MCQ
In the figure, $PR =$
- A$20\ cm$
- ✓$26\ cm$
- C$24\ cm$
- D$28\ cm$
✓
Answer
Correct option: B.
$26\ cm$
In the figure, two circles with centre $O$ and $O’$ touch each other externally $PQ$ and $RS$ are the tangents drawn to the circles.
$OQ$ and $O’S$ are the radii of these circles and $OQ = 3\ cm, PQ = 4\ cm, O’S = 5\ cm$ and $SR = 12\ cm.$
Now in right $\triangle\text{OQP}$
$OP^2 = (OQ)^2 + PQ^2$
$= (3)^2 + (4)^2$
$= 9 + 16$
$= 25$
$= (5)^2$
$OP = 5\ cm$
Similarly in right $\triangle\text{RSO}$
$(O’R)^2 = (RS)^2 + (O’S)^2$
$= (12)^2 + (5)^2$
$= 144 + 25$
$= 169$
$= (13)^2$
$O’R = 13\ cm$
Now $PR = OP + OO’ + O’R$
$= 5 + (3 + 5) + 13$
$= 26\ cm.$
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