Solids

Given: side of a square = 5cm

There are 4 isosceles triangle

Base (b) = 5cm

Height (h) = 8cm

Area of a square base (A) = side × side

= 5 × 5

Area of a square base (A) = 25 cm²

Area of one isosceles triangle

Area of one isosceles triangle = 20cm²

Area of 4 isosceles triangle = 4 × 20 = 80cm²

Total surface area = 25 + 80 = 105cm²

Given: side of a square = 16cm

Slant Height (l) = 10 cm

We know that,

Surface area of a pyramid

Where, p = perimeter of the base

l = slant height

A = area of the base

Area of a square base (A) = side × side

= 16 × 16

Area of a square base (A) = 256 cm²

Perimeter of a square base (p) = 4 × side

= 4 × 16

Perimeter of a square base (p) = 64 cm

Since, Surface area of a pyramid

⇒ Surface area of a pyramid

Total surface area of pyramid shaped toys = 500 × 576

= 288000cm²

= 28.8m²

Cost of 1m² = 80Rs.

Cost of 28.8m² = 28.8 × 80 = 2304 Rs.

Given lateral faces of square pyramid is equilateral triangle

side = 30cm

Since it is a equilateral triangle

lateral Base (b) = 30 cm

Area of a square base (A) = side × side

= 30 × 30

Area of a square base (A) = 900 cm²

Area of an equilateral triangle = 225√3 cm²

Area of an 4 equilateral triangle = 4× 225√3 = 900√3 cm²

Total surface area = 900 + 900√3 = 900(1 + √3)cm²

Let b be the base edge

And l be the lateral edge

Then we can write:

4b + 4l = 92 cm …..(1)

But 4b = 40 cm.

So we get

Substituting this value of b in (1) we get:

40 + 4l = 92

⇒ 4l = 52

⇒ l = 13cm

Altitude of one isosceles triangle = √(13² – 5²)

= √(169 – 25)

= √144 = 12cm

Now, we can calculate the total surface area

Base area = 10 × 10 = 100cm²

Area of one isosceles triangle

Area of one isosceles triangle = 60cm²

Area of 4 isosceles triangle = 4 × 60 = 240cm²

Total surface area = 100 + 240 = 340cm²

Consider pyramid as shown in figure given along.

Lateral surface are of pyramid is given by formula

As the base is given to be square, So l=w=x,

And base area = x²

So for base area to be equal to LSA,

Or, h=0

Thus, it is not possible to have a pyramid with lateral surface area equal to base area.

From the first figure we get.

Slant Height LG = 18cm

Base BC = 24cm

Height of pyramid

= √ (324 – 12²)

= √ (324 – 144)

Height AC = √180 cm = 6√5 cm

From the second diagram we get,

LH = Height of pyramid

HG = Half of a diagonal

BC = base edge = 24cm

LG = lateral edge = 30cm

BG = Full diagonal = √(24²+ 24²) = 24√2 cm

By Pythagoras theorem

HL = √(LG² – HG²)

= √(30² – (12√2)²)

= √(900 – 288)

= √612 cm

Base edge (b) = 10cm

Height (h) = 12cm

= √(144 + 5²)

= √(144 + 25)

Slant Height = √169 = 13cm

In the diagram,

e = lateral edge

s = slant height

h = height of pyramid

l = side of square base

With the help of diagram given below we get that, diagonal of square base is of length l√2

From the figures and Pythagoras theorem,

(1)

Also,

(2)

By comparing (1) and (2) we get that,

h²_, s² and e² are in AP with,

First term, a= h²

Common difference, d=

Hence, proved.

LH = Height of pyramid

HG = Half of a diagonal

BC = base edge = 30cm

LG = lateral edge = 25cm

BG = Full diagonal = √(30²+ 30²) = 30√2 cm

By Pythagoras theorem

HL = √(LG² + HG²)

= √(25² + (15√2)²)

= √(625 + 450)

= √1075 cm

LH = Height of pyramid

HG = Half of a diagonal

BC = base edge = 40cm

LG = lateral edge = 25cm

BG = Full diagonal = √(40²+ 40²) = 40√2 cm

By Pythagoras theorem

HL = √(LG² + HG²)

= √(25² + (20√2)²)

= √(625 + 800)

= √1425 cm

Given: base edge (b) = 10cm

Slant height (l) = 15cm

= √(225 – 5²)

= √(225 – 25)

Height = √200 = 10√2 cm

Base area = (base edge)²

= (10)²

Base area = 100cm²

We know that,

Let the volume of first pyramid be V₁

Let the base edge of first pyramid be a₁

Let the height of first pyramid be h₁

Let the volume of second pyramid be V₂

Let the area of first pyramid be a₂

Let the height of first pyramid be h₂

We know that,

Volume of a pyramid

Volume of a first pyramid

Volume of a second pyramid

Given: V₁ = V₂

….(1)

Given:

Substitute in (1)

⇒ h₂ = 4h₁

Let the volume of first pyramid be V₁

Let the base edge of first pyramid be a₁

Let the height of first pyramid be h₁

Let the volume of second pyramid be V₂

Let the area of first pyramid be a₂

Let the height of first pyramid be h₂

We know that,

Volume of a pyramid

Volume of a first pyramid …..(1)

Volume of a second pyramid …(2)

Given:

⇒ a₂ = 2a₁ …(a)

Given:

⇒ h₂ = 3h₁ …(b)

Consider, Volume of a second pyramid

(from a and b)

⇒ V₂ = 12×V₁

⇒ V₂ = 12×180 = 2160cm³

LH = Height of pyramid

HG = Half of a diagonal

BC = base edge = 18cm

LG = lateral edge = 18cm

BG = Full diagonal = √(18²+ 18²) = 18√2 cm

By Pythagoras theorem

HL = √(LG² – HG²)

= √(18² – (9√2)²)

= √(324 – 162)

= √162 cm

We know that,

Volume of a pyramid

Volume of a pyramid

Let the base edge be (2a) cm

Then area of one isosceles triangle

.

Area of four isosceles triangles = 4×25a = 100a cm².

Base area = (2a)² = 4a² cm².

Total surface area = (100a + 4a²)cm²

Given total surface area = 896

∴ 100a + 4a² = 896

⇒ a² + 25a – 224 = 0

Using,

We get,

⇒ a = 7 or – 32

Since, base edge cannot be negative

∴ a = 7 cm

LG = Slant height = 25cm

HG = Half of base = 7cm

Apply Pythagoras theorem

LH = √(25² – 7²)

= √(625 – 49)

= √576 = 24cm

Volume of a pyramid

Volume of a pyramid

Let all the edges of the square prism be 'a'.

The lateral edge BC is also 'a'.

HL = height of pyramid

HG = half of a diagonal of the base

LG = Lateral edge = a Cm

Full diagonal BG = √(a² + a²) = a√2 cm

Half diagonal

Applying Pythagoras theorem, we get:

LH = √(LG² – HG²)

Given: LH = 12cm

Thus

⇒ a = 12√2 cm

Volume of a pyramid

Volume of a pyramid

Base perimeter is given as 64 cm.

So base edge

Volume is given as 1280 cm³.

We know that,

⇒ h = 15 cm

Let height of pyramid (h) = 15 cm

half of base edge = 8 cm

slant height = l

Applying Pythagoras theorem, we get:

l = √(15² + 8²)

= √(225 + 64)

= √289

= 17 cm

Then area of one isosceles triangle

Area of four isosceles triangles = 4×136 = 544 cm²

Base area = 16² = 256 cm².

Total surface area = (544 + 256) = 800 cm²

Given that radius of the circle is 10 cm.

This will be same as the radius of the sector, r_s = 10

This rs will be the slant height l of the cone.

Slant height l = 10 cm

Central angle θ = 60°

the length of arc will be

Thus, the length of arc will be .

But this is same as the circumference of the base of the cone

So if r_b is the radius of the base of the cone

Given that slant height of the cone should be 25 cm.

So radius of the sector r_s = 25 cm

To find: Central angle θ

Radius of the base r_b = 10 cm

So circumference of the base = 2πr_b = 2π×10 = 20π cm ..(1)

This circumference is equal to the length of the arc

The length of arc will be

Thus, the length of arc will be . …(2)

Equating the results in (1) and (2):

⇒ θ = 144°

Here, The sector is a semicircle

Let r_s be the radius of the sector.

Then slant height of the cone will be r_s.

Central angle θ of a semi circle = 180°

The arc length of a semicircle is 'half the circumference of the full circle'

The 'circumference of the full circle' is 2πr_s.

So half of it is πrs.

This is same as the circumference of the base of the cone

So if r_b is the radius of the base of the cone,

2πr_b = πr_s

But r_s is the slant height.

Given: base radius r_b = 12cm

Slant height l = 25cm

We know that, Curved surface area of cone = πr_bl

Where r_b = base radius and l = slant height

Curved surface area of cone = π× 25× 12 = 300π cm²

height of the cone = 40 cm

Base diameter = 30cm

Base radius

l = slant height

Applying Pythagoras theorem, we get:

l = √(h² + r_b²)

= √(40² + 15²)

= √(1600 + 225) = √1825 cm

We know that, Curved surface area of cone = πr_bl

Where r_b = base radius and l = slant height

Curved surface area of cone = π× 15× √18255 = 640.8π cm²

Surface area of base = πr_b² = π × 15² = 225π cm².

Total surface area = (640.8 + 225)π = 865.8π cm²

height of the cone (h) = 12 cm

Diameter of a cone = 10cm

Base radius

l = slant height

Applying Pythagoras theorem, we get:

l = √(h² + r_b²)

= √(12² + 5²)

= √(144 + 25) = √169 cm = 13cm

We know that, Curved surface area of cone = πr_bl

Where r_b = base radius and l = slant height

Curved surface area of cone = π× 5× 13 = 65π cm²

Surface area of base = πr_b² = π × 5² = 25π cm².

Total surface area = (65 + 25)π = 90π cm² = 90× 3.14 = 282.6cm²

Surface area of 10000 fire works = 282.6 × 10000 = 2826000 cm².

So cost of colour paper = 282.6 × 2 = Rs 565.20

Let the radius of the sector be r_s

Central angle θ is 180° (∵ the sector is a semicircle)

Let the radius of the base be r_b

So circumference of the base = 2πr_b …(1)

This circumference is equal to the length of the arc

Since, the arc length of a semicircle is 'half the circumference of the

full circle'

The 'circumference of the full circle' is 2πrs.

So half of it is πrs. …(2)

Equating the results in (1) and (2):

2πr_b = πr_s

⇒ 2r_b = r_s. ..(3)

Find the area of the sector

The area of the sector is area of the semicircle which is

Let us substitute for r_s using the result in (3). We get:

= 2πr_b²

Area of the sector is same as the area of curved surface.

Area of the curved surface of the cone = 2πr_b² …(4)

To find the base area.

radius of the base of the cone = r_b.

So area of the base of the cone = πr_b². …(5)

Comparing the results in (4) and (5), we get:

Area of the curved surface of the cone = Twice the base area

Base radius = 15cm

Height of cylindrical block = 40cm

Volume of the cone =

= 5 × 15 × 40 π

= 3000π cm³

For the cylinder:

Radius of base = 12 cm

Height = 20 cm

Volume of the cone =

= 4 × 12 × 20 π

= 2880π cm³

For the cone:

Radius of base = r_b = 4 cm

Height = h = 5 cm

= 108

Given that radius of the circle is 25 cm.

This will be same as the radius of the sector r_s = 25 cm

This r_s will be the slant height of the cone

Slant height = 25 cm

Central angle θ = 216°

The length of arc will be

Thus, the length of arc will be .

But this is same as the circumference of the base of the cone

So if r_b is the radius of the base of the cone

⇒ 2πr_b = 30π

⇒ r_b = 30cm

Also,

height of the cone = h

base radius = r_b = 15 cm

slant height = l = 25 cm

Applying Pythagoras theorem, we get:

h = √(l² – r_b²)

= √(25² – 15²)

= √(625 – 225) = √400 = 20 cm

= 1500π cm³

Given that

Also given that

We know that, Volume of the cone =

Given that .

Also given that V₁ = V₂

We know that, Volume of the cone =

⇒

⇒

⇒

⇒

⇒

Let r cm be the radius of the total sphere.

Then it's surface area would be 4πr² cm²

This surface area is given as 120 cm².

we can equate the two:

= 120

⇒ = 30

…..(1)

Curved surface area of a hemisphere

Base area =

So total surface area of a hemisphere = 2 + = 3

Substitute value from (1)

So total surface area of each hemisphere

Let the volume and radius of the first sphere be V₁ and r₁

Respectively

Let the volume and radius of the second sphere be V₂ and r₂

Respective

Then volume of first sphere =

volume of second sphere

Given that

Let S₁ and S₂ be the surface areas. Then we get:

Total volume available for melting = Volume of the cylinder

=

= π×4²×10

= 160π cm³

Volume of one .

Number of spheres = 15

Total volume available for melting = Volume of the sphere

=

=

= 2304π cm³

Let 'r' be the radius of one small sphere.

Then volume of one small sphere = =

But number of spheres is given as 27.

So we can write:

R = 4cm

1. Consider the red sphere in fig (a) below.

• Two ellipses are drawn inside it: A dotted ellipse and a dashed ellipse

• The dashed ellipse represents a circle whose centre is same as the centre of the sphere

♦ Also this circle is horizontal

• So this circle divides the sphere into an upper hemisphere and a lower hemisphere

• This circle is taken as the base of the cone in fig.b.

• We can see that, the cone fits perfectly in the upper hemisphere.

• This is shown more clearly in fig.c

2. From fig.c we can see that, the height of the cone will be the height of the hemisphere, which is 10 cm

• But cone given in the question has a height of 16 cm.

• So the given cone does not fit inside the upper hemisphere alone.

♦ It will occupy some portion of the lower hemisphere also

• This is shown in fig (b) below. In that fig. we can see that the, base of the new cone is below the dashed ellipse

3. In fig (c), the measurements are given

• One half of the cone is represented by the right triangle ABC

• The distance of the apex C from the centre O will be the radius of the sphere, which is 10 cm

• So the remaining distance OA will be (16 – 10) = 6 cm

• Distance OB will also be the radius 10 cm

• Applying Pythagoras theorem to the right triangle OAB, we get:

–

⟹ = 10² – 6²

⟹ = 100 – 36

⟹ = 64

⟹ AB = 8 cm

4. Thus we have:

• Height of the cone, h = 16 cm

• Radius of the cone, r_c = 8 cm

• So Volume,

• Volume of sphere,

5. Taking ratios, we get:

• So 'volume of the cone' is of the 'volume of the sphere'

The tank has two hemispherical parts and one cylindrical part

The yellow dashed line indicates the axis of the tank

From the fig., it is clear that radius of the hemisphere is 1 m.

So it's volume

Thus volume of two hemispheres

Height of a hemisphere will be equal to it's radius.

So length of the cylindrical part = [6 – (2×1)] = 4 m

Volume of cylinder = V_c = πr²h = π×1²×4 = 4π m³

Thus total volume .

We know that 1 liter is the volume of a cube of edge 10 cm

Now, 16746666.67 cm3.

(∵ 1 m = 100 cm)

Thus the no. of liters = 16746666.67⁄1000 = 16746.67 liters

1. Consider the red hemisphere in fig (a) below.

• A dotted ellipse and a dashed curve are drawn inside it

• The dashed ellipse represents the base of the hemisphere

• For maximum possible volume, this base is taken as the base of the cone in fig. (b)

• We can see that, the cone fits perfectly in the hemisphere.

• This is shown more clearly in (c)

2. From fig.c, we have:

• Height of the cone, = r

• Radius of the cone, = r

• So Volume,

3. Consider the red hemisphere in fig (a) below. It is the same hemisphere of radius r, that we saw for the cone above

• A square is drawn in the base of the hemisphere

• This square is the base of the pyramid

4. For maximum possible volume, the diagonal of the square must be equal to the diameter of the circle

• So in fig. c, we can write:

OP = OQ = half of diameter = radius = r

5. OPQ is a right triangle. We can apply Pythagoras theorem

• Then base edge = PQ = = = = r

6. So volume of the pyramid, =

7. Now we can take the ratio:

• Thus we get:

: = 2 : π