Question

ABC and BDE are two equilateral triangles such that D is the mid point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1

Answer 1

Given: ABC and BDE are two equilateral triangles
Since, D is the mid point of BC and BDE is also an equilateral traingle.
Hence, E is also the mid point of AB.
Now, D and E are the mid points of BC and AB.
In a triangle, the line segment that joins the midpoints of the two sides of a triangle is parallel to the third side and is half of it.
$\mathrm{DE}\parallel \mathrm{CA}\mathrm{and}\mathrm{DE}=\frac{1}{2}\mathrm{CA}$
Now, In △ABC and △EBD
∠BED = ∠BAC (Corresponding angles)
∠B = ∠B (Common)
By AA-similarity criterion
△ABC ∼ △EBD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{ABC}\right)}{\mathrm{area}\left(△\mathrm{DBE}\right)}={\left(\frac{\mathrm{AC}}{\mathrm{ED}}\right)}^2={\left(\frac{2\mathrm{ED}}{\mathrm{ED}}\right)}^2=\frac{4}{1}$
Hence, the correct answer is option (d).