In ICSE Class 10 Mathematics, Chapter 17 focuses on "Mensuration." This chapter deals with the measurement of geometric figures, such as areas and volumes. It is an essential part of mathematics as it helps students understand and calculate the sizes and capacities of various shapes and objects.
In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, the chapter on "Mensuration" is a fundamental topic that deals with the measurement of geometric figures, including calculating their areas and volumes. Mensuration is a practical and essential part of mathematics with numerous real-world applications. Mensuration is not just a mathematical concept; ICSE class 10 maths mensuration is a practical tool that helps us quantify and measure objects in our everyday lives. Whether you're designing a building, calculating the area of a land plot, or estimating the volume of a container, mensuration plays a vital role in solving real-world problems accurately. Here are some area and volume class 10 ICSE questions.
Ans: Mensuration is the branch of mathematics that focuses on measuring the sizes, areas, and volumes of various geometric shapes and figures. ICSE class 10 maths mensuration provides us with the tools and formulas to quantify and understand the spatial properties of objects in both two and three dimensions.
Key Concepts and Objectives:
Area and Perimeter: Mensuration helps us calculate the area (space inside a shape) and perimeter (the boundary length) of 2D figures such as rectangles, squares, circles, and triangles.
Volume: It allows us to determine the volume (space occupied) of 3D figures like cubes, cuboids, cylinders, cones, and spheres.
Formulas: Mensuration provides us with specific formulas and methods for each type of shape, making it possible to compute measurements accurately.
Ans. (C) 660 cm3
Explanation:
volume of cone = (1/3) πr2h = 220 cm3
Volume of cylinder = πr2h = 3×220 cm3 (∵ radius and height of cylinder are as same as that of cone)
Ans. (b)
Explanation: Circumference of circle = 2πR
Perimeter of square = 4a
According to question
2πR = 4a
⇒ πR = 2a
⇒ \(\frac{\pi R}{2}\)= a
\(Area of square = a^2 = \left(\frac{\pi R}{2}\right )^2 \) =\(\frac{\pi^2 R^2}{4}\)
\(\begin{bmatrix} \because \pi = 3.14 and 3.14 \lt 4 \\ \therefore \frac{3.14}{4}\lt 1\end{bmatrix}\)
π R2>\(\frac{\pi}{4}\)π R2
Area of circle > Area of square
Explanation:
(i) Here,
r=\(\frac{4.0}{2}\) = 2m and h = 4⋅4 m
Curved surface area = 2πrh m2
=2×\(\frac{22}{7}\)×2×4⋅4 cm2
= 55.31 m2
(ii) Since \(\frac{1}{2}\) of the actual steel used was wasted, the area of the steel which has gone into the
tank =\(\begin{pmatrix}1- \frac{1}{12}\end{pmatrix}\)=\(\frac{11}{12}\)of x, where
x = total area of steel used.Steel used = (2πrh + 2πr2) m2
=(55⋅31+2×\(\frac{22}{7}\)×4) m2
= (55·31 + 25·14) m2
= 80·45 m2
∴ \(\frac{11}{12}\)x =80.45
⇒ x = 87·76 m2
Hence, the actual area of the steel used = 87·76 m2.
Explanation:
Cylindrical area=2πrh
= 2 ×\(\frac{22}{7}\)×\(\frac{105}{2}\)×3 m2
=and conical area = πrl
=\(\frac{22}{7}\)×\(\frac{105}{2}\)× 53 m2
=2×\(\frac{22}{7}\)×\(\frac{105}{2}\)×3+\(\frac{22}{7}\)×\(\frac{22}{7}\)×\(\frac{105}{2}\)×53
=15×11(2×3+53)
=15×11×59
=165×59 m2
Length of canvas =\(\frac{165×59}{5}\)m
=33×59 m
=1947 m.
Explanation:
Height of the tent = Height of cone + Height of the cylinder
Given, height of tent = 60 m
H = height of cone = 20 m
∴ Height of cylinder
= h = 60 – 20 = 40 m
and Radius of cone = Radius of cylinder
r = 10 m
∴ Volume of the tent = Volume of cylinder+ Volume of the cone
=πr2h + \(\frac{1}{3}\) πr2H
= πr2 \(\begin{pmatrix}h+ \frac{H}{3}\end{pmatrix}\)
= π(10)2 \(\begin{pmatrix} 40+ \frac{20}{3}\end{pmatrix}\)
=100 ×\(\frac{22}{7} \begin{pmatrix}\frac{140}{3}\end{pmatrix}\)
= 14666·6 m3
Slant height of the cone is
I=\(\sqrt{H^2+r^2}\)
=\(\sqrt{400+100}\)
=\(\sqrt{500}\)=\(\sqrt[10]{5}\)m
Since, curved surface area of cone
= πrl
=\(\frac{22}{7}\)×10×10\(\sqrt{5}\)m2
and curved surface area of cylinder
= 2πrh
=2×\(\frac{22}{7}\)×10×40
∴ Total surface area of the canvas in making the tent
= C.S.A. of cylinder + C.S.A. of cone
= 2πrh + πrl
= πr (2h + l)
=\(\frac{22}{7}\)×10(2×40+10\(\sqrt{5}\))m2
=\(\frac{220}{7}\) (80+10 \(\sqrt{5}\))m2
Total Surface Area = 3217·04 m2
In conclusion, the chapter on ICSE class 10 maths mensuration is an integral part of the curriculum, offering students the essential knowledge and skills to measure and calculate areas and volumes of geometric shapes. These concepts extend far beyond the classroom, finding application in various real-world scenarios such as construction, design, and everyday problem-solving. Mastery of mensuration equips students with a valuable toolkit for understanding and quantifying the spatial attributes of objects, laying the foundation for more advanced mathematical concepts and practical life applications.
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Ans: Mensuration in mathematics is the study of measuring and calculating the areas and volumes of geometric shapes and figures.
Ans: Mensuration is crucial because it helps us quantify and understand the sizes of objects in real-world situations, including construction, design, and everyday problem-solving.
Ans: Common 2D shapes include rectangles, squares, circles, triangles, and parallelograms.
Ans: The area of a rectangle is calculated by multiplying its length and width (A = length × width).
Ans: The perimeter of a square is four times the length of one of its sides (P = 4 × side length).