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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.
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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 :
$=\ \frac{Height\ of\ cylinder }{Diameter\ of\ wire}$
$=\ \frac{12 }{0.3}\ =\ 40\ rounds$
Thus the length of wire in 40 rounds will be $=\ 40\times 2\pi \times 5\ =\ 400 \pi\ cm\ =\ 12.57\ m$
And the volume of wire is: Area of cross-section $\times$ Length of wire
$=\pi \times (0.15)^2\times 1257.14$
$=\ 88.89\ cm^3$
Hence the mass of wire is. $=\ 88.89\ \times 8.88\ =\ 789.41\ gm$
The volume of the double cone will be = Volume of cone 1 + Volume of cone 2.
$=\ \frac{1}{3}\ \pi r^2h_1\ +\ \frac{1}{3}\ \pi r^2h_2$
$=\ \frac{1}{3}\ \pi \times 2.4^2\times 5$ (Note that sum of heights of both the cone is 5 cm - hypotenuse).
$=\ 30.14\ cm^3$
Now the surface area of a double cone is :
$=\ \pi rl_1\ +\ \pi rl_2$
$=\ \pi \times 2.4 \left ( 4\ +\ 3 \right )$
$=\ 52.8\ cm^2$
The total volume of the cistern is : $=\ 150\times 120\times 110\ =\ 1980000\ cm^3$
And the volume to be filled in it is $= 1980000 - 129600\ =\ 1850400\ cm^3$
Now let the number of bricks be n.
Then the volume of bricks : $= n\times 22.5\times 7.5\times 6.5 \ =\ 1096.87n\ cm^3$
Further, it is given that brick absorbs one-seventeenth of its own volume of water.
Thus water absorbed :
$=\ \frac{1}{17}\times 1096.87n\ cm^3$
Hence we write :
$1850400\ +\ \frac{1}{17}( 1096.87n)\ =\ 1096.87n$
$n\ =\ 1792.41$
Thus the total number of bricks is 1792.
Firstly we will calculate the volume of rainfall :
Volume of rainfall :
$=\ 7280\times \frac{10}{100\times 1000}$
$=\ 0.7280\ Km ^3$
And the volume of the three rivers is :
$=\ 3\left ( 1072\times \frac{75}{1000}\times \frac{3}{1000} \right )\ Km^3$
$=\ 0.7236\ Km ^3$
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 :
$=\ \sqrt{\left ( r_1\ -\ r_2 \right )^2\ +\ h^2}$
$=\ \sqrt{\left ( 9\ -\ 4 \right )^2\ +\ 12^2}$
$=\ 13\ cm$
Now, the area of the tin shed required :
= Area of frustum + Area of the cylinder
$=\ \pi \left ( r_1\ +\ r_2 \right )l\ +\ 2\pi r_2h$
$=\ \pi \left ( 9\ +\ 4 \right )13\ +\ 2\pi \times 4\times 10$
$=\ 782.57\ cm^2$
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
$\\=\ \pi r_1l_1\ -\ \pi r_2 (l_1\ -\ l)\\\\=\ \pi r_1\left ( \frac{lr_1}{r_1\ -\ r_2} \right )\ -\ \pi r_2\left ( \frac{r_1l}{r_1\ -\ r_2}\ -\ l \right )\\\\=\ \left ( \frac{\pi r_1^2 l}{r_1\ -\ r_2} \right )\ -\ \left ( \frac{\pi r_2^2 l}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( \frac{r_1^2\ -\ r_2^2}{r_1\ -\ r_2} \right )\\\\=\ \pi l\left ( r_1\ +\ r_2 \right )$
And the total surface area of the frustum is = CSA of frustum + Area of upper circle and area of lower circle.
$=\ \pi l\left ( r_1\ +\ r_2 \right )\ +\ \pi r_1^2\ +\ \pi r_2^2$
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
$=\ \frac{1}{3} \pi r^2_1 h_1\ -\ \frac{1}{3} \pi r^2_2 (h_1\ -\ h)$
$=\ \frac{\pi}{3}\left ( r_1^2 h_1\ -\ r_2^2 (h_1\ -\ h) \right )$
$=\ \frac{\pi}{3}\left ( r_1^2 \left ( \frac{hr_1}{r_1\ -\ r_2} \right )\ -\ r_2^2 \left ( \frac{hr_1}{r_1\ -\ r_2}\ -\ h \right ) \right )$
$=\ \frac{\pi}{3}\left ( \frac{hr_1^3}{r_1\ -\ r_2}\ -\ \frac{hr_2^3}{r_1\ -\ r_2} \right )$
$=\ \frac{\pi}{3}h\left ( \frac{r_1^3\ -\ r_2^3}{r_1\ -\ r_2} \right )$
$=\ \frac{1}{3} \pi h\left ( r_1^2\ +\ r_2^2\ +\ r_1r_2 \right )$
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
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On Question asked by student community
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.
Hello Aspirant,
Here's how you can find it:
School ID Card: Your registration number is often printed on your school ID card.
Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.
School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.
Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.
Your school is the best place to get this information.
Hello,
It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.
The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.
For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.
Let me know if you need imformation about any exam eligibility criteria.
good wishes for your future!!
Hello Aspirant,
"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).
For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.
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