# Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. [closed]

Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides.the length of each perpendicular is equal to a units.The distance between their feet being equal to b units.Find the area of the rhombus

• How did you try to solve this problem? Jan 22, 2017 at 19:44

## 1 Answer

Set the $A,B,C,D \,\,$ vertices of the rhombus in $(m,0)$, $(0,n)\,\,$, $(-m,0)\,\,$, and $(0, -n) \,\,$, respectively. The slope of the side $AB\,\,$ is $-n/m \,\,$. So the slope of the perpendicular to $AB\,\,$, drawn from its opposite vertex $C$, is $m/n\,\,$. Since this perpendicular line passes through $(-n,0) \,\,$, its equation is $y=m/n x +m^2n \,\, \,\,$. The $(x_0,y_0 ) \,\,$ coordinates of the crossing point between this perpendicular line and $AB$ (let us call this foot point $E$) can be found by solving

$$m/n x +m^2/n=-n/m x +n$$

which gives

$$x_0=\frac {m (n^2-m^2)}{n^2+m^2}$$

and then

$$y_0=\frac {2n m^2}{n^2+m^2}$$

Since the distance between $E$ and $C$ is $a$, we have

$$\left ( \frac {m (n^2-m^2)}{(n^2+m^2)}+m \right)^2+\frac {4n^2 m^4}{(n^2+m^2)^2}=a^2$$

which reduces to

$$\frac {4n^2 m^2}{n^2+m^2}=a^2$$

Also note that, as a result of the symmetry of the problem, the distance between $E$ and its symmetric point on side $CD\,\,$ (i.e. the foot of the perpendicular drawn from vertex $A$ to $CD\,\,$) is twice the distance between $E$ and the origin. Since this distance is given as equal to $b$, we have

$$\frac {m^2 (n^2-m^2)^2}{(n^2+m^2)^2}+\frac {4n^2 m^4}{(n^2+m^2)^2}=\frac {b^2}{4}$$

which reduces to

$$m^2=\frac {b^2}{4}$$

and then $m=b/2 \,\,$. Substituting in the equation above, this also leads to

$$\frac{4n^2 b^2/4}{n^2+b^2/4}=a^2$$

$$n=\frac {ab}{2\sqrt {b^2-a^2}}$$

Since the area $A$ of the rhombus is $2mn \,\,$, it follows

$$A= 2 \frac {b}{2} \cdot \frac{ab}{2\sqrt {b^2-a^2}}$$ $$=\frac {ab^2}{2 \sqrt {b^2-a^2}}$$

Note that, when $n=m \,\,$, that is to say the rhombus is a square, $a$ corresponds to the side of the square and $b$ to its diagonal, so in this case $a=\sqrt {2}n \,\,$ and $b=2n \,\,$. Accordingly, the formula reduces to $A=2n^2\,\,$, which is the area of a square with vertices in $(n,0)\,\,$, $(0,n) \,\,$, $(-n,0) \,\,$, $(0, -n) \,\,$.

• i too solved it Jan 23, 2017 at 6:11
• Good. Could you edit your question by showing what you have done.? Jan 23, 2017 at 6:53