Let E,F,G and H be 4 distinct points inside square ABCD whose area is 1 sq unit, such that $∠ E D C = ∠ E C D = ∠ H D A = ∠ H A D = ∠ G A B = ∠ G B A = ∠ F B C = ∠ F C B = 15^{0}$
LIST-I LIST-II
P If area of quadrilateral EFGH equal to $a - \sqrt{b}$ , where $a, b \in N$
then $a + b$ is equal to 1 1
Q $∠ A E B = \frac{π}{k},$ then k is equal to 2 3
R The radius of circle circumscribing the $Δ A H D$ is equal to 3 5
S Let the lengths of perpendiculars from vertices G, A and H to opposite sides of $Δ A H D$ be $h_{1}, h_{2}$ and $h_{3}$ , respectively. Let $\frac{1}{h_{1}^{2}} + \frac{1}{h_{2}^{2}} + \frac{1}{h_{3}^{2}} = a + \sqrt{b,}$ where $a, b \in N$ , then $\frac{b}{a}$ is equal to 4 6
5 9
The correct option is

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$P \to (3), Q \to (2), R \to (1), S \to (4)$

$P \to (2), Q \to (3), R \to (4), S \to (5)$

$P \to (3), Q \to (1), R \to (4), S \to (2)$

$P \to (3), Q \to (2), R \to (4), S \to (1)$

answer is A.

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Detailed Solution

$Δ A E M,$ $a + b = 2 + 3 = 5$ $E F G H = \frac{1}{2} F H \times E G$ We have, ABCD is square whose each side is 1 sq unit, and E,F,G,H,4 distinct such that
$∠ E D C = ∠ E C D = ∠ H D A = ∠ H A D = ∠ G A B = ∠ G B A = ∠ F B C = ∠ F C B = 15^{0}$
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P. $F H = 1 - 2 (\frac{1}{2} \tan 15^{0}) = \sqrt{3} - 1$
$E G = 1 - 2 (\frac{1}{2} \tan 15^{0}) = \sqrt{3} - 1$
Area of quadrilateral $E F G H = \frac{1}{2} F H \times E G$
$\frac{1}{2} {(\sqrt{3} - 1)}^{2} = 2 - \sqrt{3}$
$a + b = 2 + 3 = 5$
Q.EM=EG+GM= $\sqrt{3} - 1 + \frac{1}{2} \tan 15^{0} = \frac{\sqrt{3}}{2}$
In $Δ A E M,$
$\tan ∠ E A M = \frac{E M}{A M} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} ∴ ∠ E A M = 60^{0}$

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	LIST-I		LIST-II
P	If area of quadrilateral EFGH equal to $a - \sqrt{b}$ , where $a, b \in N$ then $a + b$ is equal to	1	1
Q	$∠ A E B = \frac{π}{k},$ then k is equal to	2	3
R	The radius of circle circumscribing the $Δ A H D$ is equal to	3	5
S	Let the lengths of perpendiculars from vertices G, A and H to opposite sides of $Δ A H D$ be $h_{1}, h_{2}$ and $h_{3}$ , respectively. Let $\frac{1}{h_{1}^{2}} + \frac{1}{h_{2}^{2}} + \frac{1}{h_{3}^{2}} = a + \sqrt{b,}$ where $a, b \in N$ , then $\frac{b}{a}$ is equal to	4	6
		5	9