=\frac{7.5 \cdot 4}{8.5}=\frac{60}{17}(\text{~cm})$.

Answer: $8 x-15 y+60=0(A B), 8 x-15 y-60=0$ (DC), $8 x+$ $+15 y-60=0(B C), 8 x+15 y+60=0(A D) ; \frac{60}{17}(\text{~cm})$.