Q. 285.0( 1 Vote )

# ABC is an isosceles triangle, right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ΔABE and ΔACD. Given ∆ABC is an isosceles triangle in which B is right angled i.e. 90°

AB = BC

In right angled ∆ABC, by Pythagoras theorem, we have

(Perpendicular)2 + (Base)2 = (Hypotenuse)2

(AB)2 + (BC)2 = (AC)2

(AB)2 + (AB)2 = (AC)2

[ABC is an isosceles triangle, AB =BC]

2(AB)2 = (AC)2

(AC)2 = 2(AB)2 …(i)

It is also given that ∆ABE ~ ∆ADC

And we also know that, the ratio of similar triangles is equal to the ratio of their corresponding sides.  [from (i)] ar(∆ABE) : ar(∆ADC) = 1 : 2

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 