Question

abc and bde are two equilateral triangel such that Bd =2/3BC. Find the ratio of the areas of
triangles ABC and bde

Answer 1

area of ∆ABC : area of ∆BDE
=> √3/4(BC)^2 : √3/4(BD)^2
since, BD= 2/3 BC
so, √3/4(BC)^2 : √3/4 (2/3 BC)^2
=> 1: 4/9
=> 9:4
=> 3:2

Answer 2

Answer:

9:4

Step-by-step explanation:

Area of equilateral triangle = $\frac{\sqrt{3}}{4}a^2$

where a is the side of triangle

Area of equilateral Triangle ABC = $\frac{\sqrt{3}}{4}(BC)^2$

Area of equilateral Triangle BDE = $\frac{\sqrt{3}}{4}(BD)^2$

Since we are given that BD =2/3BC.

So, Area of equilateral Triangle BDE = $\frac{\sqrt{3}}{4}(\frac{2}{3}BC)^2$

Now the ratio of areas ΔABC and ΔBDE:

= $\frac{\frac{\sqrt{3}}{4}(BC)^2}{\frac{\sqrt{3}}{4}(\frac{2}{3}BC)^2}$

= $\frac{(BC)^2}{\frac{4}{9}BC^2}$

= $\frac{1}{\frac{4}{9}}$

= $\frac{9}{4}$

Hence the ratio of the  areas of ΔABC and ΔBDE is 9:4