Question

A rhombus has an area equal to half the sum of the areas of the squares built on its four sides. The ratio of the long diagonal to the short diagonal is:

• A. $2+\sqrt{3}$
• B. $2-\sqrt{3}$
• C. $\frac{1}{2}$
• D. Can't be determined

Answer 1

Answer: (D)

Let the side of the rhombus be $x$.

Then, the area of the square on its side will be $x^2$

And, the sum of area of $4$ squares on its side will be $4x^2$

So, the area of rhombus $=\frac{4x^2}{2}=2x^2$

Let the diagonals be $d_1$ and $d_2$

Hence the area $=\frac{1}{2}(d_1\times d_2)=2x^2$

$\Rightarrow d_1\times d_2=4x^2...\left(1\right)$

And the diagonal of rhombus bisect each other at ${90}^{\circ }$, hence they form a right triangle.

So, by Pythagoras theorem

$(\frac{d_1}{2}{)}^2+(\frac{d_2}{2}{)}^2=x^2$

$\Rightarrow d_1^2+d_2^2=4x^2...\left(2\right)$

Hence, from $\left(1\right)$ and $\left(2\right)$, we cannot get a direct relation between the diagonals.

So, $\text{D}$ is the correct option.