Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid
⇔ The solids with rectangular surfaces of which the opposite surfaces are of equal measure and adjacent surfaces are perpendicular to each other, are called Rectangular parallelopiped or cuboid.
- length = AB unit
- breadth = BC unit
- height = CE unit
- diagonal = AE unit
⇔ A rectangular parallelopiped has 8 vertices, 12 edges, 6 faces and 4 diagonals.
⇔ If each of the edges of a cuboid is of equal length, then the cuboid is called a cube.
Read and Learn More WBBSE Solutions for Class 10 Maths
⇔ If the length, breadth and height of a rectangular parallelopiped (or cuboid) be a unit, b unit, and c unit respectively then
- Volume = Area of the base x height = abc cubic unit
- The length of the diagonal = \(\sqrt{(\text { length })^2+(\text { breadth })^2+(\text { height })^2}=\sqrt{a^2+b^2+c^2}\)
- Area of total surface are a = 2 (ab + bc + ca) square unit
For cube:
- Volume = a3 cubic unit (Let each edge be a unit)
- Total surface area = 6a2 sq. unit.
- Length of each diagonal = √3a unit
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Fill In The Blanks
Example 1. Dimension of a parallelopiped is ______
Solution: 3
Example 2. No. of the surface of a cube is _____
Solution: 6
Class 10 Maths Mensuration Chapter 1 Solutions
Example 3. No. of vertices of a cuboid is ______
Solution: 8.
Example 4. If the length, breadth and height of a parallelopiped are equal then the parallelopiped is called a ______
Solution: cube.
Example 5. If side surface area of a cube is 256 sq. cm then length of each edge is _______ cm.
Solution: 8 cm.
Example 6. Ratio of the length of the diagonal of each surface of a cube and the length of the diagonal of that cube is _______
Solution: √2: √3
Example 7. The no. of diagonals of a cuboid is _____
Solution: 4
Example 8. The length of the diagonal on the surface of a cube = ________ x the length of one edge.
Solution: √2
Example 9. The number of the vertices, edges and plane surfaces of a cuboid be x, y, z respectively then x + y + z = _______
Solution: 26
Class 10 Maths Mensuration Chapter 1 Solutions
Example 10. If the size of a room is 8m x 6m x 5m, then the length of the largest straight rod that can be kept inside the room is _______
Solution: 5√5 m.
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid True Or False
Example 1. If the length of each edge of a cube is twice of that 1st cube then the volume of this cube is 4 times more than that of the 1st cube.
Solution: False
Example 2. In the rainy season, the height of rainfall in 2-meter land is 5 cm, and the volume of rainwater is 1000 cubic meters.
Solution: True.
Example 3. The total surface area of a cuboid = (length x breadth + breadth x height + height x length) sq. unit.
Solution: False.
Example 4. No. of side faces of a cuboid is 6.
Solution: False
Surface Areas And Volumes Class 10 Solutions
Example 5. If the length of a edge of a cuboid is increased by a% then its length of the diagonal is also increased by a%.
Solution: True
Example 6. If two cubes are joined side by side, then it will be again a cube.
Solution: False.
Example 7. Sum of the edges of a cuboid = 4 (length + breadth + height).
Solution: True.
Example 8. Sum of the edges of cube = 12 x length of its one edge.
Solution: True.
Example 9. The intersecting point of the edges of a parallelopiped is called its vertex.
Solution: True.
Example 10. All cuboids are cubes.
Solution: False.
Class 10 Mensuration Chapter 1 Solved Examples
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Short Answer Type Questions
Example 1. If the number of surfaces of a cuboid is x, the number of edges is y, the number of vertices is z and the number of diagonals is p, then find the value of x – y + z + p.
Solution: x = 6, y – 12, z = 8, p = 4
∴ x – y + z + p = 6 – 12 + 8 + 4 = 6
The value of x – y + z = 6
Example 2. The length of dimensions of two cuboids are 4, 6, 4 units and 8, (2h – 1), 2 units respectively. If volumes of two cubes are equal find the value of h.
Solution: 4 x 6 x 4 = 8x (2h – 1) x 4
⇒ or, 2h – 1 = 6 or, h = 3.5
The value of h = 3.5
Example 3. If each edge of a cube is increased by 50% then how much the total surface area of the cube will be increased in percent?
Solution: Let each edge = a unit, total surface area = 6a2 sq. u
⇒ Increased edge = a x \(\frac{150}{100}\) unit = \(\frac{3}{2}\) a
⇒ Increased total surface area = \(6\left(\frac{3}{2} a\right)^2\)
= \(6 \times \frac{9}{4} a^2 \text { sq. } \mathrm{u}=\frac{27}{2} a^2 \text { sq. } u\)
⇒ increased % = \(\frac{\left(\frac{27}{2}-6\right) a^2}{6 a^2} \times 100 \text { sq. u }\)
= \(\frac{15}{2 \times 6} \times 100 \text { sq. } u=125 \text { sq. u }\)
∴ Total surface area is increased by 125%.
Wbbse Class 10 Mensuration Notes
Example 4. The length of two adjacent walls of a room are 12 m, and 8 m respectively. If the height of the room is 4 m, then calculate the area of the floor.
Solution: Area of the floor = length x breadth
= 12 x 8 sq. m = 96 sq. m
Example 5. The lengths of each edge of three solid cubes are 3 cm, 4 cm, and 5 cm respectively. A new solid is made by melting these three solid cubes. Write the length of each edge of the new cube.
Solution: Let the length of each edge of the new cube = a cm
∴ a3 = (33 + 43 + 53) or, a3 = 216
∴ a = 6 cm.
Example 6. No. of dimensions, vertices, edges, and surfaces of a parallelopiped are A, B, C, and D respectively, then find the value of \(\frac{2{A}+B+C-D}{B}\).
Solution: A = 3, B = 8, C = 12, D = 6
\(\frac{2 \mathrm{~A}+\mathrm{B}+\mathrm{C}-\mathrm{D}}{\mathrm{B}}=\frac{2 \times 3+8+12-6}{8}=\frac{5}{2}\)Example 7. Volume, total surface area, and length of the diagonal of a cube are V, Z, and Z respectively, then find the value of \(\frac{AZ}{V}\).
Solution: V = a3, total surface area = 6a2, length of the diagonal = a√3
∴ \(\frac{\mathrm{AZ}}{\mathrm{V}}=\frac{6 a^2 \cdot a \sqrt{3}}{a^3}=6 \sqrt{3}\)
Example 8. Find the no. of cubes of edge 2 cm made from a cube of edge 1 m.
Solution: No. of cube made = \(\frac{(1 \mathrm{~m})^3}{(2 \mathrm{~cm})^3}=\frac{1000000}{2 \times 2 \times 2}=1,25,000\)
Surface Area And Volume Formulas Class 10
Example 9. How many boxes of dimensions 5 cm x 4 cm x 2 cm are to be kept in the box of dimensions 40 cm x 25 cm x 15 cm?
Solution: No. of boxes = \(\frac{40 \times 25 \times 15}{5 \times 4 \times 2}\) = 375
Example 10. If numerical values of total surface area and volume of a cube are equal the find the length of the diagonal.
Solution: 6a2 = a3 ∴ a = 6
length of the diagonal = 6√3 cm.