-2

Given, (x² - 4) + (x² + 4) + 16
= x² - 4 + x² +4 +16
= x² + x² + 16
= 2x² + 16

 (a - b)² = a² - 2ab + b²
Here, a = pq and b = qr
(pq - qr)² + 4pq²r
= [(pq)² - 2pq²r + (qr)²] + 4pq²r
= (pq)² - 2pq²r + 4pq²r + (qr)²
= (pq)² + 2pq²r + (qr)²
(pq+qr)²

3x(x-1)(2x-1)

The first term -xy can be factorized as: (-1) × x × y and
The second term -ay can be factorized as : (-1) × a × y
The common factor for both the terms is -y
Taking out the common factor we get,
- xy - ay = -y(x + a)

6b(a+2c)

given, 992
= (100−1)²
= (100)²−2.100.1+(1)²
= 10000−200+1
= 99801

9a² - 16b²
= (3a)²−(4b)²
(3a+4b)(3a-4b)

Using standard identity: (a + b)² = a² + 2ab + b²
Here, a = 7x and b = 5
(7x + 5)² = (7x)² + [2(7x)(5)] + 5²
49x²-70x+25

4x²y²

We have, (ab−c)²+2abc
[Using the identity, (x−y)² = x²+y²−2xy]
= (ab)² + c²− (2 × ab × c) + 2abc 
= (ab)² + c² − 2abc + 2abc
a²b²+c²

(p+6)×(q−7) = p(q−7) + 6(q−7)
p²q²+4r²-4pqr

pq+6q-7p-42

28.89 cm²

\(\frac{a^2}{4}{\pi-2}\)

Perimeter of semicircular protractor = 36 cm
πr + 2r = 36
⇒ (π+2)r = 36
⇒ \((\frac{22}{7}+2)\) = 36
⇒\(\frac{36}{7}\)r = 36
⇒ r = \(\frac{36\times 7}{36}\)
​⇒ r = 7
Therefore, diameter = 2r = 2 × 7 = 14 cm 

= r²

Rs. 32.25

\(\frac{-2}{5}\)

\(-\frac { 1 }{ 4 } ,\frac { 3 }{ 2 } ,\frac { 13 }{ 4 } \)

\(\frac{4}{3}\)

\(\frac{46}{3}\)

\(\frac{33}{13}\)

\(\frac{1}{13}\)

\(\frac{3}{2}\)

\(\frac{-17}{18}\)

\(\frac{-67}{8}\)

9a²-16b² = (3a)²-(4b)²=(3a+4b)(3a-4b)

x²+xy+8x+8y = x(x+y)+8(x+y)
= (x +y)(x + 8)

12x³y⁴+16x²y5-4x⁵y² = 4x²y²(3xy²+4y³-x³)

7(2x + 5) + 3 (2x + 5) = (2x + 5)(7 + 3)

[Using (a+b)(a-b) = a²-b²] 
(2x + 3y) (2x - 3y) = (2x)²-(3y)²
= 4x²-9y²

[ using (a - b)²=a²-2ab+b²]
(2x-3y)² = (2x)²-2(2x)(3y)+(3y)²
= 4x²-12xy+9y²

[using (a + b)² = a²+2ab+b²]
(2x+3y)2 = (2x)²+2\(\times\)(2x)\(\times\)(3y)(3y)² 
= 4x²+12xy+9y²

\(\cfrac { { 6x }^{ 3 }{ y }^{ 2 }{ z }^{ 2 } }{ { 3x }^{ 2 }yz } =\cfrac { 6 }{ 3 } { x }^{ 3-2 }{ y }^{ 2-1 }{ z }^{ 2-1 }=2xyz\)

\(\cfrac { -72{ x }^{ 2 }yz }{ -12xyz } =\cfrac { -72 }{ -12 } { x }^{ 2-1 }{ y }^{ 1-1 }{ z }^{ 1-1 }=6{ xy }^{ 0 }{ z }^{ 0 }=6x\)

\(\cfrac { { 25x }^{ 3 }{ y }^{ 2 } }{ -15{ x }^{ 2 }y } =\cfrac { -25 }{ 15 } { x }^{ 3-2 }{ y }^{ 2-1 }=\cfrac { -5 }{ 3 } xy\)

\(\cfrac { -15{ a }^{ 2 }{ bc }^{ 3 } }{ 3ab } =\cfrac { -15 }{ 3 } { a }^{ 2-1 }{ b }^{ 1-1 }{ c }^{ 3 }=-5a{ b }^{ 0 }{ c }^{ 3 }=-5{ ac }^{ 3 }\)

\(\cfrac { 12{ x }^{ 3 }{ y }^{ 3 } }{ { 3x }^{ 2 }y } =\cfrac { 12 }{ 3 } { x }^{ 3-2 }{ y }^{ 3-1 }= 4xy ^ 2\)

3mn x 4np = (3 x 4) (m x n x n x p) = 12 mn²p

4a\(\times\)3a² = (4\(\times\)3) (a\(\times\)a²) = 12a³

2x²\(\times\)5y² = (2\(\times\)5)\(\times\)(x²+y²)10x²y²

10x\(\times\)5y = (10\(\times\)5) X\(\times\)Xy = 50xy

x \(\times\) y = xy

No, subtraction is not commutative for rational numbers.
Example: Let a = \(\frac{1}{2}\) and b = \(\frac{5}{6}\)
\(a-b=\frac { 1 }{ 2 } -\frac { 5 }{ 6 } =\frac { (1\times 3)-5 }{ 6 } =\frac { 3-5 }{ 6 } =\frac { -2 }{ 6 } =\frac { -1 }{ 3 } \quad \quad \quad ...(1)\)
\(b-a=\frac { 5 }{ 6 } -\frac { 1 }{ 6 } =\frac { 5 }{ 6 } -\frac { 3 }{ 6 } =\frac { 5-3 }{ 6 } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \quad \quad \quad \quad \quad \quad ...(2)\)
From (1) and (2)
a - b ≠ b - a for rational numbers

Let a = \(\frac{1}{4}\) and b \(\frac{2}{3}\)
\(a+b=\frac { 1 }{ 4 } +\frac { 2 }{ 3 } =\frac { (1\times 3)+(2\times 4) }{ 3\times 4 } \)
\(=\frac { 3+8 }{ 12 } =\frac { 11 }{ 12 } \) is in Q.
∴ Addition of rational numbers is closed.

\(\left( \frac { -7 }{ 18 } \times \frac { 15 }{ -7 } \right) -\left( 1\times \frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 2 } \times \frac { 1 }{ 4 } \right) =\left( \frac { -7\times 15 }{ 18\times -7 } \right) \left( \frac { 1\times 1 }{ 1\times 4 } \right) +\left( \frac { 1\times 15 }{ 18\times 1 } \right) -\left( \frac { 1 }{ 4 } \right) +\left( \frac { 1 }{ 8 } \right) \)
\(\\ \\ =\frac { 5 }{ 6 } -\frac { 1 }{ 4 } +\frac { 1 }{ 8 } =\frac { (5\times 4)-(1\times 6)+(1\times 3) }{ 24 } \)
\(=\frac { 20-6+3 }{ 24 } =\frac { 14+3 }{ 24 } =\frac { 17 }{ 24 } \)

In triangles \(\triangle \)RPQ and \(\triangle \) RTS, we have
\(\angle \)RPQ=\(\angle \)RTS [∵ given]
\(\angle \)PRQ =\(\angle \)TRS [∵ common]
\(\angle \)PQR =\(\angle \)RST l∵ Remaining angle]
\(\triangle \)RPQ ~ \(\triangle \)RTS [∵ By AAA similarity]

In triangles \(\triangle \)ABC and \(\triangle \)AED.
\(\angle \)ABC=\(\angle \)AED = 90°
\(\angle \)BAC=\(\angle \)EAD [Each equal to A]
∴ \(\angle \)ADE=\(\angle \)ACB [∵ Remaining angles]
∴ By AAA criteria of similarity  \(\triangle \)ABC ~\(\triangle \)AED.

In triangles \(\triangle \)AOB and \(\triangle \)COD
\(\angle \)A =\(\angle \)C ( given) 
\(\angle \)AOB =\(\angle \)COD [∵ Vertically opposite angles] 
\(\angle \)ABO =\(\angle \)CDO l∵ Remaining angles of MOB and ~COD]
∴\(\triangle \)AOB ~\(\triangle \)COD [∵ AAA similarity]
∴ \(\triangle \)AOB ~\(\triangle \)COD [∵ AAA similarity]

In \(\triangle \)AOB and \(\triangle \)COD, we have 
\(\angle \)AOB =\(\angle \)COD [∵ Vertically opposite angles]
\(\frac { AO }{ OC } =\frac { BO }{ OD } \) [given]
So by SAS criteria of similarity we have \(\triangle \)AOB ~\(\triangle \)COD
\(\frac { AO }{ OC } =\frac { BO }{ OD } =\frac { AB }{ DC } \)
\(\frac { 1 }{ 2 } =\frac { 5 }{ DC } \)  [∵ AB = 5cm]
DC = 2 x 5
DC = 10 cm.

\(\frac { 2 }{ 3 } \div \frac { -7 }{ 12 } =\frac { 2 }{ 3 } \times \frac { 12 }{ -7 } =\frac { 2\times 12 }{ 3\times -7 } =\frac { 24 }{ -21 } =-1\)

The lower triangle cannot be constructed as the sum of two sides 5 + 3 = 8 < 10 cm. So this quadrilateral cannot be constructed.

Here AD = CD
AB = CB
DB = DB (common)
\(\triangle \)DBA \(\equiv \) \(\triangle \)DBC  [∵ By SSS congruency]
Also RHS rule also bind here to say their congruency.

1-(iv), 2-(iii), 3-(i), 4-(ii)

Similar shapes:
(i) W and L
(ii) B and J
(iii) A and G
(iv) B and J
(v) B and Y
(vi) E and N
(vii) H and Q
(viii) R and T
(ix) S and T
Congruent shapes:
(i) Z and I
(ii) J and Y
(iii) C and P
(iv) B and K
(v) R and S
(vi) I and Z
You can find more.

\(1\div \frac { 1 }{ 2 } =\frac { 1 }{ 1 } \times \frac { 2 }{ 1 } =\frac { 1\times 2 }{ 1\times 1 } =2\)

\(\frac { 7 }{ 17 } -\frac { 6 }{ 17 } =\frac { 7-6 }{ 17 } =\frac { 1 }{ 17 }\)

\(\frac { 7 }{ 4 } -\frac { 3 }{ 4 } =\frac { 7-3 }{ 4 } =\frac { 4 }{ 4 } =1\)

\(\frac { 7 }{ 9 } +\frac { (-12) }{ 9 } =\frac { 7+(-12) }{ 9 } =\frac { -5 }{ 9 } \)

\(\frac { 3 }{ 5 } +\frac { 13 }{ 5 } =\frac { 3+13 }{ 5 } =\frac { 16 }{ 5 } =3\frac { 1 }{ 5 } \)

Least number of planes = 4, the solid is tetrahedron.

Radius of the sector = 7 cm
Arc length of the sector = 20 cm
Area of sector = \(\frac { lr }{ 2 } =\frac { 20\times 7 }{ 2 } ={ 70cm }^{ 2 }\)
Area of the sector = 70 cm²

Radius of the sector = 20 cm
Perimeter of the sector = 65 cm
l + 2r = 65
l + 2(20) = 65
l + 40 = 65
l = 65 - 40
l = 25 cm
Length of the sector = 25 cm

(i) Area = 6 x Area of a square of side 6 cm
= 6 x (6 x 6) cm²
= 216 cm²
(ii) Area = Area of 2 rectangles of side (8 x 6) cm² + Area of 2 rectangles of side (8 x 4) cm² + Area of 2 rectangles of side (6 x4) cm²
= (8 x 6) + (8 x 4) + (6 x 4) cm²
= 48 + 32 + 24 cm²
= 104 cm²

(i) Cube - dice, building blocks, jewel box.
(ii) Cuboid - books, bricks, containers.
(iii) Cylinder - candles, electric tube, water pipe.
(iv) Cone - Funnel, cap, ice cream cone
(v) Sphere - ball, beads, Lemmon.

Area of the mat = Area of I trapezium + Area of II trapezium
= [ \(\frac12\)x h₁ x (a₁ + b₁)] + [ \(\frac12\)x h₂ x (a₂ + b₂)]

= [ \(\frac12\)x 2 x (7 + 5)] + \(\frac12\) x 2 x (9 + 7) sq. feet
= 12 + 16 = 28 sq.feet
∴ Cost per sq.feet = Rs.20
Cost for 28 sq. feet = Rs.20 x 28 = Rs.560
∴ Total cost for the entire mat = Rs.560
Both the answers are the same.

For a regular polygon all sides and all the angles must be equal. But in a rhombus all the sides are equal, But all the angles are not equal.
∴ It is not a regular polygon.

If r = 2 r₁ ⇒  Area of the circle = πr² = π (2r₁)² = π 4r₁² = 4 πr₁²

No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle.
Here the boundaries are not radii.

There are 5 stages, Number - 1
Number - 2
Number - 3
Letter - 1
Letter - 2
There are 10 possible numbers 0 to 9
There are 26 possible letters A to Z.
∴ We have 10 x 10 x 10 x 26 x 26 = 6,76,000 possible codes.

There are 4 stages
1. Choosing a Sandwich
2. Choosing a side
3. Choosing a dessert
4. Choosing a drink
There are 4 different types of sandwich, 3 different types of side two different type of desserts and five different types of drink.
∴  The number of meal combos possible is = 4 x 3 x 2 x 5 = 120

(x-2)(x-4)

(x-8)(x+1)

(x-2)(x-10)

11 pqr

(x²+25)

4x³yz³

Area of the rectangle = Length × Breadth
Now, find the factors of the given expression:
x² - 3x + 2 = x² - 2x - 1x + 2
= x(x - 2) - 1(x - 2)
= (x - 2)(x - 1)
Hence, the possible length and breadth of the given rectangle are (x - 2) and (x - 1).

\(\frac{239}{126}\)

\(\frac{-5}{2}\)

\(\frac{37}{15}\)

\(\frac{1}{5}\)

1\(\frac{9}{10}\)

4x²-4xy+y²-9z² = (4x²-4xy+y²)-9z²
= {(2x)²-2(2x)(y)+y²}-(3z)²
= (2x-y)²-(3x)²=(2x-y+3z)(2x-y-3z)

[\(\because\) using a²-2ab+b² = (a-b)²]
4a²-4a+1 = (2a)²-2\(\times\)(2a)\(\times\)1+1²
= (2a-1)²
= (2a-1)(2a-1)

[\(\because\) using a²+2ab+b² = (a+b)²]
x²+ 8x + 16 = x²+2\(\times\)x\(\times\)4+4²
= (x+4)²
= (x+4)(x+4)

[\(\because\) using a²-b²=(a+b)²]
81a²-121b² = (9a)²-(11b)²
= (9a+11b)(9a-11b)

95 x 97 =(100 - 5) x (l00 - 3) = {100 + (-5)} x {100 +(- 3)}
100² + {(-5) + (-3)} x 100 + (-5)(-3)
10000 - 8 x 100 + 15= 10000 - 800 + 15 = 921

56 x 48 = (50+6)X(50-2)=(50+6){50+(-2)}
50² + (6 + (-2)) x 50 + 6 x (-2)
2500 + 4 x 50 - 12 = 25000 - 200 - 12
2700 - 12 = 2688

107\(\times\)103 = (100+7)\(\times\)(100+3) = 100²+(7+3)\(\times\)(100)+(7\(\times\)3)
= 10000+10\(\times\)100+21
= 1000+1000+21 = 11021

[\(\because\) (x+a)(x+b)=x²(a+b)x+ab]
(y-7)(y+3)=y²+(-7+3)y+(-7)(3)
= y²-4y+(-21)=y²-4y-21

[\(\because\) (x+a)(x+b)=x²+(a+b)x+ab]
(2x - 3) (2x + 5) = (2x)² + (-3 + 5) (2x) + (-3) (5)
= 2²x²+2X2x+(-15)
 = 4x²+4x-15

\(\cfrac { 32{ m }^{ 2 }{ n }^{ 3 }{ p }^{ 2 } }{ 4mnp } =\cfrac { 32 }{ 4 } { m }^{ 2-1 }{ n }^{ 3-1 }{ p }^{ 2-1 }={ 8m }n^{ 2 }p\)

\(\cfrac { 16{ m }^{ 3 }{ y }^{ 2 } }{ { 4m }^{ 2 }y } =\cfrac { 16 }{ 4 } { m }^{ 3-2 }{ y }^{ 2-1 }=4\quad my\)

\(\cfrac { -79{ a }^{ 4 }{ b }^{ 5 }{ c }^{ 8 } }{ -9{ a }^{ 2 }{ b }^{ 2 }{ c }^{ 3 } } =\cfrac { -72 }{ -9 } { a }^{ 4-2 }{ b }^{ 5-2 }{ c }^{ 8-3 }={ 8a }^{ 2 }{ b }^{ 3 }{ c }^{ 5 }\)

\(\cfrac { 72xy{ z }^{ 2 } }{ -9xz } =\cfrac { -72 }{ -9 } { x }^{ 1-1 }{ y }^{ 2-1 }=-8{ x }^{ 0 }{ yz }^{ 1 }-8yz\)

\(\cfrac { -21abc^{ 2 } }{ 7abc } =\cfrac { -21 }{ 7 } { a }^{ 1-1 }{ b }^{ 1-1 }{ c }^{ 2-1 }=-3{ a }^{ 0 }{ b }^{ 0 }{ c }^{ 1 }=-3{ { a }^{ 0 }{ b }^{ 0 }{ c }^{ 1 } }=-3c\)

\(\cfrac { 24{ a }^{ 3 }{ b }^{ 3 } }{ -8ab } =\cfrac { 24 }{ -8 } { a }^{ 3-1 }{ b }^{ 3-1 }=-3{ a }^{ 2 }{ b }^{ 2 }\)

\(\cfrac { 15{ m }^{ 2 }{ n }^{ 3 } }{ { 5m }^{ 3 }{ n }^{ 2 } } =\cfrac { 15 }{ 5 } ={ m }^{ 2-2 }{ n }^{ 3-2 }={ 3m }^{ 0 }n^{ 1 }=3n\)

2.1a²bcX4ab² = (2.1 x 4) x (a² x a x b x b² x c)
= 8.4 a²+1b¹⁺²c = 8.4a³b³c

\(\left( \cfrac { 3 }{ 14 } { x }^{ 2 }y \right) \left( \cfrac { 7 }{ 2 } { x }^{ 4 }y \right) =\left( \cfrac { 3 }{ 14 } \times \cfrac { 7 }{ 2 } \right) \times \left( { x }^{ 2 }\times { x }^{ 4 }\times y\times y \right) \) 
= \(\cfrac { 3 }{ 4 } { x }^{ 2+4 }{ y }^{ 1+1 }=\cfrac { 3 }{ 4 } { x }^{ 6 }{ y }^{ 2 }\)

\(\left( -\cfrac { 8 }{ 5 } { x }^{ 2 }{ yz }^{ 3 } \right) \times \left( -\cfrac { 3 }{ 4 } { xy }^{ 2 }z \right) =\left( -\cfrac { 8 }{ 5 } \times -\cfrac { 3 }{ 4 } \right) \times \left( { x }^{ 2 }\times x\times y\times z\times { z }^{ 2 } \right)\)
= \(\cfrac { 6 }{ 5 } { x }^{ 2+1 }{ y }^{ 3+1 }=\cfrac { 6 }{ 5 } { x }^{ 3 }{ y }^{ 2 }{ z }^{ 4 }\)

\(\left( { 4x }^{ 2 }yz \right) \times \left( -\cfrac { 3 }{ 2 } { x }^{ 2 }{ yz }^{ 2 } \right) =\left( { x }^{ 2 }\times { x }^{ 2 }\times y\times y\times z\times { z }^{ 2 } \right)\)
= -6x²⁺²y¹⁺¹z¹⁺² = -6x⁴y²z³

(3ab³c³)X(5a³b²c) =(3X5)(aXa³Xb²Xb²Xc³Xc)
= 15a¹⁺³,b²⁺²,c³⁺¹=15a⁴b⁴c⁴

In \(\triangle \)APB and \(\triangle \)DPC
\(\angle \)A =\(\angle \)D = 90° [given]
\(\angle \)APB =\(\angle \)DPC [Vertically opposite angles]
\(\angle \)ABP =\(\angle \)DCP [Remaining angle]
 ∴ \(\triangle \)APB ~ \(\triangle \)DPC [AAA criteria]
 \( \frac {AP }{ DP } =\frac { BP }{ PC }\) [Corresponding sides are proportional]
AP x PC = BP x DP

In  \(\triangle \)EPB and \(\triangle \)DPC,
\(\angle \)PEB =\(\angle \)PDC = 90° [given]
\(\angle \)EPB =\(\angle \)DPC [Vertically opposite angles]
\(\angle \)EPB =\(\angle \)PCD [∵ Remaining angles]
Thus, \(\triangle \)EPB ~ \(\triangle \)DPC [∵ By AAA criteria]
\(\frac { EP }{ DP } =\frac { PB }{ PC } \)
BP x PD = EP x PC

Proof:
In \(\triangle \)ABC and \(\triangle \)DAC we have,
\(\angle \)ADC =\(\angle \)BAC [ ∵ given ]
\(\angle \)L =\(\angle \)C [ ∵ common]
\(\angle \)ABC =\(\angle \)DAC [∵ Remaining angle]
\(\triangle \)ABC ~ \(\triangle \)DAC [∵  By AAA similarity]
⇒ \(\frac { AB }{ DA } =\frac { BC }{ AC } =\frac { AC }{ DC } \) [sides are proportional]
⇒ \( \frac { CB }{ CA } =\frac { CA }{ CD }\)
or CA² = CB x CD.

\(=\frac { (-12\times 12)+(-90\times 8)-(3\times 15) }{ 120 } \)
\(=\frac { -144+(-720)-45 }{ 120 } =\frac { -864-45 }{ 120 } \)
\(=\frac { -864+(-45) }{ 120 } =\frac { -909 }{ 120 } =\frac { -303 }{ 40 } \\ \)

Let the other number be x
\(\frac { 14 }{ 27 } \times x=\frac { -28 }{ 81 } \)
\(x=\frac { -28 }{ 81 } \div \frac { 14 }{ 27 } =\frac { -28 }{ 81 } \times \frac { 27 }{ 14 } =\frac { -2 }{ 3 } \times \frac { 1 }{ 1 } =\frac { -2 }{ 3 } \)
\(\therefore\) other number \(=\frac{-2}{3}\)

Let the number to be subtracted = x
\(\frac { 7 }{ 3 } -x=\frac { 5 }{ 4 } \)
\(\frac { 7 }{ 3 } -\frac { 5 }{ 4 } =x\)
\(x=\frac { (7\times 4)-(5\times 3) }{ 3\times 4 } =\frac { 28-15 }{ 12 } =\frac { 13 }{ 12 } =1\frac { 1 }{ 12 } \)
\(\frac{13}{12}\) should be subtracted

\(\frac { 4 }{ -3 } +\frac { 8 }{ 15 } =\frac { -4 }{ 3 } +\frac { 8 }{ 15 } \)
LCM of 3 and 15 is 15
\(\frac { (-4\times 5)+(8\times 1) }{ 15 } \) 

Given that r = 12cm
θ = 30°
Length of the arc l = \(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r\)
= \(\frac { { 30 }^{ o } }{ { 360 }^{ o } } \) x 2 x 3.14 x 12
= \(\frac12\) x 2 x 3.14 x 12
l = 6.28 cm
Perimeter of.a sector = l + 2r
= 6.28 + 2 (12) cm
= 6.28 + 24 cm = 30.28 cm
Perimeter of the sector = 30.28 cm

Radius of the circle = 14cm
Area of the circle = 63 cm²
Area of the circle = \(\frac { 1 }{ 2 } \)lr
\(\frac { 1 }{ 2 } \) l x r = 63
\(\frac { 1 }{ 2 } \) l x 14 = 63
l x 7 = 63
l = \(\frac{63}{7}\)
l = 9 cm
Length of the arc = 9 cm 

For a triangular prism
Faces = 5, Edges = 9, Vertices = 6
By Euler's formula F + V - E = 5 + 6 - 9
11 - 9 = 2

A pyramid has faces = 5, Vertices = 5, Edges = 8
By Euler's formula F + V - E = 5 + 5 - 8 = 10 - 8 = 2

Length of arc = \(\frac { { \theta }^{ o } }{ 360^{ o } } \times 2\pi r\) units
Given central angle θ = 60⁰
Radius of the sector r = 42 cm
l = \(\frac { { 60 }^{ o } }{ { 360 }^{ o } } \times 2\times \frac { 22 }{ 7 } \times 42\) cm = 44 cm
∴ Length of the arc = 44 cm

72

225

200

100

(2x+3y)

(x-1)²-(x-2)²={(x-1+(x-2)}{(x-1)-(x-2)}
= (2x-3)-(x-1-x+2)
= (2x-3)X1=2x-3

100 (x +y)² - 81 (a + b)² = {10(x+y)}2-{(a(a+b)}²
= {10(x+y)+9(a+b)}{10(x+y)-9(a+b)}
= (10x + 10y+.9a- 9b)} (10x +10y - 9a- 9b)

(p+q)²-(a-b)²+p+q-a+b = {(p+q)²-(a-b)²}+(p+q)-(a-b)}
= {(p+q)-(a-b)}{(p+q)-(a-b)+{(p+q)-(a-b)
={(P + q + a - b}+(p + q - a + b) + (p + q - a + b)
= (p+q -a+b)(p+q+a-b+ 1)

x¹⁶-y¹⁶+x⁸+y⁸ = {(x⁸)2-(y⁸)2}+{x⁸+y⁸}
= (x⁸-y⁸)(x⁸+y⁸)+(x⁸+y⁸)
= (x⁸ +y⁸)+ (x⁸ - y⁸ + 1)

9-a⁶+2a³-b⁶ = 9 - (a⁶ - 2a³b³ + b⁶),
= 32-{(a³)²-2Xa³Xb³+(b³)²}
= 3²-(a³-b³)²
={3 +(a³ - b³)}{3 - (a³ - b³)}
= (3 + a³ - b³) (3 - a³ + b³)
= (a³ - b³ + 3)( - a³ + b³ + 3)
 

x²+2xy+y²-a²+2ab-b² =(x²+2xy+y²)-(a²-2ab+b²)
= (x+y)²-(a-b)²
= {(x+y)+(a-b)}{(x+y)-(a-b)}
= (x+y+a-b)(x+y-a+b)

(3x + 2y)² = (3xi +(2y)² + 2 (3x) (2y)
=9x² + 41 + 12xy
12² = 9x³ + 41 + 12 x 6
144 = 9x³+ 41 + 72
144-72 = 9x²+4y²
\(\therefore\) 9x³+4y2=72

(x + y)²=x²+y²+2xy
12²=x²+y²+2X14
144=x³+y²+28
x²+y²=116

\(\cfrac { \frac { 2 }{ 3 } { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }+\cfrac { 4 }{ 3 } { ab }^{ 2 }{ c }^{ 3 }-\cfrac { 1 }{ 5 } { ab }^{ 3 }{ c }^{ 2 } }{ \frac { 1 }{ 2 } abc } =\cfrac { \frac { 2 }{ 3 } { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } }{ \frac { 1 }{ 2 } abc } +\cfrac { \frac { 4 }{ 3 } { ab }^{ 3 }{ c }^{ 3 } }{ \frac { 1 }{ 2 } abc } -\cfrac { \frac { 1 }{ 5 } a{ b }^{ 3 }{ c }^{ 3 } }{ \frac { 1 }{ 2 } abc } \)
= \(\cfrac { 2 }{ 3 } \times \cfrac { 2 }{ 1 } { a }^{ 2-1 }{ b }^{ 2-1 }{ c }^{ 2-1 }+\cfrac { 4 }{ 3 } \times \cfrac { 2 }{ 1 } { a }^{ 1-1 }{ b }^{ 2-1 }{ c }^{ 3-1 }-\cfrac { 1 }{ 5 } \times \cfrac { 2 }{ 1 } { a }^{ 1-1 }{ b }^{ 3-1 }{ c }^{ 2-1 }\)
= \(\cfrac { 4 }{ 3 } abc+\cfrac { 8 }{ 3 } { a }^{ 0 }{ bc }^{ 2 }-\cfrac { 2 }{ 5 } { a }^{ 0 }{ b }^{ 2 }c\) 
= \(\cfrac { 4 }{ 3 } abc+\cfrac { 8 }{ 3 } { bc }^{ 2 }-\cfrac { 2 }{ 5 } { b }^{ 2 }c\)

\(\cfrac { { 6x }^{ 4 }yz-{ 3xy }^{ 3 }+{ 8x }^{ 2 }{ yz }^{ 4 } }{ 2xyz } =\cfrac { { 6x }^{ 4 }yz }{ 2xyz } -\cfrac { 3xy^{ 2 }z }{ 2xyz } +\cfrac { { 8x }^{ 2 }{ yz }^{ 4 } }{ 2xyz } \)
= \(\cfrac { 6 }{ 2 } { x }^{ 4-1 }{ y }^{ 1-1 }{ z }^{ 1-1 }-\cfrac { 3 }{ 2 } { x }^{ 1-1 }{ y }^{ 3-1 }z^{ 1-1 }+\cfrac { 8 }{ 2 } { x }^{ 2-1 }{ y }^{ 1-1 }{ z }^{ 4-1 }\) 
= \(3{ x }^{ 3 }{ y }^{ 0 }{ z }^{ 0 }-\cfrac { 3 }{ 2 } { x }^{ 0 }{ y }^{ 2 }{ z }^{ 0 }+{ 4xy }^{ 0 }{ z }^{ 3 }\) 
= \({ 3x }^{ 3 }-\cfrac { 3 }{ 2 } { y }^{ 2 }-4{ xz }^{ 3 }\)

\(\cfrac { { 24x }^{ 3 }y+20{ x }^{ 2 }{ y }^{ 2 }-4xy }{ 2xy } =\cfrac { 24{ x }^{ 3 }y }{ 2xy } +\cfrac { 20{ x }^{ 2 }{ y }^{ 2 } }{ 2xy } -\cfrac { 4xy }{ 2xy } \) 
= \(\cfrac { 24 }{ 2 } { x }^{ 3-1 }y^{ 1-1 }+\cfrac { 20 }{ 2 } { x }^{ 2-1 }{ y }^{ 2-1 }-\cfrac { 4 }{ 2 } { x }^{ 1-1 }{ y }^{ 1-1 }\)
= 12x²y⁰+10xy-2x⁰y⁰=12x²+10xy-2

\(\cfrac { 9{ m }^{ 5 }+12{ m }^{ 4 }-{ 6m }^{ 2 } }{ { 3m }^{ 2 } } =\cfrac { { 9m }^{ 5 } }{ { 3m }^{ 2 } } +\cfrac { { 12m }^{ 4 } }{ { 3m }^{ 2 } } -\cfrac { { 6m }^{ 2 } }{ { 3m }^{ 2 } } \)
= \(\cfrac { 9 }{ 3 } { m }^{ 5-2 }+\cfrac { 12 }{ 3 } { m }^{ 4-2 }-\cfrac { 6 }{ 2 } { m }^{ 2-2 }\)
= 3m³+4m²-2m⁰=3m³+4m²-2

(5 -x)(3 - 2x) (4 - 3x) = {(5 -x) (3 - 2x)} x (4 - 3x) [\(\therefore\) Multiplication in association]
= {5(3-2x)-x(3-2x)}X(4-3x)
= (15 - 10x - 3x + 2x²) x (4 - 3x)
= (2x²-13x+15)(4-3x)
= 2x²X(4-3x)-13x(4-3x)+15(4-3x)
= 8x³-6³-52x+39x²+60-45x
= -6x³+47x²-97x+60

(3x - 2) (x - 1) (3x + 5) = {(3x -2) (x -1)} x (3x +5) [\(\therefore\) Multiplication in associative]
= {3x (x - 1) - 2 (x - 1)} x (3x + 5)
= (3x² - 3x - 2x + 2) x(3x + 5)
= (3x² - 5x + 2) (3x + 5)
= 3x² x (3x + 5) - 5x (3x + 5) + 2 (3x + 5)
= 9x³ + 15x² - 15x² - 25x + 6x + 10
= 9x³ -19x + 10

\(\left( \frac { 6 }{ 7 } \times \frac { 2 }{ 3 } \right) +\left( \frac { 9 }{ 11 } +\frac { 2 }{ 3 } \right) +\left( \frac { 2 }{ 3 } \times \frac { 4 }{ 9 } \right) =\left\{ \left( \frac { 6 }{ 7 } \times \frac { 2 }{ 3 } \right) +\left( \frac { 9 }{ 11 } +\frac { 2 }{ 3 } \right) +\left( \frac { 2 }{ 3 } \times \frac { 4 }{ 9 } \right) \right\} \) [∵ Associativity of addition]
\(=\left\{ \left( \frac { 9 }{ 11 } +\frac { 2 }{ 3 } \right) +\left( \frac { 6 }{ 7 } \times \frac { 2 }{ 3 } \right) \right\} \left\{ \frac { 2 }{ 3 } \times \frac { 4 }{ 9 } \right\} \) [∵Addition is commutative]
\(=\left\{ \frac { 9 }{ 11 } +\frac { 2 }{ 3 } \right\} +\left\{ \left( \frac { 6 }{ 7 } \times \frac { 2 }{ 3 } \right) +\left( \frac { 2 }{ 3 } \times \frac { 4 }{ 9 } \right) \right\} \)[∵ Addition is associative]
\(=\left\{ \frac { (9\times 3)+(2\times 11) }{ 33 } \right\} +\left\{ \left( \frac { 2 }{ 3 } +\frac { 6 }{ 7 } \right) +\left( \frac { 2 }{ 3 } \times \frac { 4 }{ 9 } \right) \right\} \)
\(=\frac { 27+22 }{ 33 } +\left\{ \frac { 2 }{ 3 } \times \left[ \frac { 6 }{ 7 } +\frac { 4 }{ 9 } \right] \right\} \) [∵ Multiplication associative]
\(\\ =\frac { 49 }{ 33 } +\left\{ \frac { 2 }{ 3 } \times \left[ \frac { (6\times 9)+(4\times 7) }{ 63 } \right] \right\} \) [∵ Distributive property of multiplication over addition]
\(=\frac { 49 }{ 33 } +\left\{ \frac { 2 }{ 3 } \times \left[ \frac { 54+28 }{ 63 } \right] \right\} =\frac { 49 }{ 33 } +\{ \frac { 2 }{ 3 } \times \frac { 82 }{ 63 } \} \)
\(=\frac { 49 }{ 33 } +\frac { 164 }{ 189 } =\frac { (49\times 63)+(164\times 11) }{ 2079 } =\frac { 3087+1804 }{ 2079 } \)
\(=\frac { 4891 }{ 2079 } \)

\(\frac { QT }{ PR } =\frac { QR }{ QS } \\ \frac { QT }{ QR } =\frac { PR }{ QS } \) [given] ...(1)
\(\angle \)1 = \(\angle \)2 
PR=PQ ...(2)  [∵ Sides opposite to equal angles are equal]
From (1) and (2)
\(\frac { QT }{ QR } =\frac { PQ }{ QS } \\ \frac { PQ }{ QT } =\frac { QS }{ QR } \)
\(\angle \)PQS = \(\angle \)TQR =\(\angle \)Q
:. By SAS criteria \(\triangle \)PQS ~\(\triangle \)TQR.

Given \(a=\frac{5}{6}, b=\frac{-3}{4},c=\frac{4}{7}\)
To verify (a + b) + c = a + (b + c)
Let \(a+b=\frac { 5 }{ 6 } +\frac { -3 }{ 4 } =\frac { (5\times 2)+(-3\times 3) }{ 12 } =\frac { 10+(-9) }{ 12 } \)
\(a+b=\frac { 1 }{ 12 } \)
Now \((a+B)+c=\frac { 1 }{ 12 } +\frac { 4 }{ 7 } \)
\(=\frac { (1\times 7)+(4\times 12) }{ 84 } =\frac { 7+48 }{ 84 } =\frac { 55 }{ 84 } \)
\(\therefore (a+B)+c=\frac { 55 }{ 84 } \quad \quad ...(1)\)
Now \(b+c=\frac { -3 }{ 4 } +\frac { 4 }{ 7 } \)
\(=\frac { (-3\times 7)+(4\times 4) }{ 28 } =\frac { -21+16 }{ 28 } =\frac { -5 }{ 28 } \)
\(x a+(b+c)=\frac { 5 }{ 6 } +\left( \frac { -5 }{ 28 } \right) =\frac { (5\times 14)+(-5\times 3) }{ 84 } =\frac { 70+(-15) }{ 84 } \)
\(a+(b+c)=\frac { 55 }{ 84 } \) ...(2)
From (1) and (2) we have (a + b) + c = a + (b + c) .
∴ Associative property is true for addition of rational numbers.

AB = AP + PB
= 3 + 6 cm = 9 cm.
AC = AQ + QC = 510 cm  =15
\(\frac { AP }{AB } =\frac { 3 }{ 9 } =\frac { 1 }{ 3 } \)
\(\frac { AQ }{AC } =\frac { 5 }{15 } =\frac { 1 }{ 3 } \)
⇒\(\frac { AP }{AB } =\frac { AQ }{AC } \)
Thus in triangles APQ and ADC we have \(\frac { AP }{AB } =\frac { AQ }{AC } \) and \(\angle \)A = \(\angle \)A.
∴ By SAS criteria of similarity \(\triangle \)APQ ~\(\triangle \)ABC
⇒ \(\frac { AP }{AB } =\frac { PQ }{ BC } =\frac { AQ }{ AC } \)
⇒ \(\frac { PQ }{BC } =\frac { AQ }{ AC} \)
\(\frac { PQ }{BC } =\frac { 5 }{15 } =\frac { 1 }{ 3 } \)
\(\frac { PQ }{BC } =\frac { 1 }{ 3 } \)
⇒ BC = 3PQ

OB = OC = 35 m
AB = CD= 14 m
OA = OD = (35 + 14) m
Area of the shaded region = Area of the rectangle ABCD + Area of the rectangle EFGH + 2 [ {Area of semi circles with radius 49m} - {Area of semi -circle with radius 35m} ]
= (14 x 120) + (14 x 120) + 2 [\(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \) x 49 x 49] - 2[\(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \) x 35 x 35]
= 1680 + 1680 + \(\frac { 22 }{ 7 } \)(49² - 35²) m²
= 3360 + \(\frac { 22 }{ 7 } \) (49 + 35) (49 - 35) m²
= 3360 + \(\frac { 22 }{ 7 } \) x 8 x 14 m² = 3360 + (44 x 84) m²
= 7056 m²

Radius of the big semicircle = 14 cm
∴ Area of big semicircle = \(\frac { 1 }{ 2 } \)πr² sq. units
= \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 14\times 14\)
= 308 cm²
Radius of small semi circles = 7cm
Area of 2 small semi circles = \(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 7\times 7\)
= 154 cm²
∴ Required area = 308 + 154 cm²
= 462 cm²

Area of quadrant AOBCA = \(\frac14\)πr²
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 3.5\times 3.5\)cm²
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \)cm² = \(\frac { 27 }{ 8 } \)cm²
= 9.625 cm²
Area of ΔAOD = \(\frac { 1 }{ 2 } \) x b x h sq. units
= \(\frac { 1 }{ 2 } \)x 3.5 x 2 cm²
= 3.5 cm²
Area of the shaded portion = Area of the quadrant - Area of triangle
= 9.625 - 3.5 cm²
= 6.125 cm²

PS = Diameter of a circle of radius 6 cm = 12 cm
PQ = QR = RS = \(\frac{12}{3}\) = 4cm
QS = QR + RS = 4 + 4 = 8 cm
∴ Perimeter of the shaded part = Arc length of semi - circle of radius 6 em + Arc length of semicircle of radius 4 cm + Arc length of semi-circle of radius 2 cm
= (π x 6) + (π x 4) + (π x 2) cm
P = 12 π cm
Area required = Area of semicircle with PS as. diameter + Area of semi circle with PQ as diameter - Area of semi-circle with QS as diameter
=\(\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times \left( { 6 }^{ 2 }+{ 2 }^{ 2 }+{ 4 }^{ 2 } \right) =\frac { 1 }{ 2 } \times \frac { 22 }{ 7 } \times 24\)
= \(\frac { 264 }{ 7 } { cm }^{ 2 }\) = 37.71 cm²
Area required = 37.71 cm²

Given arc length of the sector = 5π cm
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\pi r=\) 5π cm   .....(1)
Also area of the same sector = 20π cm²
\(\frac { { \theta }^{ o } }{ { 360 }^{ o } } \times \pi { r }^{ 2 }=20\pi { cm }^{ 2 }\).....(2)
Dividing equation (2) by equation (1), we have
\(\frac { \frac { { \theta }^{ o } }{ { 360 }^{ o } } \times \pi \times r\times r }{ \frac { { \theta }^{ o } }{ { 360 }^{ o } } \times 2\times \pi \times r } =\frac { 20\pi }{ 2\pi } \)
\(\frac { r }{ 2 } =4\)
r = 4 x 2
r = 8
Radius of the circle = 8 cm

Radius of the sector = 21 cm
Length of the arc = \(\frac { { \theta }^{ o } }{ 360^{ o } } \times 2\pi r\) units = \(\frac { { 150 }^{ o } }{ { 360 }^{ o } } \) x 2 x \(\frac{22}{7}\)
l = 55 cm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 55\times 21 }{ 2 } \) cm²
= \(\frac{1155}{2}\) = 577.5 cm²
∴ Length of the arc = 55 cm
Area of the sector = 577.5 cm²

Solid	Name	F	V	E	F + V - E
	Cube	6	8	12	6 + 8 -12 = 14 -12 = 2
	Cuboid	6	8	12	6 + 8-12 = 14-12 = 2
	Triangular Prism	5	6	9	5+6-9 = 11- 9 =2
	square Pyramid	5	5	8	5 + 5 - 8 = 10 - 8 = 2
	Triangular Pyramid	4	4	6	4 + 4 - 6 = 8 - 6 = 2

From the table F + V - E = 2 for all the solid shapes

Area of I + Area of III = Area of square - Area of two semi circles having centres as Q and S.

= \({ a }^{ 2 }-2\frac { 1 }{ 2 } \pi { r }^{ 2 }\)
= \({ a }^{ 2 }-\pi { r }^{ 2 }={ a }^{ 2 }-\pi { \left( \frac { a }{ 2 } \right) }^{ 2 }\)
= \({ a }^{ 2 }-\frac { 22 }{ 7 } \times \frac { { a }^{ 2 } }{ 4 } =\frac { { 14a }^{ 2 }-{ 1 }1a^{ 2 } }{ 14 } \)
= \(\frac { { 14a }^{ 2 }-{ 1 }1a^{ 2 } }{ 14 } \)
Area of I + Area of III = \(\frac { 3{ a }^{ 2 } }{ 14 } \)sq. units ...(1)
Area of II + Area of IV = Area of square -  Area of 2 semicircles having centres at P and R
= \({ a }^{ 2 }-2\frac { 1 }{ 2 } \pi { r }^{ 2 }={ a }^{ 2 }-\pi { \left( \frac { a }{ 2 } \right) }^{ 2 }\)
Area of II + Area of IV = \(\frac { 3{ a }^{ 2 } }{ 14 } \) sq. units ....(2)
∴  Area of unshaded region = Area of I + Area of II + Area of III + Area of IV
= \(\frac { 3{ a }^{ 2 } }{ 14 } +\frac { 3{ a }^{ 2 } }{ 14 } \) sq. units = \(\frac { { 6a }^{ 2 } }{ 14 } \) sq. units
= \(\frac { 3{ a }^{ 2 } }{ 7 } \) sq. units
∴ Area of un shaded region = \(\frac { 3{ a }^{ 2 } }{ 7 } \) sq. units
Area of the shaded portion = Area of the square - Area of the unshaded region
= \({ a }^{ 2 }-\frac { 3{ a }^{ 2 } }{ 7 } =\frac { 3{ 7a }^{ 2 }-3{ a }^{ 2 } }{ 7 } =\frac { 4{ a }^{ 2 } }{ 7 } \) sq. units
Area of the shaded part = \(\frac { 4{ a }^{ 2 } }{ 7 } \) sq. units