Area of figure A, for radius 8 cm
 =\(\pi\)r²h = \(\frac { 22 }{ 7 } \times \)8\(\times\)8=201.14 cm²
Since, there are two pipes.
∴ Area = 2\(\times\)201.14 = 402.28 cm²
Now, area of figure B, for radius 15 cm
 =\(\pi\)r² =\(\frac { 22 }{ 7 } \times \) 15\(\times\)15 = 707.14 cm²
Hence, pipe B fill the swimming pool faster.

There are two cases
Case I When the rectangular sheet is rolled along its length, then length of the sheet forms the circumference of its base and breadth of sheet becomes the height of cylinder.
Let r cm be the radius of the base and h cm be the height.
Then, h = 33 cm
and circumference of the base = 44 cm
⇒2\(\pi\)r =44   2\(\times\)\(\frac { 22 }{ 7 } \)\(\times\)r⇒  =44
⇒ r =\(\frac { 44\times 7 }{ 2\times 22 } \)⇒ r = 7 cm
∴ Volume of the cylinder r =\(\pi\)r²h =\(\frac { 22 }{ 7 } \) \(\times\)7\(\times\)7\(\times\)33
= 5082 cm³
Case II When the rectangular sheet is rolled along its breadth, then breadth of the sheet forms the circumference of its base and length of the sheet becomes the height of the cylinder.
Now,h = 44 cm and circumference = 33 cm
⇒2\(\pi\)r =33⇒ 2\(\times\)\(\frac { 22 }{ 7 } \)\(\times\)r= 33  ⇒  r=33
⇒  r=\(\frac { 33\times 7 }{ 2\times 22 } \)⇒ r =\(\frac { 21 }{ 4 } \) cm
∴ Volume of the cylinder =\(\pi\)r²h= \(\frac { 22 }{ 7 } \times \frac { 21 }{ 4 } \times \frac { 21 }{ 4 } \times 4\)
= 3811.5 cm³

In II-nd sailboat, area of sails is calculated below:
Area of ΔABE =\(\frac { 1 }{ 2 } \times \) Base\(\times\)Height
 =\(\frac { 1 }{ 2 } \times \)10.9\(\times\)19.5 = 106.3 m²
Similarly, area of ΔBCD = 23.9\(\times\)8.6
 = 102.8 m²

We know that, the dimensions of cuboid, l=70 m, b = 35 m, h = 40 m,
diameter of cylinder is 35 m and cost of painting is Rs. 2 per m²
Area of cylindrical to be painted
\(\frac { 1 }{ 2 } \)\(\times\)Total surface area =\(\frac { 1 }{ 2 } \) [2\(\pi\) r(r+h)] 
= \(\frac { 1 }{ 2 } \left[ 2\times \frac { 22 }{ 7 } \times \frac { 35 }{ 2 } \left( \frac { 35 }{ 2 } +70 \right) \right] \)=4812.5 m²
Area of cuboid to be painted = Area of three walls = lh+2bh
= 70\(\times\)40 + 2\(\times\)40\(\times\)35 = 2800 + 2800
 = 5600 m²
Total area to be painted = 4812.5 + 5600=10412.5 m²
∵ Cost of painting per m² = Rs. 2
∴ Cost of painting 10412.5 ² = Rs. 10412.5\(\times\)2 = Rs. 20825
Verification Verify your answer by adopting some other plan i.e. here in this problem instead of taking area in two steps, let us find in one step.
Area to be painted= Area of three walls + Area of cylindrical part
 =  2bh+lh +\(\frac { 1 }{ 2 } \)[2\(\pi\)rh+2\(\pi\)r²]
= h[2b+l]+ [\(\pi\)R(r+h)]
= 40 [2\(\times\)35+70]+ \(\frac { 22 }{ 7 } \frac { 35 }{ 2 } \left( \frac { 35 }{ 2 } +70 \right) \)
= 40 [140] + 55\(\times\)87.5
= 5600+ 4812.5=10412.5m²
∴  Final cost (similar as in previous method) = Rs. 20825

For box A,
Given, length of box = 20 cm
Breadth of box = 8 cm
and height of box = 3 cm
∴ Volume of the box = I \(\times\)b \(\times\)h = 20\(\times\)8\(\times\)3 cm³ =480 cm³
Total surface area of the box = 2(lb + bh + hl)
= 2(20\(\times\) 8 + 8\(\times\)3 + 3\(\times\)20) = 2\(\times\)(160+24+60) =2\(\times\)244 = 488 cm³
For box B,
Given, length of box = 12 cm
Breadth of box = 10 cm and height of box = 4 cm
∴ Volume of the box = I \(\times\) b \(\times\) h = 12 \(\times\) 10 \(\times\) 4 = 480 cm³
Total surface are of the box = 2 (lb + bh + hl)
 = 2(120 + 40 + 48) = 2\(\times\)208 = 416cm²
It is clear from above that both boxes have same volume, but the surface area of type B box is less than that of type A box.
Thus, material required for box B is less.
∴ Box B is more economical than box A.
Now, let another box of size be 12 cm\(\times\)8 crn\(\times\)5cm
Volume of this box = 12\(\times\)8\(\times\)5 = 480 cm³
and its surface area =2(lb + bh + hl) = 2(12\(\times\)8 + 8\(\times\)5 + 5\(\times\)12)
= 2(96 + 40 + 60) = 2\(\times\)196 = 392
Clearly, surface area of this box is less than that of box B
Hence, we can suggest any other box of size 12 cm\(\times\)8 cm \(\times\)5 crn, which has the same volume buts is more economical than the given boxes.

There are some cases in which we can arrange 12 cubes of equal length say b to form a cuboid, as follows:
Case I: We can arrange 12 cubes of equal length as shown in the figure.
Here, length (L) = 2b, breadth (B) = band height (H) = 6b
 ∴ Total surface area of the cuboid 
= 2(LB + BH + HL)
= 2(2b\(\times\)b + b\(\times\)6b + 6b\(\times\)2b)
= 2(2b² + 6b² + 12b²)
= 2\(\times\)20b² = 40b²

Case II We can arrange 12 cubes of equal length as shown in the figure.
Here, length (L) = 6b, breadth (B) = b and height (H) = 2b

 ∴ Total surface area of the cuboid = 2(LB + BH + HL)
= 2(6b\(\times\)b+ 2b+2b\(\times\)6b)
= 2(6b²+2b²+12b²) = 2\(\times\)20b² =40b²
Case III We can arrange 12 cubes of equal length as shown in the figure.
Here, length, L = 4 b, breadth, B = band height, H = 3b
  ∴ Total surface area of the cuboid
=2(LB+BH+HL)
= 2(4b\(\times\)b+b\(\times\)3b+3b\(\times\)4b)
=2(4b²+3b²+12b²) = 2 19b² = 38b²

Case IV
We can arrange 12 cubes of equal length as shown in the figure. Here, length (L) = 3b,breadth (B) = 2b and height (H) = 2b
 ∴ Total surface area of the cuboid 
= 2(LB + BH + HL) = 2(3b\(\times\)2b + 2b\(\times\)2b + 2b\(\times\)3b)
= 2(6b² + 4b² + 6b²) = 216b² =32b²
It is clear from the above cases that in Case IV, we get minimum surface area.
So, we arrange 12 cubes according to Case IV to get smallest surface area.

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

Given, polygon ABCDE is divided into four parts, so it is clear from given figure that 
Area of polygon ABCDE =Area of ΔAFB + Area of trapezium FBCH + Area of ΔCHD+ Area of ΔADE           ....(i)
Also, AD =8 cm, AH=6cm, AG = 4cm, AF=3 cm,BF = 2 cm, CH = 3 cm and EG = 2.5 cm
Now, area of ΔAFB = AF \(\times\)BD =\(\frac { 1 }{ 2 } \times \) 3\(\times\) 2 = 3 cm²
Area of trapezium FBCH = \(\frac { 1 }{ 2 } \times \)FH \(\times\)(BF+CH)
 = \(\frac { 1 }{ 2 } \times \)3 \(\times\)(2+3)
[∵ FH= AH - AF=6 -3 =3 cm ]
 =\(\frac { 1 }{ 2 } \times \) 3 \(\times\)5 = =7.5 cm²
Area of ΔCHD =\(\frac { 1 }{ 2 } \times \) HD \(\times\)CH
=\(\frac { 1 }{ 2 } \times \) (AD-AH)\(\times\) CH [∵ HD=AD-AH]
=\(\frac { 1 }{ 2 } \times \) (8-6)\(\times\)3 =\(\frac { 1 }{ 2 } \times \) 2\(\times\)3 = 3 cm²
Now, area of ΔADE =\(\frac { 1 }{ 2 } \times \) AD\(\times\) GE =\(\frac { 1 }{ 2 } \times \) 8\(\times\) 2.5
=  4\(\times\) 2.5 = 10 cm²
On putting all these values in Eq. (i), we get
Area of polygon ABCDE = (3 + 7.5 + 3 + 10) = 23.5 cm²

Let ABCD be a given quadrilateral, which is a parallelogram. Join the diagonal BD of the parallelogram ABCD and it divides the parallelogram into two ΔABD and ΔBCD.

Then, area of parallelogram ABCD = Area of ABD + Area of ΔBCD.
 = \(\frac { 1 }{ 2 } \)\(\times\)AB\(\times\) h + CD\(\times\) h
=  \(\frac { 1 }{ 2 } \)\(\times\) b \(\times\)h + b \(\times\)h
=  \(\frac { bh }{ 2 } +\frac { bh }{ 2 } =\frac { bh+bh }{ 2 } =\frac { 2bh }{ 2 } \) = bh sq units
We know that,
Area of parallelogram = Base \(\times\) Height
 = b \(\times\)h = bh sq units
We also know that, a parallelogram can also be a trapezium.
∴ Area of trapezium ABCD =\(\frac { 1 }{ 2 } \)  \(\times\)(Sum of parallel lines) \(\times\)(Perpendicular distance between the parallel sides)
=\(\frac { 1 }{ 2 } \)\(\times\) (b + h) \(\times\)h
=\(\frac { 1 }{ 2 } \) \(\times\)2b\(\times\)h = bh sq units
Hence, we can say that the above relation agrees with formula that we know already.

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

Given side of square plot = 25 m
∴ Area of square plot = (Side)² = 25\(\times\) 25 = 625 m²
From the figure, it is clear that constructed portion, i.e. house is in rectangular shape. It has length 20 m and breadth 15 m.
So, area of the constructed house = Length\(\times\) Breadth
= 20 \(\times\)15 = 300 m²
∴ Area of the garden = Area of square plot - Area of constructed house
= 625 - 300 = 325 m²
It is given that, the cost of developing the garden
 = Rs 55 per m²
∴  Total cost of developing the garden
 = Rs (55 \(\times\)325) = Rs.17875

(i) Total length of the fence surrounding it = Perimeter of -he park = 30m + 20m + 30m + 20m = 100 m
(ii) Land occupied by the park = Area of the park = (30 \(\times\) 20) m² = 600 m²
(iii) Area of cemented path = Area of the park - Area of the park left after cementing the path
= 600 m² - [(30 - 2) \(\times\) (20 - 2)] m² = 600 m² - (28 \(\times\) 18) m² = 600 m² - 504 m² = 96 m²
Number of cement bags = \(\frac { Area\quad of\quad the\quad path }{ Area\quad cemented\quad by\quad 1\quad bag } =\frac { 96 }{ 4 } \)=24
(iv) Area covered by the grass =Area of the park left after cementing the path - Area of rectangular beds
= 504 m² -2(1.5 2)m²= 504m²-6m²=498m²