Question

two chords AB, CD of lengths 5cm and 11cm respectively of a circle are parallel. If the distance between AB and CD is 3cm, find the radius of the circle. ​

Answer 1

Construct circle in AB parallel to CD on opposite sides.

Draw M,N perpendicular to AB and CD.

According to figure,

In triangle CNO,

${r}^{2} = \: \: \: \: ( { \frac{11}{2} })^{2} \: \: \: \: + ({3 - x})^{2} \\ r \: \: = \: \: \sqrt{ ({ \frac{11}{2} })^{2} } + \: \: ( {3 - x})^{2} \: \: \: \: \in: (equation 1) \\ in \: triangle \: amo \\ {r}^{2} = ({ \frac{5}{2}) }^{2} + {x}^{2} \\ r \: \: = \: \: \sqrt{ ({ \frac{5}{2} })^{2} } + {x}^{2} \\ from \: equation \: 1st \: and \: 2nd \\ \sqrt{ ({ \frac{11}{2} )}^{2} } + {3 - {x}^{2} x}^{2} = \: \: \sqrt{ ({ \frac{5}{2} })^{2} } + {x}^{2} \\ (squaring \: both \: sides) \\ ({ \frac{11}{2} })^{2} + ( {3 - x)}^{2} = \: \: ( { \frac{5}{2} })^{2} + {x}^{2} \\ \frac{121}{4} + {3}^{2} + {x}^{2} = \: \: ({ \frac{5}{2} })^{2} + {x}^{2} \\ \\ \frac{121}{4} + 9 + {x}^{2} - 2 \times 3 \times x = \frac{25}{4} + { x}^{2} \\ \frac{121}{4} + 9 + {x}^{2} - 6x = \frac{25}{4} + {x}^{2} \\ \frac{121 + 36}{4} + {x}^{2} - 6x = \frac{25}{4} + {x}^{2} \\ \frac{157}{4} + {x}^{2} - 6x = \frac{25}{4} + {x}^{2} \\ {x}^{2} - {x}^{2} - 6x \: = \frac{25}{4} - \frac{157}{4} \\ - 6x \: = \frac{ - 132}{4} \\ x = \frac{331}{6} \\ x = \frac{11}{2} \\ r = \sqrt{( { \frac{5}{2} })^{2} } + ({ \frac{11}{2} })^{2} \\ r = \sqrt{ \frac{25 + 121}{4} } \\ r = \sqrt{ \frac{146}{4} } \\ r = \sqrt{ \frac{73}{2} }$

Answer 2

Step-by-step explanation:

Let O be the centre.

Let distance of first chord from centre be x and that of second chord be y.

Let x be the distance from chord of 5cm and y be the distance from chord of 11 cm.

We know that,

x+y=3

i.e. y=3-x...(1)

The perpendicular drawn from the centre of the cicle bisects the chord.

By pythagorus theorem.

r^2=(2.5)^2+x^2...(2)

r^2=(5.5)^2+y^2...(3)

In eq.(2) and (3) LHS is equal therefore RHS will also be equal.

6.25+x^2=30.25+y^2

x^2-y^2=24...(4)

Substituting eq(1) in eq(4).

x^2-(3-x)^2=24

x^2-(9-6x+x^2)=24

x^2-9+6x-x^2=24

-9+6x=24

6x=33

x=5.5

Substituting x=5.5 in eq(2)

r^2=(2.5)^2+(5.5)^2

r^2=6.25+30.25

r^2=36.50

Taking square root of both sides

r=6.04 cm.(approx)