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.
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.