Pythagoras' Theorem - Solving Complex Problems: Calculate PN in triangle OMPN and find the area of a square with the same perimeter as trapezium ABCD.

Problem 1: Work out the length of \( PN \)

#### Given:
- Triangle \( OMN \) is a right triangle.
- \( MP = 7 \, \text{cm} \)
- \( OM = 25 \, \text{cm} \)
- \( ON = 30 \, \text{cm} \)

We need to find the length of \( PN \).

#### Step 1: Use Pythagoras' Theorem in \( \triangle OMP \)
In \( \triangle OMP \):
- \( OM \) is the hypotenuse.
- \( OP \) and \( MP \) are the legs.

Using Pythagoras' Theorem:
\[
OM^2 = OP^2 + MP^2
\]
Substitute the given values:
\[
25^2 = OP^2 + 7^2
\]
\[
625 = OP^2 + 49
\]
Solve for \( OP^2 \):
\[
OP^2 = 625 - 49
\]
\[
OP^2 = 576
\]
Take the square root of both sides:
\[
OP = \sqrt{576} = 24 \, \text{cm}
\]

#### Step 2: Use Pythagoras' Theorem in \( \triangle OPN \)
In \( \triangle OPN \):
- \( ON \) is the hypotenuse.
- \( OP \) and \( PN \) are the legs.

Using Pythagoras' Theorem:
\[
ON^2 = OP^2 + PN^2
\]
Substitute the known values:
\[
30^2 = 24^2 + PN^2
\]
\[
900 = 576 + PN^2
\]
Solve for \( PN^2 \):
\[
PN^2 = 900 - 576
\]
\[
PN^2 = 324
\]
Take the square root of both sides:
\[
PN = \sqrt{324} = 18 \, \text{cm}
\]

#### Final Answer for Problem 1:
\[
\boxed{18}
\]

---

Problem 2: Work out the area of the square

#### Given:
- Trapezium \( ABCD \) has bases \( AD = 7 \, \text{cm} \) and \( BC = 10 \, \text{cm} \).
- The height of the trapezium is \( 4 \, \text{cm} \).
- A square has the same perimeter as the trapezium \( ABCD \).

We need to find the area of the square.

#### Step 1: Calculate the perimeter of the trapezium \( ABCD \)
The perimeter of a trapezium is the sum of the lengths of its sides. For trapezium \( ABCD \):
- \( AD = 7 \, \text{cm} \)
- \( BC = 10 \, \text{cm} \)
- The height is \( 4 \, \text{cm} \).

To find the lengths of the non-parallel sides \( AB \) and \( CD \), we use the fact that the height forms right triangles with these sides. Let's denote the horizontal projection of \( AB \) and \( CD \) as \( x \).

From the diagram:
- The total horizontal distance between \( A \) and \( B \) is \( 10 - 7 = 3 \, \text{cm} \).
- This distance is split equally between the two right triangles formed by the height, so each projection is \( \frac{3}{2} = 1.5 \, \text{cm} \).

Using Pythagoras' Theorem in one of the right triangles (e.g., for side \( AB \)):
\[
AB^2 = 4^2 + 1.5^2
\]
\[
AB^2 = 16 + 2.25
\]
\[
AB^2 = 18.25
\]
\[
AB = \sqrt{18.25} \approx 4.27 \, \text{cm}
\]

Since \( AB = CD \), the perimeter of the trapezium is:
\[
\text{Perimeter} = AD + AB + BC + CD
\]
\[
\text{Perimeter} = 7 + 4.27 + 10 + 4.27
\]
\[
\text{Perimeter} = 25.54 \, \text{cm}
\]

#### Step 2: Determine the side length of the square
The perimeter of the square is equal to the perimeter of the trapezium:
\[
4s = 25.54
\]
Solve for \( s \):
\[
s = \frac{25.54}{4} = 6.385 \, \text{cm}
\]

#### Step 3: Calculate the area of the square
The area of a square is given by:
\[
\text{Area} = s^2
\]
\[
\text{Area} = (6.385)^2 \approx 40.77 \, \text{cm}^2
\]

#### Final Answer for Problem 2:
\[
\boxed{40.77}
\]