Let's break down and solve each part of the problem in the image:

(a)

Given:

• Base = 3 meters
• Perpendicular = 4 meters

We are asked to find the hypotenuse of this right-angled triangle. By using the Pythagorean Theorem:

$\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2$

$\text{Hypotenuse}^2 = 3^2 + 4^2 = 9 + 16 = 25$

$\text{Hypotenuse} = \sqrt{25} = 5 \, \text{meters}$

Answer (a): The hypotenuse is 5 meters.

(b)

Given:

• Base = 48 meters
• The difference between the hypotenuse and the perpendicular = 36 meters

Let the hypotenuse be $h$ and the perpendicular be $p$.

From the problem, we have: $h - p = 36 \quad \text{(Equation 1)}$ And using the Pythagorean Theorem: $h^2 = 48^2 + p^2 \quad \text{(Equation 2)}$

Let's substitute $h = p + 36$ from Equation 1 into Equation 2: $(p + 36)^2 = 48^2 + p^2$

Expanding both sides: $p^2 + 72p + 1296 = 2304 + p^2$

Simplifying: $72p + 1296 = 2304$

$72p = 1008$

$p = \frac{1008}{72} = 14 \, \text{meters}$

Now, substituting $p = 14$ into Equation 1: $h = 14 + 36 = 50 \, \text{meters}$

Answer (b): The hypotenuse is 50 meters, and the perpendicular is 14 meters.

(c)

Given:

• The sum of the hypotenuse and one side is 153 meters
• The other side is 51 meters

Let the hypotenuse be $h$ and the unknown side be $x$.

From the problem, we have: $h + x = 153 \quad \text{(Equation 1)}$ Using the Pythagorean Theorem: $h^2 = x^2 + 51^2 \quad \text{(Equation 2)}$

Substitute $x = 153 - h$ from Equation 1 into Equation 2: $h^2 = (153 - h)^2 + 51^2$

Expanding both sides: $h^2 = (153 - h)(153 - h) + 2601$

$h^2 = 23409 - 306h + h^2 + 2601$

Canceling $h^2$ from both sides: $0 = 26010 - 306h$

Solving for $h$: $306h = 26010$

$h = \frac{26010}{306} = 85 \, \text{meters}$

Now, using $h = 85$ in Equation 1: $x = 153 - 85 = 68 \, \text{meters}$

Answer (c): The hypotenuse is 85 meters, and the other side is 68 meters.

Would you like further details on any of these solutions?

Additional Questions:

1. How do we derive the Pythagorean Theorem?
2. What would happen if the base and perpendicular are swapped in part (a)?
3. How can we apply the quadratic formula in part (b)?
4. What other methods can solve part (c) without algebraic substitution?
5. What are some real-world applications of the Pythagorean Theorem?

Tip: When solving right-angled triangle problems, always start by identifying known sides and applying the Pythagorean Theorem for unknowns.