- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
About
Simple and Easy Learning
Tutorials Point originated from the idea that there exists a class of readers who respond better to online content and prefer to learn new skills at their own pace from the comforts of their drawing rooms.
The journey commenced with a single tutorial on HTML in 2006 and elated by the response it generated, we worked our way to adding fresh tutorials to our repository which now proudly flaunts a wealth of tutorials and allied articles on topics ranging from programming languages to web designing to academics and much more.
40 million readers read 100 million pages every month
Our Text Library Content and resources are freely available and we prefer to keep it that way to encourage our readers acquire as many skills as they would like to. We don't force our readers to sign up with us or submit their details either to use our Free Text Tutorials Library. No preconditions and no impediments, Just Simply Easy Learning!
We have established a Digital Content Marketplace to sell Video Courses and eBooks at a very nominal cost. You will have to register with us to avail these premium services.
Tutorialspoint
52 Views
$(a)$. Here, $r=14\ mm$Area of the circle$=\pi r^2$$=\pi\times14\times14$$=\frac{22}{7}\times14\times14$$=616\ mm^2$$(b)$. Here, diameter$=49\ m$$r=\frac{49}{2}\ m$Area$=\pi r^2$$=\frac{22}{7}\times\frac{49}{2}\times\frac{49}{2}$ $=\frac{3773}{2}\ m^2$$=1886.5\ m^2$ $(c)$. Here, radius$=5\ cm$Area$=\pi r^2$$=\frac{22}{7}\times5\times5$$=\frac{550}{7}\ m^2$$=78.57\ cm^2$
Tutorialspoint
39 Views
Given:The circumference of a circular sheet is \( 154 \mathrm{~m} \).To do:We have to find the radius and area of the sheet.Solution:Let the radius of the sheet be $r$.This implies, $2 \pi r=154\ m$$2\times\frac{22}{7}\times r=154$$r=\frac{7\times7}{2}$$r=\frac{49}{2}\ m$Area of the sheet$= \pi r^2$$=\frac{22}{7}\times\frac{49}{2}\times\frac{49}{2}$$=\frac{11\times7\times49}{2}$$=\frac{3773}{2}$$=1886.5\ m^2$The radius of the sheet is $24.5\ m$ and ... Read More
Tutorialspoint
43 Views
Diameter of the circular garden$=21\ m$Radius$=\frac{21}{2}\ m$ Circumference$=2\pi r$$=2\times\frac{22}{7}\times\frac{21}{2}$$=66\ m$Length of rope needed for 2 rounds$=2\times66\ m=132\ m$Cost of the rope$=Rs\ 4\times132=Rs\ 528$
Tutorialspoint
31 Views
Given:Radius of circular sheet$=4\ cm$.Radius of the circle removed$=3\ cm$.To do:We have to find the area of the remaining sheet.Solution:The area of the circular sheet with $4\ cm$ radius $=\pi (4)^2=16 \pi$The area of the circular sheet with $3\ cm$ radius $=\pi (3)^2=9 \pi$Area of the remaining sheet$=$Area of the ... Read More
Tutorialspoint
90 Views
From the question it is given that, Diameter of the circular table $= 1.5\ m$We know that, radius $(r)=\frac{d}{2}$$=\frac{1.5}{2}$$=0.75\ m$Then, the Circumference of the circle$=2\pi r$$=2\times 3.14\times 0.75$$=4.71\ m$So, the length of lace$=4.71\ m$Cost of $1\ m$ lace$=Rs.\ 15$ [given]Cost of $4.71\ m$ lace $=Rs.\ 15\times 4.71$$=Rs.\ 70.65$Read More
"
Tutorialspoint
57 Views
Diameter$=10\ cm$$r=\frac{10}{2}\ cm$$=5\ cm$The perimeter of a semicircle$=\pi r$$=\frac{110}{7}cm+diameter$$=\frac{110}{7}\ cm+10\ cm$ $=(\frac{110+70}{7})\ cm$$=\frac{180}{7}\ cm$$=25.71\ cm$ Hence the required perometer$=25.7\ cm\ (approx.)$
"
Tutorialspoint
53 Views
Given: ∆ABC is right angled at $A$ $(Fig\ 11.25)$. $AD$ is perpendicular to $BC$. If $AB = 5\ cm, \ BC = 13\ cm$ and $AC = 12\ cm$.To do: To find the area of $∆ABC$ and also find the length of $AD$.Solution:Area of right triangle $ABC=\frac{1}{2}\times AB\times AC$$=\frac{1}{2}\times5\times12$$=30\ cm^2$ Area of $\triangle ... Read More
"
Tutorialspoint
64 Views
To do: To find the area of each of the given parallelograms.Solution:$(a)$. Area of the parallelogram$=base\times altitude$$=7\ cm\times4\ cm$$=28\ cm^2$$(b)$. Area of the parallelogram$=base\times altitude$$=5\ cm\times3\ cm$$=15\ cm^2$$(c)$. Area of the parallelogram$=base\times altitude$$=2.5\ cm\times3.5\ cm$ $=8.75\ cm^2$$(d)$. Area of the parallelogram$=base\times altitude$$=5\ cm\times4.8\ cm$$=24\ cm^2$ $(e)$. Area of the parallelogram$=base\times altitude$$=2\ cm\times4.4\ cm$$=8.8\ ... Read More
Find the missing values:
| Base | Height | Area of Triangle |
| $15\ cm$ | $87\ cm^2$ | |
| $31.4\ mm$ | $1256\ mm^2$ | |
| $22\ cm$ | $170.5\ cm^2$ |
Tutorialspoint
56 Views
$(a)$. Area of the triangle $= 87\ cm^2$Base $= 15\ cm$Area of a triangle $= \frac{1}{2}\times Base\times Height$$87\ cm^2 = \frac{1}{2}\times 15\ cm\times Height$Height $= \frac{87\ cm^2\times 2}{15\ cm}$Height $= \frac{174\ cm^2}{15\ cm}$Height $= 11.6\ cm$$(b)$. Area of the triangle $= 1256\ mm^2$Height $= 31.4\ mm$Area of triangle $= \frac{1}{2}\times ... Read More
"
Tutorialspoint
40 Views
To do: To find the are of each of the given triangles.Solution: $(a)$. Area of the triangle$=\frac{1}{2}\times b\times h$ $=\frac{1}{2}\times4\ cm\times3\ cm$ $=6\ cm^2$$(b)$. Area of the triangle$=\frac{1}{2}\times b\times h$$\frac{1}{2}\times5\ cm\times3.2\ cm$$=8.0\ cm^2$$(c)$. Area of the triangle$=\frac{1}{2}\times b\times h$$=\frac{1}{2}\times3\ cm\times4\ cm$$=6\ cm^2$$(d)$ Area of the triangle$=\frac{1}{2}\times b\times h$$=\frac{1}{2}\times3\ cm\times2\ cm$$=3\ cm^2$Read More