AD is an altitude of an equilateral triangle ABC. On AD as base another equilateral triangle ADE is constructed. prove that area (ADE) : area (ABC) = 3 : 4
Answers
Answered by
7
Check Diagram
Given ABC and ADE are equilateral triangles.
Let AB=BC=CA = a
Recall that the altitude of an equilateral triangle is √3/2 times its side
Hence AD = √3a/2
ΔABC ~ ΔADE [By AAA similarity criterion]
ar(∆ ADE)/ar (ABC) = AD² / BC ²
[Since ratio of areas of two similar triangles is equal to the ratio of their corresponding sides]
(^^) ( Img 2...)
Hence area(ADE):area(ABC)=3:4.
^_^^_^^_^^_^
~~~
Hope it helps...
~~~
(^^)
Given ABC and ADE are equilateral triangles.
Let AB=BC=CA = a
Recall that the altitude of an equilateral triangle is √3/2 times its side
Hence AD = √3a/2
ΔABC ~ ΔADE [By AAA similarity criterion]
ar(∆ ADE)/ar (ABC) = AD² / BC ²
[Since ratio of areas of two similar triangles is equal to the ratio of their corresponding sides]
(^^) ( Img 2...)
Hence area(ADE):area(ABC)=3:4.
^_^^_^^_^^_^
~~~
Hope it helps...
~~~
(^^)
Attachments:
Similar questions