Finding the Sum of Diagonals of a Rhombus Given Side and Area

Let's analyze the properties of a rhombus to solve this problem. A rhombus is a quadrilateral where all four sides are equal in length. Its diagonals bisect each other at right angles. This property is key to relating the side length to the diagonals using the Pythagorean theorem.

We are given:

• Length of one side of the rhombus ($a$) = 61 cm
• Area of the rhombus ($A$) = 1320 cm²

We need to find the sum of the lengths of its diagonals. Let the lengths of the two diagonals be $d_1$ and $d_2$.

There are two important formulas relating the side, area, and diagonals of a rhombus:

1. Area ($A$) = $\frac{1}{2} \times d_1 \times d_2$
2. Relation between side and diagonals: Since the diagonals bisect each other at right angles, they form four right-angled triangles within the rhombus. The legs of each right triangle are half the lengths of the diagonals ($\frac{d_1}{2}$ and $\frac{d_2}{2}$), and the hypotenuse is the side length ($a$) of the rhombus. By the Pythagorean theorem: $a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2$. This simplifies to $a^2 = \frac{d_1^2}{4} + \frac{d_2^2}{4}$, which can be rewritten as $4a^2 = d_1^2 + d_2^2$.

Using the given information, we can set up equations based on these formulas.

From the area formula:

$\qquad 1320 = \frac{1}{2} d_1 d_2$

Multiplying both sides by 2, we get:

$\qquad d_1 d_2 = 2 \times 1320$

$\qquad d_1 d_2 = 2640 \quad \text{(Equation 1)}$

From the relation between side and diagonals:

$\qquad d_1^2 + d_2^2 = 4a^2$

Substitute the side length $a = 61$ cm:

$\qquad d_1^2 + d_2^2 = 4 \times (61)^2$

$\qquad d_1^2 + d_2^2 = 4 \times 3721$

$\qquad d_1^2 + d_2^2 = 14884 \quad \text{(Equation 2)}$

We need to find the sum of the lengths of the diagonals, which is $d_1 + d_2$. We know an algebraic identity related to squares and products:

$\qquad (d_1 + d_2)^2 = d_1^2 + d_2^2 + 2d_1 d_2$

Now, we can substitute the values from Equation 1 and Equation 2 into this identity:

$\qquad (d_1 + d_2)^2 = 14884 + 2 \times 2640$

$\qquad (d_1 + d_2)^2 = 14884 + 5280$

$\qquad (d_1 + d_2)^2 = 20164$

To find $d_1 + d_2$, we take the square root of both sides:

$\qquad d_1 + d_2 = \sqrt{20164}$

Calculating the square root:

$\qquad \sqrt{20164} = 142$

So, the sum of the lengths of the diagonals is 142 cm.

Let's verify the steps and calculations:

• Area formula used correctly: $A = \frac{1}{2} d_1 d_2$.
• Pythagorean relation used correctly: $a^2 = (\frac{d_1}{2})^2 + (\frac{d_2}{2})^2 \implies 4a^2 = d_1^2 + d_2^2$.
• Algebraic identity used correctly: $(d_1 + d_2)^2 = d_1^2 + d_2^2 + 2d_1 d_2$.
• Calculations: $2 \times 1320 = 2640$, $61^2 = 3721$, $4 \times 3721 = 14884$, $2 \times 2640 = 5280$, $14884 + 5280 = 20164$, $\sqrt{20164} = 142$. All seem correct.

The sum of the lengths of the diagonals is 142 cm.

Revision Table: Rhombus Properties and Formulas

Additional Information: Understanding Rhombus Geometry

A rhombus is a special type of parallelogram where all sides are equal. Because it's a parallelogram, its opposite angles are equal, and consecutive angles are supplementary. However, unlike a square (which is a rhombus with right angles), the angles of a rhombus are not necessarily 90 degrees. The diagonals of a rhombus not only bisect each other at right angles but also bisect the angles of the rhombus. This geometric understanding helps in visualizing and deriving the formulas used in this problem. The fact that the diagonals divide the rhombus into four congruent right-angled triangles is fundamental to using the Pythagorean theorem to relate the side length to the diagonals.