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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

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  • $\begingroup$ How did you try to solve this problem? $\endgroup$
    – hardmath
    Jan 22, 2017 at 19:44

1 Answer 1

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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) \,\, $.

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  • $\begingroup$ i too solved it $\endgroup$
    – Pole_Star
    Jan 23, 2017 at 6:11
  • $\begingroup$ Good. Could you edit your question by showing what you have done.? $\endgroup$
    – Anatoly
    Jan 23, 2017 at 6:53

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