Question

What will be the diameter of the circle with two parallel chords AB and CD which is 6cm and 8cm respectively and the distance between these two chords is 1cm.
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Answer 1

Topic :-

Circle

Given :-

Two chords AB and CD with length 6 cm and 8 cm respectively are there on a circle. Distance between these two chords is 1 cm.

To Find :-

Diameter of the circle.

Solution :-

Name centre of the circle as 'O'.

Jôin OA and OC.

OA = OC = Radius of the Circle

(∵ Radius of Circle remains same for a circle.)

Draw perpendicular to the chords from the centre 'O'.

Let perpendicular intersects CD at E and AB at F.

So, it is given that, EF = 1 cm.

Now, as we know,

Perpendicular to a chord from the centre of circle bisects the chord into two halves.

So,

CE = 8 cm / 2 = 4 cm

AF = 6 cm / 2 = 3 cm

From Pythagoras Theorem,

(OC)² = (OE)² + (CE)²

(OC)² = (OE)² + (4)²     . . . .  equation (1)

(OA)² = (OF)² + (AF)²

(OA)² = (OE + EF)² + (3)²

(OA)² = (OE + 1)² + (3)²  . . . . equation (2)

OA = OC (∵ It is radius of the circle)

(OA)² = (OC)²

From equation (1) and (2),

(OE + 1)² + (3)² = (OE)² + (4)²

(OE)² + 2(OE) + 1² + 3² = (OE)² + 4²

(OE)² gets cancelled from both sides,

2(OE) + 1 + 9 = 16

2(OE) + 10 = 16

2(OE) = 16 - 10 = 6

2(OE) = 6

OE = 3 cm

Now, put back value of OE in any equation,

(OA)² = (OE + 1)² + (3)²  . . . . equation (2)

(OA)² = (3 + 1)² + (3)²

(OA)² = (4)² + (3)²

(OA)² = 16 + 9 = 25

(OA)² = (5)²

OA = 5 cm

Hence, radius of the Circle is 5 cm.

Diameter of the Circle = 2(Radius of the Circle)

Diameter of the Circle = 2(5 cm)

Diameter of the Circle = 10 cm

Answer :-

So, diameter of the given circle is 10 cm.

Answer 2

$\sf\blue{Given :-}$

➪ Two parallel chords AB and CD with length 6 cm and 8 cm respectively are there on a circle. ➪ Distance between these two chords is 1 cm.

$\sf\blue{To Find :-}$

✈︎ Diameter of the circle.

$\sf\blue{Solution :-}$

✈︎ Name centre of the circle as 'O'.

✈︎ Join OA and OC.

✈︎ OA = OC = Radius of the Circle.

(∵ Radius of Circle remains same for a circle.)

✈︎ Draw perpendicular to the chords from the centre 'O'.

✈︎ Let perpendicular intersects CD at E and AB at F.

✈︎ So, it is given that, EF = 1 cm.

✈︎ Now, as we know,

✈︎ Perpendicular to a chord from the centre of circle bisects the chord into two halves.

So,

☮︎ CE = $\frac{8cm}{2}$ = 4 cm

☮︎ AF = $\frac{6cm}{2}$ = 3 cm

✈︎ From Pythagoras Theorem,

☮︎ (OC)² = (OE)² + (CE)²

☮︎ (OC)² = (OE)² + (4)²     . . . . $\sf\pink{ equation (1)}$

☮︎ (OA)² = (OF)² + (AF)²

☮︎ (OA)² = (OE + EF)² + (3)²

☮︎ (OA)² = (OE + 1)² + (3)²  . . . . $\sf\pink{equation (2)}$

☮︎ OA = OC (∵ It is radius of the circle)

☮︎ (OA)² = (OC)²

✈︎ From equation (1) and (2),

☮︎ (OE + 1)² + (3)² = (OE)² + (4)²

☮︎ (OE)² + 2(OE) + 1² + 3² = (OE)² + 4²

☮︎ (OE)² gets cancelled from both sides,

☮︎ 2(OE) + 1 + 9 = 16

☮︎ 2(OE) + 10 = 16

☮︎ 2(OE) = 16 - 10 = 6

☮︎ 2(OE) = 6

☮︎ OE = 3 cm

✈︎ Now, put back value of OE in any equation,

☮︎ (OA)² = (OE + 1)² + (3)²  . . . . $\sf\orange{By\:equation (2)}$

☮︎ (OA)² = (3 + 1)² + (3)²

☮︎ (OA)² = (4)² + (3)²

☮︎ (OA)² = 16 + 9 = 25

☮︎ (OA)² = (5)²

☮︎ OA = 5 cm

✈︎ Hence, radius of the Circle is 5 cm.

✈︎ Diameter of the Circle = 2(Radius of the Circle)

✈︎ Diameter of the Circle = 2(5 cm)

✈︎ Diameter of the Circle = 10 cm

$\therefore$ Diameter of the given circle is $\fbox\red{10 cm}$