Question

In a ΔABC , point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area (ΔADE): Area (âBCED) =
A. 3 : 4
B. 9: 16
C. 3: 5
D. 9 : 25

Answer 1

B. 9: 16

Step-by-step explanation:

Given:

In a ΔABC , point D is on side AB and point E is on side AC, such that BCED is a trapezium. DE : BC = 3 : 5

Find: Area ΔADE: Area of trapezium BCED

Solution

In ΔADE an ΔABC, ∠ADE = ∠B (corresponding angles)

∠A is comon

So ΔABC ~ ΔADE (AA similarity)

For similar triangles, the ratio of their areas will be in the ratio of the square of their corresponding sides.

So area of ΔADE /  area of ΔABC = DE² / BC² = 3² / 5² = 9/ 25

Let's assume that area of ΔADE is 9 sq. units and area of ΔABC  is 25 sq. units.

Area of trapesium BCED = Area of ΔABC - Area of ΔADE = 25 - 9 = 16 sq. units

area of ΔADE / Area of trapezium BCED = 9 / 16  

Option B is the answer.

Answer 2

Answer:In ΔADE an ΔABC, ∠ADE = ∠B (corresponding angles)

∠A is comon

So ΔABC ~ ΔADE (AA similarity)

For similar triangles, the ratio of their areas will be in the ratio of the square of their corresponding sides.

So area of ΔADE / area of ΔABC = DE² / BC² = 3² / 5² = 9/ 25

Let's assume that area of ΔADE is 9 sq. units and area of ΔABC is 25 sq. units.

Area of trapesium BCED = Area of ΔABC - Area of ΔADE = 25 - 9 = 16 sq. units

area of ΔADE / Area of trapezium BCED = 9 / 16

Step-by-step explanation: