Similar triangles and area removal:\nIn ΔABC, points D on AB and E on AC are such that DE ∥ BC. Given AD : DB = 2 : 5 and area(ΔABC) = 98 cm^2, find the area (in cm^2) of quadrilateral BDEC.

Introduction / Context:
When a segment joins the sides of a triangle parallel to the base, it forms a smaller triangle similar to the original. Areas scale by the square of the similarity ratio. We subtract to obtain the quadrilateral’s area.

Given Data / Assumptions:

• DE ∥ BC with D on AB and E on AC.
• AD : DB = 2 : 5 → AD/AB = 2/(2+5) = 2/7.
• Area(ΔABC) = 98 cm^2.

Concept / Approach:
ΔADE ~ ΔABC with linear ratio 2/7. Hence area ratio is (2/7)^2 = 4/49. Compute area(ΔADE), then subtract from 98 to get area(BDEC).

Step-by-Step Solution:

Verification / Alternative check:
Consistency: If linear scale is 2/7, then the removed top triangle is small (8 cm^2), leaving most of the 98 cm^2 for the lower trapezoid-like region BDEC, i.e., 90 cm^2.

Why Other Options Are Wrong:
98 is the whole triangle; 94, 86, 88 do not equal 98 − 8 given the similarity ratio calculation.

Common Pitfalls:
Using 2/7 for area directly (forgetting to square the linear ratio), or mixing AD:DB with AD:AB.

Final Answer:
90