This question was uploaded on 04/08/21 on social media accounts.
In this puzzle, three squares are drawn on the sides of a random triangle and three squares are drawn on the vertices of those three squares. You have to find the ratio of areas of Inner squares to Outer squares.
Give this question a try then check your answer with a solution.
Step 1:
Mark Sides of Inner squares as a, b, c
Step 2:
Mark Sider of Outer squares as x, y, z
Step 3:
Mark any angle in an inner triangle as shown in the figure given below.
Apply cosine rule at angles A and (180-A)
\[cosA=\frac{b^2+c^2-a^2}{2bc} -----(1)\]
\[cos(180-A)=\frac{b^2+c^2-x^2}{2bc} -----(2)\]
cos(180-A) = -cosA
\[\Rightarrow \frac{a^2-b^2-c^2}{2bc}=\frac{b^2+c^2-x^2}{2bc}\]
\[\Rightarrow x^2=2b^2+2c^2-a^2\]
Similarly,
\[y^2=2c^2+2a^2-b^2\]
\[z^2=2a^2+2b^2-c^2\]
Adding these three equations and we will get:
\[x^2+y^2+z^2=3(a^2+b^2+c^2)\]
\[\Rightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=3\]
Note here Green area = x^2+y^2+z^2
and Orange area = a^2+b^2+c^2
⇒ Green Area : Orange Area = 3
Here is the Image solution:
Puzzle related to Geometry, squares, ratio, area
