In the parallelogram $ABCD$, let $AD=BC=4 \text{ cm}, AB=CD=7 \text{ cm}, BD=x, AC=x+2$, and the acute angle at vertex $A$ be $\alpha$. For triangles $ABD$ and $ABC$, we apply the cosine rule:

$$x^{2}=4^{2}+7^{2}-2 \cdot 4 \cdot 7 \cdot \cos \alpha, \quad \text{and} \quad (x+2)^{2}=4^{2}+7^{2}-2 \cdot 4 \cdot 7 \cos (180^{\circ}-\alpha)$$

Since $\cos \alpha = -\cos (180^{\circ}-\alpha)$, adding the corresponding sides of the two equations, we get:

$$x^{2}+(x+2)^{2}=2(4^{2}+7^{2})$$

Rearranging the equation, we obtain the quadratic equation:

$$x^{2}+2x-63=0$$

From this, we find $x=7$ (the other root is negative) and $x+2=9$ as the lengths of the two diagonals.

Fiedler Ágnes (Esztergom, Sz. István Gimn., 11th grade)

Remarks. Equation (1) can be written directly using the so-called parallelogram theorem, which states that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four sides. Several students (rightly) used this known theorem.

2. The problem can also be solved using the Pythagorean theorem or the Heron's formula. These solutions are much more complex.

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