Question

2. Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm​

Answer 1

★ Refers to the attachments

Let AB be a chord of a circle with centre O and radius 10 cm .

From O, draw OP perpendicular to AB. Join OA .

• OP = 5 cm and OA = 10 cm

In right triangle OPA ,we have

$\sf{OA {}^{2} = OP {}^{2} + AP {}^{2} }$

$\implies \sf{{10}^{2} = {5}^{2} + {AP}^{2} }$

$\implies \sf{100 = 25 + {AP}^{2} }$

$\implies \sf{ {AP}^{2} = 100 - 25}$

$\implies \sf \: AP = \sqrt{75}$

$\implies \sf{AP = 8.66}$

Since, the perpendicular from the centre to a chord bisects the chord.

Therefore,

AB = 2AP = 2 × 8.66 = 17.32 cm

Hence, the length of a chord is 17.32 cm.

Answer 2

GIVEN:

• Radius of the circle (OA) = 10 cm
• OC = 5 cm

TO FIND:

• What is the length of the chord ?

SOLUTION:

Let O be the centre of the circle of radius 10 cm.

Let AB be a chord of the circle. Then,

$\rm{\dashrightarrow AC = CB \:and \: OC \perp AB }$

Applying Pythagoras theorem in ∆OCA

$\bf{\star \: (Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2\: \star}$

According to question:-

On putting the given values in the formula, we get

$\rm{\dashrightarrow (OA)^2 = (AC)^2 + (OC)^2 }$

$\rm{\dashrightarrow (10)^2 = (AC)^2 + (5)^2}$

$\rm{\dashrightarrow 100 = AC^2 + 25 }$

$\rm{\dashrightarrow 100 - 25 = AC^2}$

$\rm{\dashrightarrow \sqrt{75} = AC^2}$

$\bf{\dashrightarrow 8.66 \: cm = AC }$

$\rm{\dashrightarrow AB = AC + CB }$ ( AC = CB )

$\rm{\dashrightarrow AB = 8.66 + 8.66}$

$\bf{\dashrightarrow AB = 17.32 \: cm }$

❝ Hence, the length of the chord is 17.32 cm ❞

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