Math, asked by Shreem2726, 3 months ago

P is a point at a distance 12 cm from the centre O of a circle of radius 5 cm. PA and PB are the tangents to the circle A and B. If AP = 12 cm then find the value of chord AB​


amitnrw: AP can not be 12 if P is a point at a distance 12 cm from the centre O , AP will be sqrt (119)

Answers

Answered by Anonymous
25

Given:

✰ P is a point at a distance 12 cm from the centre O of a circle of radius 5 cm.

✰ PA and PB are the tangents to the circle A and B.

✰ AP = 12 cm

To find:

✠ The value of chord AB.

Solution:

We know that,

⟹ d = 12 cm

⟹ r = 5 cm

Here,

  • d is the point P at a distance 12 cm from the centre O.
  • r is of a circle of 5 cm.

Length of the tangent = l

Length of the tangent = √(d²-r²)

➤ l = √(12²-5²)

➤ l = √144-25

➤ l = √119

➤ PA = PB = √119 cm

Area of rhombus = 1/2 × product of diagonals

⟼ Area of ∆OAP = ∆OBP = 1/2 × 5 × √119

⟼ Area of rhombus OAPB = 2 × 1/2 × 5 × √119

⟼ Area of rhombus OAPB = 5√119

  • One diagonal = OP = 12 cm
  • Another diagonal = AB

Hence,

⟼ 1/2 × 12 × AB = 5√119

⟼ 6 × AB = 5√119

⟼ AB = 5√119/6 cm

The value of chord AB = 5√119/6 cm

_______________________________

Answered by amitnrw
24

Given : P is a point at a distance 12 cm from the centre O of a circle of radius 5 cm.

PA and PB are the tangents to the circle at A and B.  

To find : the value of chord AB.​

Solution:

OP = 12  cm

OA = OB = Radius = 5 cm

PA² = OP² - OA²

=> PA² = 12² - 5²  = 119

=> PA  = √119

area of ΔOAP = ΔOBP = (1/2) * 5 * √119

=> area of OAPB ( rhombus) = 2 * (1/2) * 5 * √119  = 5√119

Also area of rhombus = (1/2) * product of diagonals

one diagonal = OP = 12

another diagonal = AB

Hence = (1/2) * 12 * AB  =   5√119

=> AB = 5√119/6  cm

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