Question

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.​

Answer 1

$\huge \star \: \: \underline{ \red{ \mathfrak{solution }} : }$

⍟ Given :

• AB = 5cm
• CD = 11cm
• NM = 6cm (Distance between AB and CD
• OM ⏊ CD
• OM = x
• ON = 6 - x

⍟ To Find :

• Value of radius = r
• Value of 'x'(OM)

⍟ On Solving :

Since , OM ⏊ CD

• M is the mid point of CD

$\large \star \: \underline{ \bold{\: theorem : }}$

➜ The perpendicular from the centre of the circle ta a chord bisects the chord !

• Let OM = x cm

So, MD = MC

$\large \sf \implies \: MD = \dfrac{1}{2} (11) \\ \\ \large \therefore \: \bold{ MD = \frac{11}{2} cm}$

• According to Theorem :

N is the mid point of AB

$\large \star \: \underline{ \bold{\: theorem : }}$

➜ The perpendicular from the centre of the circle ta a chord bisects the chord !

So, NB = AN

$\large \sf \implies \: NB = \dfrac{1}{2} (5) \\ \\ \large \therefore \: \bold{ MD = \frac{5}{2} cm}$

✪ In Right angled triangle ONB :

• Pythagoras Theorem :

$\large \looparrowright \: \boxed{ \rm \: \pink{{h}^{2} = {b}^{2} + {p}^{2} }}$

$\large \implies \sf \: OB²=ON²+NB² \\ \\ \large \sf \mapsto \: {r}^{2} = (6 - x {)}^{2} + {\left( \frac{5}{2} \right)}^{2} \longrightarrow \boxed{ \rm1}$

✪ In Right angled triangle OMD

• Pythagoras Theorem :

$\large \sf \mapsto \: {r}^{2} = {x}^{2} + {\left( { \dfrac{11}{2} } \right) }^{2} \longrightarrow \boxed{2}$

$\large \star \: \: from \: \: \boxed{1} \: \: and \: \: \boxed{2} \: \: we \: \: get :$

$\\ \large \sf \looparrowright \: (6 - x {)}^{2} + { \left( \frac{5}{2} \right)}^{2} = {x}^{2} + { \left( \frac{11}{2} \right)}^{2} \\ \\ \large \sf \implies \: 36 - 12x + {x}^{2} + \frac{25}{4} = {x}^{2} + \frac{121}{4} \\ \\ \large \rm \star \: on \: \: transposing \: \: the \: \: terms \: - \\ \\ \large \sf \implies \: 12x = 36 + \frac{25}{4} - \frac{121}{4} \\ \\ \large \sf \implies \: 12x = 12 \\ \\ \large \sf \implies \: x = \frac{ \cancel{12}} {\cancel{12} } \\ \\ \large \therefore \: \: \underline{ \boxed{ \green{\sf \: x = 1}}}$

➜ Putting x = 1 in [2], we get :

$\large \sf \looparrowright \: {r}^{2} = {1}^{2} + { \left( \dfrac{11}{2} \right)}^{2} \\ \\ \large \implies \sf \: {r}^{2} = 1 + \frac{121}{4} \\ \\ \large \implies \sf \: {r}^{2} = \frac{125}{4} \\ \\ \large \implies \sf \: {r} = \sqrt{ \frac{125}{4} }$

$\\ \\ \large \therefore \: \purple{ \underline{ \boxed{ \sf \green{radius = \frac{5 \sqrt{5} }{2} }}}}$

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⍟ Formulas , identity and Theorem :

• Pythagoras Theorem

$\large \rm \mapsto \: {h}^{2} = {b}^{2} + {p}^{2}$

• Theorem

➜ The perpendicular from the centre of the circle ta a chord bisects the chord !

• Identity

$\large \mapsto \: \rm \: (a + {b)}^{2} = {a}^{2} + {b}^{2} + 2ab$