If one diagonal of a rhombus is equal to its side, determine the ratio of the diagonals of the rhombus.

Question

Accepted Answer

To solve this problem, we use the properties of a rhombus. The diagonals of a rhombus bisect each other at right angles and divide it into four congruent right-angled triangles. Let the side length of the rhombus be $$a$$, and let the diagonals be $$d_1$$ and $$d_2$$.

Given that one diagonal $$d_1$$ is equal to the side length $$a$$, we denote half of this diagonal as $$\frac{d_1}{2} = \frac{a}{2}$$. The other diagonal, $$d_2$$, is unknown, so we denote half of it as $$\frac{d_2}{2}$$.

Using the Pythagorean theorem in one of the right-angled triangles formed by the diagonals and the side:  
$$
a^2 = \left(\frac{d_1}{2}ight)^2 + \left(\frac{d_2}{2}ight)^2
$$

Substitute $$d_1 = a$$ into the equation:  
$$
a^2 = \left(\frac{a}{2}ight)^2 + \left(\frac{d_2}{2}ight)^2
$$

Simplify:  
$$
a^2 = \frac{a^2}{4} + \frac{d_2^2}{4}
$$

Multiply through by 4 to eliminate fractions:  
$$
4a^2 = a^2 + d_2^2
$$

Rearrange to solve for $$d_2^2$$:  
$$
d_2^2 = 4a^2 - a^2 = 3a^2
$$

Take the square root of both sides:  
$$
d_2 = \sqrt{3}a
$$

Now, calculate the ratio of the diagonals:  
$$
	ext{Ratio} = \frac{d_1}{d_2} = \frac{a}{\sqrt{3}a} = \frac{1}{\sqrt{3}} = 1:\sqrt{3}
$$

Thus, the ratio of the diagonals is **1:√3**, making **Option a)** correct.

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