Question : In the given figure $\mathrm{PQRS}$, a square of side $20\;\mathrm{cm}$ and $\mathrm{SR}$ is extended to point $\mathrm{T}$. If the length of $\mathrm{QT}$ is $25\;\mathrm{cm}$, then what is the distance (in $\mathrm{cm}$) between the centres, $\mathrm{O_1}$ and $\mathrm{O_2}$, of the two circles?
Option 1: $5\sqrt{10}$
Option 2: $4\sqrt{10}$
Option 3: $2\sqrt{10}$
Option 4: $16\sqrt{2}$
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Correct Answer: $5\sqrt{10}$
Solution :
We have, $\mathrm{QT}=25\;\mathrm{cm}$ and $\mathrm{QR=SR}=20\;\mathrm{cm}$
Construction: Join $\mathrm{O_1B}$ and $\mathrm{O_2C}$.
In $\triangle \mathrm{QRT}$,
$⇒\mathrm{QT^2=QR^2+RT^2}$
$⇒\mathrm{RT=15}\;\mathrm{cm}$
Inradius of triangle QRT $=\mathrm{\frac{15+20-25}{2} = 5\;\mathrm{cm}}$
$⇒\mathrm{O_2C=5\;\mathrm{cm}}$
Since, $\mathrm{O_1B=10}\;\mathrm{cm}$ and $\mathrm{AB=5\;\mathrm{cm}}$
$⇒\mathrm{O_1A= O_1B-AB=5\;\mathrm{cm}}$
$⇒\mathrm{O_2A= O_1B+AB=15\;\mathrm{cm}}$
Use Pythagoras' theorem in $\triangle\mathrm{ AO_1O_2}$
$⇒\mathrm{(O_1O_2)^2=(O_1A)^2+(O_2A)^2=15^2+5^2=250}$
$⇒\mathrm{O_1O_2}=5\sqrt{10}\;\mathrm{cm}$
Hence, the correct answer is $5\sqrt{10}$.
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