A Geometry problem

os=5-2 =3cm NOW IT IS A RIGHT ANGLE TRAINGL SO X=4CM

Raidus is known, 5cm. Using pythagoras theorem, ans is 4cm.

lol I created a function y=sqr(25-x^2) and plugged in 3 for y and solved for x.

using P.T. a=3; b= x; c= 5

just using Pythagorean theorem (5 5)=(3 3)+(x*x) , here 3= (5-2)

Radius OR = 5. Thus, OS = OR-SR= 5-2=3cm. Let P be the left end-point of line segment x. Thus, Radius OP=5. Thus, in right triangle OPS, By Pythagoras's theorem, PS^2= 5^2 -3^2 = 25-9 =16. Thus PS = 4 or -4. But length is always positive. Thus PS = 4cm

Using Similar Triangles

8/x = x/2

x^2 = 16 x = 4cm

Length of chord formula is 2under root radius square - perpendicular height from centre to chord. Subsituting these values we get 2 underroot 25 - 9 == 8, therefore length of half chord inquestion is 4 Ans

.K.K..GARG,India

Here the radius is 5 cm.

So, OR = 5 cm

SR = 2 cm and

OS = OR - SR = 5 cm - 2cm = 3cm

Draw a radius OA that starts with O and ends in the point of the line opposite to the line x.

Since, radius of circles are equal radius OA = OR = 5 cm

By using Pythagoras theorem on right angled triangle OSA we have,

     (OA) ^ {2} = (OS) ^ {2} +  (AS) ^ {2}

or, (AS) ^ {2} = (OA) ^ {2} - (OS) ^ {2}

or, (AS) ^ {2} = (5cm) ^ {2} - (3cm) ^ {2}

or, (AS) ^ {2} = 25 (cm) ^ {2} - 9(cm) ^ {2}

or, (AS) ^ {2} = 16\­( (cm) ^ {2}

or, (AS) ^ {2} = sqrt{16\­( (cm) ^ {2}}

Hence, AS = 4cm

The chord is bisected by the radius OR.

So, AS = x = 4cm

Hence, x =4cm

Its very simple. Just use the Pythagorean theorem and solve to get the answer 4 cm .