Table of Contents
Ans. (C) 660 cm^3 \\
Explanation:
volume of cone = (1/3) πr^2h = 220 cm^3 \\
\text{Volume of cylinder} = πr^2h = 3×220 cm^3 \\
(∵ 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 \\
\text{Area of square =} a^2 \\
= \left(\frac{\pi R}{2}\right )^2 \\
=\frac{\pi^2 R^2}{4} \\
\begin{bmatrix} \because \pi = 3.14 \space \space and \space \space 3.14 \lt 4 \\
\therefore \frac{3.14}{4}\lt 1\end{bmatrix} \\
π R^2> \frac{\pi}{4} π R^2 \\
Area of circle > Area of square
Explanation:
(i) Here,
r= \frac{4.0}{2} = 2m \space and \space h = 4⋅4 m \\
\text{Curved surface area =} 2πrh m^2 \\
=2× \frac{22}{7} ×2×4⋅4 cm^2 \\
= 55.31 m^2 \\
(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 + 2πr^2) m^2 \\
=(55⋅31+2× \frac{22}{7} ×4) m^2 \\
= (55·31 + 25·14) m^2 \\
= 80·45 m^2 \\
∴ \frac{11}{12}x =80.45 \\
⇒ x = 87·76 m^2 \\
Hence, the actual area of the steel used = 87·76 m^2.
Explanation:
Cylindrical area=2πrh
= 2 × \frac{22}{7} × \frac{105}{2} ×3 m^2 \\
= and conical area = πrl
=\frac{22}{7} × \frac{105}{2} × 53 m^2
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=πr^2h + \frac{1}{3} πr^2H \\
= πr^2 \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 m^3 \\
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} m^2
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})m^2\\
= \frac{220}{7} (80+10 \sqrt{5})m^2\\
\text{Total Surface Area =} 3217·04 m^2
Download Mind Map of this chapter
Download NowWant to Practice Mock Tests of this chapter
Practice NowDownload Important Questions of this chapter
Download NowChapter No. | Chapter Name |
---|---|
Chapter 1 | Goods and Service Tax (GST) |
Chapter 2 | Banking |
Chapter 3 | Shares and Dividends |
Chapter 4 | Linear inequations |
Chapter 5 | Quadratic Equations in one variable |
Chapter 6 | Ratio and proportion |
Chapter 7 | Factorization |
Chapter 8 | Matrices |
Chapter 9 | Arithmetic Progression |
Chapter 10 | Geometric Progression |
Chapter 11 | Coordinate Geometry |
Chapter 12 | Reflection |
Chapter 13 | Similarity |
Chapter 14 | Loci |
Chapter 15 | Circles |
Chapter 16 | Constructions |
Chapter 17 | Mensuration |
Chapter 18 | Trigonometry |
Chapter 19 | Statistics |
Chapter 20 | Probability |
Chapter Wise Important Questions for ICSE Board Class 10 Mathematics |
---|
Goods and Service Tax (GST) |
Banking |
Shares and Dividends |
Linear inequations |
Quadratic Equations in one variable |
Ratio and proportion |
Factorization |
Matrices |
Arithmetic Progression |
Geometric Progression |
Coordinate Geometry |
Reflection |
Similarity |
Loci |
Circles |
Constructions |
Mensuration |
Trigonometry |
Statistics |
Probability |
CBSE Important Questions Class 9
CBSE Important Questions Class 10
CBSE Important Questions Class 12
CBSE Practice Papers
CBSE Practice Papers
ICSE Important Questions Class 9
ICSE Important Questions Class 10
ICSE Practice Papers
ISC Important Questions Class 12
ISC Practice Papers
Contact Us