Question

Two circles having radii 5 cm and 3 cm intersect each other at two distinct points. If the distance between their centres is 4 cm, then what is the length of the common chord?​

Answer 1

Step-by-step explanation:

$\huge\bf\underline{\red{A}\purple{N}\pink{S}\orange{W}\blue{E}\green{R}}$

Given:-

PQ = 4cm

Now, seg PQ l Chord AB

ஃ AR = RB = 1 AB ( perpendicular from the centre to the chord, bisect the chord )

2

Let PR = xcm. So, RQ = (4 - x) cm

In ∆ ARP

${AP}^{2} = {AR}^{2} + {PR}^{2} \\ \\ {AR}^{2} = {5}^{2} - {x}^{2}$

In, ∆ ARQ

${AQ}^{2} = {AR}^{2} + {QR}^{2} \\ \\ {AR}^{2} = {3}^{2} - {(4 - \times )}^{2}$

Therefore,

$25 \: - {x}^{2} = 9 \: - (16 \: - \: 8x \: + {x}^{2} ) \\ \\ 25 \: - {x}^{2} = - 7 \: + 8x \: - \: {x}^{2} \\ \\ 32 \: = \: 8x \\ \\ x \: = \: 4$

Substituting equation,

${AR}^{2} = 25 \: - \: 16 \: = 9$

ஃ AR = 3cm

ஃ AB = 2 × AR = 2 × 3

ஃ AB = 6cm

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So, the length of the common chord AB is 6cm

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