घन आकृत्यांचे घनफळ आणि पृष्ठफळ कॅल्क्युलेटर

दंडगोल, शंकू, गोळा, पिरॅमिड, लोलक, वलयाकार घन व लंबगोलाभ अशा 14 घनांचे घनफळ, पृष्ठफळ, तिरपी उंची आणि लिटरमधील क्षमता काढा.

अद्यतनित तपासलेली उदाहरणे: १३

वापरून पाहा
घनफळ
cm³
घनफळ: 282.7433 cm³
कमाल दशांश स्थाने: ४; जवळचे मूल्य; समान अंतरावर शून्यापासून दूर
Total surface area
245.0442cm²
Lateral (curved) surface area
188.4956cm²
Base area
28.2743cm²
Capacity
0.282743L

The cylinder (r = 3 cm, h = 10 cm) holds 282.7433 cm³ — 0.2827 litres — and its outside surface is 245.0442 cm².

Cylinder (r = 3 cm, h = 10 cm)

Drag to rotate
गणना कशी केली जाते S
  1. घनफळ

    V=πr2h=π×32×10=282.743339 cm3V = \pi r^2 h = \pi \times 3^2 \times 10 = 282.743339\,\text{cm}^{3}
  2. Curved surface

    Slateral=2πrh=188.495559 cm2S_\text{lateral} = 2\pi r h = 188.495559\,\text{cm}^{2}
  3. Total surface area (with both ends)

    S=2πr(r+h)=245.044227 cm2S = 2\pi r(r + h) = 245.044227\,\text{cm}^{2}
  4. Capacity

    282.743339 cm3=0.28274334 L282.743339\,\text{cm}^{3} = 0.28274334\,\text{L}

    1 L = 1000 cm³ = 0.001 m³.

घन आकृत्यांचे घनफळ आणि पृष्ठफळ कॅल्क्युलेटर बद्दल

Choose one of 14 solids and enter its dimensions to get the volume and total surface area, plus the curved (lateral) surface, base area, slant height and space diagonal where the solid has them. The volume is also given as a capacity in litres, using 1 L = 1,000 cm³. The formulas are the standard ones, for example V = πr²h for a cylinder, V = ⅓πh(R² + Rr + r²) for a frustum and V = 2π²Rr² for a torus.

The default cylinder, 3 cm in radius and 10 cm tall, holds 282.743 cm³ (0.283 L) and has 245.044 cm² of surface, of which 188.496 cm² is the curved side. Volume sizes tanks, containers and concrete pours; surface area sizes paint, insulation and sheet material.

Every result is exact except the surface of a general ellipsoid, which uses Knud Thomsen's approximation, within about 1.06%. The 3D model is drawn to the proportions entered; drag it to rotate.

सोडवलेली उदाहरणे

Cylinder r = 3 cm, h = 10 cm

Solid
Cylinder
त्रिज्या
3 cm
उंची
10 cm
निकालाचे एकक
सेंटीमीटर (cm)
घनफळ
282.743339 cm³
Total surface area
245.044227 cm²
Lateral (curved) surface area
188.495559 cm²
Capacity
0.282743 L

पडताळणीचा स्रोत: Python 3.8 math: 90*pi, 78*pi, 60*pi; 282.743… cm³ ÷ 1000 = L

Box 6 × 4 × 10 m

Solid
Cuboid (box)
लांबी
6 m
रुंदी
4 m
उंची
10 m
निकालाचे एकक
मीटर (m)
घनफळ
240 m³
Total surface area
248 m²
Space diagonal
12.328828 m
Capacity
240,000 L

पडताळणीचा स्रोत: 6·4·10; 2(24 + 60 + 40); Python 3.8 math.sqrt(152); 1 m³ = 1000 L

Cone r = 3, h = 4 (slant 5)

Solid
Cone
त्रिज्या
3 cm
उंची
4 cm
निकालाचे एकक
सेंटीमीटर (cm)
Slant height
5 cm
घनफळ
37.699112 cm³
Lateral (curved) surface area
47.12389 cm²
Total surface area
75.398224 cm²

पडताळणीचा स्रोत: 3-4-5 slant; Python 3.8 math: 12*pi, 15*pi, 24*pi

Frustum R = 5, r = 3, h = 4

Solid
Frustum of a cone
Bottom radius (R)
5 cm
Top radius (r)
3 cm
उंची
4 cm
निकालाचे एकक
सेंटीमीटर (cm)
Slant height
4.472136 cm
घनफळ
205.25072 cm³
Total surface area
219.211186 cm²

पडताळणीचा स्रोत: Python 3.8 math: sqrt(20), pi*4*(25+15+9)/3, pi*8*sqrt(20) + 34*pi

प्रश्न

How do you calculate the volume of a cylinder?

Multiply the area of the circular end by the height: V = πr²h. A radius of 3 cm and a height of 10 cm give 90π ≈ 282.74 cm³, or 0.283 L. Use the radius, not the diameter: a tank 1 m across and 1.5 m tall has r = 0.5 m and holds π × 0.25 × 1.5 ≈ 1.178 m³, which is 1,178 L.

How do you find the surface area of a cylinder?

Add the curved side, 2πrh, to the two circular ends, 2πr²: S = 2πr(r + h). For r = 3 cm and h = 10 cm the curved side is 60π ≈ 188.50 cm² and the total is 78π ≈ 245.04 cm². An open-topped tank has only one end, so subtract πr², here 28.27 cm², to get 216.77 cm².

How many litres are in a cubic metre?

1,000. A litre is exactly one cubic decimetre, 1,000 cm³, a definition fixed by the General Conference on Weights and Measures (CGPM) in 1964. Divide cubic centimetres by 1,000 to get litres. In US units, one gallon is exactly 231 in³, or 3.785411784 L, and one cubic foot is 28.316846592 L.

What is the formula for the volume of a sphere?

V = ⁴⁄₃πr³, which is two thirds of the cylinder that just encloses the sphere, as Archimedes showed. A sphere of radius 3 cm holds 36π ≈ 113.10 cm³, and a ball 1 m across holds 0.5236 m³, or 523.6 L. Volume grows with the cube of the radius, so doubling the diameter multiplies the volume by 8.

How do you calculate the volume of a cone or a pyramid?

Take one third of the base area times the height: V = ⅓Bh. A cone of radius 3 and height 4 has V = ⅓π × 9 × 4 = 12π ≈ 37.70, and a square pyramid with a 6 × 6 base and height 4 has V = ⅓ × 36 × 4 = 48. A cone holds exactly one third of the cylinder with the same base and height.

“घन आकृत्यांचे घनफळ आणि पृष्ठफळ कॅल्क्युलेटर” किती अचूक आहे?

अचूकता तुमच्या इनपुटवर आणि पद्धतीच्या गृहीतकांवर अवलंबून असते. दशांश गणना 50 सार्थ अंक वापरते, पण अंदाज, संख्यात्मक पद्धती आणि मूळ डेटा कमी अचूक असू शकतात; दाखवलेल्या मूल्यांचे पूर्णांकन केल्याने या मर्यादा दूर होत नाहीत. स्वतंत्र स्रोतांतील सोडवलेल्या उदाहरणांशी पडताळणी: १३. उदाहरणार्थ, “Cylinder r = 3 cm, h = 10 cm” ची पडताळणी Python 3.8 math: 90*pi, 78*pi, 60*pi; 282.743… cm³ ÷ 1000 = L याच्याशी केली आहे.

या पद्धतीचा स्रोत कोणता?

Zwillinger, D. (ed.) CRC Standard Mathematical Tables and Formulas, 33rd ed., §4.6 (solids); Weisstein, E. W. “Torus”, “Spherical Cap”, “Ellipsoid” — MathWorld; Thomsen, K. — ellipsoid surface approximation, as quoted in “Ellipsoid § Surface area”, Wikipedia.

या कॅल्क्युलेटरबद्दल

Vsphere=43πr3,Vcone=13πr2hVfrustum=13πh(R2+Rr+r2),Vtorus=2π2Rr2\begin{gathered} V_\text{sphere} = \tfrac{4}{3}\pi r^3,\quad V_\text{cone} = \tfrac{1}{3}\pi r^2 h \\ V_\text{frustum} = \tfrac{1}{3}\pi h(R^2 + Rr + r^2),\quad V_\text{torus} = 2\pi^2 R r^2 \end{gathered}

स्रोत

  1. Zwillinger, D. (ed.) CRC Standard Mathematical Tables and Formulas, 33rd ed., §4.6 (solids)
  2. Weisstein, E. W. “Torus”, “Spherical Cap”, “Ellipsoid” — MathWorld
  3. Thomsen, K. — ellipsoid surface approximation, as quoted in “Ellipsoid § Surface area”, Wikipedia

संदर्भांशी पडताळलेले

या कॅल्क्युलेटरमध्ये स्वतंत्र स्रोतांमधील उत्तरांसह १३ सोडवलेली उदाहरणे आहेत. ती चाचणी संचात चालतात आणि तुम्ही ती येथेही चालवू शकता.

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