Question

how to solve this.......​

Answer 1

★ AnswEr :-

• → 4 cm

★ GiVEn :-

• In a circle, the length of chord is 6 cm
• The radius is 5 cm

★ To Find :-

• The distance from the centre of the circle to the chord

★ Diagram :-

$\setlength{\unitlength}{1mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\qbezier(6,25)(10,25)(44,25)\qbezier(5,25)(5,25)(30,45)\qbezier(30,45)(30,45)(30,25)\put(5.2, 25){\circle*{1}} \put(44.8, 25){\circle*{1}}\put(30, 45.2){\circle*{2}}\put(26 ,45){ \sf O }\put(14,40){ \textsf{\textbf{ 5 cm}} }\put(14 ,21){ \textsf{ \textbf{ 3 cm }} }\put(2 ,25){ \sf A }}\put(45.5 ,25){ \sf B }\put(28.5 ,21){ \sf P }\put(28,25){\dashbox{0.01}(2,2)}\end{picture}$

★ Solution :-

→ Let the centre of the circle be O

→ And the distance of the chord to the centre is OP

→ Now, the chord, AB = 6 cm and the radius, OA = 5 cm

→ We know the theorem,

★ The perpendicular from the centre of a circle to a chord bisects the chord.

→ Therefore, AP = PB = 3 cm

→ As OP is perpendicular upon AB,

∠OPA = 90°

→ Therefore,

$\sf \triangle \: OAP \: is \: right \: angle \: triangle$

In a right angle triangle,

$\implies \boxed{ \sf{Hypotenuse^{2} = base^{2} + height^{2} }}$

Where,

• Hypotenuse = 5 cm
• Base = 3 cm
• Height = OP cm

$\implies \boxed{ \sf{5^{2} = 3^{2} + OP^{2} }}$

$\implies \boxed{ \sf{ OP^{2} = {5}^{2} - {3}^{2} }}$

$\implies \boxed{ \sf{ OP^{2} = 25 - 9 }}$

$\implies \boxed{ \sf{ OP^{2} = 16}}$

$\implies \boxed{ \sf{ OP = \sqrt{16} }}$

$\implies \boxed{ \red{ \sf{ \therefore \: OP= 4cm}}}$

The distance between the centre of the circle and chord is 4 cm