x18 cm = 9 cm.2Two Marks Questions4. AABC is an equilateraltriangle. D

1.	x18 cm = 9 cm.2Two Marks Questions4. AABC is an equilateraltriangle. D and E are mid-points of sides BC and ABrespectively. If BC = 4 cm,find ariABED).
Answer» 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 )AndD 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÷4So,Area of ∆ BED = 4 cm2 ( Ans )

x18 cm = 9 cm.2Two Marks Questions4. AABC is an equilateraltriangle. D and E are mid-points of sides BC and ABrespectively. If BC = 4 cm,find ariABED).

Answer»

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 )AndD 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÷4So,Area of ∆ BED = 4 cm2 ( Ans )