What is the approximate area of the triangle ABC?

Given:

Perimeter of triangle ABC = 105 cm

Ratio of sides AB : BC : CA = 5 : 10 : 6

Formula used:

Heron's Formula for the area of a triangle:

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

Where a, b, c are the lengths of the sides of the triangle, and

s is the semi-perimeter s = \((\dfrac{\text{a + b + c}}{2})\)

Calculations:

From the previous problem, we established

The ratio of the sides AB : BC : CA = 5 : 10 : 6.

Let the sides be AB = 5x, BC = 10x, and CA = 6x.

The perimeter is the sum of the sides:

Perimeter = AB + BC + CA = 5x + 10x + 6x = 21x

Given Perimeter = 105 cm.

⇒ 21x = 105

⇒ x = \((\dfrac{105}{21})\)

⇒ x = 5

a (BC) = 10x = 10 × 5 = 50 cm

b (CA) = 6x = 6 × 5 = 30 cm

c (AB) = 5x = 5 × 5 = 25 cm

The semi-perimeter (s): s = \((\dfrac{\text{Perimeter}}{2})\) = \((\dfrac{105}{2})\) = 52.5 cm

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

Area = \((\sqrt{52.5 \times (52.5 - 50) \times (52.5 - 30) \times (52.5 - 25)})\)

Area = \((\sqrt{52.5 \times 2.5 \times 22.5 \times 27.5})\)

Area = \((\sqrt{99738.28125})\)

Area ≈ 284.975 cm²

Area ≈ 285 cm²

∴ The correct answer is option 4.