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The distance between two parallel lines is unity. A point P lies between the lines at a distance a from one of them. Find the length of a side of an equilateral triangle $\mathrm{P Q R}$ , vertex $\mathrm{Q}$ of which lies on one of the parallel lines and vertex $\mathrm{R}$ lies on the other line.

Option: 1
$\mathrm{ r=2 \sqrt{a^2-a+1} }$

Option: 2
$\mathrm{r=\frac{1}{\sqrt{3}} \sqrt{a^2-a+1} }$

Option: 3
$\mathrm{r=\frac{2}{\sqrt{3}} \sqrt{a^2-a+1} }$

Option: 4
$\mathrm{r=\frac{2}{\sqrt{3}} \sqrt{a^2+a+1}}$

Answers (1)

best_answer

$\mathrm{Let}$ $\mathrm{ \mathrm{PQ}=\mathrm{QR}=\mathrm{RP}=\mathrm{r} }$

$\mathrm{ \text { and } \quad \angle \mathrm{PQL}=\theta }$

$\mathrm{ \text { then } \quad \angle \mathrm{XQR}=\theta+60^{\circ} }$

$\mathrm{ \text { Given } P L=a \text { and } R N=1 \text { unit } }$

$\text { In } \quad \triangle \mathrm{PQL}, \sin \theta=\frac{\mathrm{PL}}{\mathrm{QP}}=\frac{\mathrm{a}}{\mathrm{r}}$

$\therefore \quad \mathrm{a}=\mathrm{r} \sin \theta$ .............(i)

$\mathrm{ \text { and in } \triangle Q R N \text {, } }$

$\mathrm{\sin \left(\theta+60^{\circ}\right)=\frac{\mathrm{RN}}{\mathrm{QR}}=\frac{1}{\mathrm{r}} }$
$\mathrm{\therefore \quad r \sin \left(\theta+60^{\circ}\right)=1 }$

$\mathrm{ r\left(\sin \theta \cos 60^{\circ}+\cos \theta \sin 60^{\circ}\right\}=1 }$

$\mathrm{ \Rightarrow \quad r\left\{\frac{1}{2} \sin \theta+\frac{\sqrt{3}}{2} \cos \theta\right\}=1 }$

$\mathrm{\Rightarrow \quad r\left\{\frac{1}{2} \times \frac{a}{r}+\frac{\sqrt{3}}{2} \times \sqrt{1-\frac{a^2}{r^2}}\right\}=1 }$

$\Rightarrow \quad \frac{a}{2}+\frac{\sqrt{3}}{2} \sqrt{\left(\mathrm{r}^2-\mathrm{a}^2\right)}=1$
$\Rightarrow \quad \frac{\sqrt{3}}{2} \sqrt{\left(\mathrm{r}^2-\mathrm{a}^2\right)}=1-\frac{\mathrm{a}}{2}$
$\text { or } \quad \frac{3}{4}\left(\mathrm{r}^2-\mathrm{a}^2\right)=1+\frac{\mathrm{a}^2}{4}-\mathrm{a}$
$\Rightarrow \quad 3 \mathrm{r}^2-3 \mathrm{a}^2=4+\mathrm{a}^2-4 \mathrm{a}$
$\Rightarrow \quad 3 \mathrm{r}^2=4 \mathrm{a}^2-4 \mathrm{a}+4=4\left(\mathrm{a}^2-\mathrm{a}+1\right)$
$\therefore \quad \mathrm{r}=\frac{2}{\sqrt{3}} \sqrt{\left(\mathrm{a}^2-\mathrm{a}+1\right)} \text { units. }$

Hence option 3 is correct.

Posted by

Pankaj

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