Question : ABC is a right-angled triangle with AB = 6 cm and BC = 8 cm. A circle with centre O has been inscribed inside $\triangle ABC$. The radius of the circle is:
Option 1: 1 cm
Option 2: 2 cm
Option 3: 3 cm
Option 4: 4 cm
Correct Answer: 2 cm
Solution :
Let x be the radius of the circle.
In the right-angled $\triangle$ ABC,
AC
2
= AB
2
+ BC
2
(by Pythagoras Theorem)
⇒ AC
2
= 6
2
+ 8
2
⇒ AC
2
= 36 + 64
⇒ AC
2
= 100
$\therefore$ AC = 10
Now in quadrilateral OPBR, $\angle$B = $\angle$P = $\angle$R = 90°
Hence, $\angle$ROP = 90° (sum of all angles of a quadrilateral is 360°) and also OP = OR (each radius).
Let OPBR be a square with each side x cm.
So, BP = RB = x cm
Therefore, CR = (8 − x) and PA = (6 − x)
Since the tangents from an external point to a circle are equal in length,
$\therefore$ AQ = AP = (6 − x) and CQ = CR = (8 − x)
Now, AC = AQ + CQ
⇒ 10 = 6 − x + 8 – x
⇒ 10 = 14 − 2x
$\therefore$ x = 2 cm
Hence, the correct answer is 2 cm.
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