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To solve this problem, let's analyze the information provided.

We have:
- An isosceles right-angled triangle, where the two shorter sides are equal.
- The perimeter of the triangle is given as \(7 + 7\sqrt{2}\).

For an isosceles right-angled triangle with equal shorter sides \(a\):
1. The hypotenuse \(c\) can be calculated using the Pythagorean theorem:
   \[
   c = a\sqrt{2}
   \]

2. The perimeter \(P\) is the sum of all sides:
   \[
   P = a + a + a\sqrt{2} = 2a + a\sqrt{2}
   \]

Since we know the perimeter \(P\) is \(7 + 7\sqrt{2}\), we can set up the equation:
\[
2a + a\sqrt{2} = 7 + 7\sqrt{2}
\]

To solve for \(a\):
1. Separate terms involving \(a\):
   \[
   a(2 + \sqrt{2}) = 7 + 7\sqrt{2}
   \]

2. Divide both sides by \(2 + \sqrt{2}\):
   \[
   a = \frac{7 + 7\sqrt{2}}{2 + \sqrt{2}}
   \]

To simplify this, we can rationalize the denominator by multiplying the numerator and denominator by \(2 - \sqrt{2}\):
\[
a = \frac{(7 + 7\sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})}
\]

Expanding the numerator and simplifying the denominator:
\[
a = \frac{14 - 7\sqrt{2} + 14\sqrt{2} - 7 \cdot 2}{4 - 2} = \frac{14 + 7\sqrt{2}}{2} = 7
\]

Thus, the length of each shorter side is:
\[
\boxed{7}
\]

Would you like further details or have any questions?

Here are some related questions you might explore:
1. What would be the hypotenuse if the perimeter of the triangle were given differently?
2. How do we calculate the area of an isosceles right triangle given only its perimeter?
3. How does the Pythagorean theorem apply specifically in isosceles right triangles?
4. What happens if the perimeter is given in terms of a different irrational number?
5. How would this problem change if it were not an isosceles right triangle?

**Tip:** When simplifying expressions with radicals, rationalizing the denominator is a useful technique to make the expression easier to work with.

The perimeter of an isosceles right-angled triangle is given as 7 + 7√2. Then find the length of the shorter side.

Explore how to find the length of the shorter side of an isosceles right-angled triangle given its perimeter as 7 + 7√2. This solution uses key concepts in geometry and algebra, including the Pythagorean theorem and rationalizing radicals. Suitable for Grades 9-10.

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