Find the area of a rectangle if its diagonal is 26 cm, and the half-perimeter is 34 cm.

1. The perimeter of the rectangle is:
P = 2a + 2b,
where a and b are the length and width of the rectangle.
The semi-perimeter of the rectangle is:
p = P / 2;
(2a + 2b) / 2 = 34.
Thus:
a + b = 34.
2. The diagonal of the rectangle (d) forms a right-angled triangle with two of its sides, in which it is the hypotenuse. By the Pythagorean theorem:
d² = a² + b²;
a² + b² = 26²;
a² + b² = 676.
3. We have obtained a system of equations with two unknowns:
a + b = 34;
a² + b² = 676.
In the first equation, we express a:
a = 34 – b.
Substitute the expression into the second equation:
(34 – b) ² + b² = 676;
34² – 2 * 34 * b + b² + b² – 676 ​​= 0;
2b² – 68b + 1156 – 676 ​​= 0;
2b² – 68b + 480 = 0;
b² – 34b + 240 = 0.
Discriminant:
D = b² – 4ac,
where a, b, c are coefficients at b.
D = (-34) ² – 4 * 1 * 240 = 1156 – 960 = 196.
b = (-b +/- √D) / 2a.
b₁ = (- (- 34) + √196) / 2 * 1 = (34 + 14) / 2 = 48/2 = 24.
b₂ = (- (- 34) – √196) / 2 * 1 = (34 – 14) / 2 = 20/2 = 10.
Then:
a₁ = 34 – b₁ = 34 – 24 = 10.
a₂ = 34 – b2 = 34 – 10 = 24.
Thus:
a = 10, b = 24.
4. The area of ​​a rectangle is equal to the product of its length and width:
S = ab = 10 * 24 = 240 (cm²).
Answer: S = 240 cm².