Given, the shape of a garden is rectangular in middle and semi-circular at both ends.
Also, diameter of circular part = 7 m
∴ Radius of circular part = \(\frac { 7 }{ 2 } \)= 3.5 m
Then, length of rectangular part = 20m -(3.5+3.5)m and breadth of rectangular part = 7 m
Now,area of one semi-circular part = \(\frac { 1 }{ 2 } { \pi r }^{ 2 }\)
where, r is the radius of semi-circle.
∴ Area of both semi-circular parts = \(2\times \frac { 1 }{ 2 } { \pi r }^{ 2 }={ \pi r }^{ 2 }\)
 =  \(\frac { 22 }{ 7 } \times \left( \frac { 7 }{ 2 } \right) ^{ 2 }\)    \(\left[ \because \pi =\frac { 22 }{ 7 } \right] \)
= \(\frac { 22 }{ 7 } \times \frac { 49 }{ 7 } =\frac { 77 }{ 2 } \) m²
Area of rectangular part = Length \(\times\) Breadth
= 13m\(\times\) 7m = 91m²
∴ Area of garden = Area of both circular parts + Area of rectangular part
=\(\frac { 77 }{ 2 } \) + 91 m²
=\(\left( \frac { 77+182 }{ 2 } \right) { m }^{ 2 }\)= 129.5 m²
Now, Perimeter of one semi-circle part = \(\frac { 2\pi r }{ 2 } =\pi r\)
∴ Perimeter of both semi-circle parts = \(2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \) =22m
Here, rectangular part is in between two semi-circular parts, so for perimeter of garden, we will take only length of rectangular part.
∴  Perimeter of garden = Perimeter of both circular parts + 22 \(\times\) Length of rectangular part
= 22 + 2 \(\times\)13 = (22+26) m = 48m

The diagram of the picture frame is as shown in the figure: The picture frame has four trapezium shaped sections, which are ABCD, BEFC, EHGF and HADG.
Out of these four sections, the area of opposite sections will be same
i.e  area of section ABCD = Area of section EHGF and area of section BEFC  = Area of section HADG
Width of section ABCD =\(\frac { (AH-DG) }{ 2 } =\frac { (28-20) }{ 2 } =\frac { 8 }{ 2 }\) = 4cm
[∵ width of EFGH and ABCD is same]
Then, width of each section is 4 cm.
Now, area of trapezium ABCD
=\(\frac { 1 }{ 2 } \) (AB+CD)\(\times\) Width of frame
= \(\frac { 1 }{ 2 } \)\(\times\)(24+16)\(\times\) 4 =\(\frac { 1 }{ 2 } \) 40\(\times\) 4 = 80 m²
and area of trapezium BEFC 
 = \(\frac { 1 }{ 2 } \)(BE + CF) \(\times\)Width of frame
 = \(\frac { 1 }{ 2 } \)(28 + 20)\(\times\) 4 =\(\frac { 1 }{ 2 } \) \(\times\)48\(\times\) 4 = 96m²
∴ Area of section ABCD =Area of section EHGF = 80 cm² and area of section BEFC = Area of section HADG = 96 cm²

(i)(48)²=(50-2)²
Since, (a - b)² = a² - 2ab + b²
(50-2)²=(50)²-2x50x2+(2)²
= 2500 - 200 + 4   (a-b)2=a²-2ab+b²
= 2504 - 200 = 2304
(ii) 181² -19² =(181-19)(181+ 19)
where, a = 50 and b = 2
= 162 x 200 = 32400
(iii) 497 x 505 =(500 - 3)(500 + 5)
= 5002 + (-3 + 5) x 500 + (-3)(5)
[.: (x + a)(x + b) = x² + (a +. b) x + ab]
= 250000 + 1000 -15 = 250985
(iv) 2.07 x1.93 = (2 + 0.07)(2 - 0.07)
= 2² -(0.07)²
[.: where, a = 50 and b = 2]
= 3.9951

\(\frac{1}{4}\)

(a) 9 bags
(b) 440 kg

Here, as the base increased, the required number of green pigments will also increase.
\(\therefore\) The quantity vary directly:
i.e.,\(\frac { 1 }{ 6 } =\frac { 4 }{ { x }_{ 1 } } =\frac { 5 }{ { x }_{ 2 } } =\frac { 6 }{ { x }_{ 3 } } \) 
\(\therefore\) \(\frac { 4 }{ { x }_{ 1 } } =\frac { 1 }{ 6 } \Rightarrow 1\times { x }_{ 1 }=4\times 6\Rightarrow { x }_{ 1 }=24\)
\(\frac { 1 }{ 6 } =\frac { 5 }{ { x }_{ 2 } } \Rightarrow 1\times { x }_{ 2 }=5\times 6\)
\(\Rightarrow { x }_{ 2 }=\frac { 5\times 6 }{ 1 } =30\)
\(\frac { 1 }{ 6 } =\frac { 6 }{ { x }_{ 3 } } \Rightarrow 1\times { x }_{ 3 }=6\times 6\)
\(\Rightarrow \ { x }_{ 3 }=\frac { 6\times 6 }{ 1 } =36\)
Thus, the required unknown quantities are: x₁ = 24, x₂ = 30, and x₃ = 36.

Scale:
On horizontal axis: 2 units = 2 cm
On vertical axis: 1 unit = 2 cm
1. Mark side of the square (in em) on horizontal axis.
2. Mark area (in cm²) on vertical axis.
3. Plot the points (2, 4) (3, 9), (4, 16), (5, 25) and (6, 36).
4. Join the points.
The graph we get is not linear.

289 cm²

We have, 3x x (13x + 12z - 2y) = 3xx13x + 3x x 12z - 3xx2y = 39x²+ 36xz -6xy

We know that, a prism is a polyhedron whose base and top are congruent polygons and whose other faces, i.e lateral faces are parallelograms in shape. Here, top and base of a prism and of a cylinder are congruent and parallel to each other and a prism becomes a cylinder if the number of sides of its base becomes larger and larger.

∵ 998 = 1000 - 2
∴ (998i= (1000 - 2)²
= (1000)² - 2(1000)(2) + (2)²   [Using (a - b)² = a² - 2ab + b²]
= 1000000 - 4000 + 4 = 996004

\({ \left\{ { \left( \frac { -2 }{ 3 } \right) }^{ -2 } \right\} }^{ 2 }\) \(\because \quad ({ a }^{ m }{ ) }^{ n }={ a }^{ mn }\)
\(\therefore \quad { \left\{ { \left( \frac { -2 }{ 3 } \right) }^{ -2 } \right\} }^{ 2 }={ \left( \frac { -2 }{ 3 } \right) }^{ (-2)\times 2 }={ \left( \frac { -2 }{ 3 } \right) }^{ -4 }\)
=\(\frac { { (-2) }^{ -4 } }{ { (3) }^{ -4 } } \left[ \because { \left( \frac { a }{ b } \right) }^{ m }=\frac { { a }^{ m } }{ { b }^{ m } } \right] \)
\(=\frac { { 3 }^{ 4 } }{ { (-2) }^{ 4 } } =\frac { 3\times 3\times 3\times 3 }{ (-2)\times (-2)\times (-2)\times (-2) } =\frac { 81 }{ 16 } \) 

32

300

To check the divisibility of 1052896
1+ 0 + 5 + 2 + 8 + 9 + 6 = 31
Since, 31 is not divisible by 9. Hence, this number is not divisible by 9.

Given, (7, 3)
In this point, x-coordinate (abscissa) is 7 i.e.
x =7.

Given, (3, 5)
In this point, y-coordinate (ordinate) is 5 i.e. y = 5.

x = 10

ax + bx - ay - by
= x[a + b] + (-y)[a + b]
=(x - y)[a + b]

First expression = 3l(l - 4m + 5n)
= (3l) x (l) - (3l) x (4m) + (3l) x (5n)
= 3l² -12lm + 15ln
Second expression = 4l(10n - 3m + 2l)
= (4l)x (I0n) - (4l) x (3m) + (4l) x (2l)
= 40ln -12lm + 8l²
On subtracting first expression from second expression, we get
40ln -12lm + 8l²
15ln - 12lm + 3l²
(-)      (+)     (-)
___________________
25ln + 0 + 5l²
___________________
= 25ln + 5l²

Let ABCDEFGH be the regular octagonal surface as shown below in the figure:

Given, each side of octagon = 5m,AD = 11 m
perpendicular from B to AD i.e. BP =4m .Join HE
Since, it is a regular octagon, so trapeziums ABCD and HEFG will be congruent and ADEH is a rectangle.
Now, area of trapezium ABCD = Area of trapezium HEFG
 = (AD+BC)BP
 = \(\frac { 1 }{ 2 } \)(11+5)\(\times\) 4 =\(\frac { 1 }{ 2 } \) \(\times\)16\(\times\)4
 = 8 \(\times\)4 =32 m²
and area of rectangle ADEH = AD\(\times\) DE =11 \(\times\)5 =55m²
Now, area of octagon ABCDEFGH
= Area of trapezium ABCD +Area of rectangle ADEH +Area of trapezium HFGH
= 32 + 55 + 32= 119 m²
Hence, required area of octagonal surface is 119 m²

According to the question, on cutting the cylinder along its height, we get a rectangular sheet of width 33 cm which is height of cylinder. So, the height of cylinder, h = 33 cm. Let radius of cylinder be r cm.
 Given, lateral surface area of cylinder = 4224 cm²
∴ 2\(\pi\)r = 4224 ⇒ \(2\times \frac { 22 }{ 7 } \times r\times 33\)=4224
⇒  \(r=\frac { 7\times 4224 }{ 2\times 22\times 33 } =\frac { 7\times 96 }{ 33 } =\frac { 7\times 32 }{ 11 } cm\)
Also, on cutting the hollow cylinder, the perimeter of base becomes the length of the rectangular sheet 
∴ Length of rectangular sheet = Perimeter of circular base
  = 2\(\pi\)r
 \(2\times \frac { 22 }{ 7 } \times \frac { 7\times 32 }{ 11 } \)  = 128 cm
and breadth of rectangular sheet = 33 cm
∴  Perimeter of rectangular sheet = 2(1 + b) = 2(128 +33) = 2\(\times\)161 =322 cm

Given, \(\frac{5^{m}\times5^{3}\times5{-2}}{5^{-5}}=5^{12}\)
⇒ \(\frac{5^{m+3-2}}{5^{-5}}=5^{12}\)
⇒ 5^m+1+5 = 5¹²
⇒ (5)^m+6 = (5)¹²
⇒ m + 6 = 12   [ ∵  bases are same]
⇒ m= 12 - 6 = 6

Let the number of boxes filled be x.
Then, we have the following table:

Here, more the number of bottles in a box, less would be the number of boxes. Therefore, it is a case of inverse proportion.
So, \(12\times 25=20\times x\Rightarrow x=\frac { 12\times 25 }{ 20 } =\frac { 12\times 5 }{ 4 } \)
[dividing numerator and denominator by 5]
x = 15
Hence, 15 boxes would be filled, if the same batch is packed using 20 bottles in each box.

8pq

Here, 5P²-25P+20 = 5(p²-5P + 4)
= 5[p² +(-4-1)p+4]
= 5(p²-4p-P+ 4)    [∵ ab = 4 and a + b= -5, ஃ a= -4, b= -1]
= 5[p(p-4)-1(p-4)]
= 5(p-4)(p-1)
Now, (5p²-25p + 20) ÷ (p-1)
=\(\frac{5p^{2}-25p+20}{(p-1)}\)
=\(\frac{5(p-4)(p-1)}{(p-1)}=5(p-4)\)

If a polyhedron is having number of faces as F, number of edges as E and the number of vertices as V, then the relationship
F+ V=E+2
is known as Euler's formula. Following figure is a solid pentagonal prism.

It has:
Number of faces (F) = 7
Number of edges (E) = 15
Number of vertices (V) = 10
Substituting the values of F, E and V in the relation,
F+ V=E+2
We have
7 + 10 = 15 + 2
⇒ 17 = 17
which is true, the Euler's formula is verified.

Since 125 = 5 \(\times\)5\(\times\)5=5³
6^-5=(2\(\times\)3)^-5 = 2^-5 \(\times\) 3^-5
\(\therefore \frac { { 2 }^{ -5 }\times { 3 }^{ -5 }\times 125 }{ { 5 }^{ -4 }\times { 6 }^{ -5 } } =\frac { { 2 }^{ -5 }\times { 3 }^{ -5 }\times { 5 }^{ 3 } }{ { 5 }^{ -4 }\times { 2 }^{ -5 }\times { 3 }^{ -5 } } \)
\(=\frac { { 2 }^{ -5 } }{ { 2 }^{ -5 } } \times \frac { { 3 }^{ -5 } }{ { 3 }^{ -5 } } \times { 5 }^{ 3+4 }=1\times { 5 }^{ 7 }\)
=5 \(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5\(\times\)5 = 78125.

We have 50x² - 98 = 2(25x² - 49)
= 2[(5x)² - (7)²] == 2[(5x + 7)(5x - 7)] [Using a² - b² = (a + b)(a - b)]
= \(\frac { 81x^{ 3 }[50x^{ 2 }-98] }{ 27x^{ 2 }[5x+7] } \)
=\(\frac { { 81x }^{ 3 } }{ { 27x }^{ 2 } } \left[ \frac { 2(5x-7)(5x+7) }{ (5x+7) } \right] =3x[2(5x-7)]\)
= 3x\(\times \)2\(\times \)(5x- 7) = 6 x (5x - 7)

\({10\over 1000}={4\over ?}⇒?=400\)

Area = \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \)x7x7 = 77 cm²

Total surface area = 2\(\pi\)r(h+r)
=2\(\pi\)5(10+5) = 150\(\pi\) cm²