Question

For $△ABC$ whose vertices are given by $A(3,6),B(-4,7)$ and $C(10,-7),$ if $A_1,B_1,C_1$ represent the mid points of sides opposite to vertices $A,B,C$ respectively, of $△ABC$ and $A_2,B_2,C_2$ represent the mid points of sides opposite to vertices $A_1,B_1,C_1$ respectively, of $△A_1{B}_1{C}_1$ and so on, then which of the following is/are correct?

• A. The centroid of $△A_2{B}_2{C}_2$ is $(3,2)$
• B. The centroid of $△A_3{B}_3{C}_3$ is $(3,2)$
• C. $\frac{\text{Area (}△A_2{B}_2{C}_2\text{)}}{\text{Area (}△ABC\text{)}}=\frac{1}{16}$
• D. $\frac{\text{Area (}△A_3{B}_3{C}_3\text{)}}{\text{Area (}△ABC\text{)}}=\frac{1}{16}$

Answer 1

Answer: (A), (B), (C)

$\frac{\text{Area (}△A_2{B}_2{C}_2\text{)}}{\text{Area (}△ABC\text{)}}=\frac{1}{16}$

Given vertices of $△ABC$ are $A(3,6),B(-4,7)$ and $C(10,-7)$

We know that,
Centroid of triangle $=$ Centroid of triangle formed by midpoints.
So, centroid of triangles $△ABC=△A_1{B}_1{C}_1=△A_2{B}_2{C}_2\dots =(\frac{3-4+10}{3},\frac{6+7-7}{3})=(3,2)$


Area of triangle formed by midpoints $=\frac{1}{4}\times$Area of original triangle.

So, $\frac{\text{Area (}△A_2{B}_2{C}_2\text{)}}{\text{Area (}△A_1{B}_1{C}_1\text{)}}=\frac{1}{4}$
and $\frac{\text{Area (}△A_1{B}_1{C}_1\text{)}}{\text{Area (}△ABC\text{)}}=\frac{1}{4}$
$\therefore \frac{\text{Area (}△A_2{B}_2{C}_2\text{)}}{\text{Area (}△ABC\text{)}}=\frac{1}{16}$

Similarly, $\frac{\text{Area (}△A_3{B}_3{C}_3\text{)}}{\text{Area (}△ABC\text{)}}=\frac{1}{64}$