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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
A square is given with side 10 and a circle drawn inside square (Now touching sides of squares). 3 tangents are drawn from circle to corner of square of length 7, 3, 8. Find radius of Circle.
Step 1:
Give names to each point and center of circle.
Step 2:
For each Tangent, draw radius and join center of circle with corner of circle.
Step 3:
Calculate distance between center of circle and corner of square.
\[AO = \sqrt{R^2+9}\]
\[BO = \sqrt{R^2+64}\]
\[DO = \sqrt{R^2+49}\]
Step 4:
Now consider Triangle ADO.
Let ∠DAO = θ
AD is 10 and we calculated AO and DO
Apply cosine rule:
\[cos\theta=\frac{10^2+(R^2+9)-(R^2+49)}{2\times10\times\sqrt{R^2+9}}\]
\[cos\theta=\frac{60}{20\sqrt{R^2+9}}\]
\[cos\theta=\frac{12}{4\sqrt{R^2+9}}\]
Step 5:
Now consider Triangle ABO.
Let ∠BAO = 90-θ
AB is 10 and we calculated AO and BO
Apply cosine rule:
\[cos(90-\theta)=\frac{10^2+(R^2+9)-(R^2+64)}{2\times10\times\sqrt{R^2+9}}\]
\[sin\theta=\frac{45}{20\sqrt{R^2+9}}\]
\[sin\theta=\frac{9}{4\sqrt{R^2+9}}\]
Step 6:
Apply:
\[sin^2\theta+cos^2\theta=1\]
Now put all values:
\[\Rightarrow\left(\frac{9}{4\sqrt{R^2+9}}\right)^2+\left(\frac{12}{4\sqrt{R^2+9}}\right)^2=1\]
\[\Rightarrow\frac{225}{16(R^2+9)}=1\]
\[\Rightarrow16(R^2+9)=225\]
\[\Rightarrow16R^2+144=225\]
\[\Rightarrow16R^2=81\]
\[\Rightarrow R^2=\frac{81}{16}\]
\[\Rightarrow R=\frac{9}{4}\]
So we get:
R=9/4
or
R=2.25
