CBSE BOARD X, asked by Madhavsrivatsavk, 1 year ago

AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP.

Answers

Answered by gurudeep87
177
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Answered by HanitaHImesh
20

Given,

The length of the chord, AB = 8 cm

The radius of the circle = 5 cm

The tangents to the circle at A and B intersect at P.

To find,

The length of AP.

Solution,

The length of AP will be 20/3 cm.

We can easily solve this problem by following the given steps.

Now, let's take the centre of the circle to be O.

Let's draw a perpendicular from the centre on the chord at a point C such that AC = CB = 4 cm.

Join OP.

OA = OB = 5 cm ( radius)

OA = OB = 5 cm ( radius)AC = CB = 4 cm

∆ OCA is a right-angled triangle.

Using the Pythagoras theorem in ∆ OCA,

OA² = AC² + OC²

(5)² = (4)² + OC²

25 = 16 + OC²

OC² = 25 - 16

OC² = 9

OC = √9

OC = 3 cm

Then, OP = (CP + 3) cm

Now, let's take AP to be b cm and CP to be a cm.

Then, OP = (a+3) cm

∆ ACP is a right-angled triangle.

Using the Pythagoras theorem in ∆ ACP,

AP² = AC² + CP² --- equation 1

b² = (4)² + a²

b² = 16 + a² --- equation 2

Now, we know that the tangents to the circle meet its radius at a right angle.

So, angle OAP is a right angle and ∆ OAP is a right-angled triangle.

Using the Pythagoras theorem in ∆ OAP,

OP² = AP² + OA²

OP² = (AC² + CP²) + OA² [ Putting the value of AP² from equation 1]

(a+3)² = (4)² + a² + (5)²

a² + (3)² + 2×a×3 = 16 + a² + 25 [ Using the identity, (a+b)² = a² + b² + 2ab]

a² + 9 + 6a = a² + 41

a² - a² + 9 + 6a = 41 ( Moving a² from the right-hand side to the left-hand side will result in the change of the sign from plus to minus.]

9 + 6a = 41

6a = 41-9

6a = 32

a = 32/6

a = 16/3 cm ( Converting 32/6 into the simplest form by dividing both the numerator and denominator by 2.)

Putting the value of a in equation 2,

b² = 16 + a² --- equation 2

b² = 16 + (16/3)²

b² = 16 + 256/9

b² = (16×9 + 256)/9 [Taking the LCM of 9 and 1 and adding the two fractions.]

b² = (144 + 256)/9

b² = 400/9

b =√400/√9

b = 20/3 cm

Hence, the value of AP is 20/3 cm.

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