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
If an isosceles triangle is inscribed in a circle of radius 10 cm then the area of the triangle is 23.46 cm².
Step-by-step explanation:
It is given that,
ΔPQR is an isosceles triangle inscribed in a circle with center O where PQ = PR = 8 cm
The radius of the circle PO = OQ = OR = 10 cm (refer to the figure attached below).
Required formula:
(1) Area of the triangle = ½ * [product of two sides] * [sine of the included angle]
(2) The cosine rule: a² = b²+ c² – (2bc cos A) ← this rule holds true for all respective sides and angles.
Step 1:
Based on formula in (1), we have
The area of ∆ PQR as,
= ½ * [PQ * PR ] * sin ∠P
= ½ * 8 * 8 * sin ∠P
= ½ * 64 * sin ∠P
= 32 * sin ∠P …… (i)
Step 2:
Now, considering the side QR of triangle PQR and applying the cosine rule for the formula in (2), we get
QR² = PQ² + PR² - [2PQ * PRcos∠P]
⇒ QR² = 8² + 8² - [2 * 8²*cos∠P]
⇒ QR² = 64 + 64 – 128 [cos∠P]
⇒ QR² = 128* [1 - cos∠P] …… (ii)
Similarly, considering the side QR of triangle QOR and applying the cosine rule for the formula in (2), we get
QR² = QO² + RO² - [2 QO*OR*cos∠QOR]
⇒ QR² = 10² + 10² - [2*10²cos∠2P] …… [since angle subtended by chord QR at the centre is twice the angle subtended at the arc, therefore ∠QOR = 2∠P )
⇒ QR² = 100 + 100 – 200[ 1 - cos∠2P]
⇒ QR² = 200 * [ 1 - cos∠2P] ……. (iii)
Step 3:
Comparing eq. (ii) & (iii), we get
128* [1 - cos∠P] = 200 * [ 1 - cos∠2P]
⇒ 16* [1 - cos∠P] = 25 * [ 1 - cos∠2P]
⇒ 16 * [1 - cos∠P] = 25 * [1 – (2cos²∠P-1)]
⇒ 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][25cos∠P + 17] = 0
⇒ cos∠P = 1 or cos∠P = -17/25
[For cos∠P = 1, the angle ∠P = 0, so this is not possible ]
∴ cos∠P = -17/25
⇒ ∠P = cos⁻¹ [-17/25] = 132.84°
∴ sin∠P = sin 132.84° = 0.733
Thus, substituting the value of sin∠P in eq. (i), we get
The Area of triangle PQR,
= 32 * 0.733
= 23.46 cm²
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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
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