MCQ
In the figure, if $AP = 10\ cm,$ then $BP =$
- A$\sqrt{91}\text{ cm}$
- ✓$\sqrt{127}\text{ cm}$
- C$\sqrt{119}\text{ cm}$
- D$\sqrt{109}\text{ cm}$
✓
Answer
Correct option: B.
$\sqrt{127}\text{ cm}$
In the figure,
$OA = 6\ cm, OB = 3\ cm$ and $AP = 10\ cm$
$OA$ is radius and $AP$ is the tangent
$OA \perp AP$
Now in right $\triangle\text{OAP}$
$O P^2=A P^2+O A^2=(10)^2+(6)^2$
$=100+36=136$
Similarly $BP$ is tangent and $OB$ is radius
$ O P^2=O B^2+B P^2 $
$ 136=(3)^2+B P^2$
$136=9+B P^2$
$\Rightarrow BP^2= 136 – 9 = 127$
$BP = \sqrt{127}\text{ cm}$
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