Class 10 Compulsory Mathematics – क्षेत्रफल र आयतन (Mensuration – Area and Volume)
यो अध्यायले कक्षा १० को अनिवार्य गणितको लागि क्षेत्रफल र आयतन (Mensuration – Area and Volume) को पूर्ण विवरण प्रस्तुत गर्दछ। यसले विभिन्न आकारहरूको क्षेत्रफल, पृष्ठीय क्षेत्रफल र आयतन गणना गर्ने सूत्रहरू समावेश गर्दछ।
This chapter provides a complete overview of Mensuration – Area and Volume for Class 10 Compulsory Mathematics. It covers formulas for calculating area, surface area, and volume of various geometric shapes.
Chapter Information
Chapter: क्षेत्रफल र आयतन (Mensuration – Area and Volume)
Subject: Compulsory Mathematics, Class 10 SEE
Writer: D.R. Simkhada
Publisher: Readmore Publishers & Distributors
Description: Complete notes on Mensuration covering area and volume formulas for triangles, quadrilaterals, prisms, cylinders, cones, spheres, pyramids with detailed examples and solutions as per SEE Nepal curriculum.
Table of Contents
Detailed Chapter Notes
क्षेत्रफलका सूत्रहरू (Area Formulas)
त्रिभुजका क्षेत्रफल सूत्रहरू (Triangle Area Formulas)
| S.N. | Name | Key Points & Area Formula |
|---|---|---|
| 1. | Triangle (त्रिभुज) | Area, \( A = \frac{1}{2} \times b \times h \) |
| 2. | Triangle (त्रिभुज) | \( s = \frac{a+b+c}{2} \), Area, \( A = \sqrt{s(s-a)(s-b)(s-c)} \) |
| 3. | Equilateral Triangle (समबाहु त्रिभुज) | Area, \( A = \frac{\sqrt{3}}{4}a^2 \) |
| 4. | Isosceles Triangle (समद्विबाहु त्रिभुज) | Area, \( A = \frac{b}{4}\sqrt{4a^2-b^2} \) |
| 5. | Right Angled Triangle (समकोणी त्रिभुज) | Area, \( A = \frac{1}{2} \times b \times p \) |
चतुर्भुजका क्षेत्रफल सूत्रहरू (Quadrilateral Area Formulas)
| S.N. | Name | Area Formula |
|---|---|---|
| 6. | Parallelogram (समानान्तर चतुर्भुज) | Area, \( A = b \times h \) |
| 7. | Rhombus (समबाहु चतुर्भुज) | Area, \( A = \frac{1}{2} \times d_1 \times d_2 \) |
| 8. | Quadrilateral (चतुर्भुज) | Area, \( A = \frac{1}{2} \times d \times (h_1 + h_2) \) |
| 9. | Trapezium (समलम्ब चतुर्भुज) | Area, \( A = \frac{1}{2} \times h \times (a + b) \) |
| 10. | Arrow-head (एरोहेड) | Area, \( A = \frac{1}{2} \times d_1 \times d_2 \) |
| 11. | Kite (चङ्गा) | Area, \( A = \frac{1}{2} \times d_1 \times d_2 \) |
प्रिज्म (Prism)
Area of cross-section = A
Perimeter of base = P
Length of prism = h (or l)
Lateral Surface Area (LSA) = \( P \times h \)
Total Surface Area (TSA) = \( 2A + Ph \)
Volume (V) = \( A \times h \)
विशेष प्रिज्महरू (Special Prisms):
त्रिभुजाकार प्रिज्म (Triangular Prism):
\( s = \frac{a+b+c}{2} \), Area of base (A) = \( \sqrt{s(s-a)(s-b)(s-c)} \)
Perimeter of base (P) = \( a+b+c \)
समबाहु त्रिभुज आधार भएको प्रिज्म (Prism with Equilateral Triangle Base):
Area of base (A) = \( \frac{\sqrt{3}}{4}a^2 \)
Perimeter of base (P) = \( 3a \)
समकोणी त्रिभुज आधार भएको प्रिज्म (Prism with Right-angled Triangle Base):
Area of base (A) = \( \frac{1}{2} \times p \times b \)
Perimeter of base (P) = \( p + b + h \)
बेलना (Cylinder)
Area of base (A) = \( \pi r^2 \)
Circumference of base (C) = \( 2\pi r \)
Curved Surface Area (CSA) = \( 2\pi rh \)
Total Surface Area (TSA) = \( 2\pi r(r+h) \)
Volume (V) = \( \pi r^2 h \)
खोक्रो बेलना (Hollow Cylinder):
External Radius = R, Internal Radius = r
Area of cross-section (A) = \( \pi(R^2 – r^2) \)
Internal CSA = \( 2\pi rh \)
External CSA = \( 2\pi Rh \)
TSA = \( 2\pi rh + 2\pi Rh + 2\pi(R^2 – r^2) \)
Volume of material (V) = \( \pi(R^2 – r^2)h \)
गोला (Sphere)
Surface Area (A) = \( 4\pi r^2 \)
Volume (V) = \( \frac{4}{3}\pi r^3 \)
अर्धगोला (Hemisphere)
Curved Surface Area (CSA) = \( 2\pi r^2 \)
Area of flat surface = \( \pi r^2 \)
Total Surface Area (TSA) = \( 3\pi r^2 \)
Volume (V) = \( \frac{2}{3}\pi r^3 \)
सोली (Cone)
Slant height (l) = \( \sqrt{h^2 + r^2} \)
Curved Surface Area (CSA) = \( \pi r l \)
Total Surface Area (TSA) = \( \pi r(r + l) \)
Volume (V) = \( \frac{1}{3}\pi r^2 h \)
पिरामिड (Pyramid)
Area of base = A
Perimeter of base = P
Lateral Surface Area (LSA) = \( \frac{1}{2} \times P \times l \) (where l = slant height)
Total Surface Area (TSA) = LSA + A
Volume (V) = \( \frac{1}{3} \times A \times h \)
वर्गाकार आधार भएको पिरामिडका लागि महत्वपूर्ण सम्बन्धहरू (Important relations for a square-based pyramid):
\( l^2 = h^2 + \left(\frac{a}{2}\right)^2 \)
\( e^2 = l^2 + \left(\frac{a}{2}\right)^2 \)
\( e^2 = h^2 + \left(\frac{d}{2}\right)^2 \)
\( d = \sqrt{2}a \)
संयुक्त ठोसहरू (Combined Solids)
बेलना र अर्धगोला (Cylinder and Hemisphere):
TSA = \( 2\pi rh + \pi r^2 + 2\pi r^2 = \pi r(2h + 3r) \)
Volume = \( \pi r^2 h + \frac{2}{3}\pi r^3 = \pi r^2 \left(h + \frac{2}{3}r\right) \)
बेलना र सोली (Cylinder and Cone):
TSA = \( 2\pi rh + \pi r^2 + \pi r l \)
Volume = \( \pi r^2 h + \frac{1}{3}\pi r^2 H \)
सोली र अर्धगोला (Cone and Hemisphere):
TSA = \( \pi r l + 2\pi r^2 = \pi r(l + 2r) \)
Volume = \( \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 (h + 2r) \)
अभ्यास (Exercises)
१. प्रिज्मको क्षेत्रफल र आयतन (Prism Area and Volume)
१. (क) Find the area of the cross-section, LSA, and TSA of the given prism:
Here, the base is a right-angled triangle with p = 8 cm, b = 6 cm. Length of the prism (h) = 12 cm.
Solution:
(i) Area of cross-section (A) = \( \frac{1}{2} \times p \times b = \frac{1}{2} \times 8 \times 6 = 24 \text{cm}^2 \)
Hypotenuse = \( \sqrt{p^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{cm} \)
Perimeter of base (P) = \( p + b + h’ = 8 + 6 + 10 = 24 \text{cm} \)
(ii) Lateral Surface Area (LSA) = \( P \times h = 24 \times 12 = 288 \text{cm}^2 \)
(iii) Total Surface Area (TSA) = \( Ph + 2A = 288 + 2 \times 24 = 288 + 48 = 336 \text{cm}^2 \)
१. (ख) Find the volume of the following prism:
Here, the base is a right-angled triangle. b = 8 cm, p = 6 cm, and length of prism (h) = 15 cm.
Solution:
Area of base (A) = \( \frac{1}{2} \times b \times p = \frac{1}{2} \times 8 \times 6 = 24 \text{cm}^2 \)
Volume of prism (V) = \( A \times h = 24 \times 15 = 360 \text{cm}^3 \)
२. बेलनाको क्षेत्रफल र आयतन (Cylinder Area and Volume)
२. (क) Find the LSA, TSA and Volume of the cylinder:
Here, radius (r) = 7 cm, height (h) = 10 cm.
Solution:
LSA = \( 2\pi rh = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{cm}^2 \)
TSA = \( 2\pi r(h + r) = 2 \times \frac{22}{7} \times 7(10 + 7) = 44 \times 17 = 748 \text{cm}^2 \)
Volume (V) = \( \pi r^2 h = \frac{22}{7} \times 7^2 \times 10 = 22 \times 7 \times 10 = 1540 \text{cm}^3 \)
३. गोलाको क्षेत्रफल (Sphere Surface Area)
३. (क) Find the CSA and TSA of the given solid sphere:
Here, radius (r) = 7 cm.
Solution:
Surface Area = \( 4\pi r^2 = 4 \times \frac{22}{7} \times 7^2 = 4 \times 22 \times 7 = 616 \text{cm}^2 \)
४. अर्धगोलाको क्षेत्रफल (Hemisphere Surface Area)
४. (क) Find the CSA and TSA of the given hemisphere:
Here, radius (r) = 7 cm.
Solution:
CSA = \( 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308 \text{cm}^2 \)
TSA = \( 3\pi r^2 = 3 \times \frac{22}{7} \times 7^2 = 3 \times 22 \times 7 = 462 \text{cm}^2 \)
५. सोलीको क्षेत्रफल र आयतन (Cone Area and Volume)
५. (क) The total surface area of a cone is 704 cm² and the radius of the base is 7 cm. Find its slant height.
Solution:
TSA of cone = \( \pi r(l + r) \)
\( 704 = \frac{22}{7} \times 7(l + 7) \)
\( 704 = 22(l + 7) \)
\( \frac{704}{22} = l + 7 \)
\( 32 = l + 7 \)
\( l = 32 – 7 = 25 \text{cm} \)
५. (ख) The volume of a cone is \( 100\pi \text{cm}^3 \) and its height is 12 cm. Find the radius of the base.
Solution:
Volume of cone = \( \frac{1}{3}\pi r^2 h \)
\( 100\pi = \frac{1}{3}\pi r^2 \times 12 \)
\( 100 = 4r^2 \)
\( r^2 = \frac{100}{4} = 25 \)
\( r = 5 \text{cm} \)
६. पिरामिडको क्षेत्रफल र आयतन (Pyramid Area and Volume)
६. (क) Find the value of the unknowns in the following square-based pyramids:
Here, length of the side of the base (a) = 10 cm, vertical height (h) = 12 cm.
Solution:
Slant height (l) = \( \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{12^2 + \left(\frac{10}{2}\right)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{cm} \)
६. (ख) Find the total surface area of the given square-based pyramid:
Here, length of the side of the base (a) = 12 cm, slant height (l) = 10 cm.
Solution:
TSA = \( a^2 + 2al = 12^2 + 2 \times 12 \times 10 = 144 + 240 = 384 \text{cm}^2 \)
७. संयुक्त ठोसहरूको आयतन (Combined Solids Volume)
७. (क) A solid object is formed by combining a cube and a square-based pyramid. Find the volume of the solid object.
Here, side of cube = 12 cm, total height = 18 cm.
Solution:
Volume of cube = \( (12)^3 = 1728 \text{cm}^3 \)
Height of pyramid = \( 18 – 12 = 6 \text{cm} \)
Area of square base = \( (12)^2 = 144 \text{cm}^2 \)
Volume of pyramid = \( \frac{1}{3} \times 144 \times 6 = 288 \text{cm}^3 \)
Total volume = \( 1728 + 288 = 2016 \text{cm}^3 \)
७. (ख) A tent is in the shape of a cylinder surmounted by a cone. Find the area of the canvas required.
Height of cylindrical part = 10 m, radius = 24 m, slant height of cone = 26 m.
Solution:
Area of canvas = CSA of cone + CSA of cylinder
= \( \pi r l + 2\pi r h = \pi r(l + 2h) \)
= \( \frac{22}{7} \times 24 \times (26 + 2 \times 10) \)
= \( \frac{22}{7} \times 24 \times 46 = 3469.71 \text{m}^2 \)
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