Question

An isosceles triangle LMN is inscribed in a circle where LM=LN= 40 cm and MN=48 cm. Find the radius of the circle.

Answer 1

Answer:

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refer the reference diagram uploaded above

Step-by-step explanation:

$so \: let \: the \: centre \: of \: circle \: in \: which \: isosceles \: triangle \: LMN \: is \: inscribed \: be \: O$

$now \: for \: isosceles \: triangle \: LMN \\ LM = LN = 40 \: cm \\ MN = 48 \: cm$

$so \: here \: MO \: is \: radius \: of \: adjoining \: circle \\ so \: as \: isosceles \: triangle \: LMN \: is \: inscribed \: in \: this \: circle \\ we \: can \: say \: that \: the \: given \: circle \: circumscribes \: triangle \: LMN \: and \: thus \: MO \: can \: be \: called \: as \: circumradius$

$so \: circumradius \: is \: calculated \: using \: given \: formula \\ ie \: circumradius = abc \div \sqrt{(a + b + c)(a - b + c)(a + b - c)(b + c - a)} \\ wherein \: a \: b \: and \: c \: are \: sides \: of \: inscribed \: triangle$

$so \: hence \: using \: \\ MO = LM \times LN \times MN \div \sqrt{ \:(LM + MN +LN)(LM - MN +LN)( LM +MN -LN)(MN +LN - LM)} \\ = 48 \times 40 \times 40 \div \sqrt{(40 + 40 + 48)(40 - 48 + 40)(40 - 40 + 48)(40 + 48 - 40)} \\ = 1600 \times 48 \div \sqrt{(128 \times 32 \times 48 \times 48)} \\ = 1600 \times 48 \div 48 \times 8 \sqrt{2} \times 4 \sqrt{2} \\ = 1600 \div 8 \times 4 \times 2 \\ = 1600 \div 64 \\ thus \: MO = 25.cm$

$hence \: radius \: of \: circle \: is \: 25.cm$