Pythagoras' Theorem: Question 1

Worked solution

Part (a): Length of the ladder

The wall is vertical and the ground is horizontal, so the wall, the ground and the ladder form a right-angled triangle. The right angle is where the wall meets the ground.

The ladder is the side opposite the right angle, so the ladder is the hypotenuse, the longest side. We use Pythagoras’ theorem:

$h^2 = a^2 + b^2$

where $h$ is the hypotenuse (the ladder) and $a$, $b$ are the two shorter sides (the distance along the ground and the height up the wall).

Substitute $a = 1.4$ and $b = 4.8$:

$h^2 = 1.4^2 + 4.8^2$

$h^2 = 1.96 + 23.04$

$h^2 = 25$

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

$h = \sqrt{25} = 5$

$\boxed{\text{The ladder is } 5\ \text{m long.}}$

Alternative check (recognising a Pythagorean triple). Notice that $1.4 : 4.8 : 5$ is just the well-known triple $7 : 24 : 25$ scaled down by a factor of $5$ (since $7\times0.2=1.4$, $24\times0.2=4.8$, $25\times0.2=5$). This confirms the answer of $5$ m without further calculation.

Part (b): Length of the guy rope

The flagpole is vertical and the ground is horizontal, so the pole, the ground and the rope again form a right-angled triangle, with the right angle at the foot of the pole.

The rope runs from the top of the pole to the peg, so the rope is the hypotenuse. Let its length be $L$. By Pythagoras’ theorem:

$L^2 = 4^2 + 2.5^2$

$L^2 = 16 + 6.25$

$L^2 = 22.25$

$L = \sqrt{22.25} = 4.71699\ldots$

Rounding to 2 decimal places (the third decimal digit is $6$, so we round the second decimal up):

$\boxed{L \approx 4.72\ \text{m}}$

Key idea

In both parts the unknown side is the hypotenuse (the longest side, opposite the right angle), so we add the squares of the two shorter sides and then take the square root.