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In the given figure ABC is a right angled triangle with AB=6 cm, and BC=8 cm. A circle with centre O has been inscribed inside the triangle. Find the radius of the circle. |
Answer» Solution :Let x be the radius of the circle. In right triangle ABC, `AC^(2)=AB^(2)+BC^(2)""`(by Pythagoras Th.) `implies""AC^(2)=6^(2)+8^(2)` `implies""AC^(2)=36+64` `implies""AC^(2)=100` `implies""AC=10` Now in QUADRILATERAL OPBR `angleB=angleP=angleR=90^(@)` each Hance, `""angleROP=90^(@)""`(sum of all angles of a quadrilateral is 360^(@)) and also `""OP=OR""`(each radii) HENCE, OPBR is a square with each side x cm. Hence, `""BP=RB=xcm` Therefore, CR=(8-x) and PA=(6-x) Since, the tangents from an EXTERNAL point to a circle are equal in length `:.AQ=AP=(6-X) and CQ=CR=(8-x)` Now,`""AC=AQ+CQ` `implies""10=6-x+8x` `implies""10=14-2x` `implies""x=2cm` |
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