Question

The sides of a triangle are 9cm, 40cm and 41cm. The area of the triangle is

• A. 180 sq.cm
• B. 184 sq.cm
• C. 190 sq.cm
• D. 240 sq.cm

Hint:

Familiarity with standard Pythagorean triplets is a major shortcut in exams.
Common triplets include \((3, 4, 5)\), \((5, 12, 13)\), \((8, 15, 17)\), and \((9, 40, 41)\).
Identifying the triplet immediately avoids the complex calculations of Heron's formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This is a geometry question requiring us to calculate the area of a triangle when the lengths of its three sides are provided.

Step 2: Key Formulas and approach:
First, check if the given side lengths (\(9\text{ cm}\), \(40\text{ cm}\), and \(41\text{ cm}\)) form a right-angled triangle.
A triangle is right-angled if its side lengths satisfy the Pythagorean theorem:
\[ a^2 + b^2 = c^2 \] If it is a right-angled triangle, the area can be calculated using the simple formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] Otherwise, we would need to apply Heron's formula.

Step 3: Detailed Explanation:

• Test the side lengths for a Pythagorean triplet, where the longest side (\(41\text{ cm}\)) is the hypotenuse:
\[ 9^2 + 40^2 = 81 + 1600 = 1681 \] \[ 41^2 = 1681 \]

• Since \(9^2 + 40^2 = 41^2\), the sides satisfy the Pythagorean theorem and form a right-angled triangle.

• In this right-angled triangle, the two shorter sides perpendicular to each other are the base and the height:
\[ \text{base} = 40\text{ cm} \] \[ \text{height} = 9\text{ cm} \]

• Now calculate the area of the triangle:
\[ \text{Area} = \frac{1}{2} \times 40 \times 9 \] \[ \text{Area} = 20 \times 9 = 180\text{ sq. cm} \]

Step 4: Final Answer:
The area of the triangle is \(180\text{ sq. cm}\), which corresponds to Option (A).