5. In an isosceles trapezoid $A B C D$ with lateral sides $A B$ and $C D$, the lengths of which are 10, perpendiculars $B H$ and $D K$ are drawn from vertices $B$ and $D$ to the diagonal $A C$. It is known that the bases of the perpendiculars lie on segment $A C$ and $A H: A K: A C=5: 14: 15$. Find the area of trapezoid $A B C D$.
Solution. Let $x = BH$, $y = DK$. From the similarity of right triangles $DKA$ and $BHC$, since $\angle BHC = \angle DAK$, we get

$$
\frac{x}{y} = \frac{CH}{AK} = \frac{10}{14} = \frac{5}{7}, \quad 5y = 7x, \quad y = \frac{7x}{5}.
$$

By the condition $AH : AK : AC = 5 : 14 : 15$, therefore $AH : CK = 5 : 1$. By the Pythagorean theorem,

$$
AH^2 = AB^2 - BH^2 = 100 - x^2, \quad CK^2 = CD^2 - DK^2 = 100 - y^2
$$

Thus, $\frac{25}{1} = \frac{AH^2}{CK^2} = \frac{100 - x^2}{100 - y^2}$. Since $y = \frac{7x}{5}$, substituting $y$ into the last equation, we find that $x = \sqrt{50}$, $y = \frac{7\sqrt{50}}{5}$, $AH = \sqrt{50}$, $AC = 3\sqrt{50}$, $S_{ABCD} = AC \cdot (DK + BH) / 2 = AC \cdot (x + y) / 2 = 180$.

Answer: 180. 6. Can the number $\left(x^2 + x + 3\right)^2 + \left(y^2 + y + 3\right)^2$ be a perfect square for some integers $x$ and $y$?

Solution. Since $x^2 + x + 3 = x(x + 1) + 3$, the value of each expression in parentheses is an odd number for any integers $x$ and $y$. The square of an odd number, when divided by 4, leaves a remainder of 1, since $(2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1$. Thus, the value of the given expression is an even number and leaves a remainder of 2 when divided by 4.

Suppose it is a perfect square, then it is the square of an even number. But the square of any even number is divisible by 4 - a contradiction.

Answer: No, it cannot.

Answer:

Attention! The problem is considered solved if, in addition to the correct answer, necessary explanations are provided.

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