(i) False, because zero is a whole number but not a natural number.
(ii) True, because every integer m can be expressed in the form m/1, and so it is a rational number.
(iii) False because 3/5 is a rational number but not an integer.

2

7

Since, AB = BC
AX + BX = BY + CY
Since, BX = BY
AX + BX - BX = BY + CY - BY
AX = CY
Axiom : If equals are subtracted from the equals, the remainders are equal.

1(\(\neq 0\)) is a rational number and \(\sqrt { 2 } \)(\(\neq 0\)) is an irrational number.
\(\frac { 1 }{ \sqrt { 2 } } \) is an irrational number.
The quotient of a non-zero rational number with an irrational number is irrational.

\(p(x)=2{ x }^{ 2 }+kx+\sqrt { 2 } \)
If x-1 is a factor of p(x), then p(1)=0 | By factor Theorem
\(\Rightarrow \ 2{ (1) }^{ 2 }+k(1)+\sqrt { 2 } =0\)
\(\Rightarrow \ 2+k+\sqrt { 2 } =0\)
\(\Rightarrow \ k=-(2+\sqrt { 2 } ).\)

I

Let ABCD be the plot of land of the shape of a quadrilateral. Let the portion ADE be taken over by the Gram Panchayat of the village from one corner D to construct a Health Centre. Join AC. Draw a line through D parallel to AC to meet BC produced in P. Join EP. Then Itwaari must be given the land ECP adjoining his plot so as to form a triangular plot ABP as then
\(ar(\Delta ADE)=ar(\Delta PEC)\)
  
Proof: \(\Delta \)DAP and \(\Delta \)DCP are on the same base DP and between the same parallels DP and AC.
ar(\(\Delta \)DAP) = ar(\(\Delta \)DCP)
| Two triangles on the same base (or equal bases) and between the same parallels are equal in area
ar(\(\Delta \)DAP) - ar(\(\Delta \)DEP)
= ar(\(\Delta \)DCP) - ar(\(\Delta \)DEP)
I Subtracting ar(\(\Delta \) DEP) from both sides
ar(\(\Delta \) ADE) = ar(\(\Delta \) PCE)
ar(\(\Delta \)DAE) + ar(\(\Box \) ABCE)
= ar(\(\Delta \)PCE) + ar(\(\Box \) ABCE)
I Adding are \(\Box \) ABCE) to both sides
ar(\(\Box \) ABCD) = ar(\(\Delta \) ABP).

(a) 114.89 cm²  
(b) \(40\sqrt { 3 } \)cm²  
(c) \(24\sqrt { 3 } \)cm²  
(d) 330 m²

(i) For herbarium 
l = 30 cm, b = 25 cm,
h = 25 cm
\(\therefore \)  Area of the glass = 2 (lb + bh + hl)
= 2[(30)(25) + (25)(25) + (25)(30)]
= 2[750 + 625 + 750] = 4250 cm².
(ii) The tape needed for all the 12 edges
= 4 (l + b + h)
= 4(30 + 25 + 25) = 320 cm.

(i) Surface area of the sphere = \(4\pi { r }^{ 2 }\)
(ii) For cylinder
Radius of the base = r
Height  = 2r
\(\therefore\) Curved surface area of the cylinder
\(=2\pi \left( r \right) \left( 2r \right) =4\pi { r }^{ 2 }\)
(iii) Ratio of the areas obtained in (i) and (ii)
 \(=\frac { Surface\ area\ of\ the\ sphere }{ Curved\ surface\ area\ of\ the\ cylinder } \)
\(\\ =\frac { 4\pi { r }^{ 2 } }{ 4\pi { r }^{ 2 } } =\frac { 1 }{ 1 } =1:1.\)

\((a)\frac { 7 }{ 10 }\)  
\((b)\frac { 7 }{ 40 } \)

Given, A(-1,-1) and C(1,1)
Then, B(1,-1) and D(-1,1)
Also, equations of sides of square are,
AB: Y=-1
BC: X=1
CD: Y=1
DA: X=-1

In \(\triangle\)ABC 
AB = BC
\(\angle\)ABC = \(\angle\)ACB = 8
\(\angle\)B = \(\angle\)C = 8
\(\angle\)A + \(\angle\)B + \(\angle\)C = 180⁰
50° + 8 + 8 = 180⁰
28 = 180° - 50°
= 130°
8 = 65°
\(\angle\)B = \(\angle\)C = 65°
\(\therefore\) LMII BC 1
\(\therefore\) \(\angle\)LMC + \(\angle\)BCM = 180°
[\(\therefore\) \(\angle\)LMC and \(\angle\)BCM]
\(\angle\)LMC + 65° = 180°
\(\angle\)LMC = 180° - 65°
\(\angle\)LMC = 115°

Number of terms = 3
Terms: x⁴, -3x², 2

\(\therefore \) PO \(\bot \) AB
\(\therefore \angle AOP=90^{ 0 }\)
x:y:z=1:3:5=9
\(\therefore \angle x=\frac { 1 }{ 9 } x90^{ 0 }=10^{ 0 }\)
\(y=\frac { 3 }{ 9 } x90^{ 0 }=30^{ 0 }\)
\(y=\frac { 5 }{ 9 } x90^{ 0 }=50^{ 0 }\)

Given: ABCD is a cyclic trapezium with
AD || BC. \(\angle B\) = 70°.
To determine: Other three angles of the trapezium.
Determination:
 \(\angle B+\angle D=180°\) | ∵ Opposite angles of a cyclic quadrilateral are supplementary

⇒ ​​​​​\(70°\angle D=180°\)
⇒ \(\angle D=180°-70°\)
⇒ \(\angle D=110°\)​​​​​​
Again, ∵ AD || BC and transversal AB intersects them
∴ \(\angle A+\angle B=180°\) | ∵ The sum of the consecutive interior angles on the same side of a transversal is 180°
⇒ \(\angle A+70°=180°\)
⇒ \(\angle A=180°-70°\)
⇒ ​​​​​​ \(\angle A=110°\)
Also, \(\angle A+\angle C=180°\)| ∵ Opposite angles of a cyclic quadrilateral are supplementary
⇒ \(110°+\angle C=180°\)
⇒ \(\angle C=180°-110°\)
⇒ ​​​​​​​\(\angle C=70°\). 

b + b + 6 = 24
 \(\Rightarrow \)   b = 9 cm
\(\therefore \) Area \(=\frac { 9 }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } =\frac { 6 }{ 4 } \sqrt { 4{ (9) }^{ 2 }-{ (6) }^{ 2 } } \)
 \(=\frac { 3 }{ 2 } \sqrt { 288 } =\frac { 3 }{ 2 } .12\sqrt { 2 } \)
\( 18\sqrt { 2 } \)  cm²

\(\therefore\)   Mean \(=\frac {{\sum fx}}{\sum f}= \frac {750}{30}=25\)

\({2\over3}x-y=2\)
\(\Rightarrow 2x-3y=6\)
\(\Rightarrow 2x=3y+6\)
\(\Rightarrow x=\frac{3y+6}{2}\)
(i) When the line cuts x-axis then put y=0
i.e., 2x=6
x=3
Hence point is (3,0)
(ii) When the line cuts y-axis then put x=0
i.e., 3y+6=0
y=-2
Hence point is (0,-2)

Steps of construction:
i) Draw a line segment BC = 4.7 cm. and at point B construct an angle of 45^o i.e ㄥXBC=45^o
ii) Cut the line segment BD = 2 cm (equal to AB-AC)on ray BX
iii) Join DC and draw the perpendicular bisector PQ of DC
iv) The perpendicular bisector intersects BXat point A. Jon AC ΔABC is the required triangle.