SATHEE: Properties of Triangles 3 Question 10

Properties of Triangles 3 Question 10

10. In a $△ X Y Z$ , let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$ respectively and $2 s = x + y + z$ . If $\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$ and area of incircle of the $△ X Y Z$ is $\frac{8 π}{3}$ , then

(2016 Adv.)

(a) area of the $△ X Y Z$ is $6 \sqrt{6}$

(b) the radius of circumcircle of the $△ X Y Z$ is $\frac{35}{6} \sqrt{6}$

$\begin{array}{ll} (c) \sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} = \frac{4}{35} & (d) \sin^{2} \frac{X + Y}{2} = \frac{3}{5} \end{array}$

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Answer:

Correct Answer: 10. $(a, c, d)$

Solution:

Given a $△ X Y Z$ , where $2 s = x + y + z$

and

$\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$

$∴ \frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$

$= \frac{3 s - (x + y + z)}{4 + 3 + 2} = \frac{s}{9}$

$\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2} = \frac{s}{9} = λ (let)$

$\Rightarrow s = 9 λ, s = 4 λ + x, s = 3 λ + y$

and $s = 2 λ + z$

$∴ s = 9 λ, x = 5 λ, y = 6 λ, z = 7 λ$

Now, $Δ = \sqrt{s (s - x) (s - y) (s - z)}$

[Heron’s formula]

$= \sqrt{9 λ \cdot 4 λ \cdot 3 λ \cdot 2 λ} = 6 \sqrt{6} λ^{2}$

Also, $π r^{2} = \frac{8 π}{3}$

$\Rightarrow r^{2} = \frac{8}{3}$

and

$R = \frac{x y z}{4 Δ} = \frac{(5 λ) (6 λ) (7 λ)}{4 \cdot 6 \sqrt{6} λ^{2}} = \frac{35 λ}{4 \sqrt{6}}$

Now, $r^{2} = \frac{8}{3} = \frac{Δ^{2}}{S^{2}} = \frac{216 λ^{4}}{81 λ^{2}}$

$\Rightarrow \frac{8}{3} = \frac{8}{3} λ^{2}$

[from Eq. (ii)]

$\Rightarrow λ = 1$

(a) $△ X Y Z = 6 \sqrt{6} λ^{2} = 6 \sqrt{6}$

$∴$ Option (a) is correct.

(b) Radius of circumcircle $(R) = \frac{35}{4 \sqrt{6}} λ = \frac{35}{4 \sqrt{6}}$

$∴$ Option (b) is incorrect.

$r = 4 R \sin \frac{X}{2} \cdot \sin \frac{Y}{2} \cdot \sin \frac{Z}{2}$

$\begin{array}{ll} \Rightarrow & \frac{2 \sqrt{2}}{\sqrt{3}} = 4 \cdot \frac{35}{4 \sqrt{6}} \sin \frac{X}{2} \cdot \sin \frac{Y}{2} \cdot \sin \frac{Z}{2} \\ \Rightarrow & \frac{4}{35} = \sin \frac{X}{2} \cdot \sin \frac{Y}{2} \cdot \sin \frac{Z}{2} \end{array}$

$∴$ Option (c) is correct.

(d) $\sin^{2} \frac{X + Y}{2} = \cos^{2} \frac{Z}{2}$

$as \frac{X + Y}{2} = 90^{\circ} - \frac{Z}{2} = \frac{s (s - z)}{x y} = \frac{9 \times 2}{5 \times 6} = \frac{3}{5}$

$∴$ Option (d) is correct.

Properties of Triangles 3 Question 10

10. In a $△ X Y Z$ , let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$ respectively and $2 s = x + y + z$ . If $\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$ and area of incircle of the $△ X Y Z$ is $\frac{8 π}{3}$ , then

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Properties of Triangles 3 Question 10

10. In a △XYZ, let x,y,z be the lengths of sides opposite to the angles X,Y,Z respectively and 2s=x+y+z. If s−x4=s−y3=s−z2 and area of incircle of the △XYZ is 8π3, then

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10. In a $△ X Y Z$ , let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$ respectively and $2 s = x + y + z$ . If $\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$ and area of incircle of the $△ X Y Z$ is $\frac{8 π}{3}$ , then