Pythagoras' Theorem: Question 2

Worked solution

Part (a): Finding the shorter side (distance along the ground)

The ladder, the wall and the ground form a right-angled triangle. The right angle is at the base of the wall, where the ground meets the vertical wall.

• The ladder is the side opposite the right angle, so it is the hypotenuse: $4.5\text{ m}$.
• The height up the wall is one of the shorter sides: $4.1\text{ m}$.
• The distance along the ground, call it $d$, is the other shorter side; this is what we want.

Because we are finding a shorter side, we subtract the squares. Pythagoras’ theorem states:

$\text{(hypotenuse)}^2 = \text{(side)}^2 + \text{(side)}^2$

$4.5^2 = 4.1^2 + d^2$

Rearranging to make $d^2$ the subject:

$d^2 = 4.5^2 - 4.1^2$

$d^2 = 20.25 - 16.81 = 3.44$

Take the positive square root (a length cannot be negative):

$d = \sqrt{3.44} = 1.85472\ldots$

Rounding to 2 decimal places:

$\boxed{d = 1.85\text{ m}}$

Check: $1.85^2 + 4.1^2 = 3.4225 + 16.81 = 20.2325 \approx 20.25 = 4.5^2$. ✓ The horizontal distance ($1.85\text{ m}$) is sensibly shorter than the ladder ($4.5\text{ m}$).

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Part (b): Distance between two points (2D problem)

To find the distance between two points we build a right-angled triangle whose hypotenuse is the line segment $AB$. The two shorter sides are the horizontal and vertical gaps between the points.

Horizontal change (difference in $x$-coordinates):

$\Delta x = 4 - (-3) = 7$

Vertical change (difference in $y$-coordinates):

$\Delta y = -3 - 2 = -5$

Now $AB$ is the hypotenuse, so we add the squares of the two shorter sides:

$AB^2 = (\Delta x)^2 + (\Delta y)^2 = 7^2 + (-5)^2$

Squaring removes the sign of the $-5$:

$AB^2 = 49 + 25 = 74$

$AB = \sqrt{74} = 8.6023\ldots$

Rounding to 1 decimal place:

$\boxed{AB \approx 8.6\text{ km}}$

Alternative: the distance formula

This is exactly Pythagoras written as a single formula:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{\bigl(4-(-3)\bigr)^2 + \bigl(-3-2\bigr)^2} = \sqrt{49+25} = \sqrt{74} \approx 8.6\text{ km}.$

Both methods give the same result because the distance formula is Pythagoras’ theorem applied to the horizontal and vertical separations.