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⇒ radius of CONE(r) = 12 CM

⇒ HEIGHT of cone(h) = 48 cm

⇒ Let radius of sphere be R

⇒ Given volume of cone = volume of sphere

⇒ (1/3)πr²h = 4/3πR³

⇒ r²h = 4 × R³

⇒ 12 × 12 × 48 = 4 × R³

⇒ 12 × 12 × 12 = R³

⇒ R = 12 cm

∴ radius of the sphere is 12 cm

Initial Area of a circle, A = π r², where r is the radius.

New area, A’ = A - 36% A = 64%A = 0.64 A

∴ A’/A = 0.64 ⇒ (r’/r)² = 0.64 ⇒ r’/r = 0.8 ⇒ r’ = 0.8 r

∴ New perimeter, P = 2πr’ = 2π (0.8)r = 0.8(2πr) = 0.8 P

∴ percentage DECREASE in perimeter is

$(\frac{{P - P'}}{P} \times 100 = 0.2 \times 100 = 20\% )$

Total surface AREA of HOLLOW sphere = 4πr₁² + 4πr₂²

Where, r₁ = radius of OUTER surface

r₂ = radius of inner surface

Here, r₁ = 18 cm, thickness = 10 mm = 1 cm

⇒ r₂ = 17 cm

Solution:

Area of an equilateral triangle with side $a = [ \sqrt{3} a ^2 ] / 4$

Perimeter of an equilateral triangle $= 3 a$

Semi-perimeter of an equilateral triangle $= 3 a /2$

Radius of the incircle = Area / Semi-perimeter 

$r =[ \sqrt{3} a ^2 ] × 2 / 4 × 3a$

$r =[ \sqrt{3} a ] / 6$

Diameter of the incircle

= DIAGONAL of the SQUARE inscribed in the incircle 

$2r =$ Diagonal of the square 

Diagonal of the square $= 2 × [ \sqrt{3} a] / 6$

Diagonal of the square $= [ \sqrt{3}) a] / 3$

Area of the square $=(diagonal )^2 / 2$

 $= [ \sqrt{3} a]^2 / 3^2 ×2$

$= 3 a^2 / ( 9 × 2)$

$= 3 a^2 / 18$

$= a^2 / 6$

Therefore the area of the square inscribed in the circle is $a^2 / 6$.

The correct option is 3). $a^2 / 6$

⇒ Total NUMBER of squares over the length 15.4 cm = 7

⇒ Total number of squares over the breadth 13.2 cm = 6

⇒ Side of each SQUARE is = 15.4/7 or 13.2/6 = 2.2 cm

⇒ Area of one square = (2.2) ² = 4.84

⇒ Total number of RED squares = 42/2 = 21

∴ Area of all red squares = 21 X 4.84 = 101.64 cm²

Radius of SMALLER ball = 2 cm

Diameter of bigger ball = 16 cm

∴ The radius of the bigger ball = 8 cm

Volume of Sphere of bigger ball = (4/3)πr³, where r is the radius of bigger ball

Volume of sphere of smaller ball = (4/3)πr₁³ , where r₁ is the radius of smaller ball

As, each angle of a RECTANGLE is 90°

Length = √{(diagonal)² – (BREADTH)²} = √{(26)² – (10)²} = 24 cm

AREA of rectangle = LENGTH × Breadth

⇒ 60 = Length × 5

⇒ Length = 50/5 = 12 cm

? Diagonal of rectangle = √(length² + breadth²)

Solution:

Let us take ABCD as the rhombus with obtuse angle at D and B. Let us take E as the point where perpendicular from D MEETS AB and we know that DE is the perpendicular height of the rhombus.

Let the two diagonals of the rhombus meet at the point O.

We have two right angle triangles ∆AOB and ∆DEB. These two triangles are SIMILAR as they have the same angle B and the perpendicular of rhombus bisect each other at right angle.

So,∆AOB ~ ∆DEB (so, their sides are PROPORTIONAL)

AO/DE = AB/DB 

So, DE(perpendicular height) = AO × (DB/AB)

It is given that DB= 48CM and AC=55cm so, AO= 55/2cm

Using PYTHAGORAS theorem in ∆AOB we get,

AB² = AO² + OB²

AB² = (55/2)² + (48/2)²

AB² = 27.5² + 24²

AB² = 756.25 + 576

AB² = 1332.25

AB = √1332.25

AB = 36.5cm

We know DE(p) = AO × (DB/AB)

DE(p) = 27.5 × (48/36.5)

DE(p) = 27.5 × 1.315

DE(p) = 36.15 cm

So,the correct option is 1). 36 cm < p < 37 cm

Surface area of a SPHERE = 4πr²

Volume of a sphere $(= \FRAC{4}{3}\pi {r^3})$

Ratio of surface area = (ratio of radius)²

Given, surface area of TWO spheres are in the ratio 9 : 16.

⇒ ratio of radius = 3 : 4

Ratio of volumes = (ratio of radius)³

Formula: Volume of a right circular cone $(= \;\frac{{\pi {r^2}h}}{3})$

And area of curved SURFACE of right circular cone = πrl

Given: volume = 1232 cm³

Height = 24 cm

$(\therefore \frac{{\pi {r^2}h}}{3} = 1232)$

$(\Rightarrow \frac{{22}}{7} \times {r^2} \times \frac{{24}}{3} = 1232)$

$(\Rightarrow {r^2} = \frac{{1232 \times 7}}{{22 \times 8}} = 49)$

⇒ r = 7 cm

By Pythagoras theorem, $(L\; = \;\sqrt {{r^2} + {h^2}} )$

$(\therefore l\; = \sqrt {{7^2} + {{24}^2}} )$

l = 25 cm

Now, area of curved surface = π r l

Area of TRAPEZIUM =$({\rm{\;}}\FRAC{1}{2}\left( {Sumofparallelsides} \right){\rm{\;}} \times {\rm{\;}}height)$

$(\begin{array}{l} \Rightarrow 175 = \frac{1}{2}\left( {20 + 15} \right) \times h\\ \Rightarrow h = \frac{{175 \times 2\;}}{{35}} = 10\;cm \end{array})$

⇒ Let radius be R, height be h and VOLUME be V

⇒ Then initial volume of cone is

⇒ V = (1/3) × (π × r² × h)

⇒ Now if radius is DECREASED by 20% then new radius will be = r – 20% of r

⇒ New radius = 0.8r

⇒ Suppose new height INCREASED by x %

⇒ Then new height = h + x% of h

⇒ New height = (100h + x × h)/100

⇒ New height = (h/100) × (100 + x)

⇒ New volume = (1/3) × (π × 0.8r2 × (h/100) × (100 + x)

⇒ As volume remains the same so ratio of volume will be same

⇒ (Initial volume)/(New Volume) = 1 ----- 1

⇒ Putting the value of each in 1 we get

⇒ {(1/3) × (π × r2 × h) }/{(1/3) × (π × 0.8r2 × (h/100) × (100 + x) } = 1

⇒ 100/(0.64) × (100 + x) = 1

Let the POLYGON has N SIDES. We also know that sum of interior ANGLE of a polygon with side n is (n - 2) × 180°

According to the question,

⇒ (n - 2) × 180° = 540°

⇒ n - 2 = 3

Perimeter of a square = 4 × side

Given, side of the square = 22 cm

∴ Perimeter of the square = 88 cm

Now, the same wire is USED to FORM a circle.

∴ The circumference of the circle formed = perimeter of the square

Circumference of a circle = 2πr

⇒ 2πr = 88

⇒ r = 44/π = 14 cm

Area of a circle = πr²

⇒ given: RADIUS of cone, r = 16 CM, Height, h = 12 cm

⇒ Radius of cylindrical jar, r₁ = 8 cm, height, h₁ = x

⇒ Volume remains the same in this case

⇒ Volume of cone = Volume of Cylinder

⇒ (1/3) × (π × r² × h) = (π × r₁² × h₁)

⇒ (1/3) × (π × 16² × 12) = (π × 8² × x)

⇒ 16 × 16 × 4 = 8 × 8 × x

Let the height of the cylinder be ‘h’ cm

Let the radius of the cross section be ‘r’ cm

the total SURFACE area of cylinder = 616 cm²

Since, the curved surface area and the total surface area of a cylinder are in the ratio 1 : 2

The surface area of the FLAT SURFACES= 2πr² = (616/2) cm²

⇒ r² = (154/π)

⇒ r = 7 cm

The curved surface area = 2πrh = 616/2 = 308 cm²

⇒ h = 308/(2πr)

⇒ h = 7 cm

For a triangle ABC with vertices A(A_X, A_y), B(B_X, B_y) and C(C_X, C_y), the COORDINATES of the centroid of triangle are obtained as,

(O_X, O_y) = [(A_X + B_X + C_X)/3], [(A_y + B_y + C_y)/3]

GIVEN, A(-3, -1), B(3, 5) and O(-1, 4)

Computing x-coordinate,

⇒ -1 = (-3 + 3 + C_X)/3

⇒ C_X = -3

Computing y-coordinate,

⇒ 4 = (-1 + 5 + C_y)/3

⇒ C_y = 12 - 4 = 8

∴ Coordinates of vertex C is (-3, 8)

Let the radius of the sphere and the RIGHT CIRCULAR cylinder be ‘r’ and the height of the cylinder be ‘H’.

Radius of sphere = r = Radius of right circular cylinder

CURVED surface area of sphere = Curved surface of the right circular cylinder

⇒ 4πr² = 2πrh

⇒ h = 2r

Ratio of volumes of sphere and right circular cylinder = [(4/3) πr³] / (πr²h) = (4r/3) : 2r = 2 : 3

PERIMETER of the CIRCLE = Perimeter of the rectangle

Perimeter of the circle = 2 × (length + BREADTH)

Perimeter of the circle = 2 × (18 + 34) = 104 cm

2πr = 104

r = 16.5 cm

Area of the circle = π × r² = π × 16.5² = 856 cm²

VOLUME of sphere = 4/3 × πr³

⇒ 179.67= 4/3 × 22/7 × r³

⇒ r³ = 7 × 7 × 7

∴ r = 6.54 CM

∴ r = 7 cm

∴ d = 14 cm

LET the BREADTH be ‘b’ and LENGTH be ‘l’, ACCORDING to question

Length = 2 × breadth ⇒ l = 2b

Perimeter of a rectangle = 2 (length + breadth) = 2(2b + b) = 2 × 3b = 6b

Perimeter of the given rectangle = 780 cm

⇒ 6b = 780 cm

⇒ b = 130 cm

⇒ l = 2 × 130 = 260 cm

We know that area of Rectangle = Length × Breadth

⇒ VOLUME of CUBE = (edge length)³ = (10)³ = 1000 cm³

Now,

⇒ Volume of a square pyramid = (1/3) × AREA of square base × perpendicular HEIGHT = (1/3) × 15 × 15 × 30 = 2250 cm³

When the cube is placed inside the pyramid,

⇒ Remaining volume of pyramid = 2250 – 1000 = 1250 cm³

Total surface area before the cut = 2πrh + 2πr²

⇒ 2πr(r + h)

⇒ 2 × 22/7 × 7 × 25 = 1100 cm²

After CUTTING in 3 EQUAL parts by 2 cuts PARALLEL to the base, curved surface area will not increase only base area will increase,

⇒ Increase in base area = 4πr² = 4 × 22/7 × 7 × 7 = 616 cm²

We know that,

Diagonal of cube = √3 × SIDE

⇒ 8√3 = √3 × Side

⇒ Side = 8

Inner volume of cube = 216

⇒ Side³ = 216

⇒ Side = 6

Thickness = t = (8 - 6)/2 = 1 cm

Let r = radius of base and h = height

Volume of CYLINDERS = πr²h

Given,

R/r = 2/3 and H/h = 5/3

Ratio of their VOLUMES =

= (2/3)² × (5/3)

= 20/27

LET the LENGTH be x and BREADTH be y

Area = xy = 160

New length = x + 10

Area = (x + 10) × y = 160 + 40 = 200

⇒ xy + 10y = 200

⇒ 160 + 10y = 200

⇒ y = 4

∴ x = 160/y = 160/4 = 40 m

Height of the cone = Total height – Height of the cylinder = 15 – 10 = 5 cm

RADIUS of the cone = Radius of the cylinder = 50 m

Total volume of tent = Volume of cylinder + Volume of cone

⇒ πr²H + 1/3 × πr²h

⇒ π50² × 10 + 1/3 × π50² × 5

⇒ 50² × π × (10 + 5/3)

⇒ 2500 × 35/3 × π

The radius of a hemispherical BOWL is 8 cm.

We KNOW,

Volume of a HEMISPHERE = (2πr³)/3 , where R = radius.

The RATIO of curved surface AREA of a right CIRCULAR CYLINDER to the total area of its two bases is 2 : 1;

Since total surface area of cylinder is 23100 cm²

∴ Curved surface area = 2πrh = 2/3 × 23100 = 15400 cm²

And Area of the base = 1/2 × 1/3 × 23100 = 3850 cm²

⇒ πr² = 3850

⇒ R² = 1225

⇒ r = 35 cm

And

2πrh = 15400

h = (15400 × 7) / (22 × 35 × 2)

h = 70 cm

Cylinder is made by rolling RECTANGLE, so a circle will be CREATED having circumference equal to BREADTH of rectangle.

Circumference = 2πr = 2 × (22/7) × r

7 = (44/7) × r

r = 49/44 m

And height of cylinder = 11 meter

∴ VOLUME of cylinder = πr²h, where r is RADIUS of cylinder and ‘h’ is height of cylinder.

= (22/7) × (49/44)² × 11

= (22/7) × (49/44) × (49/44) × 11

= (7/2) × (49/4)

= 343/8

LET, the radius of the RESULTANT CIRCLE = R cm.

∴ Area of the resultant circle = πR² cm²

Radius of the 1^st circle = 7cm.

∴ Area of the 1^st circle = π × 7² = 49π cm²

Radius of the 2^nd circle = 8 cm.

∴ Area of the 2^nd circle = π × 8² = 64π cm²

Radius of the 3^rd circle = 11cm.

∴ Area of the 3^rd circle = π × 11² = 121π cm²

According to problem,

⇒ πR² = 49π + 64π + 121π

⇒ R² = 234

⇒ R ≈ 15

∴ Radius of the circle = 15 cm.

A.1089 cm²

Is the ANSWER!!!!!!!!!!!!!!

Let, vertical distance between parallel lines = H cm.

∴ HEIGHT of the TRIANGLE = H cm.

Base of PARALLELOGRAM = base of the triangle = 35 cm.

∴ Area of triangle,

⇒ ½ × base × height

⇒ ½ × 35 × H

⇒ 35H/2

According to PROBLEM,

⇒ 35H/2 = 630

⇒ H = 630 × (2/35)

⇒ H = 36

∴ Vertical distance between the parallel lines = 36 cm.

Let the length and WIDTH of the given rectangle be l and b respectively.

We know that, perimeter of rectangle = 2 × (length + width)

⇒ 160 = 2 × (l + b)

⇒ l + b = 80 m …(1)

Given, difference of two sides = 48 m

∴ l – b = 48 …(2)

SOLVING above two EQUATIONS we get,

l = 64 meters & b = 16 meters

We know that, Area of rectangle = l × b

Let the side of the square be ‘a’ meters. Thus, according to the question

⇒ a² = l × b [?Area of square = side²]

⇒ a² = 64 × 16

The garden is 26 m LONG and 14 m wide.

So, the area of the garden = 26 × 14 = 364 m²

There is a path 2 m wide OUTSIDE the garden along its sides.

Then, the NEW length of the garden with the path = 26 + 2 + 2 = 30 m.

The new width of the garden with the path = 14 + 2 + 2 = 18 m.

The area of the garden with the path along its sides = 30 × 18 = 540 m²

So, the area of the 2m wide path along the sides of the garden = 540 – 364 = 176 m²

The path is to be CONSTRUCTED with square marble tiles 40 cm × 40 cm = 0.40 m × 0.40 m.

The area of one square marble tile = 0.40 × 0.40 = 0.16 m²

• Given a cube is placed inside the CONE of radius 20 CM. Let the height PQ of the cube inside the cone be p m = 10 m and let poq be the total height where o is the centre of pq.
• Let oq = x and so oa be the side = x / 2
• Now po = 10 – x
• Also the base qb = 20 cm which is the radius.
• Triangle poa is CONGRUENT to pqb
• Po / pq = oa / qb
• 10 – x / 10 = x/2 / 20
• 10 – x / 10 = x / 40
• 400 – 40 x = 10 x
• 50 x = 400
• Or x = 400 / 50
• Or x = 8 cm
• option (C) is correct

We know that,

All sides of rhombus are EQUAL in length.

Let us find the area of triangle formed by 2 sides of rhombus and 1 diagonal.

By Heron’s formula,

Semi-perimeter, S = (10 + 10 + 12)/2 = 32/2 = 16

Area of triangle $( = \sqrt {s \TIMES \left( {s - a} \right) \times \left( {s - b} \right) \times \left( {s - C} \right)} )$

⇒ Area of triangle $( = \sqrt {16 \times \left( {16 - 10} \right) \times \left( {16 - 10} \right) \times \left( {16 - 12} \right)} )$

⇒ Area of triangle $( = \sqrt {16 \times 6 \times 6 \times 4}= 48)$

∴ Area of rhombus = 48 × 2 = 96 cm²

First we will find QUANTITY A,

Quantity A:

Let the SIDE of the cube be ‘a’.

We know that,

Perimeter of ONE face of a cube = 4a

⇒ 24 = 4a

⇒ a = 24/4 = 6 cm

We know that,

Volume of the cube = a³ = 6³ = 216 cm³

∴ Volume of the cube = 216 cm³

Now,

Quantity B:

Let the side of the cube be ‘a’.

We know that ,

Volume of the cube = a³

⇒ 1728 = a³

⇒ a = 1728^1/3 = 12

We know that,

A cube consists of 12 EDGES.

∴ The total length of the edges of the cube = 12 × 12 = 144 cm

Detailed SOLUTION:

Area of TRAPEZIUM can be written as (1/2) × (sum of PARALLEL SIDES) × height

⇒ Area = (1/2) × (21 + 9) × 10

⇒ Area = 150 cm²

∴ the area of trapezium is 150 cm²

Let the BASE of right angle triangle be ‘5X’ cm and height be ‘4X’ cm

Area of the right angled triangle = 1/2 × base × height

⇒ 160 = 1/2 × 5x × 4x

⇒ 160 = 10x²

⇒ x² = 160/10 = 16

⇒ x = √16 = 4 cm

∴ Height of triangle = 4x = 4 × 4 = 16 cm

PERIMETER of a rectangle = 2(length + BREADTH)

⇒ 48 = 2(length + 10)

⇒ 24 = length + 10

⇒ Length = 24 - 10 = 14 cm

∴ Area of rectangle = length × breadth = 14 × 10 = 140 cm²

The BASE of a regular hexagonal pyramid has 6 EQUAL sides

∴ Side = base perimeter/6 = 24/6 = 4 m

HEIGHT of pyramid = 18 m

The $VOLUME of a $pyramid is = 1/3 × (the area of the base) × (the height)$

Area of the base = $ (3$( {\sqrt{3} })$/2) × (Side)²

Volume of regular hexagonal pyramid =(1/3)×(3$( {\sqrt{3} })$/2) × (Side)² × height = 

($( {\sqrt{3} })$/2) × (Side)² × height

⇒ ($( {\sqrt{3} })$/2) × (4)² ×18 = 249.42 m³

Let the base of the triangle be B, Altitude of first triangle is X and altitude of SECOND triangle is Y

Area of first triangle is = (1/2) × B × X = 15 cm²----(1)

Area of second triangle = (1/2) × B × Y = 20 cm²----(2)

DIVIDING equation (1) by (2), we get

⇒ 15/20 = X/Y

⇒ 3Y = 4X

⇒ Y = (4/3)X = (1 + 1/3)X

Difference in two lengths = (1/3)X

Wire is in the shape of a right circular CYLINDER with height 300000 cm and DIAMETER 0.4 cm(radius 0.2 cm).

Volume of cube = Volume of wire

Let SIDE length of cube be T cm.

⇒ T³ = 3.14 X 0.2 x 0.2 x 300000

⇒ T³ = 37680

⇒ T = 33.525

Area of circle = Area of each face of cube = 33.525 x 33.525 = 1123.926 square cm

Area of a circle with radius of 6CM = π × R² = 36π

Area of each piece = 36π/8 = 4.5π

Another circle having area 70% of 1 piece = 0.7 × 4.5π

If the radius of this new circle is R.

⇒ πR² = 3.15π

⇒ R = √3.15 = 1.7741.8 cm