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The ratio of radius and slant height = \(\frac{120}{360}=\frac{1}{3}\) = 1 : 3

curved surface area (πrl) = 108π

rl = 108; r × 3 r = 108

3 × r² = 108; r² = 36

radius (r) = 6 cm; slant height (l) = 3r = 3 × 6 = 18cm

Area of the sector with central angle 120° is one-third of the area of the circle.

Curved surface area = 108 π

Ares of die; sector = 108 π,

which is one-third of the area of the circle

Area of the circle = 108 π × 3

πr² = 324 π

Radius of the circle r = 18 cm

Slant height = 18 cm

Radius of the cone = 6 cm

Radius = \(\frac{30}{2}\) = 15 cm = r

Height = h = 40 cm

Total surface area = Base area + Curved surface

area = πr² + πrl

Slant height = \(\sqrt{h^2+x^2}=\sqrt{40^2+15^2}\)

= \(\sqrt{1825}\approx\) 42.72

πr² + πrl = 225π + π x 15 x 42.72

= 865 πr = 2718.6 m²

a. Central angles are in the ratio 2 : 3 , so, let the perimeter of the two clones which made by rolling up this two sectors be \(\frac{2}{5}\) part and \(\frac{3}{5}\) part of the perimeter of the circle

That is perimeter of each sector be \(2\pi r\times\frac{2}{5}\) and \(2\pi r\times\frac{3}{5}\)

Ratio between base perimeters of cone = \(2\pi r\times\frac{2}{5}\) : \(2\pi r\times\frac{3}{5}\) = 2 : 3

b. Base perimeter of the cones will be \(\frac{2}{5}\) and \(\frac{3}{5}\) parts of the circumference of the 5 circle.

Ratio between the perimeters of die cones \(=\pi r^2\times\frac{2}{5}\) : \(\pi r^2\times\frac{3}{5}\) = 2; 3

r₁ : r₂ = 4 : 5

Ratio of the heights : V₁ = V₂

\(\frac{1}{3}\pi \times4^2\times h_1\) = \(\frac{1}{3}\pi \times5^2\times h_2\)

16h₁ = 25h₂, \(\frac{h_1}{h_2}=\frac{25}{16}\)

h₁ : h₂ = 25 : 16

r₁ : r₂ = 3 : 5

h₁ : h₂ = 2 : 3

Ratio of the volumes

\(=\frac{1}{3}\pi \times3^2\times2:\frac{1}{3}\pi \times5^2\times3\)

\(=\frac{1}{3}\pi\times9\times2:\frac{1}{3}\pi\times25\times3\)

= 18 : 75 = 6 : 25

Volume of Meera’s pyramid = \(\frac{1}{3}\) x Baste edge x height 3

= \(\frac{1}{3}\) x 10² x 6 = 200 cm³

Volume of Manu’s pyramid = \(\frac{1}{3}\) x 52 x 24 = 200 cm³

Volumes are equal

Perimeter of base of a cone is equal to half of perimeter of large cone.

Radius of pyramid = R/2

Perimeter of base

\(=\pi(\frac{R}{2})^2=\pi\frac{R^2}{4}\)

\(=\frac{1}{2}\pi \,R^2=\frac{\pi\,R^2}{2}=2\times\frac{\pi\,R^2}{4}\)

= 2 × Base area , that is twice

Ratio of volumes

\(=\frac{4}{3}\pi r^3_1:\frac{4}{3}\pi r^3_2=27:64\)

\(r^3_1:r^3_2=3^3:4^3\)

\(r_1:r_2=3:4\)

r = 5 cm

h = 12 cm 

t = 13 cm 

Total surface area of one firework = Base area + Curved surface area

= π x 5² + π x 5 x 13 = 25π + 65π

= 90π cm² = 90 x 3.14

= 282.6 cm²

Surface area of 10000 firework

= 282.6 x 10000 = 2826000 cm²

\(\frac{2826000}{100\times100}=282.6\) m²

Total cost = 282.6 x 2 = Rs 565

Surface area of the solid sphere = \(4\pi r^2\)

\(4\pi r^2=120\)

\(r^2=\frac{120}{4\pi}=\frac{30}{\pi}\)

Surface area of one hemisphere = 3πr²

3πr² = 3 π × \(\frac{30}{\pi}\) = 90 cm²

Total surface area of cone = base area + curved surface area

= 25 π +165 π = 190 cm²

Slant height =10 cm, height = 6 cm

Base edge/2 = \(\sqrt{(10)^2-(6)^2}\) 

= \(\sqrt{100-36}\)

= \(\sqrt{64}\)

= 8 cm

Base edge = 16 cm

lateral edge = \(\sqrt{100+64}=\sqrt{100+64}\)

= \(\sqrt{164}\)

= \(2\sqrt{41}\) cm

Length of total edges =  4 x 16 + 4 x \(2\sqrt{41}\)

= \(64+8\sqrt{41}\) cm

Radius = 12cm

Volume of bigger sphere = \(\frac{4}{3}\pi R^3=\frac{4}{3}\pi \times12^3\)

If the radius of smaller sphere is ‘r’ Volume of 27 smaller spheres = Volume of the bigger sphere

\(27\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\)

\(27\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi \times12^3\)

27r³ = 12³

\(r^3=\frac{12^3}{27}=\frac{1728}{27}=64\)

r³ = 64

Radius of smaller sphere = 4 cm

slant height of cone = 16 cm

\(\frac{90}{360}\) of radius of sector = 16 x \(\frac{90}{360}\) = 4 cm

Volume of cone = \(\frac{1}{3}\) × base perimeter × height 

= \(\frac{1}{3}\) × 81 π × 12 = 324 π cm³

Base Radius = 30 cm

Height = 40 cm

Slant height = \(\sqrt{40^2+30^2}\) = 50

Surface area of the cone = πr² + πrl = π × 302 + π × 30 × 50

= 900π + 1500π = 2400π

Total cost to paint 10 cones = \(\frac{2400\pi \times10\times50}{10000}\)

= \(\frac{2400\times3.14\times10\times50}{10000}\) = 377Rs

v₁ : v₂ = 1 : 8

v₁ = \(\frac{1}{3}\) a₂ h

v₂ = \(\frac{1}{3}\)(2a)² x 2h, v₁ : v₂ = 1 : 8

Volume of the second pyramid = 800 cm²

a. Central angle of the small sector = 60°

\(\frac{1}{6}\) part of the area of the circle Base radius of cone formed from above sector = 5 cm

Radius of the circle = 5 × 6 = 30

Similarly, area of the sector of central angle 120° = \(\frac{1}{3}\) of the area of the circle Base radius of the bigger cone = \(30\times\frac{1}{3}\) = 10

Base area of the bigger cone = π × 102 = 100 ×

b. Curved surface area ofthe bigger cone = π × 10 × 30 = 300π

Surface area = 100π + 300π = 400π cm²

Radius of sector = slant height of cone = 20 cm

Central angle of the sector = radius of cone \(\times\frac{360}{R}=10\times\frac{360}{20}=18^\circ\)

Height of cone = 4 cm

Slant height = 5 cm

Radius = \(\sqrt{(5)^2-(4)^2}=\sqrt{25-16}\)

= \(\sqrt{9}=3\) cm

Surface area = π (3)² + π x 3 x 5 = 9 π + 15 π = 24 π cm²

a. r² + I² - h²

r² = 41² - 40² = 81

r = 9

Diameter = 18 cm

b. Volume = \(\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\times9^2\times40\)

= 3394.3 cm³

Volume of square pyramid = \(\frac{1}{3}\) a³

Volume of cone

= \(\frac{1}{3}\pi (\frac{a}{2})^2\times a=\frac{1}{3}\times\pi\frac{a^3}{4}\) = \(\frac{1}{12}\pi a^3\)

Volume of sphere = \(\frac{4}{3}\pi\times(\frac{a}{2})^3\)

= \(\frac{4}{3}\pi\times\frac{a^3}{8}=\frac{1}{6}\pi a^3\)

Radius of sector = slant height of cone = \(4\times\frac{5}{2}=10\) cm

Central angle of the sector = radius of cone

\(\times \frac{360}{R}=4\times\frac{360}{10}=144^\circ\)

Central angle = \(360\times\frac{2}{5}=144^\circ\)

a₁ : a₂ = 1: 2 = x : 2x,

h₁ : h₂ = 1:3 = y : 3y

Volume of 1^st pyramid = \(\frac{1}{3}\times x^2\times y=180\)

Volume of 2^nd pyramid

= \(\frac{1}{3}\times(2x)^2\times 3y=\frac{1}{3}\times 12\times x^2y\)

= \(12\times\frac{1}{3}x^2y=180\times12=2160\) cm²

r = 35 cm

l = 37 cm

CSA = πrI = π(35 x 37) = 4070 cm²

TSA = πr (1 + r )

\(=\frac{22}{7}\times35\times(37+35)=7920\) cm²

360 x \(\frac{4}{5}\) = 288

∴ Radius of the cone 4 = \(\frac{4}{5}\) x radius of the big circle

∴ If r is the radius of the circle Radius of the Radius of the cone = \(\frac{4}{5}\) r

But radius of the circle = slant height of cone

i.e., 1 = r

∴ Ratio between the radius of the cone and slant height

\(=\frac{4}{5}r:r=\frac{4}{5}:1=4:5\)

Ratio between the radius and slant height 2 : 3

Area length of the sector is equal to \(\frac{2}{3}\) part of the circle perimeter.

Central angle of the sector = 360 x \(\frac{2}{3}\) = 240°

 Length of base edge a = 18 cm

Volume = \(\frac{1}{3}a^2h\)

Slant height, \(l=\sqrt{18^2-9^2}\)

= \(\sqrt{324-81}\)

= \(\sqrt{243}\) = \(9\sqrt{3}\)

\(h=\sqrt{l^2-(a/2)^2}\)

= \(\sqrt{243-81}\)

= \(\sqrt{162}\) = \(9\sqrt{2}\) 

Volume = \(\frac{1}{3}a^2h\)

= \(\frac{1}{3}\) x 18² x \(9\sqrt{2}\)

= 1371.61 cm³

Central anglejof the sector

(x) = \(\frac{r}{l}\times360\),

r = 10 cm, l = 25 cm

= \(\frac{10}{25}\times360\)

= 144°

Curved surface area = 2al

Area of base = a²

a² = 2al

a = 21 ⇒ 1 = a/2

For making a square pyramid first we must determine its base, one side of the lateral will be the base. Other two sides make half of base by reducing the angle. That is angle at apex will be less than 90°.

Base edge = 42 cm

Slant edge = 29 cm

Slant height = \(\sqrt{29^2-(\frac{42}{2})^2}\)

= \(\sqrt{29^2-21^2}\)

= \(\sqrt{841-441}\)

= \(\sqrt{400}\)

= 20 cm

Slant height is less than the half of the base edge 21 cm so not possible for making a square pyramid

l = 25 cm

Surface area = 896 cm

a² + 2al = 896

a² + 2a × 25 = 896

a² + 50a – 896 = 0

a = \(\frac{50\pm\sqrt{2500+3584}}{2}\)

= \(\frac{50\pm\sqrt{6084}}{2}\) = \(\frac{50\pm78}{2}\)

a = \(\frac{50+78}{2}\) or \(\frac{50-78}{2}\)

= \(\frac{28}{2}=14\) or \(\frac{-128}{2}=-64\)

Length of base edge = 14 cm

height h = \(\sqrt{1^2-(\frac{a}{2})^2}\)

= \(\sqrt{25^2-7^2}\)

= \(\sqrt{625-49}\) = \(\sqrt{576}\) = 24

Volume = \(\frac{1}{3}a^2h\) = \(\frac{1}{3}\) x 14² x 24 = 1568 cm³

Base perimeter 4a = 64

a = 16 cm

Volume = \(\frac{1}{3}a^2h=1280\)

a²h = 1280 x 3

h = \(\frac{1280\times3}{a^2}\) = \(\frac{1280\times3}{16^2}\)

= \(\frac{3840}{256}\) = 15 cm

\(l=\sqrt{h^2+(\frac{a}{2})^2}\) = \(\sqrt{15^2+8^2}\) = \(\sqrt{289}\) = 17

Surface area = a² + 2al

= 162 + 2 × 16 × 17 = 256 + 544 = 800 cm²

Surface area = Curved surface area + base area

4 × 1/2 × 16 × 10 + 162 = 320 + 256 = 576cm

Total surface area of 500 square prism is = 500 × 576 = 28800 cm²

1 cm² = \(\frac{1}{10,000}\) m²

288000 cm² = \(\frac{288000}{10,000}\) = 28.8 m²

Cost of painting 1m² is = 80

Total cost = 28.8 x 80 = 2304 rupees

Slant height = \(\sqrt{25^2-15^2}\) = \(\sqrt{625-225}\) = \(\sqrt{400}\) = 20 cm

Height = \(\sqrt{20^2-15^2}\) = 13.2 cm

Let base edge be 40 cm

Slant height = \(\sqrt{20^2-20^2}\) = \(\sqrt{625-400}\) = \(\sqrt{225}\) = 15 cm

Height = \(\sqrt{15^2-15^2}\) = 0

It is impossible to make a square pyramid of base edge 40cm.

a = 10, l = 15

Volume of pyramid = \(\frac{1}{3}a^2h\)

= \(\frac{1}{3}\times10^2\times10\sqrt{2}\)

= \(\frac{1000\sqrt{2}}{3}\) cm³

\(h=\sqrt{l^2-(a/2)^2}\)

= \(\sqrt{15^2-5^2}\)

= \(\sqrt{225-25}\)

= \(\sqrt{200}\) = \(10\sqrt{2}\)

a. Slant height = 10 √3 cm

b. Lateral surface area

= \(20^2+4\times\frac{1}{2}\times20\times10\sqrt{3}\)

= 400 + \(400\sqrt{3}\)

= 400(1 + \(\sqrt{3}\)) cm²

c. Height = \(\sqrt{(10\sqrt{3})^2-10^2}\) = \(10\sqrt{2}\) cm

Volume = \(\frac{1}{3}400\times10\sqrt{2}\) = \(4000\frac{\sqrt{2}}{3}\) cm³

Base edge = 10cm

The given figure can be divided into n two right-angled triangles their angles are 30°, 60°, 90°, so ratio of their side will be 1: √3: 2.

2x = 10, x = 5

Slant height = 5 √3 cm³

Let breadth of the room be bm and length of the room is l = 112 m given 

\(\therefore\) Area of rectangular room = lb m²

But given that area of rectangular room is 784 m².

\(\therefore\) lb = 784

⇒ 112 b = 784

⇒ b = 784/112 = 7

\(\therefore\) Breadth of the rectangular room is 7m.

(2) Statement I is correct but statement II is incorrect

According to Hardy Schulze Rule statement 1 is correct. (Generally, the greater the valence of the flocculating ion added, the greater is its power to cause precipitation)

According to Hardy Schulze Rule statement 2 is incorrect