Question

Let $ABC$ be a triangle with $\angle A={45}^{\circ }$. Let $P$ be a point on the side $BC$ with $PB=3$ and $PC=5$. If ${}^{\prime }{O}^{\prime }$ is the circumcentre of the triangle $ABC$ then the length $OP$ is equal to

• A. $\sqrt{15}$
• B. $\sqrt{17}$
• C. $\sqrt{18}$
• D. $\sqrt{19}$

Answer 1

Answer: (B)

$\sqrt{17}$

$Concept-$ If in a $△ABC$, Circumradius is $R$ and $\angle A=\theta$ then side $BC=2Rsin\theta$

$\Rightarrow BC=2R\sin {45}^o$

$\Rightarrow BC=\sqrt{2}R$

But $BC=BP+PC=3+5=8$

$\Rightarrow R=4\sqrt{2}$

Now from the figure

$MO=NO=R$

$PO=x$

$PB=3$

$PC=5$

where $O$ is circumcentre.

$\Rightarrow PM=MO-PO=R-x$

$\Rightarrow PN=NO+PO=R+x$

Now applying the property of chords of circle,we get

$\Rightarrow PB\times PC=PM\times PN$

$\Rightarrow 5\times 3=(R+x)(R-x)$

$\Rightarrow 15=R^2-x^2$

$\Rightarrow x^2=(4\sqrt{2}{)}^2-15$

$\Rightarrow x=\sqrt{17}$

Hence, the correct answer is option $B$