Perimeter of triangle = 540 cm The ratio of sides of triangle = 12: 17: 25 Let, one of the sides of triangle a = 12 x Therefore, remaining two sides are b = 17 x and c = 25x. We know that the perimeter of triangle = a + b + c ⇒ 540 = 12 x + 17 x + 25 x ⇒ 540 = 54x ⇒ x = 540/54 = 10 So, the sides ofRead more
Perimeter of triangle = 540 cm
The ratio of sides of triangle = 12: 17: 25
Let, one of the sides of triangle a = 12 x
Therefore, remaining two sides are b = 17 x and c = 25x.
We know that the perimeter of triangle = a + b + c
⇒ 540 = 12 x + 17 x + 25 x
⇒ 540 = 54x
⇒ x = 540/54 = 10
So, the sides of triangle are a = 12 × 10 = 120 cm, b = 17 × 10 = 170 cm and c = 25 × 10 = 250 cm.
So, the semi- perimeter of triangle is given by
S = (a + b + c /2) = 540/2 = 270 cm
Therefore, using Heron’s formula, the area of triangle = √s(s – a)(s – b)(s – c)
= √270(270 – 120)(270 – 170)(270 – 250)
= √270(150)(100)(20)
= √81000000
= 9000 cm²
Hence, the area of triangle is 9000 cm².
Perimeter of triangle = 30 cm Two sides of triangle b = 12 cm and c = 12 cm. Let, the third side = a cm We know that the Parimeter of triangle = a + b + c ⇒ 30 = a + 12 + 12 ⇒ 30 - 24 = a ⇒ a = 6 So, the Semi- perimeter of triangle is given by S = a + b + c /2 = 30/2 = 15 cm Therefore, using Heron'sRead more
Perimeter of triangle = 30 cm
Two sides of triangle b = 12 cm and c = 12 cm.
Let, the third side = a cm
We know that the Parimeter of triangle = a + b + c
⇒ 30 = a + 12 + 12
⇒ 30 – 24 = a
⇒ a = 6
So, the Semi- perimeter of triangle is given by
S = a + b + c /2 = 30/2 = 15 cm
Therefore, using Heron’s formula, the area of triangle = √s(s – a)(s – b)(s – c)
= √15(15 -6)(15 -12)(15 -12)
= √15(9)(3)(3)
= 9√15 cm²
Hence, the area of triangle is 9√15 cm².
Radius of capsule r = 3.5/2 = 1.75 mm Volume of medicine to fill the capsule = = 4/3πr³ = 4/3 × 22/7 × 1.75 × 1.75 × 1.75 = 4/3 × 22 × 0.25 × 1.75 × 1.75 = 22.46 mm³ (approx) Hence, 22.46 mm³ medicine is required to fill this capsule.
Radius of capsule r = 3.5/2 = 1.75 mm
Volume of medicine to fill the capsule = = 4/3πr³
= 4/3 × 22/7 × 1.75 × 1.75 × 1.75
= 4/3 × 22 × 0.25 × 1.75 × 1.75
= 22.46 mm³ (approx)
Hence, 22.46 mm³ medicine is required to fill this capsule.
Surface area of sphere A = 154 cm² Let, the radius of sphere = r cm We know that the surface area of sphere = 4πr² ⇒ 154 = 4 × 22/7 × r² ⇒ r² = (154 × 7)/(22×4) = 49/4 ⇒ r = √49/4 = 7/2 Volume of sphere = 4/2πr³ = 4/3 × 22/7 × (7/2)³ = 4/3 × 22/7 × 7/2 × 7/2 × 7/2 = 539/3 = 179(2/3) cm³ Hence the voRead more
Surface area of sphere A = 154 cm²
Let, the radius of sphere = r cm
We know that the surface area of sphere = 4πr²
⇒ 154 = 4 × 22/7 × r²
⇒ r² = (154 × 7)/(22×4) = 49/4
⇒ r = √49/4 = 7/2
Volume of sphere = 4/2πr³
= 4/3 × 22/7 × (7/2)³
= 4/3 × 22/7 × 7/2 × 7/2 × 7/2
= 539/3 = 179(2/3) cm³
Hence the volume of sphere is 179(2/3) cm³.
Let the internal radius of dome = r m Internal surface area of dome = 2πr² cost of white washing at the rate of Rs 2 = 2πr² × Rs 2 = Rs 4πr² ⇒ Rs 4πr = Rs 498.96 ⇒ 4 × 22/7 × r² = 498.96 ⇒ r² = (498.96×7)/4×22) = 39.69 ⇒ r = √39.69 = 6.3 m Therefore, the internal surface of dome = 2πr² = 2 × 22/7 ×Read more
Let the internal radius of dome = r m
Internal surface area of dome = 2πr²
cost of white washing at the rate of Rs 2 = 2πr² × Rs 2 = Rs 4πr²
⇒ Rs 4πr = Rs 498.96
⇒ 4 × 22/7 × r² = 498.96
⇒ r² = (498.96×7)/4×22) = 39.69
⇒ r = √39.69 = 6.3 m
Therefore, the internal surface of dome = 2πr²
= 2 × 22/7 × (6.3)2
= 2 × 22/7 × 6.3 × 6.3
= 2 × 22 × 0.9 × 6.3
= 249.98 m²
Hence, the inside surface area of the dome is 249.48 m².
(II) Volume of the air inside the dome = 2/3πr³
= 2/3 × 22/7 × (6.3)³
= 2/3 × 22/7 × 6.3 × 6.3 × 6.3
= 2 × 22 × 0.3 × 6.3 × 6.3
= 523.9 cm³
Hence, the volume of the air inside the dome is 523.9 cm³.
(i) Given that the radius of solid sphere is r and the radius of new sphere is r'. Volume of solid sphere = 4/3 πr³ Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³ According to question : Volume of new sphere = volume of 27 solid spheres ⇒ 4/3π (r')³ = 27 × 4/3πr³ ⇒ (r')³ = 27 × r³ ⇒ r' = 3Read more
(i) Given that the radius of solid sphere is r and the radius of new sphere is r’.
Volume of solid sphere = 4/3 πr³
Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³
According to question :
Volume of new sphere = volume of 27 solid spheres
⇒ 4/3π (r’)³ = 27 × 4/3πr³
⇒ (r’)³ = 27 × r³
⇒ r’ = 3 × r
Hence, the radius of new sphere is 3r.
(ii) Surface area of new sphere S’ = 4πr²
Surface area of new sphere S’ = 4π(r’)² = 4π(3r)² = 36πr²
Therefore,
s/s’ = 4πr²/36πr² = 1/9
Hence, the ratio of S and S’ is 1:9.
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m Therefore, the outer radius R = 1 + 0.01 = 1.01 m Volume of hemispherical tank = 2/3π(R³ - r³) = 2/3 × 22/7 × [(1.01)³ -1³] = 2/3 × 22/2 × (1.030301 - 1) = 2/3 × 22/7 × 0.030301 = 0.06348 m³ Hence, the volume of the iron uRead more
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m
Therefore, the outer radius R = 1 + 0.01 = 1.01 m
Volume of hemispherical tank = 2/3π(R³ – r³)
= 2/3 × 22/7 × [(1.01)³ -1³]
= 2/3 × 22/2 × (1.030301 – 1)
= 2/3 × 22/7 × 0.030301
= 0.06348 m³
Hence, the volume of the iron used to make the tank is 0.06348 m³.
(I) Radius of spherical ball r = 28/2 = 14 cm Volume of water displaced by spherical ball = 4/3πr³ = 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³ Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³. (II) Radius of spherical ball r = 0.21/2 = 0.105 m VoluRead more
(I) Radius of spherical ball r = 28/2 = 14 cm
Volume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³
Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³.
(II) Radius of spherical ball r = 0.21/2 = 0.105 m
Volume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 0.105 × 0.105 × 0.105 = 4 × 22 × 0.005 × 0.63 × 0.63 = 0.004861 m³
Hence, the volume of water displaced by spherical ball is 0.004861 m³.
Radius of metallic ball r = 4.2/2 = 2.1 cm Therefore, the volume of metallic ball = 4/3πr³ = 4/3 × 22/7 × 2.1 × 2.1 × 2.1 = 4 × 22 × 0.1 × 2.1 × 2.1 = 38.808 cm² Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx) Hence, the mass of the ball is 345. 39 gram.
Radius of metallic ball r = 4.2/2 = 2.1 cm
Therefore, the volume of metallic ball = 4/3πr³
= 4/3 × 22/7 × 2.1 × 2.1 × 2.1
= 4 × 22 × 0.1 × 2.1 × 2.1
= 38.808 cm²
Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx)
Hence, the mass of the ball is 345. 39 gram.
Let, the radius of Earth be R Therefore, the diameter of Earth = 2R According to question, diameter of moon = 1/4(2R) So, the radius of moon = 1/4(2R)/2 = (1/4)R Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64 ⇒ Volume of moon = 1/64 × volume of Earth Hence, the volume of moRead more
Let, the radius of Earth be R
Therefore, the diameter of Earth = 2R
According to question, diameter of moon = 1/4(2R)
So, the radius of moon = 1/4(2R)/2 = (1/4)R
Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64
⇒ Volume of moon = 1/64 × volume of Earth
Hence, the volume of moon is 1/64 the volume of Earth.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Perimeter of triangle = 540 cm The ratio of sides of triangle = 12: 17: 25 Let, one of the sides of triangle a = 12 x Therefore, remaining two sides are b = 17 x and c = 25x. We know that the perimeter of triangle = a + b + c ⇒ 540 = 12 x + 17 x + 25 x ⇒ 540 = 54x ⇒ x = 540/54 = 10 So, the sides ofRead more
Perimeter of triangle = 540 cm
See lessThe ratio of sides of triangle = 12: 17: 25
Let, one of the sides of triangle a = 12 x
Therefore, remaining two sides are b = 17 x and c = 25x.
We know that the perimeter of triangle = a + b + c
⇒ 540 = 12 x + 17 x + 25 x
⇒ 540 = 54x
⇒ x = 540/54 = 10
So, the sides of triangle are a = 12 × 10 = 120 cm, b = 17 × 10 = 170 cm and c = 25 × 10 = 250 cm.
So, the semi- perimeter of triangle is given by
S = (a + b + c /2) = 540/2 = 270 cm
Therefore, using Heron’s formula, the area of triangle = √s(s – a)(s – b)(s – c)
= √270(270 – 120)(270 – 170)(270 – 250)
= √270(150)(100)(20)
= √81000000
= 9000 cm²
Hence, the area of triangle is 9000 cm².
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Perimeter of triangle = 30 cm Two sides of triangle b = 12 cm and c = 12 cm. Let, the third side = a cm We know that the Parimeter of triangle = a + b + c ⇒ 30 = a + 12 + 12 ⇒ 30 - 24 = a ⇒ a = 6 So, the Semi- perimeter of triangle is given by S = a + b + c /2 = 30/2 = 15 cm Therefore, using Heron'sRead more
Perimeter of triangle = 30 cm
See lessTwo sides of triangle b = 12 cm and c = 12 cm.
Let, the third side = a cm
We know that the Parimeter of triangle = a + b + c
⇒ 30 = a + 12 + 12
⇒ 30 – 24 = a
⇒ a = 6
So, the Semi- perimeter of triangle is given by
S = a + b + c /2 = 30/2 = 15 cm
Therefore, using Heron’s formula, the area of triangle = √s(s – a)(s – b)(s – c)
= √15(15 -6)(15 -12)(15 -12)
= √15(9)(3)(3)
= 9√15 cm²
Hence, the area of triangle is 9√15 cm².
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?
Radius of capsule r = 3.5/2 = 1.75 mm Volume of medicine to fill the capsule = = 4/3πr³ = 4/3 × 22/7 × 1.75 × 1.75 × 1.75 = 4/3 × 22 × 0.25 × 1.75 × 1.75 = 22.46 mm³ (approx) Hence, 22.46 mm³ medicine is required to fill this capsule.
Radius of capsule r = 3.5/2 = 1.75 mm
See lessVolume of medicine to fill the capsule = = 4/3πr³
= 4/3 × 22/7 × 1.75 × 1.75 × 1.75
= 4/3 × 22 × 0.25 × 1.75 × 1.75
= 22.46 mm³ (approx)
Hence, 22.46 mm³ medicine is required to fill this capsule.
Find the volume of a sphere whose surface area is 154 cm².
Surface area of sphere A = 154 cm² Let, the radius of sphere = r cm We know that the surface area of sphere = 4πr² ⇒ 154 = 4 × 22/7 × r² ⇒ r² = (154 × 7)/(22×4) = 49/4 ⇒ r = √49/4 = 7/2 Volume of sphere = 4/2πr³ = 4/3 × 22/7 × (7/2)³ = 4/3 × 22/7 × 7/2 × 7/2 × 7/2 = 539/3 = 179(2/3) cm³ Hence the voRead more
Surface area of sphere A = 154 cm²
See lessLet, the radius of sphere = r cm
We know that the surface area of sphere = 4πr²
⇒ 154 = 4 × 22/7 × r²
⇒ r² = (154 × 7)/(22×4) = 49/4
⇒ r = √49/4 = 7/2
Volume of sphere = 4/2πr³
= 4/3 × 22/7 × (7/2)³
= 4/3 × 22/7 × 7/2 × 7/2 × 7/2
= 539/3 = 179(2/3) cm³
Hence the volume of sphere is 179(2/3) cm³.
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, find the
Let the internal radius of dome = r m Internal surface area of dome = 2πr² cost of white washing at the rate of Rs 2 = 2πr² × Rs 2 = Rs 4πr² ⇒ Rs 4πr = Rs 498.96 ⇒ 4 × 22/7 × r² = 498.96 ⇒ r² = (498.96×7)/4×22) = 39.69 ⇒ r = √39.69 = 6.3 m Therefore, the internal surface of dome = 2πr² = 2 × 22/7 ×Read more
Let the internal radius of dome = r m
See lessInternal surface area of dome = 2πr²
cost of white washing at the rate of Rs 2 = 2πr² × Rs 2 = Rs 4πr²
⇒ Rs 4πr = Rs 498.96
⇒ 4 × 22/7 × r² = 498.96
⇒ r² = (498.96×7)/4×22) = 39.69
⇒ r = √39.69 = 6.3 m
Therefore, the internal surface of dome = 2πr²
= 2 × 22/7 × (6.3)2
= 2 × 22/7 × 6.3 × 6.3
= 2 × 22 × 0.9 × 6.3
= 249.98 m²
Hence, the inside surface area of the dome is 249.48 m².
(II) Volume of the air inside the dome = 2/3πr³
= 2/3 × 22/7 × (6.3)³
= 2/3 × 22/7 × 6.3 × 6.3 × 6.3
= 2 × 22 × 0.3 × 6.3 × 6.3
= 523.9 cm³
Hence, the volume of the air inside the dome is 523.9 cm³.
Twenty-seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the:
(i) Given that the radius of solid sphere is r and the radius of new sphere is r'. Volume of solid sphere = 4/3 πr³ Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³ According to question : Volume of new sphere = volume of 27 solid spheres ⇒ 4/3π (r')³ = 27 × 4/3πr³ ⇒ (r')³ = 27 × r³ ⇒ r' = 3Read more
(i) Given that the radius of solid sphere is r and the radius of new sphere is r’.
Volume of solid sphere = 4/3 πr³
Therefore, the volume of 27 solid spheres = 27 × 4/3 πr³
According to question :
Volume of new sphere = volume of 27 solid spheres
⇒ 4/3π (r’)³ = 27 × 4/3πr³
⇒ (r’)³ = 27 × r³
⇒ r’ = 3 × r
Hence, the radius of new sphere is 3r.
(ii) Surface area of new sphere S’ = 4πr²
See lessSurface area of new sphere S’ = 4π(r’)² = 4π(3r)² = 36πr²
Therefore,
s/s’ = 4πr²/36πr² = 1/9
Hence, the ratio of S and S’ is 1:9.
A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m Therefore, the outer radius R = 1 + 0.01 = 1.01 m Volume of hemispherical tank = 2/3π(R³ - r³) = 2/3 × 22/7 × [(1.01)³ -1³] = 2/3 × 22/2 × (1.030301 - 1) = 2/3 × 22/7 × 0.030301 = 0.06348 m³ Hence, the volume of the iron uRead more
The internal radius of hemispherical tank r = 1 m and thickness 1 cm = 0.01 m
See lessTherefore, the outer radius R = 1 + 0.01 = 1.01 m
Volume of hemispherical tank = 2/3π(R³ – r³)
= 2/3 × 22/7 × [(1.01)³ -1³]
= 2/3 × 22/2 × (1.030301 – 1)
= 2/3 × 22/7 × 0.030301
= 0.06348 m³
Hence, the volume of the iron used to make the tank is 0.06348 m³.
Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m
(I) Radius of spherical ball r = 28/2 = 14 cm Volume of water displaced by spherical ball = 4/3πr³ = 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³ Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³. (II) Radius of spherical ball r = 0.21/2 = 0.105 m VoluRead more
(I) Radius of spherical ball r = 28/2 = 14 cm
Volume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 14 × 14 × 14 × = 4/3 × 22 × 2 × 14 × 14 = 11498(2/3) cm³
Hence, the volume of water displaced by spherical ball is 11498(2/3) cm³.
(II) Radius of spherical ball r = 0.21/2 = 0.105 m
See lessVolume of water displaced by spherical ball = 4/3πr³
= 4/3 × 22/7 × 0.105 × 0.105 × 0.105 = 4 × 22 × 0.005 × 0.63 × 0.63 = 0.004861 m³
Hence, the volume of water displaced by spherical ball is 0.004861 m³.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?
Radius of metallic ball r = 4.2/2 = 2.1 cm Therefore, the volume of metallic ball = 4/3πr³ = 4/3 × 22/7 × 2.1 × 2.1 × 2.1 = 4 × 22 × 0.1 × 2.1 × 2.1 = 38.808 cm² Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx) Hence, the mass of the ball is 345. 39 gram.
Radius of metallic ball r = 4.2/2 = 2.1 cm
See lessTherefore, the volume of metallic ball = 4/3πr³
= 4/3 × 22/7 × 2.1 × 2.1 × 2.1
= 4 × 22 × 0.1 × 2.1 × 2.1
= 38.808 cm²
Here, the mass of 1 cm³ = 8.9 × 38.808 = 345.39g(approx)
Hence, the mass of the ball is 345. 39 gram.
The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Let, the radius of Earth be R Therefore, the diameter of Earth = 2R According to question, diameter of moon = 1/4(2R) So, the radius of moon = 1/4(2R)/2 = (1/4)R Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64 ⇒ Volume of moon = 1/64 × volume of Earth Hence, the volume of moRead more
Let, the radius of Earth be R
See lessTherefore, the diameter of Earth = 2R
According to question, diameter of moon = 1/4(2R)
So, the radius of moon = 1/4(2R)/2 = (1/4)R
Volume of moon/Volume of Earth = 4/3π(1/4R)³/4/3π(R)³ = 1/64R³/R³ = 1/64
⇒ Volume of moon = 1/64 × volume of Earth
Hence, the volume of moon is 1/64 the volume of Earth.