Question

The radius of a circle is 8 cm and the distance from the center to chord is 6 cm. Find the length of the  chord.​

Answer 1

Let O be the center of the circle .

From O draw OM⊥AB and ON⊥CD

In △AOM

AM=21AB=21×8=4cm

OM2=AO2−AM2

⇒OM=(5)2−(4)2

⇒OM=25−16

⇒OM=9=3cm

In △CON

CN=21CD=21×6=3cm

ON2=CO2−CN2

⇒ON=(5)2−(3)2

⇒ON=25−9

⇒ON=16=4cm

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Answer 2

Answer:

≈  10.6 cm

Step-by-step explanation:

Given radius, r = 8cm

distance from the cenre to the chord = 6cm

=> a right angled triangle is formed with hypotenuse 8cm (radius) and one of the sides as 6cm (distance from centre)

=> other side of the triangle = $\sqrt{r^2 - d^2}$    units    (by pythagorous theorem)

                                              = $\sqrt{8^{2} - 6^{2} }$ cm

                                              = $\sqrt{64 - 36 }$ cm

                                              = $\sqrt{28 }$ cm

                                              = $\sqrt{4 \times 7 }$ cm

                                              = $\sqrt{2^2 \times 7 }$  cm

                                              = $2 \sqrt{7 }$ cm  

which is half of the length of the chord

=> length of the chord = 2 x $2 \sqrt{7 }$ cm

                                     = 4$\sqrt{7 }$ cm

                                     ≈  4 x 2.646 cm

                                     ≈  10.584 cm

                                     ≈  10.6 cm