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NCERT Solutions for Class 10 Maths exercise 13.5- The region covered by its six rectangular faces is called a cuboid. The sum of the areas of a cuboid's six rectangular faces equals its surface area. The area of all sides apart from the top and bottom faces is known as lateral surface area (LSA). The space occupied by a cuboid's six rectangular faces is known as its volume.
The formula for cuboid with length l, breadth b, and height h :
Total surface area
Lateral surface area
Volume
NCERT solutions Class 10 Maths exercise 13.5- A cylinder is a solid shape with two circular bases that are joined together by a lateral surface. As a result, a cylinder has three faces: two circular and one lateral.
Formulas of cylinder having radius r and height h:
Curved surface area
Total surface area of Cylinder
Volume of cylinder
For find the area of complex solid: Complex figure areas can be disassembled and analysed as simpler known shapes. We can determine the required area of the unknown figure by calculating the areas of these known shapes.
Along with NCERT book Class 10 Maths chapter 13 exercise 13.5 the following exercises are also present.
A number of rounds are calculated by :
Thus the length of wire in 40 rounds will be
And the volume of wire is: Area of cross-section Length of wire
Hence the mass of wire is.
The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.
(Note that sum of heights of both the cone is 5 cm - hypotenuse).
Now the surface area of a double cone is :
The total volume of the cistern is :
And the volume to be filled in it is
Now let the number of bricks be n.
Then the volume of bricks :
Further, it is given that brick absorbs one-seventeenth of its own volume of water.
Thus water absorbed :
Hence we write :
Thus the total number of bricks is 1792.
Firstly we will calculate the volume of rainfall :
Volume of rainfall :
And the volume of the three rivers is :
It can be seen that both volumes are approximately equal to each other.
From this, we can write the values of both the radius (upper and lower) and the height of the frustum.
Thus slant height of frustum is :
Now, the area of the tin shed required :
= Area of frustum + Area of the cylinder
In the case of the frustum, we can consider:- removing a smaller cone (upper part) from a larger cone.
So the CSA of frustum becomes:- CSA of bigger cone - CSA of the smaller cone
And the total surface area of the frustum is = CSA of frustum + Area of upper circle and area of lower circle.
Similar to how we find the surface area of the frustum.
The volume of the frustum is given by-
=Volume of the bigger cone - Volume of the smaller cone
Surface area and volume of Frustum of a cone: When a solid is cut with a plane, a new type of solid is created. The frustum of a cone, which is formed when a plane cuts a cone parallel to the base, is an example of such a solid. A Frustum is the part of a right circular cone with two circular bases that are sliced by a plane parallel to its base.
Geometry includes a wide range of shapes and sizes, including the sphere, cube, cuboid, cone, cylinder, and so on. There is a volume and a surface area for each shape.
Also Read| Surface Areas and Volumes Class 10 Notes
Conversion of solid from one shape to another in exercise 13.5 Class 10 Maths teaches us about shape conversion due to melting and recasting by posing mischievous difficulties.
NCERT syllabus Class 10 Maths chapter 13 exercise 13.5 also covers the important concepts of exercise 13.2. and the previous exercise.
Class 10 Maths chapter 13 exercise 13.5 also contains concepts related to curved surface area and volume
Also, See:
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This exercise deals with surface area, volume of different types of solids, combinations of solids. Definition of frustum, volume and its surface area.
We need to simply divide the complex shape into simpler ones like cube cones etc. then we need to either add their volume or subtract in order to get the desired shape.
The volume of the object remains unchanged.
The volume of the object remains unchanged.
The volume of the object remains unchanged.
The volume of the object remains unchanged.
Hello
The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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