Question

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.

Answer 1

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  

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 * 8²cos∠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*ORcos∠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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