Question

∆ABC and ∆ADE are two equilateral
triangles such that D is the midpoint of BC. Then find the ratio of areas of ∆ABC and ∆ BDE. Please experts answer only. Answer with figure​

Answer 1

LET AB=BC=AC=2a

D is the mid point of BC

So BD=a

In right angles triangle ADC

AD=

$\sqrt{2a { }^{2} } - {a}^{2} = \sqrt{3 \: } a$

Now in trangle ABC

area of ABC=

$\sqrt{3} \div 4 {(2a)}^{2} = \sqrt{3} {a}^{2}$

Again

area of ADE=

$\sqrt{3} \div 4 \times {( \sqrt{3} }a)^{2} = \sqrt{3} \div 4 \times {3a}^{2}$

=

$3 \sqrt{3} \div 4 \times {a}^{2}$

so ratio of triangle ABC and ADE=4/3