Exercise 7.1 Surface Area and Volume of Triangular Prisms
The above-mentioned topic is mentioned in the vedanta Excel in Mathematics
book in Unit 7 - the third part of Mensuration where we deal with the surface
area and volumes of Triangular Prisms, Pyramids and Cones, along with basic
Household Mensuration exercises.
Here is a complete solution of exercise 7.2 from vedanta's book for Class 10
students. All solutions are accurate but there might be bad handwriting or
extra crossing in many solutions. In case you do not understand the
handwriting, numerical or language, do let us know in the comment section so
that we can make it clear to you.
We have almost all solutions from vedanta Excel in Mathematics Book 10 for
Grade 10 students. We also have notes and other unit tests in it. If you want
to see it, visit here:
Get solutions of other exercises from this unit by visiting the links
mentioned just below:
- Ex 7.1 - Surface Area and Volume of Triangular Prisms Class 10 Solutions | vedanta Excel in Mathematics
- Exercise 7.2 Surface Area and Volume of Pyramids Class 10 Solutions | vedanta Excel in Mathematics
- Exercise 7.3 Surface Area and Volume of Cones Class 10 Solutions | vedanta Excel in Mathematics
- Ex 7.4 - Mensuration in Household Activities Class 10 Solutions | vedanta Excel in Mathematics
Now, let us get into the solutions of Exercise 7.1 Surface Area and Volume
of Triangular Prisms Class 10.
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About vedanta EXCEL in MATHEMATICS Book 10
Author: Hukum Pd. Dahal
Editor: Tara Bahadur Magar
Vanasthali, Kathmandu, Nepal
+977-10-4382404, 01-4362082
vedantapublication@gmail.com
About this page:
Ex 7.1 - Surface Area and Volume of Triangular Prisms Class 10 Solutions
| vedanta Excel in Mathematics is a collection of the solutions related
to calculation of surface area and volume of triangular prisms from
Mensuration (III): Prisms and Pyramids chapter for Nepal's Secondary
Education Examination (SEE) appearing students.
#Class10
4 Comments
k ho yr base in right tespaxi
ReplyDeleteCould you explain?
DeleteIn 3 c how is h=14.14???
ReplyDeleteh is the hypotenuse. From Pythgaoras theore, h^2 = p^2 + b^2 => h = $\rm{\sqrt{p^2 + b^2}}$
DeleteYou can let us know your questions in the comments section as well.