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Question

In the figure below, the rectangle at the corner measures $$10 cm \times 20 cm.$$ The corner A of the rectangle is also a point on the circumference of the circle. What is the radius of the circle in cm?

103342.jpg


A
10 cm
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B
40 cm
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C
50 cm
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D
30 cm
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Solution

The correct option is C 50 cm
$$ We\quad take\quad the\quad corner\quad O\quad of\quad the\quad given\quad square\quad OPQR,\quad containing\quad the\quad 10cm\times 20cm\\ rectangle,\quad as\quad the\quad origine\quad and\quad the\quad sides\quad OP\quad \& \quad OR\quad as\quad the\quad coordinate\quad axes.\\ Let\quad the\quad sides\quad of\quad OPQR\quad be\quad 2a.\\ So\quad OPQR\quad lies\quad in\quad the\quad third\quad quadrant.\quad \quad The\quad inscribe\quad circle's\quad radius\quad is\quad r=a\\ \& \quad centre\quad is\quad C(a,a).\\ The\quad circle\quad passes\quad through\quad A(-10,-20).\\ \therefore \quad The\quad equation\quad of\quad the\quad circle\quad is\\ { \left( x-h \right)  }^{ 2 }+{ \left( y-k \right)  }^{ 2 }={ r }^{ 2 }..........(i)\\ Here\quad \left( h,k \right) =\left( a,a \right) \quad and\quad r=a\\ \therefore \quad equation\quad (i)\quad becomes\\ { \left( x-a \right)  }^{ 2 }+{ \left( y-a \right)  }^{ 2 }={ a }^{ 2 }..........(ii)\\ The\quad circle\quad passes\quad through\quad A(-10,-20).\\ i.e\quad \left( x,y \right) =\left( -10,-20 \right) \\ So\quad equation\quad (ii)\quad reduces\quad to\\ { \left( -10-a \right)  }^{ 2 }+{ \left( -20-a \right)  }^{ 2 }={ a }^{ 2 }.\\ On\quad solving\quad for\quad a\quad we\quad get\\ |a|=50\quad units\quad and\quad 10\quad units.\quad \\ (a\quad is\quad a\quad length.\quad So\quad we\quad do\quad not\quad consider\quad the\quad negative\quad sign).\\ Also\quad A(-10,-20)\quad lie\quad on\quad the\quad circle.midway\quad between\quad (a,0)\quad \& \quad (0,a).\\ So\quad a>10\quad units\quad \& \quad a>20\quad units.\\ \therefore \quad we\quad reject\quad a=10.\\ so\quad a=50units\\ Ans-\quad Option\quad C\\ \\ \\  $$
131788_103342_ans.png

Mathematics

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