Q. 215.0( 1 Vote )

# In a , 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 ( ): Area (âBCED) =A. 3 : 4B. 9: 16C. 3: 5D. 9 : 25

Given in ΔABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium and DE: BC = 3: 5. ACB = AED [corresponding angles]

A = A [common angle]

We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

We know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Let ar (ΔADE) = 9x sq. units and ar (ΔABC) = 25x sq. units

ar (trap BCED) = ar (ΔABC) – ar (ΔADE)

= 25x – 9x

= 16x sq. units

Now, ar (ΔADE): ar (trap BCED) = 9: 16

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Basic Proportionality Theorem42 mins  A Peep into Pythagoras Theorem43 mins  R.D Sharma | Solve Exercise -4.2 and 4.3FREE Class  NCERT | Strong Your Basics of Triangles39 mins  RD Sharma | Imp. Qs From Triangles41 mins  Quiz | Criterion of Similarity of Triangle45 mins  How to Ace Maths in NTSE 2020?36 mins  Know About Important Proofs in Triangles33 mins  Master BPT or Thales Theorem39 mins  R.D Sharma | Solve Exercise-4.545 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 