Download App

Circles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsCircles1 Mark
MCQ
In the given figure,$ \triangle\text{ABC}$ is right$-$angled at $B,$ such that $BC = 6\ cm$ and $AB = 8\ cm$. A circle with centre $O$ has been inscribed in the triangle. $\text{OP} \perp \text{AB}, \text{OQ} \perp \text{BC} $ and $\text{OR} \perp \text{AC}.$ If $\text{OP = OQ = OR} = x \ cm,$ then $x =?$
  • ✓
    $2\ cm$
  • B
    $2.5\ cm$
  • C
    $3\ cm$
  • D
    $3.5\ cm$
✓

Answer

Correct option: A.
$2\ cm$
In right $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2 ....($By Pythagoras theorem$)$
$\Rightarrow\text{AC}^2=8^2+6^2$
$\Rightarrow\text{AC}^2=64+36$
$\Rightarrow\text{AC}=100$
$\Rightarrow\text{AC}=10\text{cm}$
We know that tangent from an external poit to the circle are equal.
$\Rightarrow\text{CR}=\text{CQ}=\text{BC}-\text{BQ}=(6-\text{x})\text{cm}$
$\Rightarrow\text{AR}=\text{AP}=\text{AB}-\text{BP}=(8-\text{x})\text{cm}$
$\text{AC}=(\text{AR}+\text{CR})=(8-\text{x})+(6-\text{x})=(14-2\text{x})\text{cm}$
$\Rightarrow14-2\text{x}=10$
$\Rightarrow2\text{x}=4$
$\Rightarrow\text{x}=2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Circles→Circles — all question types→Maths STD 10 question papers→Maharashtra Board STD 10 question papers→Maharashtra Board question paper generator→

Similar questions

While computing the mean of the grouped data, we assume that the frequencies are:
→
If NAV of one unit is ₹ 35 then the number of units allotted for investment of ₹ 4270 is _______ .
→
In $\triangle\text{ABC}$ it is given that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}.$ If $\angle\text{B}=70^\circ$ and $\angle\text{C}=50^\circ$ then $\angle\text{BAD}=?$
→
$\triangle\text{ABC}\sim\triangle\text{PQR}$ such that $\text{ar}(\triangle\text{ABC})=4\ \text{ar}(\triangle\text{PQR}).$ If $\text{BC} = 12\ cm,$ then $\text{QR} =$
→
$\frac{1+\cot ^2 A }{1+\tan ^2 A }=?$
→
A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is:
→
In the given figure, the length of $BC$ is:
→
A right circular cylinder of radius $r$ and height $h (h > 2r)$ just encloses a sphere of diameter:
→
cosec45° = ?
→
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=$
→