An isosceles triangle has two equal sides of length 20 cm each, and the third side is 30 cm. What is the area (in square centimetres) of this triangle?

Introduction / Context:
This is a mensuration problem involving the area of a triangle when all three sides are known. The triangle is isosceles with two equal sides, but the most straightforward method here is to use Heron's formula, which applies to any triangle given its side lengths.

Given Data / Assumptions:

• Triangle is isosceles with sides: 20 cm, 20 cm and 30 cm.
• We must find the area in square centimetres.
• All side lengths are positive and form a valid triangle.

Concept / Approach:
Heron's formula states that the area of a triangle with side lengths a, b and c is:
Area = √(s * (s - a) * (s - b) * (s - c)),where s is the semi perimeter: s = (a + b + c) / 2. We compute s for the given sides and then substitute into Heron's formula, simplifying step by step.

Step-by-Step Solution:
Step 1: Let the side lengths be a = 20 cm, b = 20 cm and c = 30 cm.Step 2: Compute the semi perimeter: s = (a + b + c) / 2 = (20 + 20 + 30) / 2 = 70 / 2 = 35 cm.Step 3: Apply Heron's formula: Area = √(35 * (35 - 20) * (35 - 20) * (35 - 30)).Step 4: Simplify inside the square root: 35 - 20 = 15 and 35 - 30 = 5, so Area = √(35 * 15 * 15 * 5).Step 5: Factor the product: 35 = 7 * 5, so 35 * 15 * 15 * 5 = (7 * 5) * 15 * 15 * 5 = 7 * 25 * 225 = 7 * 5625.Step 6: Thus, Area = √(7 * 5625) = √7 * √5625.Step 7: Since √5625 = 75, the area becomes 75√7 square centimetres.

Verification / Alternative check:
Another way is to treat the triangle as two right triangles by drawing an altitude from the vertex between the equal sides to the base of length 30 cm, splitting the base into two segments of 15 cm each. Using Pythagoras theorem with hypotenuse 20 cm and one leg 15 cm, the altitude h satisfies 20^2 = 15^2 + h^2, so h^2 = 400 - 225 = 175, and h = √175 = 5√7. The area is then (1 / 2) * base * height = (1 / 2) * 30 * 5√7 = 75√7, confirming the result.

Why Other Options Are Wrong:
The values 50√5, 100, 175 and 25√7 do not match the area computed either by Heron's formula or the altitude method. Only 75√7 is consistent with both approaches and the given side lengths.

Common Pitfalls:
Common errors include miscalculating the semi perimeter, making arithmetic mistakes inside the square root, or forgetting to take the square root at the end. When using the altitude method, students sometimes split the base incorrectly or misapply Pythagoras theorem. Careful stepwise calculation avoids these mistakes.

Final Answer:
The area of the isosceles triangle is 75√7 square centimetres.