# 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 = 90°

Need to find the ratio between the areas of Δ ABE and Δ ACD

AB = BC

By Pythagoras theorem, we have AC2 = AB2 + BC2

since AB = BC

AC2 = AB2 + AB2

AC2 = 2 AB2 …..eq(1)

it is also given that Δ ABE Δ ACD

(ratio of areas of similar triangles is equal to ratio of squares of their corresponding sides)

=

= from 1

=

ar(Δ ABC):ar(Δ ACD) = 1:2

Hence the ratio is 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
Champ Quiz | Applications of Basic Proportionality Theorem24 mins
Basic Proportionality Theorem44 mins
Basic Proportionality Theorem34 mins
Basics of Similarity of Triangles39 mins
Basic Proportionality Theorem42 mins
NCERT | Strong Your Basics of Triangles39 mins
RD Sharma | Imp. Qs From Triangles41 mins
A Peep into Pythagoras Theorem43 mins
Predominant Proof of Triangles52 mins
Quiz | Criterion of Similarity of Triangle45 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