To solve the tasks from the 15th to the 20th problems, I'll transcribe and solve each one based on the text in the image.

Task 15:

"Dijagonala pravougaonika je 15 cm, a stranica a = 10 cm. Odredi drugu stranicu, obim i površinu."

1. The diagonal $d = 15 \, \text{cm}$ and one side $a = 10 \, \text{cm}$.
   • Using the Pythagorean theorem: $d^2 = a^2 + b^2 \implies 15^2 = 10^2 + b^2$ $225 = 100 + b^2 \implies b^2 = 125 \implies b = \sqrt{125} = 5\sqrt{5} \, \text{cm}$
   • Perimeter: $P = 2(a + b) = 2(10 + 5\sqrt{5}) \, \text{cm}$
   • Area: $A = a \cdot b = 10 \cdot 5\sqrt{5} \, \text{cm}^2$

Task 16:

"Dijagonala kvadrata je 12 cm. Odredi stranicu, obim i površinu."

1. The diagonal of a square $d = 12 \, \text{cm}$ relates to the side $a$ by: $d = a\sqrt{2} \implies a = \frac{d}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \, \text{cm}$
2. Perimeter: $P = 4a = 4 \cdot 6\sqrt{2} = 24\sqrt{2} \, \text{cm}$
3. Area: $A = a^2 = (6\sqrt{2})^2 = 72 \, \text{cm}^2$

Task 17:

"Osnovna jednakokrakog trougla je 10 cm, a krak je 13 cm. Odredi obim i površinu trougla."

1. Perimeter: $P = 10 + 13 + 13 = 36 \, \text{cm}$
2. To find the area, drop the height $h$ from the vertex to the base:
   • The height divides the base into two equal parts ($5 \, \text{cm}$ each).
   • Use the Pythagorean theorem: $h^2 + 5^2 = 13^2 \implies h^2 + 25 = 169 \implies h^2 = 144 \implies h = 12 \, \text{cm}$
   • Area: $A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 10 \cdot 12 = 60 \, \text{cm}^2$

Task 18:

"Osnovna jednakokrakog trougla je 25 cm. Odredi obim i visinu ako je krak 13 cm."

1. Perimeter: $P = 25 + 13 + 13 = 51 \, \text{cm}$
2. To find the height $h$, the height divides the base into two equal parts ($12.5 \, \text{cm}$ each).
   • Use the Pythagorean theorem: $h^2 + 12.5^2 = 13^2 \implies h^2 + 156.25 = 169 \implies h^2 = 12.75 \implies h = \sqrt{12.75} \approx 3.57 \, \text{cm}$

Task 19:

"Odredi visinu u jednakostraničnom trouglu ako je a = 6 cm."

1. The formula for the height $h$ of an equilateral triangle is: $h = \frac{\sqrt{3}}{2} \cdot a$
   • Substituting $a = 6 \, \text{cm}$: $h = \frac{\sqrt{3}}{2} \cdot 6 = 3\sqrt{3} \, \text{cm}$

Task 20:

"Odredi stranicu romba ako su dijagonale $d_1 = 12 \, \text{cm}$ i $d_2 = 16 \, \text{cm}$."

1. The side $a$ of a rhombus is calculated using the diagonals: $a = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}$
   • Substituting $d_1 = 12$ and $d_2 = 16$: $a = \sqrt{\left(\frac{12}{2}\right)^2 + \left(\frac{16}{2}\right)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{cm}$

Would you like detailed explanations or step-by-step derivations for any specific task? Here are five relative questions to consider:

1. How does the Pythagorean theorem apply to diagonals in rectangles and squares?
2. Can you derive a general formula for a square's side based on its diagonal?
3. Why is the height of an equilateral triangle proportional to the square root of three?
4. How do diagonal lengths uniquely determine a rhombus's side length?
5. What other properties of a right triangle can be derived from the Pythagorean theorem?

Tip: Always double-check the square roots and simplify where possible to keep answers precise!