This question was uploaded on 01/09/21 on social media accounts.
One circle and one semicircle are inscribed in a semicircle such that the inscribed semicircle is having a radius half of large semicircle and both having the same base. A small circle is touching both semicircles (One from internally and one from externally) and also touches the baseline. Find the area small circle.
Step 1:
Draw lines to join centers of all circles.
Draw a perpendicular from the center of the small circle to the baseline.
Let the radius of the small circle is 'r'.
Step 2:
Now focus on the figure ABCD,
AB = 3 + r;
BD = 6 - r;
CD = r;
AB = 3;
And also CD is perpendicular to AC.
Step 3:
Triangle BCD is a right-angled triangle,
\[\Rightarrow BC^2=BD^2-CD^2\]
\[\Rightarrow BC^2=(6-r)^2-r^2\]
\[\Rightarrow BC^2=36-12r\]
\[\Rightarrow BC=\sqrt{36-12r}\]
Also,
Triangle ACD is a right-angled triangle,
\[\Rightarrow AC^2=AD^2-CD^2\]
\[\Rightarrow AC^2=(3+r)^2-r^2\]
\[\Rightarrow AC^2=6r+9\]
\[\Rightarrow AC=\sqrt{6r+9}\]
Now AC = AB + BC:
\[\Rightarrow \sqrt{6r+9} = 3 + \sqrt{36-12r}\]
\[\Rightarrow (6r+9) = 9 + (36-12r) + 6(\sqrt{36-12r})\]
\[\Rightarrow \sqrt{36-12r}=3r-6\]
\[\Rightarrow 36-12r=9r^2-36r+36\]
\[\Rightarrow 9r^2=24r\]
\[\Rightarrow r=\frac{8}{3}\]
Now the area of the small circle is πr²
Area = π(64/9)
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Puzzle related to Geometry, Circle, Area
