Two sides of a rhombus OABC lying in the first quadrant along the lines $y=x/\sqrt 3$ and $y=x\sqrt 3$. If the area of the rhombus is 2 sq units then

Question

Prove that length of longer diagonal is $\sqrt 2(\sqrt 3+1)$ and

Length of the side of rhombus is 2 unit

The lines will form an angle of 60 and 30 with the positive x axis

B lies opposite to O (origin) while A lies lower wrt C (hope this clears up the figure I have in mind)

According to the question $$OB.AC=4$$

Let P be the point of intersection of the diagonals

OB will bisect the angle AOC and will form 45 with the positive x axis

Also angle COP will be 15

Then $\tan 15=\frac{PC}{OP}$ $$\tan 15=\frac{AC}{OB}$$

Putting this in the above equation $$OB^2=\frac{4}{\tan 15}$$

$$OB=\frac{2}{\sqrt{2-\sqrt 3}}$$

This isn’t matching. What is going wrong?

Answer 1

Note that

$$OB=\frac{2}{\sqrt{2-\sqrt 3}}=\frac{2\sqrt2}{\sqrt{4-2\sqrt 3}} =\frac{2\sqrt2}{\sqrt{(\sqrt3-1)^2}} =\frac{2\sqrt2}{\sqrt3-1}=\sqrt2(\sqrt3+1)$$

which does match.

Answer 2

Let OABC be the rhombus with point A on line $y=\sqrt{3} x$ is A$(a , \sqrt{3}a)$ and from symmetry about y=x other vertex is C$(\sqrt{3}a,a)$ and diagonals are OB : y-x=0 and AC: $x+y=(\sqrt{3}+1)a$.let diagonals intersect at M $$ $$ Length OM $=\vert \frac{(\sqrt{3}+1)a}{\sqrt{2}} \vert$ $$ $$ length AM $=\vert \frac{(\sqrt{3}-1)}{\sqrt{2}} \vert$ $$ $$ Area of triangle OAM =$\frac{1}{2}(OM)(AM)=1$ $$a^2=2\,\,\,\,\therefore a=\sqrt{2}$$ Hence longest diagonal OB is $a(\sqrt{3}+1)=\sqrt{2}(\sqrt{3}+1)$