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Surface AREA of cuboid = 48 cm²

⇒ 2(lb + bh + hl) = 48 ----- (1)

⇒ Sum of 12 EDGES = 48

⇒ 4(l + b+ h) = 48 cm

⇒ l + b + h = 12 cm ----- (2)

Diagonal of cuboid = $(\sqrt {{l^2} + {b^2} + {h^2}})$ ----- (3)

Squaring (2) :

(l + b + h)² = 12²

L² + b² + h² + 2(lb + bh + hl) = 144

⇒ Substituting from (1) and (3)

Diagonal² + 48 = 144

Diagonal² = 144 – 48 = 96

SURFACE Area of TWO bases = 2πr² 

Curved Surface Area of CYLINDER = 2πrh 

According to the question,

⇒ 2πrh/2πr² = 2/1

⇒ 2R = h

Total surface area the cylinder

⇒ 23100 = 2πr × (r + h)

⇒ 23100 = 2 × (22/7) × r(r + 2r)

⇒ 3r² = 23100 × 7/44 

⇒ r = 35

Volume of the right circular cylinder = πr²h

= (22/7) × 35² × 70 = 269500 cm³

 Number of REVOLUTIONS MADE by one wheel = 1000

⇒ Distance covered = 120 km

⇒ Distance covered in one revolution = 120/1000 = 0.12 km

⇒ Distance covered in one revolution = 0.12 ? 1000 = 120 m

⇒ Thus, 2$(\pi )$R = 120 m

⇒ 2 $(\times \FRAC{{22}}{7})$(\times )$ r = 120

⇒ r = 19.09 m

∴ Radius of wheel of the cycle = 19.09 m

We know that, volume of a cube = a³

Volume of cubes of 3 cm, 4 cm and 5 cm sides = 3³ + 4³ + 5³ cm³

= 27 + 64 + 125 = 216 cm³

Surface area of the smaller cubes = 6(9 + 16 + 25) = 6 × 50 = 300 cm²

Since the large cube is obtained by melting the three smaller cubes,

THUS, volume of larger cube = sum of VOLUMES of three smaller cubes = 216 cm³

Side of larger cube $(a)$ = ?216 cm = 6 cm

Surface area of larger cube = 6 × a² cm², where a = length of each side of a cube

= 6 × 6² = 216 cm²

$(\therefore \frac{{surface\;area\;of\;smaller\;cubes}}{{surface\;area\;of\;larger\;cube\;}}\; = \;\frac{{300}}{{216}} = \;\frac{{25}}{{18}})$

⇒ Volume of LEAD = (4/3) π × 2³

⇒ Volume of GOLD = Total volume – Volume of lead

⇒ (4/3) π × (2 + x)³ – (4/3) π × 2³

According to question,

⇒ (4/3) π × 2³ = (4/3) π × (2 + x)³ – (4/3) π × 2³

⇒ (2 + x)³ = 2 × 2³

⇒ 2 + x = 2 × ³√2

⇒ x = (2 × ³√2) – 2

⇒ 2(³√2 – 1)

⇒ 2(1.259 – 1)

⇒ Outside VOLUME of the box = 42 x 30 x 20 cm³

⇒ Outside volume of the box = 25200 cm³

Since the thickness is 1 cm, the dimensions of inner CUBOID are 2 less than the dimensions of Outer cuboid

⇒ Inner volume of the box = 40 x 28 x 18 cm³

⇒ Inner volume of the box = 20160 cm³

⇒ Volume of wood = 25200 – 20160 cm³

LET the length and BREADTH of the rectangle are 3x and 7x RESPECTIVELY. 

Perimeter of the rectangle = 80

⇒ 2(l + B) = 80

⇒ 2(3x + 7x) = 80

⇒ 10x = 40

⇒ x = 4

∴ The area of the rectangle = lb = 7x × 3x = 21x² = 21 × 16 = 336 m² 

Volume of cylinder = πr²h?

where, 

r = RADIUS of cylinder

h = height of cylinder

Curved Surface Area = 2πrh

Volume/Curved Surface Area = 616/352

⇒ (πr²h)/(2πrh) = 28/16 = 7/4

⇒ r/2 = 7/4

⇒ r = 7/2

Again,

Curved Surface Area = 352

⇒ 2 × 22/7 × 7/2 × h = 352

⇒ h = 16

Again

⇒ TOTAL Surface Area = 2πr(r + h)

⇒ 2 × 22/7 × (7/2)(7/2 + 16)

∴ Total Surface Area = 22 × 39/2 = 429 m²

Let the BREADTH be x then length be 5x

⇒ AREA = 1805

⇒ x × 5x = 1805

⇒ 5x² = 1805

⇒ x² = 1805/5 = 361

Length cannot be negative

⇒ x = 19

⇒ Length = 5x = 5 × 19 = 95 m

⇒ Breadth = x = 19 m

⇒ Length will remain same for racing track

⇒ Area of racing track = 1805 - [(95 - 6) × (19 - 6)] sq metre

⇒ Area of racing track = 648 sq metre

Solution:

Given : A circle circumscribes a rectangle with,

Length 'l' = 4 cm and BREADTH 'B' = 3 cm.

To find : Difference between the area of the circle and rectangle.

Area of rectangle = Length × Breadth

= 4 × 3 = 12 cm²

To find the area of the circle we have to find the diagonal of the rectangle which gives US the diameter of the circle.

To find the diagonal we have to use Pythagoras Theorem.

Diagonal (d) = √( l² + b² )

d= √( 4² + 3² )

d= √( 16 + 9 )

d = √( 25 )

d = 5 cm 

We KNOW that,diagonal of rectangle = diameter of circle

Diameter (D) = 5cm

Radius (r) = 5/2 = 2.5 cm 

Area of the circle = πr² cm²

Area of the circle = 3.14 × 2.5 × 2.5 cm²

Area of the circle = 19.625 cm²

Difference between the two areas = (19.625 - 12)cm² = 7.625 cm²

So, the correct option is 3). 7.625 cm²

Quantity A:

Let the number of hemispheres that can be formed = x

Radius of hemisphere to be formed = R_h

⇒ Volume of cylinder = π × r² × h = π × (6/2)² × 4 = π × 3² × 4 = 36π

Now,

 Volume of sphere = volume of cylinder

⇒ (4/3) × π × R³ = 36 × π

⇒ R = 3 cm

⇒ Radius of hemisphere forms = R_h = R/2 = 3/2

Now, since hemisphere is formed by melting sphere,

⇒ Volume of all the hemisphere formed = volume of sphere

⇒ x × {(2/3) × π × R_h ³} = 36 × π

⇒ x × {(2/3) × π × (3/2) ³} = 36 × π

⇒ x × {(2/3) × π × (27/8)} = 36 × π

⇒ x = 16

So, 16 hemispheres can be formed from melting the sphere

⇒ Quantity A = 16

Quantity B:

Let number of ice cream cones that can formed = x

Now, Ice cream container is a cylinder having radius 6 cm and height 5 cm,

⇒ TOTAL volume of ice cream in container = volume of cylinder = π × r² × h = π × 6² × 5 = 180 × π

Now, it is given that ice cream formed should also have a hemisphere built on it,

⇒ Total volume of 1 ice cream cone = volume of cone + volume of hemisphere

Given,

⇒ height of cone given = h_c= 4 cm

⇒ Slant height of cone = L = 5 cm

We know, r_c² + h_c² = l²

⇒ r_c² + 16 = 25

⇒ r_c² = 25 – 16 = 9

⇒ r_c = 3

⇒ Volume of cone = (1/3) ×π × r² × h = (1/3) × π × 3² × 4 = 12 × π

⇒ Volume of hemisphere = (2/3) × π × r_h³

As it is given that radius of hemisphere = radius of cone

⇒ r_h = 3

⇒ Volume of hemisphere = (2/3) × π × 3³ = 18 × π

⇒ Total volume of one ice cream cone = (12 × π) + (18 × π) = 30 × π

⇒ Total number of cones = total volume of ice cream/volume of one ice cream cone

⇒ Number of cones = (180 × π)/(30 × π) = 6

⇒ Number of cones that can be made = 6

⇒ Quantity B = 6

Let the length and the breadth of a rectangle be 4x and 3x.

We know that the area of rectangle = Lenght × Breadth

⇒ Area of rectangle = (4x) × (3x) = 12x²

⇒ 108 = 12x²

⇒ X² = 9

⇒ x = 3

∴ Length = 4x = 12 and Breadth = 3x = 9

It is given that, SIDE of SQUARE = Breadth of rectangle = 9

Area of square = Side² = 9² = 81

Hence, the area of square = 81 m²

Let the side of the EQUILATERAL triangle be ‘X’.

Perimeter of the triangle = 3x

Let the side of the REGULAR hexagon be ‘y’

⇒ Perimeter of regular hexagon = 6y

∴ 6y = 3x

2y = x

y = x/2 ----- (1)

Area of an equilateral triangle = √3x²/4

Area of regular hexagon = 3√3y²/2 = 3√3x²/8 

From (1)

VOLUME of Hemisphere = (2/3) × π × r³, where r is the radius of hemisphere.

In case of 1^st hemisphere,

Inner radius = 3 cm

∴ Volume of water in it = (2/3) × π × 3³ = 18π SQ. cm.

In 2^nd hemisphere,

Inner radius = 6 cm.

∴ Volume of water it can contain = (2/3) × π × 6³ = 144π sq. cm.

∴ Empty volume = 144π - 18π = 126π sq. cm.

∴ Percentage empty = (126π/144π) × 100 = 87.5%

Let ‘l’,’B’ be the LENGTH and breadth of the rectangular plot as shown in the figure.

Area of 1 metre BROAD pathway = Area of outer RECTANGLE – Area of inner rectangle

⇒ Area of 1 metre broad pathway = lb – (l – 2)(b – 2)

⇒ Area of 1 metre broad pathway = lb – lb + 2L + 2b – 4

Area of 1 metre broad pathway = 2(l + b) – 4

? We don’t know the values of l and b, the given data is inadequate.

Area of the playground covered = LATERAL surface area of roller × number of revolutions

Lateral surface area of a roller = 2πrh

Given, diameter of a roller is 91 CM and its LENGTH 100 cm.

∴ Radius = 91/2 = 45.5 cm

Lateral surface area of the roller = 2 × (22/7) × 45.5 × 100 = 28600 cm²

⇒ Lateral surface area of the roller = 2.86 m²

Given,

Radius of the BASE of the cone (R) = 10 m

Height of the Cone (H) = 24 m

? SLANT height (l) = $(\sqrt {\left( {{R^2}{\rm{\;}} + {\rm{\;}}{H^2}} \right)})$

∴ Slant Height = $(\sqrt {{{10}^2} + {{24}^2}}= 26{\rm{\;}}m)$

We know,

Curved Surface AREA of the cone = πRl

∴ Curved Surface Area = π x 10 x 26 = 260π m²

Area of the cloth USED = Curved Surface Area of the cone = 260π m²

Given, width of the cloth = 5.2 m

Length of the cloth used = Area of Cloth used/Width of Cloth

Diagonal of the square = 15 cm = √2 × Side[? √2 = 1.414]

∴ Side of the square =15/√2 cm

Area of the square = (Side)²  = 112.5 sq. cm

Diagonal of the square = 22 cm = √2 × Side

∴ Side of the square = 22√2 cm

Area of the square = (Side)²  = 242 sq. cm

DIFFERENCE = 242 – 112.5 = 129.5 sq. cm

Given that,

The radius of a garden roller = 1.4 m

Length of garden roller = 2 m

According to the QUESTION,

One revolution of the roller is EQUAL to the surface area of roller which is in CYLINDRICAL shape.

So, Area of roller = 2πrl

Where l is length of roller.

Area of roller = 2 × (22/7) × 1.4 × 2

⇒ Area of roller = 17.6 m2

So, area covered in 10 revolution = 10 × area of roller = 176 m2

∴ Area covered by garden roller in 10 revolution is 176 m2.

For the CYLINDER the CIRCUMFERENCE = length of the RECTANGULAR sheet

i.e. 2πr = 16

⇒ r = 8/π and h = 5 cm

Now volume of cylinder = πr²h = π(8/π)²(5)

Volume of a cylinder = πr²h

Given, its RADIUS is decreased by 10% and height increased by 20%.

R = r – 10% of r = 0.9r

H = h + 20% of h = 1.2h

Volume of the new cylinder = πR²H

⇒ Volume of the new cylinder = π(0.9r)² × 1.2 h

⇒ Volume of the new cylinder = 0.972 πr²h

Decrease in volume = πr²h – 0.972 πr²h = 0.028 πr²h

Area covered by the cylinder in TOTAL,

⇒ 2 × π × 0.7 × 8.4 × 400

⇒ 14784 m²

∴ Area of the square field = 14784 m²

∴ SIDE of the square

⇒ √14784 = 121.59 m

Given that,

DIMENSIONS of room are 24 m long, 16 m broad and 18 m high.

The longest rod that can be placed in a room has length EQUAL to the diagonal of the room.

We know that diagonal of CUBOID = √(length² + breadth² + height²)

Let the length, BREADTH and height of cuboid be l, b and H respectively

Length of diagonal = (l² + b² + h²)^1/2 = 5√2

Total SURFACE area of cuboid = 2(LB + bh + lh)

(l + b + h) = 12

Squaring both sides

(l² + b² + h²) + 2(lb + bh + lh) = 144

According to the condition given in the PROBLEM,

Area of circle = 154

⇒ 22/7 × r² = 154

⇒ r² = 49,

⇒ r = 7

The wire is bent in the form of square, then

Length of wire = perimeter of circle

Length of wire = 2 πr = 2π × 7 = 44 cm = perimeter of square

⇒ 4a = perimeter of square (assuming a = side of square)

⇒ 4a = 44

⇒ a = 11

Area of square = a² = 11² = 121 cm²

Sum of the INTERNAL ANGLES of a polygon with ‘n’ sides = (n - 2) × 180°

∴ For a hexagon, sum of the internal angles = (6 - 2) × 180° = 720°

Let the measure of each of the other 5 angles of the given hexagon be ‘x’

∴ ACCORDING to the given conditions,

⇒ 90° + 5X = 720°

∴ x = 126°

∴ The measure of each of the other angles is 126°

We KNOW that, Volume of Sphere = $(\frac{4}{3}\pi {R^3})$

Where R is the radius of sphere

⇒ 38808 $(= \;\frac{4}{3}\pi {R^3})$

⇒ R³ $(= \frac{{38808\; \TIMES \;3\; \times \;7}}{{4\; \times \;22}})$

∴ R = 21 m

Surface area of Sphere = 4π R²

∴ Surface Area = 4π × 21² $(= \;\frac{{4\; \times \;22\; \times \;\;21\; \times \;21}}{7})$

∴ Surface area = 5544 m²

Let the radius of hemisphere be R CM.

⇒ VOLUME of shell = Volume of hemisphere

⇒ π × (4/3) × cube of 15 - π × (4/3) × cube of 13 = π × (2/3) × cube of R

⇒ 2 × 1178 = cube of R

⇒ R = cube ROOT of 2356 = 13.3 cm

As we know, DIAGONAL of a square = √2 × SIDE of square

Let, side of square be ‘a’ cm

Diagonal = a√2 = a + √2

⇒ a√2 – a = √2

⇒ a(√2 – 1) = √2

⇒ a = √2/(√2 – 1)

⇒ a = √2/(√2 – 1) × (√2 + 1)/(√2 + 1)

⇒ a = (2 + √2)/(2 – 1)

⇒ a = 2 + √2

∴ AREA of square = a² = (2 + √2)² = 4 + 2 + 4√2 = 6 + 4√2

We know that,

AREA of a triangle = ½ × Base × Height

Let us assume that the height and base of a given triangle are `H’ and `b’ respectively

∴ Area of the triangle = ½ × b × h----(1)

Given that height of the triangle is decreased by 40% and base is INCREASED by 40%

⇒ NEW height of the triangle = h – (40h/100) = 3h/5

⇒ New base of the triangle = b + (40b/100) = 7b/5

New area of the triangle = ½ × (3h/5) × (7b/5) = 21bh/50---(2)

∴ It is clear from (1) and (2) that Area of the new triangle is decreased

Decrease in area = (bh/2) – (21bh/50) = 2bh/25

We know that,

Percentage decrease in area = $(\frac{{{\rm{Decrease\ in\ area}}}}{{{\rm{Original\ area}}}} \TIMES 100)$

∴ Percentage decrease in area = $(\frac{{\frac{{2bh}}{{25}}}}{{\frac{{bh}}{2}}} \times 100{\rm{}} = {\rm{}}16)$

Quantity A:

Let height of the cylinder = a cm, RADIUS = (a + 2) cm and Curved surface area = 220 cm²

⇒ 2πrh = 220

⇒ 2 × 22/7 × a × (a + 2) = 220

⇒ a² + 2a – 35 = 0

⇒ (a + 7) (a – 5) = 0

⇒ a = 5 or -7 (? height can’t be NEGATIVE)

⇒ Height = 5 cm and radius = 7 cm

⇒ Volume of cylinder = πr²h = 22/7 × 7 × 7 × 5 = 770 cm³

Quantity B:

⇒ Volume of the pyramid = 1/3 × (Area of base) × (Height) = 1/3 × (15 × 15) × 10 = 750 cm³

Given,

Perimeter of square = 60 cm

We KNOW that,

The perimeter of square = 4 × side

⇒ 60 = 4 × side

⇒ side = 15 cm

DIAGONAL of square = √2 × side

⇒ Diagonal of square = 15√2 = Lenght of rectangle

We know that, 

AREA of rectangle = LENGTH × Breadth

Area of rectangle = 15√2 × 10√2 = 300 cm²

Number of Revolutions = Total DISTANCE/Distance covered in 1 revolution

Distance covered in 1 revolution = Circumference of wheel = 2πr = 2 × π × 30 = 60π

⇒ Speed = Distance/Time

⇒ 45 = Distance/(8/60)

⇒ Distance = 90/15 = 6 km

⇒ Distance = 6 × 1000 m = 6000 × 100 CM = 600000 cm

Since the amount of material used for both the sphere and the material are the same, their volumes will be equal.

Let the RADIUS of the sphere be r, and that of the cylinder be R.

It is given that the height of the cylinder is 2 ½ times its radius,

⇒ H = 5/2 × R

Volume of the sphere = $(\frac{4}{3}\pi {r^3})$

Volume of the cylinder = $(\pi {R^2}H = \pi {R^2} \times \frac{5}{2}R = \frac{5}{2}\pi {R^3})$

Since we KNOW that these volumes are the same,

$(\begin{array}{L} \frac{4}{3}\pi {r^3}\; = \;\frac{5}{2}\pi {R^3}\\ \Rightarrow \;\frac{{{r^3}}}{{{R^3}}} = \frac{15}{8}\\ \Rightarrow \;r\;:\;R\; = \;\sqrt[3]{15}:2 \end{array})$

RADIUS of circle = 3 m²

Square is INSCRIBED in a circle, which means the diameter of circle is the diagonal of square.

⇒ Diameter of circle = 2 × radius = 2 × 3 = 6

Diagonal of square = 6

We know,

⇒ Side² + Side² = Diagonal²

⇒ 2 × side² = 6²

⇒ 2 × side² = 36

⇒ Side² = 18

⇒ We know, area of square = side²

Let, the radius of the sphere = r cm.

∴ Surface AREA of the sphere = 4πr² cm²

According to Problem,

⇒ 4πr² = 100π

Or, r² = 25

⇒ r = 5

∴ DIAMETER of the sphere = 2R = 2 × 5 = 10CM

Let the SIDE of the SQUARE be X cm

Area of the square = X²

CHANGED side of the square = X + 0.18 × X = 1.18X

Area of the changed square = 1.18X × 1.18X = 1.3924X²

Change in the area = 1.3924X² – X² = 0.3924X²

Percentage change in area = 39.24%

Area of a TRIANGLE = ½ × Base × Height

Let, Height and base of the triangle are 6x and 7X respectively

Given,

Area of the triangle = 3549 m²

3549 = ½ × 7x × 6x

⇒ 42x² = 7098

⇒ X² = 169

⇒ x = 13 [? x can’t be negative]

Area of circle = 1386

Πr² = 1386 ⇒ r = 21 cm

Length of RECTANGLE = 21 + (33.33/100) × 21 = 28 cm

Circumference of circle = 2πr = 132 cm

Circumference of rectangle = 220 – 132 = 88 cm

2(L + B) = 88 ⇒ b = 16 cm

The radius of base of a right circular CONE is 4 cm. Its HEIGHT is 1 cm LESS than the radius of base.

So, height will be 3 cm.

⇒ SLANT height = √ (4 × 4 + 3 × 3) = 5 cm

We know, 1 hectare = 10000 square meter

Total COST to fill it with sand = Rs. 1440

Cost to fill PER hectare is Rs. 160

∴ Area of the field = 

$(\frac{{1440}}{{160}})$ hectare = 9 hectare = 9 × 10000 square meter

If length of playground is L meter then its area is L² square meter

∴ L² = 9 × 10000

⇒ L = √ (9 × 10000) = 300 meter

Perimeter of the square playground is 4L meter = 4 × 300 =1200 meter

The cost of putting a FENCE around it at the rate of 75 paise per meter is

= Rs. (1200 × 0.75) = Rs. 900