Find the Relation between the Area of a Circle and the Semicircle Inside a Square
From the figure, ABCD is a square and CT is the tangent of the semicircle and circle. Then find the Relation between the area of the semicircle and the area of the circle inside the square.
![Geometry Math Problem: Find the Relation between the Area of a Circle and the Semicircle Inside the Square](https://educatemath.com/wp-content/uploads/2023/12/Find-the-Relation-Between-Area-of-Circle-and-Semicircle-Inside-a-Square-1.jpg)
Solution
Connect C and the center of the semicircle
![Geometry Math Problem with Solution: Find the Relation between the Area of a Circle and the Semicircle Inside the Square](https://educatemath.com/wp-content/uploads/2023/12/Find-the-Relation-Between-Area-of-Circle-and-Semicircle-Inside-a-Square-2.jpg)
Let’s assume the radius of the semicircle is R and the radius of the circle is r, then
Area of red semicircle = πR²/2
Area of blue circle = πr²
Now, the area of the blue circle/area of the red semicircle = (πr²)/(πR²/2)
⇒ Area of blue circle/area of red semicircle = 2r²/R²
Apply Pythagoras’ theorem in the triangle OBC
OC² = OB² + BC²
⇒ OC² = R² + (2R)²
⇒ OC² = R² + 4R² = 5R²
Thus, OC = R√5
From the figure triangle OBC and triangle PQC are similar, so
OB/PR = OC/PC
⇒ R/r = (R√5)/(R√5 – (R + r))
⇒ R(R√5 – (R + r)) = (R√5)r
so, R√5 – (R + r) = r√5
⇒ R(√5 – 1) = r(√5 +1)
⇒ r/R = (3 – √5)/2
Square both sides, then
r²/R² = (3 – √5)² / 4 = (14 – 6√5) / 4
⇒ r²/R² = (7 – 3√5) / 2
⇒ 2r²/R² = 7 – 3√5
Area of blue circle/area of red semicircle = 7 – 3√5