Hello, snigdha!
Isosceles $\displaystyle \Delta PQR$ is inscribed in a circle with center $\displaystyle O$
such that: .$\displaystyle PQ=PR=13\text{ cm and }QR=10\text{ cm.}$
Find the radius of the circle. Code:
P
* o *
* /|\ *
* / | \ *
* / |r \ *
/ | \ 13
* / | \ *
* / O* r \ *
* / | * \ *
/ | * \
Q o- - - - o - - - -o R
* S 5 *
* *
* * *
In right triangle $\displaystyle PSR\!:\;PS^2 + 5^2 \:=\:13^2 \quad\Rightarrow\quad PS = 12$
. . Then: .$\displaystyle OS \,=\,12-r$
In right triangle $\displaystyle OSR\!:\;\;OS^2 + SR^2 \,=\,OR^2 \quad\Rightarrow\quad (12-r)^2 + 5^2 \:=\:r^2$
And we have: .$\displaystyle 144 - 24t + r^2 + 25 \:=\:r^2 \quad\Rightarrow\quad -24r \:=\:-169$
. . Therefore: .$\displaystyle r \;=\;\frac{169}{24}\text{ cm}$