The decimal expansion is terminating.
0.3796 is a rational number.

a=5, b=4

Given: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
To Prove: \(ar(\Delta APB)\times ar(\Delta CPD)=ar(\Delta APD)\times ar(\Delta BPC)\)
Construction: From A and C, draw perpendiculars AE and CF respectively to BD.

\(ar(\Delta APB)\times ar(\Delta CPD)\)=\(\frac { (PB)(AE) }{ 2 } \times \left( \frac { DP\times CF }{ 2 } \right) \)
|Area of \(\Delta\)=\(\frac { Base\times Corresponding\ altitude }{ 2 } \)
\(=\frac { 1 }{ 4 } (PB)(AE)(DP)(CF)\quad \quad ....(1)\)
\(\\ ar(\Delta APD)\times ar(\Delta BPC)\)
\(\\ =\frac { (DP)(AE) }{ 2 } \times \frac { (PB)(CF) }{ 2 }\)
\( \\ =\frac { 1 }{ 4 } (PB)(AE)(DP)(CF)\quad \quad ....(2)\)
From (1) and  (2),
\(ar(\Delta APD)\times ar(\Delta BPC)\)

(i) Each edge of the cubical box (a) = 10 cm
\(\therefore \)  Lateral surface area of the cubical box
= 4a² = 4(10)² = 400 cm².
For cuboidal box
l = 12.5 cm, b = 10 cm,
h = 8 cm
\(\therefore \)  Lateral surface area of the cuboidal box 
= 2(l + b)h
= 2(12.5 + 10)(8) = 360 cm².
Cubical box has the greater lateral surface area than the cuboidal box by (400 - 360)cm²,
i.e.,  40 cm².
(ii) Total surface area of the cubical box = 6a²
= 6(10)² = 600 cm²
Total surface area of the cuboidal box
= 2(lb + bh + hl)
= 2[(12.5)(10) + (10)(8) + (8)(12.5)] 
= 2[125 + 80 + 100] = 610 cm².
Cubical box has the smaller total surface area than the cuboidal box by (610 - 600) cm², i.e., 10 cm².

The linear equation is 2x+ky=8
At x=2,y=1,
2(2)+k(1)=8
\(\Rightarrow\)4+k=8
k=4
If x=4, then
\(\Rightarrow\) 2(4)+4y=8
\(\Rightarrow\) 8+4y=8
\(\Rightarrow\) 4y=0
\(\therefore y=0\)

Given PQ||Rs are cut by a transversal t at A and B respectively. AC and BD are the trisector of a pair of alterna int \(\angle S,\angle PAB\ and\angle ABS\) respectively.

To prove AC||BD
Prove since PQ||RS and t is a traversal we have
\(\angle PAB=\angle ABS [Alt. Int\angle S]\)
\(\Rightarrow \frac{1}{2}\angle PAB=\frac{1}{2}\angle ABS\)
\(\Rightarrow \angle CAB=\angle ABD\)
But these are alternate interior angles formed when the transversal AB cuts AC and BD.
\(\therefore AC||BD\)

Since from a point \({ \bot }^{ r }\) line is the shortest. 
CF\({ \bot }\) AB
\(\therefore\) CF < AC and CF < BC ...(1) 
Similarly, BCis a line segment and A does not lie on
it. AD \({ \bot }\) BC
\(\therefore\) AD < AB and AD < AC ...(2) 
Also, AC a line segment and B does not lie on it.
BE\({ \bot }\)AC
\(\therefore\) BE < AB and BE < BC ...(3)
Adding (I), (2) and (3), we get
2(AD + BE + CF) < 2(AB + BC + CA)
\(\therefore\) AB + BC + CA > AD + BE + CF 
i.e., Perimeter is greater than the sum of three altitudes. Proved.

\(\left( \sqrt { 3 } +\sqrt { 5 } \right) \)

\(a^4y^4(a^2+y^2)(a-y)(a+y)(a^2+y^2-\sqrt{2}ay(a^2+y^2+\sqrt{2}ay)\)

\(\left( 0,4\sqrt { 3 } \right) \)

\(270^{ 0 }\)

6 cm

168 cm²

440 cm²

7

Equation 2x-3y=12
or 3y=2x-12
\(\therefore \ y=\frac{2x-12}{3}\)
On x-axis y=0
\(\Rightarrow x=6 \)
At point(6,0) the given line cuts the x-axis
On y-axis x=0
\(\therefore \ y=\frac{2\times0-12}{3}\)
\(\Rightarrow y=-4\)
At point(0,-4) the given line cuts the y-axis

\(a=2+\sqrt { 3 } +\sqrt { 5 } \quad \)
\(\\ a-2=\sqrt { 3 } +\sqrt { 5 } \)
\(\\ b=3+\sqrt { 3 } -\sqrt { 5 } \)
\(\\ b-3=\sqrt { 3 } -\sqrt { 5 } \)
\(\\ { \left( a-2 \right) }^{ 2 }+{ \left( b-3 \right) }^{ 2 }={ \left( \sqrt { 3 } +\sqrt { 5 } \right) }^{ 2 }{ \left( \sqrt { 3 } -\sqrt { 5 } \right) }^{ 2 }\)
\(\\ =(3+5+2\sqrt { 3 } \sqrt { 5 } )+(3+5-2\sqrt { 3 } \sqrt { 5 } )\)
\(\\ =16\)

Through O draw OE || AB || CD 
Now y= \(\angle FOG\)
=\(\angle \)FOE + \(\angle GOE\)
=\(\angle \)CFO+\(\angle \) AGO
=\(\angle \)FOE ==\(\angle \) CFO (Alternate Interior angles)
=\(\angle \)GOE=\(\angle \)AGO (Alternate Interior Angles)
 =\(45^{ 0 }\)+\(40^{ 0 }\)=\(85^{ 0 }\)

Let the base radius and height of the cylinder be r cm and h cm respectively.
Then,
\(2 \pi r=22\)
\(\Rightarrow 2\times \frac{22}{7}\times r=22\)
\(\Rightarrow r=\frac{7}{2}cm\)
h = 22 cm
ஃ  Volume of the cylinder 
\(=\pi r^2h\)
\(=\frac{22}{7}. \frac{7}{2}.\frac{7}{2}.22\)
= 847 cm³

Factor of 6 = (\(\pm \)1,\(\pm \)2,\(\pm \)3,\(\pm \)6)
p(x)=x³+2x²-5x-6
p(-1)=(-1)³+2(-1)²-5(-1)-6
=-1+2+5-6
=7-7=0
\(\because\) x = -1 is zero of p(x) of (x+1) is a factor of p(x)
\(\therefore\) x³+2x²-5x-6
=x²(x+1)+x(x+1)-6(x+1)
=(x+1)[x²+x-6]
=(x+1)[x²+3x-2x-6]
=(x+1)[x(x+3)-2(x+3)]
=(x+1)(x+3)(x-2)

Let ABCD is a parallelogram
To show LMNO is a rectangle,
ㄥA+ㄥD=180°
\(\frac{1}{2}\)ㄥA+\(\frac{1}{2}\)ㄥD=90°
ㄥOAD+ㄥODA=90°
In ΔOAD,
ㄥOAD+ㄥADO+ㄥDOA=180°
⇒ ㄥDOA = 90°
⇒ ㄥLON = 90°
Similarly, ㄥOLM =ㄥLMN =ㄥMNO = 90°

ஃ A quadrilateral with all angles 90° is a rectangle. Also opposite angles are equal. It is rectangle.

\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)is an irrational number.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } =\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 2-1 } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 1 } } =\sqrt { 2 } -1\)
which is an irrational number.
Let, there is a number x such that x³ is an irrational number but x⁵ is a rational number.
Let, x =\(\sqrt[5]{7}\) be the number.
⇒ x³ = (5√7)³ = (7)^3/5
is an irrational number.
But x⁵ = (\(\sqrt[5]{7}\))⁵=(7)^5/5  = 7
=7 is a rational number.
(ii) Accepting own mistakes gracefully, co-operative learning among the classmates.

Proof: In \(\triangle\) PQR,
PQ =PR
\(\angle\) PQR = \(\angle\)PRQ
(Angles opp. to equal sides are equal) ...(i) 
In \(\triangle\)PQS, \(\angle\)PQR > \(\angle\)PSQ 
(Ext. angle of a 6 is greater than each of interior opp. angle)
\(\angle\)PRQ > \(\angle\)PSQ, using (i)
\(\Rightarrow\) \(\angle\)PRS >\(\angle\)PSR \(\Rightarrow\) PS > PR 
PS>PQ (\(\because\) PR = PQ)
(Side opp. to greater angle is larger)