Shortcut Trick
Since DE || AC, ΔBDE is similar to ΔBAC.
Ratio of corresponding sides: BD/BA = 4 / (4 + 2√2) = 2 / (2 + √2) = √2 / (√2 + 1).
Ratio of areas: Area(ΔBDE) / Area(ΔBAC) = (BD/BA)2 = (√2 / (√2 + 1))2 = 2 / (3 + 2√2).
Area of trapezium ADEC = Area(ΔBAC) − Area(ΔBDE) = (3 + 2√2) − 2 = 1 + 2√2.
Ratio of Area(ΔBDE) to Area(ADEC) = 2 : (1 + 2√2).
∴ The correct answer is 2 : (1 + 2√2).
Alternate Method
Given:
In ΔABC, DE || AC, where D and E are points on sides AB and BC respectively.
BD = 4 cm
AD = 2√2 cm
Formula Used:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides:
Area(ΔBDE) / Area(ΔBAC) = (BD / AB)2
Calculations:
⇒ AB = BD + AD
⇒ AB = 4 + 2√2 cm
⇒ Since DE || AC, in ΔBDE and ΔBAC:
⇒ ∠BDE = ∠BAC (Corresponding angles)
⇒ ∠BED = ∠BCA (Corresponding angles)
⇒ ∴ ΔBDE ∼ ΔBAC (By AA similarity criterion)
⇒ Area(ΔBDE) / Area(ΔBAC) = (BD / AB)2
⇒ Area(ΔBDE) / Area(ΔBAC) = [4 / (4 + 2√2)]2
⇒ Area(ΔBDE) / Area(ΔBAC) = [2 / (2 + √2)]2
⇒ Area(ΔBDE) / Area(ΔBAC) = [√2 / (√2 + 1)]2
⇒ Area(ΔBDE) / Area(ΔBAC) = 2 / (√2 + 1)2
⇒ Area(ΔBDE) / Area(ΔBAC) = 2 / (2 + 1 + 2√2) = 2 / (3 + 2√2)
⇒ Let Area(ΔBDE) = 2k and Area(ΔBAC) = (3 + 2√2)k
⇒ Area of trapezium ADEC = Area(ΔBAC) − Area(ΔBDE)
⇒ Area of trapezium ADEC = (3 + 2√2)k − 2k = (1 + 2√2)k
⇒ Ratio of Area(ΔBDE) to Area of trapezium ADEC = 2k / [(1 + 2√2)k] = 2 / (1 + 2√2)
⇒ Ratio = 2 : (1 + 2√2)
∴ The correct answer is 2 : (1 + 2√2).
Additional Information
Basic Proportionality Theorem (Thales Theorem)
If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. Thus, BD/DA = BE/EC.
Area of Similar Triangles
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides, altitudes, medians, or angle-bisectors.