Question

Two circle at of radii 5 cm and 3cm intersent at two points the distance between their centres is 4cm . find the lenth of common chord​

Answer 1

$\huge \green{given : }$

• common chord AB in two circles having P and Q as the center.
• AP is the radius of the first circle having length 5cm and AQ is the radius of the other circle having length 3cm.
• Now, segment PQ is perpendicular to the chord AB using the theorem "The perpendicular from center to the chord, bisect the chord".
• Therefore, the length of AR is the half of the length AB using the theorem.
• After that using Pythagoras Theorem in the both triangle ARP and triangle ARQ. We get the value of length of AR, so the length of AB is half of the length of AR.

$\huge \green{solution : }$

$\sf \: Let \: the \: common \: chord \: be \: AB. \: P \: and \: Q \: are \: the \: \: centers \: of \: circles.$

$\sf{The \: length \: of \: AP \: is \: given \: in \: the \: question, } \bf \: AP = 5cm.$

$\sf \: The \: length \: of \: PQ \: is \: also \: given \: in \: the \: question,$

$PQ = 4cm.$

$\sf \: Using \: the \: theorem \: "The \: perpendicular \: from \: center \: to \: the \: chord, \: bisect \: the \: chord".$

$\therefore \: AR= RB \: = \frac{1}{2} AB$

$\sf \: Let \: the \: length \: of \: PR \: be \: x \: cm. \\ \\ \sf \: \: So, \: the \: length \: of \: RQ = (4x)cm.$

$\sf \blue{Now, \: consider \: the \: triangle \: ARP \: and \: apply \: Pythagoras \: Theorem \: in \: it:}$

$\sf \red{In \: triangle \: ARP,}$

$\bf \: AP² = AR² + PR$

$\bf \: AR² = {(5)}^{2} - {x}^{2}$

$\bf \: AR² = 25 - {x}^{2} \: \: \: \: \: - - (1)$

$\sf \: Applying \: the \: Pythagoras \: Theorem \: in \: the \: triangle \: ARQ,$

$\bf \: AQ² = AR² + QR²$

$\bf \: AR² = {(3)}^{2} - {(4 - x)}^{2} \: \: \: \: \: - - (2)$

$\sf \: Using \: equation \: (1) \: and \: equation \: (2), \: we \: get:$

$\therefore \: \: \: \: \: \bf 5² - x² = 3² - (4-x)²$

$\bf \: \: \: \: \: \: 25-x² =9-(16-8x+x²)$

$\sf we \: expand \: the \: \bf (4-x)² \sf \: using \: the \: identity$

$\: \: \: \: \: \: \: \: \: \: \: \bf \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

$\sf \red{Solving \: further, }$

$\implies\bf \: 25 - \cancel{{x}^{2}} = - 7 + 8x - \cancel{{x}^{2} }$

$\implies \bf \: 25 + 7 = 8x$

$\implies \bf \: 32 = 8x$

$\implies \bf \: x = \frac{ \cancel{32}}{ \cancel{8}}$

$\boxed { \implies \bf \: x = 4}$

$\sf \: Putting \: the \: value \: of \: x, in \: equation \: number \: 1, \: we \: get:$

$\bf \: \: \: \: AR² = 25-16$

$\bf \: \: \: \: \therefore \: \: \: \: AR² = 3cm$

$\bf \: \: \therefore \: \: \: AB = 2 × AR$

$\bf \: \: \: \therefore \: \: \: AB = 2 × 3$

$\bf \: \: \: \therefore \: \: \: AR = 6cm$

$\boxed{ \underline{ \red{Hence, the \: length \: of \: the \: common \: chord \: AB \: is \: 6cm}}}$