To solve this problem, we use the formula for the area of a triangle:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.$

Step 1: Define Variables

Let the height of the triangle be $h$ cm.
The base is given as $2h + 4$ cm.
The area is 53 cm².

Step 2: Substitute into the Area Formula

$53 = \frac{1}{2} \times (2h + 4) \times h.$

Step 3: Simplify the Equation

Multiply through by 2 to eliminate the fraction: $106 = (2h + 4)h.$

Expand the right-hand side: $106 = 2h^2 + 4h.$

Step 4: Rearrange into Standard Quadratic Form

$2h^2 + 4h - 106 = 0.$

Divide through by 2 to simplify: $h^2 + 2h - 53 = 0.$

Step 5: Solve the Quadratic Equation

We use the quadratic formula: $h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$ Here, $a = 1$, $b = 2$, and $c = -53$.

$h = \frac{-2 \pm \sqrt{2^2 - 4(1)(-53)}}{2(1)}.$

Simplify under the square root: $h = \frac{-2 \pm \sqrt{4 + 212}}{2}.$ $h = \frac{-2 \pm \sqrt{216}}{2}.$ $h = \frac{-2 \pm 14.7}{2}.$

Step 6: Calculate the Two Solutions

1. $h = \frac{-2 + 14.7}{2} = \frac{12.7}{2} = 6.35$ cm.
2. $h = \frac{-2 - 14.7}{2} = \frac{-16.7}{2} = -8.35$ cm (not possible because height cannot be negative).

Thus, $h = 6.35$ cm.

Step 7: Calculate the Base

$\text{Base} = 2h + 4 = 2(6.35) + 4 = 12.7 + 4 = 16.7 \, \text{cm}.$

Final Answer:

• Height: $6.35 \, \text{cm}$
• Base: $16.7 \, \text{cm}$

Let me know if you would like more details about this solution.

Here are five related questions for further exploration:

1. How do you derive the quadratic formula?
2. What are the conditions for a quadratic equation to have real roots?
3. Can you explain how to graph a quadratic equation like $h^2 + 2h - 53 = 0$?
4. How does changing the base or height affect the area of a triangle?
5. Could you calculate the dimensions of a triangle with an area of 100 cm², if the base is twice the height?

Tip: Always check the units in your problem to ensure consistency and correctness!