Area of ABCDE=ar(\(\triangle\) AFB)+ar(trap.FBCG)+ar(\(\triangle\)CGD)+ar(\(\triangle\)DEK)+ar(\(\triangle\)AKE)

Area of the rhombus = \(\frac{1}{2}\) d₁ d ₂ where d ₁, d ₂ are lengths of diagonals.
\(=\frac{1}{2} \mathrm{x}\) 10 x 8.2 cm² = 41 cm².

250 m²

Let the two parallel sides be a and b.
 According to the question,
a-b = 8    ....(i)
Area of trapezium
= \(\frac { 1 }{ 2 } \times \)h(a+b)
⇒ \(\frac { 1 }{ 2 } \) (a+b)\(\times\)19 =760
⇒ a+b =40\(\times\)2
⇒ a+b = 80                  ...(ii)
From Eqs. (i) and (ii), we get
2a =88 ⇒ \(a=\frac { 88 }{ 2 } =44\)
From Eq. (i), we get
a-b=8
⇒  b = a-8 = 44-8 = 36 cm
∴ 8 = 44 cm and b = 36 cm

Here,r = m and h = 14 m
∴ Volume of iron rods = \(\pi\)r²h
 = \(\frac { 22 }{ 7 } \times \frac { 2 }{ 100 } \times \frac { 2 }{ 100 } \times 14\)
= \(22\times \frac { 1 }{ 50 } \times \frac { 1 }{ 50 } \times 2\)
= \(\frac { 44 }{ 2500 } =\frac { 11 }{ 625 } { m }^{ 3 }\)
=0.88 m³
Number of iron rods =\(88\times \frac { 625 }{ 11 } \)=5000

Volume of water = 160 m³
Area of rectangular field = 800 m²
Let h be the height of water level in the field.
Now, volume of water = Volume of cuboid formed on the field by water
⇒ 160= Area of base x Height
⇒ 160 = 800\(\times\)h
∴ h = \(\frac { 160 }{ 800 } \)=0.2 m
Hence, the required height is 0.2 m.

Edge of a small cube = 30 cm
Surface area of each small cube = 6\(\times\)(Edge)²
=  6\(\times\)30\(\times\)30 = 5400 cm²
Edge of the cubical box = 3 m = 300 cm
Surface area of cubical box = 6\(\times\)(Edge)²
= 6\(\times\)(300)² = 6\(\times\)300\(\times\)300 = 540000 cm²
∴ Number of small cubes accomodated in the cubical box of edge 3 m or 300 cm 
= \(\frac { Surface\quad area\quad of\quad the\quad cubical\quad box }{ Surface\quad area\quad of\quad each\quad small\quad cubes } \)
= \(\frac { 54000 }{ 5400 } =100\)
Hence, 100 small cubes can be accomodated ina cubical box of 3 m edge.

Given, road roller is in cylindrical shape.
Length of road roller = 1m =100 cm  [∵1 m= 100 cm]
 and diameter of road roller = 84 cm
∴ Radius of road roller =\(\frac { 84 }{ 2 } \) = 42 cm
The curved surface area of cylindrical road roller = 2\(\pi\)rh
= \(2\times \frac { 22 }{ 7 } \times 42\times 10\) =44\(\times\)6\(\times\)100 = 26400 cm²
∵ Area covered by road roller in 1 revolution = Curved surface area of road roller = 26400 cm² = \(\frac { 26400 }{ 100\times 100 } { m }^{ 2 }\)
∴  Area covered by road roller in 750 revolutions
 = 750\(\times\)2.64 = 1980 m²
Hence, the required area of road is 1980 m²

Given, length of wall (I) = 15 m
Breadth of wall (b) = 10m and height of wall (h) = 7 m
∴  Area to be painted = Area of 4 walls + Area of ceiling
= 2h(l+b) l\(\times\)b
= 2\(\times\)7(15 + 10) + 15\(\times\)10
= 2 \(\times\)7 x 25 + 150 m²
Given, one can of paint covers 100 m² area
∴  Number of cans needed = \(\frac { Area\ to\ be\ painted }{ Area\ painted\ by\ one\ can } \)
 =\(\frac { 500{ m }^{ 2 } }{ 100{ m }^{ 2 } } \) = 5
Hence, she will need 5 cans of paint to painted the room.

Given, length of suitcase (I) = 80 cm
Breadth of suitcase (b) = 48 cm and height of suitcase (h) = 24 cm
∴ Total surface area of one suitcase = 2(1b + bh + hl)
 = 2(80\(\times\)48 + 48\(\times\)24 + 24\(\times\)80)
= 2 (3840 + 1152 + 1920) = 2\(\times\)6912 = 13824 cm²
Also, width of tarpaulin = 96 cm
Since, area of tarpaulin required to cover of suitcase will be equal to the total surface area of suitcase.
So, area of tarpaulin required to cover one suitcase =  13824 cm²
⇒ Length\(\times\)Width = 13824 [∵ tarpaulin is in the shape of a rectangle]
⇒  Length\(\times\)96 = 13824 ⇒ Length =\(\frac { 13824 }{ 96 } \) = 144 cm
Thus, length of tarpaulin required of cover 100 suitcase
= 100\(\times\)144 = 14400 cm=\(\frac { 1440 }{ 100 } m\) = 144 m
 [∵ 1m = 100 cm⇒1 cm = \(\frac { 1 }{ 100 } \) m]
Hence, 144 m of tarpaulin with width 96 cm is required to cover 100 suitcases.

Here, the amount of material to make a cuboidal box will be equal to its total surface area
For first cuboidal box, Length (l) = 60 cm
breadth (b) = 40 cm and height (h) = 50 cm
∴ Total surface area of cuboid = 2(1b + bh + hI)
 = 2(60\(\times\)40 + 40\(\times\)50 + 50\(\times\)60)
= 2(2400 + 2000 + 3000) = 2 \(\times\) 7400 = 14800 cm²
For second cuboidal box, I = b = h = 50 cm
∴ Total surface area of cuboid = 2(1b + bh + hI)
= 2(50 \(\times\) 50 + 50 \(\times\) 50 + 50 \(\times\) 50)
= 2(2500 + 2500 + 2500) = 2 \(\times\) 7500 = 15000 cm²
Since, total surface area of cuboid (a) is less, so the box (a) requires the lesser amount of material to make.

According to the question, the side along the river is parallel to and twice the side along the road.
Let the length of side along the road be x m
Then, the length of side along the river = 2x m
Given, perpendicular distance between the two parallel sides =100 m
and area of the trapezium shaped field = 10500 m²
⇒\(\frac { 1 }{ 2 } \times \) (Sum of parallel sides) Heigth = 10500
⇒\(\frac { 1 }{ 2 } \times \)  (2x+2) 100= 10500 ⇒ 3\(\times\) 50 =10500
⇒ 3x = \(\frac { 10500 }{ 50 } =\frac { 1050 }{ 5 } \)
∴ x =\(\frac { 1050 }{ 3\times 5 } \) = 70 and 2x = 2 0 =140
Hence, the length of the side along the river is 140 m.

Given, fence of the trapezium shaped field ABCD = 120 m, BC = 48 m, CD = 17 m and AD = 40 m.
Here, length of the fence of a trapezium shaped field means perimeter of the trapezium.
∴ Perimeter of trapezium shaped field ABCD = AB+BC+CD+DA
⇒ AB+BC+CD+DA =120
⇒   AB + 48 + 17 + 40 = 120
⇒ AB+105=120 
⇒ AB =120-105 = 15 m
∴ Area of the field ABCD =\(\frac { 1 }{ 2 } \times \) (BC+AD)\(\times\) AD
 =\(\frac { 1 }{ 2 } \times \) (48+40)\(\times\) 15
=\(\frac { 1 }{ 2 } \times \) 88\(\times\)15 = 660m²
Hence, area of the trapezium shaped fill is 660 m²

Given, a polygon EFGHI and FI is a diagonal of it. To find the area, we have to divide this polygon into triangle and trapezium. So, firstly draw perpendicular from opposite vertical on FI i.e. from points G, Hand E to Fl. Thus, we get perpendiculars GM, HN and EP respectively on FI and polygon is divided into 5 parts, out of which four are triangles and one is trapezium.

Area of polygon EFGHI =Area of ΔGMF + Area of trapezium GMNH + Area of ΔHNI + Area of ΔEPI + Area of ΔEPF
= \(\left( \frac { 1 }{ 2 } \times FM\times GM \right) +\left[ \frac { 1 }{ 2 } (GM+HN)\times MN \right] \) +\(\frac { 1 }{ 2 } \times NI\times HN+\left( \frac { 1 }{ 2 } \times PI\times EP \right) +\left( \frac { 1 }{ 2 } \times PF\times EP \right) \)
\([ \because area\ of\ triangle=\frac { 1 }{ 2 } \times base\times height\ and\)
\(\\ area\ of\ trapezium\ =\frac { 1 }{ 2 } \times (sum\ of\ parellel\ sides)\times height \)

Given,h = 10 cm, c = 6 cm, b = 12 cm and d = 4 cm

Area of ΔWLZ =\(\frac { 1 }{ 2 } \) \(\times\)c\(\times\) b = 6 \(\times\)10 = 30cm²
Area of rectangle LMYZ = b \(\times\)h = 12\(\times\) 10 = 120 cm²
and area of ΔXMY = \(\frac { 1 }{ 2 } \)\(\times\)d \(\times\)h = \(\frac { 1 }{ 2 } \)\(\times\) 4 \(\times\)10 = 20cm²
Now, area of trapezium WXYZ = Area of ΔWLZ + Area of rectangle LMYZ + Area of ΔXMY
=(30+120+20)cm² =170cm²
By using the formula,
Area of trapezium WXYZ = \(\frac { h(a+b) }{ 2 } =\frac { 10(22+12) }{ 2 } \)
[∵ a = c + b + d = 6 + 12 + 4 = 22 cm] = 5 \(\times\)34 =170 cm²
which is same as above.
Hence, area of trapezium is verified.