Question

. If radii of two concentric circles are 4cm and 5cm, then what will be the length
of each chord of one circle which is tangent to the other circle.

Answer 1

${\large{\underline{\pmb{\sf{Given...}}}}}$

★ The radii of two concentric circles are 4cm and 5cm.

${\large{\underline{\pmb{\sf{To\; Find...}}}}}$

★ The length  of each chord of one circle which is tangent to the other circle.

${\large{\underline{\pmb{\sf{Understanding\; the \; concept...}}}}}$

☀️ Concept : Here, we have been provided with the measure of the radii of the two concentric circles which means two circles with the same centre and that their radii are 4cm and 5cm respectively. where we are said to find out the length  of each chord of one circle which is tangent to the other circle.

❍ Now, let's draw an imaginary diagram [ something like the diagram which is given in the attachment by me ] for a better reference and that it would be easy to understand which formulae need to be used

${\large{\underline{\pmb{\sf{Using \; the \; concpt...}}}}}$

✪ Formula related to pythagoras therom :

$\tt Hypotenuse^{2} = Base^{2} + Side^{2}$

${\large{\underline{\pmb{\sf{Solution...}}}}}$

★ The length of the chord which is tangent to the other circle is 6cm

${\large{\underline{\pmb{\sf{Full \; Solution...}}}}}$

~ Here, wee can see that chord which is the tangent of the other circle forms a right triangle. so, let's use pythagoras theorem to find the length of the chord [ unknown side ]

${\bigstar\;{\underline{\purple{\underline{\sf{Formula...}}}}}}$

$\bf Hypotenuse^{2} = Base^{2} + Side^{2}$

${\bigstar\;{\underline{\purple{\underline{\sf{Where...}}}}}}$

• Base = chord
• Side = 4cm
• Hypotenuse = 5cm

~ Now let's apply the theorem for triangle OPB and find the lenght of PB and later double it to find the length of the chord...

$: \implies \sf H^{2} = S^{2} + B^{2}$

$: \implies \sf (5cm)^{2} = (4cm)^{2} + (base)^{2}$

$: \implies \sf 25cm^{2} = 16cm^{2} + (base)^{2}$

$: \implies \sf (base)^{2} = 25cm^2 - 16cm^2$

$: \implies \sf (base)^2 = 9cm^2$

$: \implies \sf base = \sqrt{9cm^2}$

$: \implies \sf base = 3cm$

~ Now let's find the length of the chord by doubling length of the base of the triangle OPB

$: \implies \sf Length \;of \;the\; chord = PB \times 2$

$: \implies \sf Length \;of \;the\; chord = 3cm \times 2$

$: \implies \sf Length \;of \;the\; chord = 6cm$

• Henceforth the length of the chord is 6cm

Answer 2

Answer:

ToFind...

ToFind...

★ The length of each chord of one circle which is tangent to the other circle.

{\large{\underline{\pmb{\sf{Understanding\; the \; concept...}}}}}

Understandingtheconcept...

Understandingtheconcept...

☀️ Concept : Here, we have been provided with the measure of the radii of the two concentric circles which means two circles with the same centre and that their radii are 4cm and 5cm respectively. where we are said to find out the length of each chord of one circle which is tangent to the other circle.

❍ Now, let's draw an imaginary diagram [ something like the diagram which is given in the attachment by me ] for a better reference and that it would be easy to understand which formulae need to be used

{\large{\underline{\pmb{\sf{Using \; the \; concpt...}}}}}

Usingtheconcpt...

Usingtheconcpt...

✪ Formula related to pythagoras therom :

\tt Hypotenuse^{2} = Base^{2} + Side^{2}Hypotenuse

2

=Base

2

+Side

2

{\large{\underline{\pmb{\sf{Solution...}}}}}

Solution...

Solution...

★ The length of the chord which is tangent to the other circle is 6cm

{\large{\underline{\pmb{\sf{Full \; Solution...}}}}}

FullSolution...

FullSolution...

~ Here, wee can see that chord which is the tangent of the other circle forms a right triangle. so, let's use pythagoras theorem to find the length of the chord [ unknown side ]

{\bigstar\;{\underline{\purple{\underline{\sf{Formula...}}}}}}★

Formula...

\bf Hypotenuse^{2} = Base^{2} + Side^{2}Hypotenuse

2

=Base

2

+Side

2

{\bigstar\;{\underline{\purple{\underline{\sf{Where...}}}}}}★

Where...

Base = chord

Side = 4cm

Hypotenuse = 5cm

~ Now let's apply the theorem for triangle OPB and find the lenght of PB and later double it to find the length of the chord...

: \implies \sf H^{2} = S^{2} + B^{2}:⟹H

2

=S

2

+B

2

: \implies \sf (5cm)^{2} = (4cm)^{2} + (base)^{2}:⟹(5cm)

2

=(4cm)

2

+(base)

2

: \implies \sf 25cm^{2} = 16cm^{2} + (base)^{2}:⟹25cm

2

=16cm

2

+(base)

2

: \implies \sf (base)^{2} = 25cm^2 - 16cm^2:⟹(base)

2

=25cm

2

−16cm

2

: \implies \sf (base)^2 = 9cm^2:⟹(base)

2

=9cm

2

: \implies \sf base = \sqrt{9cm^2}:⟹base=

9cm

2

: \implies \sf base = 3cm:⟹base=3cm

~ Now let's find the length of the chord by doubling length of the base of the triangle OPB

: \implies \sf Length \;of \;the\; chord = PB \times 2:⟹Lengthofthechord=PB×2

: \implies \sf Length \;of \;the\; chord = 3cm \times 2:⟹Lengthofthechord=3cm×2

: \implies \sf Length \;of \;the\; chord = 6cm:⟹Lengthofthechord=6cm

Henceforth the length of the chord is 6cm