An isosceles triangle pqr, in which pq= pr= 8 cm, is inscribed in a circle of radius 10 cm , find the area of the triangle
Answers
Area of triangle = 23.46 cm²
Step-by-step explanation:
Area of triangle = (PQ * PR * Sin ∠P )/2
= (8 * 8 * Sin ∠P )/2
= 32Sin ∠P
QR² = PQ² + PR² - 2PQ * PRCos∠P
=> QR² = 8² + 8² - 2*8²Cos∠P
=> QR² = 128(1 - Cos∠P)
QR² = QO² + RO² - 2 QO*ORCos∠QOR
∠QOR = 2∠P ( angle subtended by chord QR at cneter & arc)
=> QR² = 10² + 10² - 2*10²Cos∠2P
=> QR² = 200 ( 1 - Cos∠2P)
128(1 - Cos∠P) = 200 ( 1 - Cos∠2P)
=> 16(1 - Cos∠P) = 25 ( 1 - Cos∠2P)
=> 16(1 - Cos∠P) = 25 (2 - 2Cos²∠P)
=> 8(1 - Cos∠P) = 25 (1 - Cos²∠P)
=> 25Cos²∠P - 8Cos∠P - 17 = 0
=> 25Cos²∠P - 25Cos∠P + 17Cos∠P - 17 = 0
=> 25Cos∠P(Cos∠P - 1) + 17(Cos∠P - 1) = 0
=> Cos∠P = 1 or Cos∠P = -17/25
Cos∠P = 1 is not possible as angle would be zero
Cos∠P = -17/25
Sin²∠P = 1 - Cos²∠P
Sin∠P = 0.733
Area of triangle = 32Sin ∠P = 23.46 cm²
Learn more:
An equilateral triangle is inscribed in a circle of radius 7 cm. Find the ...
https://brainly.in/question/10153152
a circle with Centre P is inscribed in a triangle ABC side a b side BC ...
https://brainly.in/question/8756941
