Understanding Parallelogram Properties and Diagonals

The question asks for the sum of the squares of the diagonals of a parallelogram given the lengths of its two adjacent sides. A parallelogram is a quadrilateral with two pairs of parallel sides. In a parallelogram, opposite sides are equal in length, and opposite angles are equal. The diagonals of a parallelogram bisect each other.

Applying the Parallelogram Law for Diagonals

To find the sum of the squares of the diagonals, we can use a property specific to parallelograms known as the Parallelogram Law (also sometimes called the parallelogram identity or parallelogram rule). This law relates the lengths of the sides of a parallelogram to the lengths of its diagonals.

The Parallelogram Law states that the sum of the squares of the lengths of the two diagonals of a parallelogram is equal to the sum of the squares of the lengths of all four sides of the parallelogram.

Let the lengths of the adjacent sides of the parallelogram be $a$ and $b$. Since opposite sides are equal, the lengths of the four sides are $a, b, a, b$. Let the lengths of the diagonals be $d_1$ and $d_2$.

According to the Parallelogram Law, the relationship is:

$$d_1^2 + d_2^2 = a^2 + b^2 + a^2 + b^2$$

This simplifies to:

$$d_1^2 + d_2^2 = 2(a^2 + b^2)$$

Calculating the Sum of Squares of Diagonals

We are given the lengths of the two sides of the parallelogram as 3 cm and 10 cm. Let's set $a = 3$ cm and $b = 10$ cm.

Now, we can substitute these values into the Parallelogram Law formula:

$$d_1^2 + d_2^2 = 2(a^2 + b^2)$$

$$d_1^2 + d_2^2 = 2((3 \text{ cm})^2 + (10 \text{ cm})^2)$$

First, calculate the squares of the side lengths:

• $(3 \text{ cm})^2 = 3 \times 3 \text{ cm}^2 = 9 \text{ cm}^2$
• $(10 \text{ cm})^2 = 10 \times 10 \text{ cm}^2 = 100 \text{ cm}^2$

Now, substitute these values back into the formula:

$$d_1^2 + d_2^2 = 2(9 \text{ cm}^2 + 100 \text{ cm}^2)$$

Add the squares of the side lengths:

$$d_1^2 + d_2^2 = 2(109 \text{ cm}^2)$$

Finally, multiply by 2:

$$d_1^2 + d_2^2 = 218 \text{ cm}^2$$

The sum of the squares of the diagonals of the parallelogram is 218 cm$^2$. This value represents the combined squared lengths of the two diagonals.

Summary of Calculation Steps

The calculated sum of the squares of the diagonals is 218 cm$^2$. This corresponds to one of the given options.

Revision Table: Parallelogram Properties

Additional Information: The Parallelogram Law in Detail

The Parallelogram Law is a geometric theorem that is a direct consequence of the law of cosines. Consider a parallelogram with sides of length $a$ and $b$, and diagonals of length $d_1$ and $d_2$. Let one angle of the parallelogram be $\theta$ and the adjacent angle be $180^\circ - \theta$.

Applying the law of cosines to the triangle formed by sides $a$, $b$, and diagonal $d_1$ (which connects the vertices with angle $\theta$ between sides $a$ and $b$):

$$d_1^2 = a^2 + b^2 - 2ab \cos(\theta)$$

Applying the law of cosines to the triangle formed by sides $a$, $b$, and diagonal $d_2$ (which connects the vertices with angle $180^\circ - \theta$ between sides $a$ and $b$):

$$d_2^2 = a^2 + b^2 - 2ab \cos(180^\circ - \theta)$$

Since $\cos(180^\circ - \theta) = -\cos(\theta)$, the second equation becomes:

$$d_2^2 = a^2 + b^2 - 2ab (-\cos(\theta))$$

$$d_2^2 = a^2 + b^2 + 2ab \cos(\theta)$$

Now, sum the squares of the diagonals:

$$d_1^2 + d_2^2 = (a^2 + b^2 - 2ab \cos(\theta)) + (a^2 + b^2 + 2ab \cos(\theta))$$

$$d_1^2 + d_2^2 = a^2 + b^2 - 2ab \cos(\theta) + a^2 + b^2 + 2ab \cos(\theta)$$

The terms $-2ab \cos(\theta)$ and $+2ab \cos(\theta)$ cancel out:

$$d_1^2 + d_2^2 = a^2 + b^2 + a^2 + b^2$$

$$d_1^2 + d_2^2 = 2a^2 + 2b^2$$

$$d_1^2 + d_2^2 = 2(a^2 + b^2)$$

This confirms the Parallelogram Law formula used in the solution. It's a fundamental property in Euclidean geometry and is particularly useful when dealing with side and diagonal lengths of parallelograms, including special cases like rectangles, rhombuses, and squares.