Question

a circle of radius 5 cm . in this circle the two chord AB and CD length same 6 cm . find length BC.​

Answer 1

Answer:

$\sf\huge\boxed{BC= 9.6\:cm}$

Given that :-

• Circle of radius = 5cm.

• Two cords AB and CD length is 6 cm.

To find :-

• Length of BC

$\tt\huge{EXPLANATION-}$

∴ P divides BC in the ratio = 6 : 6 = 1 : 1.

⇒ P is mid-point of BC.

⇒ OP ⊥ BC.

In ΔABP, by pythagoras theorem,

AB2 = AP2 + BP2

⇒ BP2 = 62 - AP2 ........(1)

In right triangle OBP, we have

OB2 = OP2 + BP2

⇒ 52 = (5 - AP)2 + BP2

⇒ BP2 = 25 - (5 - AP)2 .......(2)

Equating (1) and (2), we get

62 - AP2 = 25 - (5 - AP)2

⇒ 11 - AP2 = -25 - AP2 + 10AP

⇒ 36 = 10AP

⇒ AP = 3.6 cm

putting AP in (1), we get

BP2 = 62 - (3.6)2 = 23.04

⇒ BP = 4.8 cm

⇒ BC = 2BP = 2 × 4.8 = 9.6 cm

Answer 2

ANSWER:-

Given:

A circle of radius 5cm, in this circle the two chord AB & CD length same 6cm.

To find:

Find the length of BC.

Solution:

Attachment a figure;

Here,

⚫️AB= AC =6cm

⚫️radius,(DA) =5cm

⚫️OD= y

⚫️AD= (5-y)cm

BC = 2x

BD= DC= x cm

Now,

In ∆ABD & ∆ACD,

AB= AC

AD= AD 【common】

angle ADB = angle ADC= 90°

angle BAD = angle CAD

BD= DC = 1/2BC

Therefore,

In ∆BOD,

Using Pythagoras Theorem;

=) BD² + OD²= BO²

=) x² + y² = (5)²

=) x² + y² = 25.............(1)

In ∆ABD,

angleADB = 90°

=) AD² +BD² =AB²

=) (5-y)² + x² = (6)²

=) 25+ y² -10y +x² = 36

=) x² + y² -10y= 36 -25

=) x² +y² -10y =11

=) 25 - 10y = 11 [using eq.(1)]

=) -10y = 11 -25

=) -10y = -14

=) y = 14/10

=) y = 7/5cm

So,

Putting the value of y in equation(1), we get;

$= > {x}^{2} + ( \frac{7}{5} ) {}^{2} = 25 \\ \\ = > {x}^{2} + \frac{49}{25} = 25 \\ \\ = > {x}^{2} = 25 - \frac{49}{25} \\ \\ = > {x}^{2} = \frac{625 - 49}{25} \\ \\ = > {x}^{2} = \frac{576}{25} \\ \\ = > {x}^{2} = ( \frac{24}{5} ) {}^{2} \\ \\ = > x = \frac{24}{5} \\ \\ = > x = 4.8cm \\ \\ = > BD = 4.8cm$

$\frac{1}{2} BC = BD \\ \\ = > BC = 2BD\\ \\ = > BC = 2 \times 4.8 \\ \\ = > BC = 9.6cm$

Thus,

The length of BC is 9.6cm.

Hope it helps