SATHEE: Properties of Triangles 3 Question 7

Properties of Triangles 3 Question 7

7. The radius of the circumcircle of the $△ P R S$ is

(a) 5

(b) $3 \sqrt{3}$

(d) $2 \sqrt{3}$

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Answer:

Correct Answer: 7. (b)

Solution:

Equation of circumcircle of $△ P R S$ is

$(x + 1) (x - 9) + y^{2} + λ y = 0$

It will pass through $(1, 2 \sqrt{2})$ , then

$- 16 + 8 + λ \cdot 2 \sqrt{2} = 0$

$\Rightarrow λ = \frac{8}{2 \sqrt{2}} = 2 \sqrt{2}$

$∴$ Equation of circumcircle is

$x^{2} + y^{2} - 8 x + 2 \sqrt{2} y - 9 = 0$

Hence, its radius is $3 \sqrt{3}$ .

Alternate Solution

Let $∠ P S R = θ \Rightarrow \sin θ = \frac{2 \sqrt{2}}{2 \sqrt{3}}$

$∴ \sin θ = \frac{P R}{2 R}$

$\Rightarrow P R = 6 \sqrt{2} = 2 R \cdot \sin θ$

$\Rightarrow R = 3 \sqrt{3}$

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Properties of Triangles 3 Question 7

7. The radius of the circumcircle of the △PRS is

Answer:

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7. The radius of the circumcircle of the $△ P R S$ is