Question

triangle ABC is an equilateral triangle . D and E are mid points of side BC and AB respectively . If BC =4cm , find area of triangle BED.

Answer 1

Here D and E are mid points of BC and AB , We join DE and from " converse of mid-point theorem " we get AC | | ED

And BC = 4 cm , SO

AB = BC = CA = 4 cm ( As given ABC is a equilateral triangle )
And
D and E are mid points of BC and AB , So

BE = EA = BD = DC = 2 cm

Now In ∆ BAC and ∆ BED
∠ ABC = ∠ EBD ( Same angles )

∠ BAC = ∠ BED ( As we know AC | | ED and take AB as transversal line , So these angles are Corresponding angles )

And

∠ BCA = ∠ BDE ( As we know AC | | ED and take CB as transversal line , So these angles are Corresponding angles )

Hence ∆ BAC ~ ∆ BED ( By AAA rule )

So we know

BA÷BE = AC÷ED⇒4÷2 = 4÷ED⇒ED = 2

And

Area of BAC÷Area of BED = AC²÷ED²⇒Area of BAC÷Area of BED = 4²÷2²⇒Area of BAC÷Area of BED = 16÷4
So,
Area of ∆ BED = 4 cm2 ( Ans )

Answer 2

Draw a triangle with vertex C on top.
Now mid point of BC is D and mid point of AB is E.
By converse of mid point theorem, AC parallel to ED.
And BC=4cm
Therefore, BC=AB=AC (Equilateral triangle)
Since D and E are midpoints of BC and AB, CD=DB=CE=EA= 2cm.
Now in triangle BAC and CED,
Angle ACB= angle ECD (common)
Angle EAB= angle CED (corresponding angles)
Angle CDE= angle DBA(corresponding angles)
Therefore both triangles are congruent.
Now BA/BE=AC/ED
4/2=4/ED
Therefore, ED=2
Now, ar(BAC)/ar(BED)=AC^2/ED^2
=4^2/2^2
=16/4
=4cm
Therefore ar(BED)= 4cm.
Hope it helps you.
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