Yes; 31, 37

8

Given point P(x,y) is equidistant from the points A(5,1) and B(1,5).
So, AP=BP
\(\Rightarrow A P^{2}=B P^{2}\) [squaring both sides]
\(\Rightarrow (x-5)^{2}+(y-1)^{2}=(x-1)^{2}+(y-5)^{2}\)  \(\left[\because \text { distance }=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]\) 
\(\Rightarrow x^{2}+25-10 x+y^{2}+1-2 y=x^{2}+1-2 x+y^{2}+25-10 y\) \(\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]\) 
\(\Rightarrow -10 x+2 x=-10 y+2 y\) 
\(\Rightarrow-8 x=-8 y \Rightarrow x=y\) 
Hence proved.

No

From part (i),
OA² + OB²  + OC² - OD² - OE² - OF² = AF² + BD² + CE²   ......(i)
Similarly, we can prove that
OA² + OB²  + OC² - OD² - OE² - OF² = BF² + CD² + AE²  ...... (ii)
From Eqs. (iv) and (v), 
AF² + BD² + CE² = AE² + CD² + BF²

\(\sin { \theta } =\frac { 3 }{ 4 } \) (Given)
\(\Rightarrow \quad \sin ^{ 2 }{ \theta } =\frac { 9 }{ 16 } \)
\(\because \quad \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } =1\)
\(\Rightarrow \quad \cos ^{ 2 }{ \theta } =1-\sin ^{ 2 }{ \theta } \)
\(=1-\frac { 9 }{ 16 } =\frac { 7 }{ 16 } \)
\(\cos { \theta } =\frac { \sqrt { 7 } }{ 4 } \)
and \(\tan { \theta } =\frac { \sin { \theta } }{ \cos { \theta } } \)
\(=\frac { \frac { 3 }{ 4 } }{ \frac { \sqrt { 7 } }{ 4 } } \)
\(=\frac { 3 }{ \sqrt { 7 } } \)
\(\therefore \quad \sqrt { \frac { { cosec }^{ 2 }\theta -\cot ^{ 2 }{ \theta } }{ \sec ^{ 2 }{ \theta -1 } } } +2\cot { \theta } =\frac { \sqrt { 7 } }{ x } +\cos { \theta } \)
\(\Rightarrow \quad \sqrt { \frac { 1 }{ \tan ^{ 2 }{ \theta } } } +2\times \frac { \sqrt { 7 } }{ 3 } =\frac { \sqrt { 7 } }{ x } +\frac { \sqrt { 7 } }{ 4 } \)
\(\Rightarrow \quad \frac { 1 }{ \tan { \theta } } +\frac { 2\sqrt { 7 } }{ 3 } =\frac { \sqrt { 7 } }{ x } +\frac { \sqrt { 7 } }{ 4 } \)
\(\Rightarrow \quad \frac { \sqrt { 7 } }{ 3 } +\frac { 2\sqrt { 7 } }{ 3 } -\frac { \sqrt { 7 } }{ 4 } =\frac { \sqrt { 7 } }{ x } \)
\(\Rightarrow \quad \frac { 4\sqrt { 7 } -\sqrt { 7 } }{ 4 } =\frac { \sqrt { 7 } }{ x } \)
\(\Rightarrow \quad \frac { 3\sqrt { 7 } }{ 4 } =\frac { \sqrt { 7 } }{ x } \)
\(\because \quad x=\frac { 4 }{ 3 } \)

\(\frac {22}{7}\) cm²

In right angled delta BAC, 
\({ BC }^{ 2 }={ AB }^{ 2 }+{ AC }^{ 2 }\)
\(\Rightarrow { AC }^{ 2 }={ BC }^{ 2 }-{ AB }^{ 2 }\\ \Rightarrow { AC }^{ 2 }={ (10) }^{ 2 }-{ (6) }^{ 2 }\quad [\because BC=10cmandAB=6cm]\\ \Rightarrow { AC }^{ 2 }=100-36\\ \Rightarrow { AC }^{ 2 }=64\Rightarrow AC=8cm\)
Therefore, Area of \(\Delta ABC=\frac { 1 }{ 2 } \times AB\times AC\)
 \(=\frac { 1 }{ 2 } \times 6\times 8={ 24cm }^{ 2 }\)
Area of deltaABC = Area of deltaIBC + Area of deltaICA + Area of deltaIAB
 \(\Rightarrow 24=\frac { 1 }{ 2 } (BC\times r)+\frac { 1 }{ 2 } (CA\times r)+\frac { 1 }{ 2 } (AB\times r)\)
\(\Rightarrow 24=\frac { 1 }{ 2 } r(BC+CA+AB)\\ \Rightarrow 24=\frac { 1 }{ 2 } \times r\times (10+8+6)\\ \Rightarrow 24=12r\\ \Rightarrow \quad r=2cm\)
Therefore, Area of the shaded region = Area of deltaABC - Area of incircle
\(=24-{ \pi r }^{ 2 }=24-\frac { 22 }{ 7 } \times { (2) }^{ 2 }=24-\frac { 88 }{ 7 } \\ =\frac { 168-88 }{ 7 } =\frac { 80 }{ 7 } { cm }^{ 2 }\)

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

If Assertion is incorrect but Reason is correct.

(i) (c): 3
(ii) (a): 2, 0, -2

(i) (b): Here \(\sqrt{8}\) = 2\(\sqrt{2}\) = product of rational and irrational numbers = irrational number
(ii) (c): Here, \(\sqrt{9}\) = 3 So, 2 + 2\(\sqrt{9}\)= 2 + 6 = 8 , which is not irrational.
(iii) (b): Here.\(\sqrt{15}\) and \(\sqrt{10}\) are both irrational and difference of two irrational numbers is also irrational.
(iv) (c): As \(\sqrt{5}\) is irrational, so its reciprocal is also irrational.
(v) (d): We know that \(\sqrt{6}\) is irrational. So, 15 + 3.\(\sqrt{6}\) is irrational.
Similarly, \(\sqrt{24}\) - 9 = 2.\(\sqrt{6}\) - 9 is irrational.
And 5\(\sqrt{150}\) = 5 x 5.\(\sqrt{6}\) = 25\(\sqrt{6}\) is irrational.