ABCD is an isosceles trapezium with AD ∥ BC, AB = 5 cm, AD = 8 cm and BC = 14 cm. What is the area (in sq cm) of trapezium ABCD?

Introduction / Context:
This question is about finding the area of an isosceles trapezium using side lengths. An isosceles trapezium has one pair of parallel sides and equal non parallel sides. The problem requires using right triangles formed by dropping perpendiculars from the ends of the shorter base to the longer base, and then applying the area formula for a trapezium once the height is known.

Given Data / Assumptions:

• ABCD is an isosceles trapezium.
• AD ∥ BC, so AD and BC are the parallel bases.
• AB = 5 cm, AD = 8 cm and BC = 14 cm.
• AB and CD are equal non parallel sides because the trapezium is isosceles.
• We need the area of trapezium ABCD.

Concept / Approach:
For a trapezium with parallel sides a and b and height h, area = (1 / 2) * (a + b) * h. Here the difference between the longer and shorter base is shared equally as horizontal offsets on each side because the trapezium is isosceles. This gives a right triangle on each end, whose hypotenuse is the leg of the trapezium. From the leg and horizontal offset, we can find the height using Pythagoras theorem. Then we substitute into the area formula.

Step-by-Step Solution:
Step 1: Identify the parallel sides: AD = 8 cm (shorter base) and BC = 14 cm (longer base). Step 2: The difference in base lengths is 14 − 8 = 6 cm. Step 3: Since ABCD is isosceles, this 6 cm is split equally on both sides, so each horizontal offset is 6 / 2 = 3 cm. Step 4: Each non parallel side, for example AB, forms the hypotenuse of a right triangle with horizontal leg 3 cm and vertical leg h (the trapezium height). Given AB = 5 cm. Step 5: Apply Pythagoras theorem: 5^2 = h^2 + 3^2, so 25 = h^2 + 9. Step 6: Solve for h: h^2 = 25 − 9 = 16, so h = 4 cm. Step 7: Use trapezium area formula: Area = (1 / 2) * (AD + BC) * h = (1 / 2) * (8 + 14) * 4 = (1 / 2) * 22 * 4 = 44 sq cm.

Verification / Alternative check:
We can imagine constructing the trapezium as a central rectangle of width 8 cm and height 4 cm, plus two right triangles of base 3 cm and height 4 cm on the sides. The rectangle area is 8 * 4 = 32 sq cm. Each right triangle has area (1 / 2) * 3 * 4 = 6 sq cm, and there are two such triangles, contributing 12 sq cm. The total area is 32 + 12 = 44 sq cm, which matches the earlier calculation, confirming the result.

Why Other Options Are Wrong:
An area of 36 sq cm would correspond to a smaller height or incorrect base sum. An area of 88 sq cm or 144 sq cm would require a much larger height than allowed by the leg length of 5 cm. The value 52 sq cm also does not match any correct geometric decomposition of the trapezium. Only 44 sq cm is consistent with both the right triangle dimensions and the standard trapezium area formula.

Common Pitfalls:
Learners sometimes confuse which sides are parallel and may try to use AB and CD as bases. Another error is forgetting that the difference in base lengths is split equally in an isosceles trapezium. Neglecting to square the side lengths correctly in Pythagoras theorem can also lead to wrong heights. Always confirm the shape type, identify the correct bases and use a clear diagram before doing calculations.

Final Answer:
The area of trapezium ABCD is 44 sq cm.