A straight support cable runs from the top of a vertical flagpole to a fixing point on level ground. The cable is 17 m long, and the fixing point on the ground is 8 m from the base of the pole. The pole meets the ground at a right angle. What is the height of the flagpole?

Pythagoras' Theorem: Find a Missing Side (IGCSE 0580 MCQ)

The flagpole, the ground, and the cable form a right-angled triangle. The right angle is where the pole meets the ground.

The cable is the side opposite the right angle, so it is the hypotenuse: $17\text{ m}$ .
The distance along the ground is one shorter side (a leg): $8\text{ m}$ .
The height of the pole is the other leg; this is what we want. Call it $h$ .

For a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the two legs:

$\text{hypotenuse}^2 = \text{leg}^2 + \text{leg}^2$

$17^2 = h^2 + 8^2$

Because we are looking for a leg (not the hypotenuse), we subtract:

$h^2 = 17^2 - 8^2$

$h^2 = 289 - 64$

$h^2 = 225$

Take the positive square root (a length must be positive):

$h = \sqrt{225} = 15$

Verify the three sides satisfy the theorem:

$8^2 + 15^2 = 64 + 225 = 289 = 17^2 \checkmark$

(In fact $8,\,15,\,17$ is a well-known Pythagorean triple.)

$18.8\text{ m}$ comes from adding the squares, $\sqrt{17^2+8^2}=\sqrt{353}\approx 18.8$ . This is wrong because the $17\text{ m}$ cable is the longest side (the hypotenuse), so the missing side must be shorter than $17\text{ m}$ , not longer.
$25\text{ m}$ is simply $17+8$ , and $9\text{ m}$ is simply $17-8$ ; neither uses the squares of the sides.

$\boxed{h = 15\text{ m}}$