This question was uploaded on 19/10/21 on social media accounts.
A triangle is inscribed in a triangle such that vertices of the inner triangle divide side of the outer triangle in the ratio 5:3, 3:2, 1:1. Find the fraction of the area of the inner triangle to the outer triangle.
Solution:
Let the area of the outer triangle be 'A'.
\[A1=\frac{5\cdot3}{8\cdot5}=\frac38\]
\[A2=\frac{3\cdot1}{8\cdot2}=\frac3{16}\]
\[A3=\frac{2\cdot1}{5\cdot2}=\frac15\]
Now,
\[\frac {A'}A=\frac{A-A1-A2-A3}{A}=\frac{19}{80}\]
Fraction = 19:80
To calculate ratios of areas, I used the formula for the area of triangle:
\[A=a\cdot b\cdot\sin\theta\]
sine term is cancelled due to one angle being same in triangles.
Puzzle related to hard math puzzle, geometry, triangle, area, ratio.
