Introduction to Perimeter and Areas of Plane Figures Solutions Class 10 Chapter 17 from Vedantu
FAQs on RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures
1. From Where Can I Download RS Aggarwal Solutions Class 10 Chapter 17?
Students can download the RS Aggarwal solutions from the Vedantu website. RS Aggarwal Solutions are created for the students to help easily solve their problems. By downloading the RS Aggarwal solutions students can get help in solving the questions easily. By downloading RS Aggarwal solutions you can solve all the similar questions. Make sure to download the solutions while you are preparing for the exam to get an overview of how you can solve the questions in your exams. Make sure to download the solutions while you are preparing for the exam to get an idea of how you can solve the questions in your exams.
2. What Topics are Covered in Mathematics Class 10th Chapter 17th?
Mathematics class 10th chapter 17 is about Volume and surface area of solids. Under this chapter various topics like volume and surface area of cuboid, volume and surface area of a cube, conversion of solids from one shape to another and other mixed problems, volume and surface area of a frustum of a cone, derivation of volume and surface area of a frustum of a cone, volume and surface area of sphere and hemisphere, volume and surface area of a combination of solids. You can master all these topics by downloading RS Aggarwal class 10 ch 17 solutions. RS Aggarwal solutions will help you get your concepts clear and you can answer any type of question.
3. How to prepare RS Aggarwal Class 10 Solutions Chapter 17 Perimeter and Areas of Plane Figures?
Chapter 17 is a crucial chapter taught in grade 10. Every student has to study this specific chapter with care to secure good marks. Now, you might be wondering how to prepare well. Read and practice every chapter thoroughly. Make sure to cover everything to strengthen your base and be able to approach any question with confidence. Try to solve all the questions provided in RS Aggarwal. Prepare revision notes for all the formulas and comprehend the derivations. Apart from this, you must not forget to concentrate on being clear on the formulas and concepts of each chapter.
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5. How to resolve the question to explore the hypotenuse length of an isosceles right-angled triangle which owns an area of 200 cm2. And, find the given perimeter 2–√ = 1.41?
In the right isosceles triangle, the base=height=a
So,
triangle area is= 12×base×height=12×a×a=12a2
Moreover, provided that isosceles area right triangle = 200 cm2
⇒12a2=200⇒a2=400or, a=400−−−√ =20 cm
In the isosceles right triangle, both sides are equal ('a') as well as the third side remains the hypotenuse, i.e. 'c'
Thus, c = a2+a2−−−−−−√
= 2a2−−−√= a2–√= 20×1.41= 28.2 cm
Triangle perimeter = a+a+c
=20+20+28.2
= 68.2 cm
Hypotenuse length is 28.2 cm as well as the perimeter of the triangle results as 68.2 cm.
