One leg of a right-angled triangle is 1 in less than the other. The area of the triangle is 10dm ^ 2. Find the hypotenuse.

1. A, B, C – the vertices of the triangle. ∠С = 90 °.

2. We take the length of the BC leg as x (dm). The length of the AC leg (x + 1), since according to the condition of the problem one of the legs is 1 dm larger than the other.

3. Make an equation using the formula for calculating the area (S) of a triangle:

S = x (x + 1) / 2;

10 = (x² + x) / 2;

x² + x = 20;

x² + x – 20 = 0;

The first value x = (- 1 + √1 + 80) = (- 1 + 9) / 2 = 4 dm.

The second value is x = (- 1 – 9) / 2 = – 5. Not accepted.

The length of the BC leg is 4 dm.

The length of the AC leg is 4 + 1 = 5 dm.

4. Calculate the length of the hypotenuse AB using the Pythagorean theorem:

AB = √BC² + AC² = √16 + 25 = √41 dm.

Answer: the length of the hypotenuse AB of a given triangle is √41 dm.