Area of the irregular field = Area of ΔAHF + Area of trapezium FHIE + Area of  + triangle EID + Area of ΔJDC + Area of rectangle BGJC + Area of ΔAGB
Area of the triangle = (\(\frac12\)x base x height) sq. units
Area of ΔAHF = \(\frac12\) (80 + 60) x 50 m² = 25 x 140 m²
= 3500 m²   ....(1)
Area of trapezium FHIE = \(\frac12\) h (a + b) sq. units =( \(\frac12\) x 35 x (50 + 40)) m²
= \(\frac12\)(35 x 90) = 1575 m² .....(2)
Area of ΔEID = \(\frac12\) x (65 + 25) x 40 m² = 90 x 20 m² = 1800 m² ....(3)
Area of ΔJDC = \(\frac12\) x 25 x 50 m² = 625 m² ....(4)
Area of rectangle BGJC = (60 + 35 + 65) x 50
= 160 x 50 = 8000 m² ....(5)
Area of the triangle AGB = \(\frac12\) 80 x 50 = 2000 m²
∴ Area of the field = (1) + (2) + (3) + (4) + (5) + (6)
= 3500 + 1575 + 1800 + 625 + 8000 + 2000 m²
= 17,500 m²
∴ Area of the field = 17,500 m²

(i) Central angle 45⁰, r = 16 cm
Length of the are l = \(\frac{θ^o}{360^o}\) x 2πr units
l = \(\frac{45^o}{360^o}\) x 2 x 3.14 x 16 cm
l = \(\frac18\) x 2 x 3.14 x 16 cm
l = 12.56 cm
Area of the sector = \(\frac{θ^o}{360^o}\) x  πr² sq.units
A = \(\frac{45^o}{360^o}\) x 3.14 x 16 x 16
A = 100.48 cm²
Perimeter of the sector P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm
(ii) Central angle 120⁰, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3 cm
Length of the are l = \(\frac{θ^o}{360^o}\) x  2πr units
l = \(\frac{120^o}{360^o}\) x 2 x 3.14 x 6.3 cm
l = 13.188 cm
l = 13.19 cm.
Area of the sector A = \(\frac{θ^o}{360^o}\) x 2πr units
A = \(\frac{120^o}{360^o}\) x 3.14 x 6.3 x 1.2 cm²
A = 3.14 x 6.3 x 2.1 cm²
A = 41.54 cm²
Perimeter of the sector P = l+ 2r cm
P = 13.19 + 2 (6.3) cm
= 13.19 + 12.6 cm
P = 25.79 cm

\(\frac { -6 }{ 4 } =\frac { -6\div 2 }{ 4\div 2 } =\frac { -3\times 11 }{ 2\times 11 } =\frac { -33 }{ 22 } =\frac { 33 }{ -22 } \)
\(\therefore \frac { -6 }{ 4 } \) equivalent to \(\frac { 33 }{ -22 } \)

\(\frac { 125 }{ 200 } ,\frac { 125\div 25 }{ 200\div 25 } =\frac { 5 }{ 8 } \)

Distance walked by A = \(\frac{7}{4}km\)
Distance jogged by A = \(\frac{3}{5}km\)
Total distance covered = \(\frac{7}{4}+\frac{3}{5}km\)
\(\frac{(7\times 5)+(3\times 4)}{20}km\)
= \(\frac{35+12}{20}km\)
\(=\frac{47}{20}km\)
More distance walked than Jogged = \(\frac{7}{4}-\frac{3}{5}km\)
= \(\frac{(7\times 5)-(3\times 4)}{20}km\)
= \(\frac{35-12}{20}km=\frac{23}{20}km\)

Let the total weight of a box of apple = x kg.
Weight of \(\frac{3}{4}\) of a box apples = 3 kg 225 gm.
= 3.225 kg
\(\frac{3}{4}\times\)x = 3225
x = \(\frac{3.225\times 4}{3}kg\)
= 1..075 x 4 kg
= 4.3 kg
= 4 kg 300 gm
Weight of the box of apples = 4 kg 300 gm

\(\frac { 15 }{ -4 } =-3\frac { 3 }{ 4 } \)
\(\frac { 15 }{ -4 } \) lies between -3 and - 4.

-2 and 0 
i,e. \(\frac{-2}{1}\) and \(\frac{0}{1}\)
\(\frac { -2 }{ 1 } =\frac { -2\times 10 }{ 1\times 10 } =\frac { -20 }{ 10 } \)
\(\frac { 0 }{ 1 } =\frac { 0\times 10 }{ 1\times 10 } =\frac { 0 }{ 10 } \)
∴ Five rational numbers between \(\frac{-20}{10}\) ( = -2) and \(\frac{0}{10}\)( = 0)are
\(\frac { -20 }{ 10 } ,\frac { -19 }{ 10 } ,\frac { -18 }{ 10 } ,\frac { -7 }{ 10 } ,\frac { -6 }{ 10 } ,\frac { -5 }{ 10 } ,\frac { 0 }{ 10 } (=0)\)

x²+ 6x + 8= x² + 2x + 4x + 8
= x(x + 2) + 4(x + 2)
x² + 6x + 8 = (x + 2)(x + 4)
Given length of the room = x + 4
width of the room = x + 2

(52)³ = (50 + 2)³
Comparing (50 + 2)³ with (a + b)³we have a = 50 and b = 2
(a + b)³ = a³ + 3a²b + 3ab² + b³
(50 + 2)³ = 50³ + 3 (50)²2 + 3 (50)(2)² + 2³
52³ = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
52³ = 1,40,608

Net for the cuboid is:

One of the possible nets for a cuboid of length = 5 cm, breadth = 4 cm, height = 3 cm is given above
Area of the cuboid = 20 cm² + 15 cm² + 20 cm² + 15cm² + 12cm² + 12cm²
= 94cm²
Using formula,
Surface area of a cuboid = 2 (lb + bh + lh) unit²
= 2(5 x 4 + 4 x 3 + 5 x 3) cm² = 2 (20 + 12 + 15) cm²
= 94cm²

Given that r = 150 m.
In 9 minutes the person covers the circumference of the circle.
He covers of the distance of the circumference in 3 minutes.
That is, he covers the distance in 3 minutes
\(=\frac{1}{3} \times 2 \pi \mathrm{r}=\frac{1}{3} \times 2 \times 3.14 \times 150 \)
= 314 m.

By Euler's formula F + V - E = 2 for a polyhedron
Here F = 12, V = 17, E = 22
F + V - E = 12 + 17 - 22
= 29 - 22
= 7≠ 2
∴ The polyhedron cannot have 12 faces 22 edges and 17 vertices.

Cylinder

Cube

Area of the card = Area of the rectangle + area of 2 semicircles
Length of the rectangle I = 30 cm
Breadth b = 21 cm
Radius of the semicircle = \(\frac{21}{2} \) cm
∴ Area of the card = (l x b) + \(\frac { 1 }{ 2 } \times 2\pi { r }^{ 2 }\)
= 30 x 21 + \(\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \) cm² = 630 + 346.5
= 976.5cm² (approximately)
∴ Area of the Invitation card = 976.5 cm²

6xy x a = −12x³y
\(a=-\frac{12 x y}{6 x y}\)
a = -2x²

The given figure is a combination of a semicircle and a triangle.
diameter = 6 cm
radius = 3 cm
base = 6 cm
height = 9 cm
The shaded area of the figure = Area of the semicircle + Area of the triangle
\(=\frac{1}{2} \pi \mathrm{r}^{2}+\frac{1}{2} \mathrm{bh} \)
\(=\frac{1}{2} \times 3.14 \times 3 \times 3+\frac{1}{2} \times 6 \times 9 \)
= 14.13 + 27 = 41.13 cm²

From this figure
Perimeter = 2l + 2 x 6 + 2r
\(=2 \times \frac{\theta}{360} \times 2 \pi \mathrm{r}+12+2 \times 3.5 \)
\(=2 \times \frac{90}{360} \times 2 \times \frac{22}{7} \times 3.5+12+7 \)
= 11+ 12 + 7
= 30 cm
The area of shaded part
= Area of two circular quadrants + Area of the rectangle
\(=\left(2 \times \frac{\theta}{360} \times \pi r^{2}\right)+(l \times b) \)
\(=\left(2 \times \frac{90}{360} \times \frac{22}{7} \times 3.5 \times 3.5\right)+(6 \times 3.5) \)
= 19. 25 + 21
= 40.25 cm²

-10p³+6p²-14p

6x \(\times\) 4= (6 \(\times\) 4)(x) = 24x

The teacher is going to select one student from class VI out of 10 students in 10 ways from class VII out of 15 students in 15 ways and from class VIII out of 20 students in 20 ways.
ஃ Number of ways 3 students can be selected = 10 + 15 + 20 = 45 ways

Given that r = 70 cm ,n = 5
Since the circle is divided into 5 equal sectors, the area of each of the sector
\(=\frac{1}{n} \pi \mathrm{r}^{2}=\frac{1}{5} \times \pi \times 70 \times 70=980 \pi \mathrm{cm}^{2}\)

Closure property for addition.
Let a = \(\frac{-5}{7}\) and b = \(\frac{8}{9}\) be the given rational numbers.
\(a+b=\frac { -5 }{ 7 } +\frac { 8 }{ 9 } \)
\(=\frac { (-5\times 9)+(8\times 7) }{ 7\times 9 } =\frac { -45+56 }{ 63 } =\frac { 11 }{ 63 } \) is in Q.
i.e a + b = \(\frac { -5 }{ 7 } +\frac { 8 }{ 9 } =\frac { 11 }{ 63 } \) is in Q.
∴ Closure property is true for addition of rational numbers.
Closure property for multiplication
Let a = \(\frac{-5}{7}\) and b=\(\frac{8}{9}\)
\(a\times b=\frac { -5 }{ 7 } \times \frac { 8 }{ 9 } =\frac { -40 }{ 63 } \) is in Q.
i.e. \(a\times b=\frac { -5 }{ 7 } \times \frac { 8 }{ 9 } =\frac { -40 }{ 63 } \) is in Q.
∴ Closure property is true for multiplication of rational numbers.