If ABC and BDE are two equilateral triangles such that D is the mid-po

1.	If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar (ΔABC) : ar (ΔBDE)
Answer» ΔABC and ΔBDE are equilateral triangles We know that, Area of equilateral triangle = \(\frac{\sqrt3}{4}a^2\) D is the mid-point of BC then, Area (ΔBDE) = \(\frac{\sqrt3}{4}\) x \((\frac{a}{2})^2\) = \(\frac{\sqrt3}{4}\) x \(\frac{a\times a}{2}\) Now, Area (ΔABC) : Area (ΔBDE) \(\frac{\sqrt3}{4}\) x a²: \(\frac{\sqrt3}{4}\) x \(\frac{a\times a}{2}\) 1 : \(\frac{1}{4}\) 4: 1 Hence, Area (ΔABC) : Area (ΔBDE) is 4: 1

If ABC and BDE are two equilateral triangles such that D is the mid-point of BC, then find ar (ΔABC) : ar (ΔBDE)

Answer»

ΔABC and ΔBDE are equilateral triangles

We know that,

Area of equilateral triangle = \(\frac{\sqrt3}{4}a^2\)

D is the mid-point of BC then,

Area (ΔBDE) = \(\frac{\sqrt3}{4}\) x \((\frac{a}{2})^2\)

= \(\frac{\sqrt3}{4}\) x \(\frac{a\times a}{2}\)

Now,

Area (ΔABC) : Area (ΔBDE)

\(\frac{\sqrt3}{4}\) x a²: \(\frac{\sqrt3}{4}\) x \(\frac{a\times a}{2}\)

1 : \(\frac{1}{4}\)

4: 1

Hence,

Area (ΔABC) : Area (ΔBDE) is 4: 1