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Now let B be the base of the cuboidal box whose area is 35 sq. cm and let A be the area of one of the FACE which is 60 sq. Cm

We can see that there is a COMMON SIDE to face A and face B.

∴ We will find the factors of the area of both faces and the common number will be length of the common side

⇒ 35 = 5 × 7

⇒ 60 = 2 × 2 × 3 × 5

Now we can see that the length of the common side = 5 cm

∴ Length of the other sides of the cuboid will be remaining factors of the area

∴ other sides are 7 and 2 × 2 × 3 i.e. 7 and 12

Now

Volume of cuboid = l × b × h

 Where l = length, b = breadth, h = height

Solution;

Given : ABCD is a parallelogram.

AB = CD = 20 cm 

AD = BC = 10 cm 

It is ALSO given that

Let us take DM with M on the AB line as the perpendicular with height 'h'.

Now in the triangle DAM,

SINE teta = DM / AD

sine 30 degree = h / 10

1/2 = h / 10

By cross multiplication we get 

2h = 10

h = 5 cm

Area of the parallelogram

= perpendicular height × perpendicular base (AB) sq.units.

= 5 × 20

= 100 sq.cm.

The correct option is 1). 100 sq.cm.

Volume of a CYLINDER = πr² × h

Hence the ratio of their volumes

= (πr₁² × h₁)/(πr₂² × h₂)

= (9² × 3)/(4² × 7)

Let ABCD be the given rhombus. And AC be the bigger diagonal.

So, AB = BD = 4 cm

So, ΔABD is an EQUILATERAL triangle

⇒ Ar (ΔABD) = √3 (side)²/4

⇒ Ar (ΔABD) = 4√3

Now, Ar (rhombus ABCD) = 2 [Ar (ΔABD)]

⇒ Ar (rhombus ABCD) = 8√3

Also, Ar (rhombus ABCD) = ( 1/2) × AC ×BD

⇒ (1/2) × AC × BD = 8√3

⇒ AC = (8√3 × 2)/4 = 4√3

Now, AC is the side of new equilateral Δ

⇒ Area of new triangle = [√3 × (4√3)²]/4

⇒ REQUIRED area = 12√3 cm²

∴ the correct option is 4) 

Volume of SPHERE = (4/3) π r³

Given Diameter (d) = 42 cm

Radius = Diameter/2 = 42/2 = 21 cm

Required Volume = (4/3) × (22/7) × 21³ = (88/21) × 21³ = (88 × 21²) = 88 × 441 = 38808

R = 32 cm, r = 26 cm and V = 55.704 litre

55.704 litre = 55704 cm³

Suppose the height of the frustum is ‘h’ cm;

We KNOW that:

Volume of the frustum = π/3 × [R² + r² + RR] × h

⇒ 22/7 × 1/3 × [32² + 26² + 32 × 26] × h = 55704

⇒ 22/21 × [1024 + 676 + 832] × h = 55704

⇒ 22/21 × 2532 × h = 55704

⇒ 22h/21 = 22

⇒ h = 21

As we know, in a rectangle,

(DIAGONAL)² = (length)² + (breadth)²

⇒ (26)² = (other SIDE)² + (10)²

⇒ (other side)² = 676 - 100 = 576

⇒ Other side = √576 = 24 cm

∴ Area of rectangle = 24 × 10 = 240 cm²

⇒ Volume of THREE spheres = {(4/3) π 6³} + {(4/3) π 8³} + {(4/3) π 10³} = (4/3) π × (1728)

⇒ Volume of bigger sphere will be same to that of volume of three spheres

⇒ Volume of bigger sphere = (4/3) π × (1728)

⇒ (4/3) π R³ = (4/3) π × (1728)

⇒ R = 12

Radius of the BASE = 5m

Height = 12m

L² = r² + h²

⇒ l² = 5² + 12² = 25 + 144 = 169 ⇒ l = 13m

We KNOW that,

The surface area of SPHERE = 4πr²

The curved surface area of CYLINDER = 2πrh

Diameter of sphere = HEIGHT of cylinder

⇒ 2r = h

Ratio $(= \frac{{4{\rm{\pi }}{r^2}}}{{2{\rm{\pi rh\;}}}} = \frac{{4\pi {r^2}}}{{4\pi {r^2}}} = \frac{1}{1})$

Let the radius of the WHEEL be ‘r’.

DISTANCE covered by the wheel in one revolution = 13.2/5000 km = (13200 × 100)/5000 CM = 1320/5cm = 264 cm (? 1 km = 1000 m = 1000 × 100 cm)

⇒ Circumference of the wheel = 264 cm

⇒ 2πr = 264

⇒ r = 264/2π = 264/(2 × 22/7) = 42 cm

∴ Diameter of the wheel = 2R = 2 × 42 = 84 cm

Radius of conical tent, r = 7 m

HEIGHT of conical tent, h = 24 m.

The slant height of the conical tent, l = √(r² + h²) = √(7² + 24²) = √(49 + 576) = √625 = 25 m.

Then, the surface area of the conical tent = π × radius × slant height = (22/7) × 7 × 25 = 22 × 25 = 550 m²

The area of the cloth used to make the tent = 550 m²

The breadth of the cloth = 11 m.

The LENGTH of that cloth required = 550/11 = 50 m

Let r and A be the RADIUS and area of the circle.

⇒ A = πr²----(1)

When r is INCREASED by 35%, new area is:

⇒ A’ = π(1.35r) ²

⇒ A’ = 1.8225πr²----(2)

Let the diagonals be x and 2x

In a rhombus diagonals are PERPENDICULAR to each other

So,

(2x/2)² + (x/2)² = (10)²

x² + x²/4 = 100

5x²/4 = 100

x = √80

Area of rhombus = Half of Product of diagonals = 2x²/2 = x² = 80 cm²

AREA of rectangular sheet = 25 × 15 = 375 cm²

Area of square = 5 × 5 = 25 cm²

LET the sides of the TRIANGLE be a, b & c RESPECTIVELY.

$(\because a:b:c = \;\frac{1}{2}:\frac{1}{3}:\frac{1}{4})$

⇒ a = x/2, b = x/3, c = x/4

We KNOW that,

Perimeter of the triangle = a + b + c

$(\Rightarrow 78 = \frac{x}{2} + \frac{x}{3} + \frac{x}{4})$

$(\Rightarrow 78 = \frac{{6x + 4X + 3x}}{{12}} = \frac{{13x}}{{12}})$

⇒ x = 72 cm

∴ Length of the smallest side = c = x/4 = 72/4 = 18 cm

Let ‘b’ be the base and ‘a’ is the adjacent side

⇒ b = a + 3……eq1

It is given that a × b = 108m

⇒ a = 108/b

Put value of a in eq1

⇒ b² – 3b – 108 = 0

⇒ (b – 12)(b + 9) = 0

⇒ b = 12 or - 9

b = - 9 is ignored as length cannot be negative

Area of PARALLELOGRAM = b × H = 72

⇒ 12 x h = 72

∴ h = 6M

Solution:

The length of the TWO tangents from point P(outside the circle) to points on circle Q and R is given by ,

PQ = PR = 20 cm 

It is given that centre of the circle is 'M' and the radius is 5 cm.

The length of any point on the circle to the centre equals the radius of the circle. So, MQ = MR = 5 cm.

We have to find the area of the quadrilateral PQMR.

We now know all the sides of the quadrilateral PQMR,that is 

PQ = 20 cm = a

QM = 20 cm = B

MR = 5 cm = c 

RP = 5 cm = d

Brahmagupta's formula to find the area of quadrilateral with all four sides a ,b , c and d is given by

A = [(s - a)(s - b)(s - c)(s - d)]^1/2

where s =( a + b + c + d)/2

s = (20 + 20 + 5 + 5)/2

s = 50/2

s = 25

So, A = [(25 - 20)(25 - 20)(25 - 5)(25 - 5)]^1/2

A = [ 5 × 5 × 20 × 20 ]^1/2

A = (10,000)^1/2

A = 100 sq.cm 

Therefore the correct option is 3).100 sq.cm.

TRICK:

If length is increased by 40% and breadth is decreased by 50% then % change WOULD be 30%. Now again the PROCESS is repeated and now %change would be 21%. Overall percentage change = 51%

Detailed SOLUTION:

Let the initial length and breadth be x and y respectively

Area = xy

Now length is increased by 40% and breadth is decreased by 50%

⇒ New length = x + (40% of x) = (140/100)x

⇒ New breadth = y – (50% of y) = (50/100)y

Now after repeating the same process

⇒ New length = (140/100)x + {40% of (140/100)x}

⇒ (140/100)x + (56/100)x = (196/100)x

⇒ New breadth = (50/100)y – {50% of (50/100)y}

⇒ (50/100)y – (25/100)y = (25/100)y

⇒ New Area = {(196/100)x} × {(25/100)y} = (49/100)xy

The ratio of length and breadth of a cuboid is 5 : 4. The ratio of breadth and height of this cuboid is 3 : 2.

LCM of 4 and 3 is 12. So, ratio of length, breadth and height will be 15 : 12 : 8.

Let length, breadth and height be 15T, 12T and 8T, respectively.

⇒ 15T × 12T × 8T = 1440

⇒ cube of T = 1

⇒ T = 1

So, length, breadth and height are 15 cm, 12 cm and 8 cm, respectively.

∴ Total surface AREA = 2(15× 12 + 12 × 8 + 8 × 15) = 792 SQUARE cm

ACTUAL VOLUME of CYLINDER = πr²h = 22/7 × 2.14 × 2.14 × 10.15 = 146.089 cm³

Computed volume of cylinder = πr²h = 22/7 × 2.1 × 2.1 × 10 = 138.6 cm³

Error in volume = 146.089 – 138.6 = 7.489 cm³

⇒ Height of the Trapezium (h) = 25 CM

⇒ Area of the Trapezium (A) = 475 cm²

⇒ Let assume, line PARALLEL to the highway = X cm,

Line parallel to bridge = y cm

⇒ From given CONDITION, x = y + 4 (I)

⇒ Area of the trapezium = 1/2 (x + y) × h

⇒ 475 = 1/2(x + y) × 25

⇒ 475 × 2 = (x + y) × 25

⇒ 950/25 = (x + y)

⇒ (x + y) = 38

From the (I),

⇒ y + 4 + y = 38

⇒ 2Y + 4 = 38

⇒ 2y = 34

GIVEN, slant height L = 12 cm

Let the BASE RADIUS be r. We have

Curved Surface AREA = πrl

⇒ 226.08 = 3.14 × r × 12

⇒ r = 6 cm

We know that,

The measure of an EXTERIOR angle of a regular n - SIDED POLYGON is 360^o/n.

⇒ The NUMBER of sides of a regular polygon whose exterior angles are each 40^o = 360/40 = 9

Surface area = 2 × base area + base PERIMETER × length

Surface area = 2 × 50 + (48 × 24)

Surface area = 100 + 1152 = 1252 m²

LET the altitude of the triangle be ‘H’ centimeters.

We KNOW that, Area of a triangle = ½ × BASE × height

And, area of a square = side²

According to question, Area of the GIVEN triangle = Area of the given square

$( \Rightarrow \frac{1}{2} \times 15 \times h = {15^2})$

⇒ h = 30 cm

Hence, the height of the triangle is 30 centimeters.

Area of a RHOMBUS = ½ × Product of the diagonals

∴ Area of the GIVEN rhombus = ½ × 21 × 32 = 336 sq.cm

∴ COST of tiling = 2.75 × 336 = RS. 924

The AREA of the circular park = 12,474 m²

∴ π × (RADIUS)² = 12474

⇒ (Radius)² = 3969 m

⇒ Radius = 63 m

Now, Perimeter of circular park = 2π × (radius)

∴ Perimeter = 2 × π × 63

⇒ Perimeter = 396 m

Now, COST of fencing = Rs. 250 per metre

∴ Total cost of fencing = 250 × Perimeter = 250 × 396

We KNOW that, Area of triangle = ½ × BASE × height

⇒ Hence, the area of GIVEN triangle = ½ × 20 × 24

LET the sides of triangle be a, b and c respectively.

a + b + c = 32

⇒ 11 + b + c = 32

⇒ b + c = 21----(i)

And given that, b - c = 5----(II)

From eq. (i) and (ii), we get

⇒ b + c + b - c = 21 + 5

⇒ 2b = 26

∴ b = 13

∴ c = 8

Given? PERIMETER of triangle ‘2s’ = 32

∴ s = 16

And, a = 11, b = 13, c = 8

Area of triangle = √{s(s - a)(s - b)(s - c)}

⇒ Area of triangle = √{16(16 - 11)(16 - 13)(16 - 8)} = √(16 × 5 × 3 × 8)

Let the initial RADIUS be ‘r’.

⇒ Area of CIRCLE = πr²(1)

Now, radius is increased by 15%

NEW radius = r’ = r + 0.15r = 1.15r

∴ Area = πr’² = π(1.15r)² = 1.3225πr²(2)

From (1) and (2)

RATIO of the sides of a cuboid is 2 ? 3 ? 6

Let the LENGTH, breath and height of a cuboid be 2x, 3x and 6x

Volume of SOLID cylinder = Volume of cuboid

⇒ πr²h = 2x × 3x × 6x

⇒ (22/7) × 7 × 7 × 28 = 36x³

⇒ x³ = (22 × 7 × 7) /9

⇒ x = [(22 × 7 × 7) /9]^1/3

Total surface area of cuboid = 2(lb + bh + HL)

⇒ Total surface area of cuboid = 2(2x × 3x + 3x × 6x + 6x × 2x)

⇒ Total surface area of cuboid = 2(6x² + 18x² + 12x²)

⇒ Total surface area of cuboid = 2(36x²)

⇒ Total surface area of cuboid = 2 × 36 × [(22 × 7 × 7) /9]^2/3

∴ Total surface area cuboid = 72 × (1078/9)^2/3

⇒ Given, curved SURFACE AREA of the CONE = 13200 sq.cm

⇒ π × r × L = 13200

⇒ π × 60 × l = 13200

⇒ l = 70 cm

We KNOW that height = √(l² – r²) = √(70² – 60²) = 36.05 cm

Since they have the same base (i.e. same radius of the base) and they have same height.

Thus the ratio of their volumes in the given ORDER = πr²H : πr²h/3 : 2/3 πr³

The radius R of the hemisphere is equal to h, hence = πr³ : πr³/3 : 2/3 πr³

= 1 : 1/3 : 2/3

Total SURFACE area of a part = Area of one face of the cube + 2 × 1/4 × Area of TOP face + 2 × (SIDE of cube) × (1/2 × face DIAGONAL)

⇒ 8 × 8 + 1/2 × 64 + 2 × 8 × ½ × 8√2

⇒ 64 + 32 + 64√2

Volume of a tetrahedron = $(V = \;\FRAC{{{a^3}}}{{6\sqrt 2 }})$

Where, a = length of each side

V = (24 × 24 × 24)/6√2

∴ V = (4 × 576)/√2

Given:

Initial AREA, A = 64m²

⇒ Length of each side of square, L = √(64) = 8 m

So as to double the PERIMETER, the length of the side of the square has to be doubled

⇒ New length of each side, l = 2 × 8 = 16 m

We know that surface area of a sphere = 4πr²

Where r = radius

Let the surface area of sphere A = x

But given that surface area of sphere B = x + 800% of surface area of sphere A = x + 800% of x = 9x

⇒ Surface area of sphere A / surface area of sphere B = 4πr²/4πR² = x/9x

⇒ r/R = 1/√ 9

⇒ R = 3r

We know that VOLUME of sphere = (4/3)πr³

⇒ Volume of sphere A is found to be K% LOWER than the volume of B = (4/3πR³ - 4/3πr³)/4/3πR³ × 100

Solution :

Given: $AD : DB = 2 : 3$ and 

Area of $ABC = 100$ sq.cm.

From the given information, we can easily find out that triangle ADE and triangle ABC are similar. 

So, by EQUATING we get

Area of ADE / Area of ABC $= AD^2 / AB^2$

Area of ADE / Area of ABC $= AD^2 / AB ^2$

Area of ADE / 100( given) $= 4/ 25$( given)

Area of ADE $= (4×100)/25$

Area of ADE $= 400/25 =16 $sq.cm.

Now,

Area of QUADRILATERAL BDEC

= Area of triangle ABC - Area of triangle ADE

$= 100 - 16$

$= 84$ sq.cm.

The correct OPTION is 1). 84

Solid sphere:

RADIUS of the solid sphere = r = 6 cm

Volume of the solid sphere = (4/3) πr³ = (4/3) × (22/7) × 6³ = (6336/7) CM3

Weight of the solid sphere = 4.78 kg

⇒ (6336/7) cm³ volume of the given material weighs 4.78 kg.

Hollow sphere:

Outer DIAMETER of the hollow sphere = 16 cm

⇒ Outer radius of the hollow sphere = r₁ = 16/2 = 8 cm

Inner diameter of the hollow sphere = 12 cm

⇒ Inner radius of the hollow sphere = r₂ = 12/2 = 6 cm

Volume of the hollow sphere = (4/3) × π × (r₁³ – r₂³) = (4/3) × (22/7) × (8³ - 6³) = (26048/21) cm3

Weight of the hollow sphere with volume (26048/21) cm3 = [ (26048/21) × 4.78]/(6336/7) ≈ 6.55 kg

∴ Weight of the hollow sphere MADE with same material = 6.55 kg

Formula for Area of a semicircle = πr²/2

Given πr²/2 = 308

⇒ (22/7) × R² = 308 × 2

⇒ r² = 308 × 2 × (7/22) = 14 × 14

⇒ r = √(14 × 14) = 14 cm

Diameter (d) = 2 × 14 = 28 cm

PERIMETER of semicircle = [(1/2) πd] + d = [(1/2) × (22/7) × 28] + 28 = 44 + 28 = 72

We know that, the RATIO of the length of the line segment BD to the length of segment DC is equal to the ratio of the length of side AB to the length of side AC

⇒ AC/AB = CD/BD

⇒ CD/BD = 4/3

ALTITUDE of ?ACD = Altitude of ?ADB = Altitude of ?ABC = h

So, their areas would be in the ratio of their bases

⇒ Ar(?ACD)/ Ar(?ABD) = 4/3

Let Ar(?ACD) = 4a and Ar(?ABD) = 3A

⇒ Ar(?ACD) + Ar(?ABD) = Ar(?ABC)

⇒ 4a + 3a = 350

⇒ 7a = 350

⇒ a = 50

A cistern 5 m long and 8 m wide, CONTAINS water up to depth of 2 m 45 cm.

Here, cistern is in cuboid form

Length of cistern = 5 m

Breadth of cistern = 8 m

HEIGHT of cistern upto which its WET = 2 m 45 cm = 2.45 m

Total area of wet SURFACE = {2(length + breadth) × height} + (length × breadth)

= {2(5 + 8) × 2.45} + (5 × 8)

= 63.7 + 40

As we know, a cone with RADIUS ‘R’ and height ‘h’, has a volume = 1/3 πr²h

Total volume of liquid = volume of FIRST conical vessel = (1/3) × (22/7) × (7)² × 30 = 1540 cm³

Volume of liquid poured in one vessel = volume of second conical vessel

= (1/3) × (22/7) × (5)² × 21 = 550 cm³

Total number of vessels filled = 1540/550 = 2.8

Hence, the third vessel is only filled = 0.8 = 80%

∴ The LAST vessel remained empty by = 100 – 80 = 20% = 1/5

Given length of the Diagonal = 17 cm, Length of one SIDE (L) = 8 cm

In a RECTANGLE, (Diagonal)² = (Length)² + (Breadth)²

⇒ (Breadth)² = (Diagonal)² - (Length)² = (17)² - (8)² = 289 - 64 = 225

⇒ (Breadth) = √225 = 15 cm

PERIMETER of the Rectangle = 2 (L + B) = 2 (8 + 15) = 2 (23) = 46 cm