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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