The problem asks for the area of a triangle with sides 25, 34, and 39 cm². To solve this, we can use Heron's formula for the area of a triangle when the lengths of all three sides are known:

$A = \sqrt{s(s-a)(s-b)(s-c)}$

where:

• $a = 25 \, \text{cm}$
• $b = 34 \, \text{cm}$
• $c = 39 \, \text{cm}$
• $s$ is the semi-perimeter: $s = \frac{a + b + c}{2} = \frac{25 + 34 + 39}{2} = 49 \, \text{cm}$

Now, plugging values into Heron's formula:

$A = \sqrt{49(49 - 25)(49 - 34)(49 - 39)}$ $A = \sqrt{49 \times 24 \times 15 \times 10}$ $A = \sqrt{49 \times 3600} = \sqrt{176400} = 420 \, \text{cm}^2$

Thus, the area of the triangle is $420 \, \text{cm}^2$, corresponding to the last option in your image.

Would you like a more detailed explanation or have any questions?

Here are five related questions:

1. How do you derive Heron's formula?
2. What other methods exist for calculating the area of a triangle?
3. How does Heron’s formula relate to right triangles?
4. Can Heron’s formula be used for any type of triangle?
5. How can you simplify the calculation if the triangle is isosceles or equilateral?

Tip: When using Heron's formula, make sure the triangle inequality theorem is satisfied to ensure valid triangle side lengths.