NCERT MCQ Solutions for Class 9 Maths Chapter 11 Surface Areas and Volumes

The curved surface area of a cone is derived from unfolding its surface and finding the area of the resulting sector. The formula is πrl, where r is the base radius and l is the slant height. This only accounts for the curved part without including the base.

The total surface area of a cone includes both the curved surface area and the circular base area.
The curved surface area is πrl and the base area is πr².
So, the total surface area is πrl + πr² which simplifies to πr(l + r) when factored.

The curved surface area of a cone is πrl.
Substituting π = 22/7, r = 7 cm and l = 10 cm, we get πrl = (22/7) × 7 × 10 = 220 cm².
The 7 in the denominator and numerator cancel out for easy calculation.

A cone’s volume is one-third that of a cylinder with the same base and height.
The formula for a cylinder is πr²h.
So, the cone’s volume becomes (1/3)πr²h, where r is the base radius and h is the perpendicular height of the cone.

The slant height of a cone is the hypotenuse of the right triangle formed by the radius and height.
Using the Pythagoras theorem, l = √(r² + h²), where r is the base radius and h is the vertical height from base to vertex.

The surface area of a cuboid is calculated by summing up the areas of all six rectangular faces. A cuboid has three pairs of opposite faces, and the area of each pair is as follows:
1. Two faces with area lb (length × breadth),
2. Two faces with area bh (breadth × height),
3. Two faces with area lh (length × height).
Thus, the total surface area is given by:
Surface Area = 2(lb) + 2(bh) + 2(lh)
Factoring out the common factor of 2, we get:
Surface Area = 2(lb + bh + lh)

The perimeter of the floor of the rectangular hall is given as 250m. This means:
Perimeter = 2(l + b) = 250m,
where l is the length and b is the breadth of the floor.
From this, we can calculate:
l + b = 250 / 2 = 125m.
The lateral surface area (LSA) of the four walls is given by:
LSA = 2h(l + b),
where h is the height of the room.
Substituting l + b = 125 into the formula, we get:
LSA = 2h(125) = 250h.
The total cost of whitewashing is ₹15000
The cost per square meter is assumed to be ₹10 (as it is a standard rate in such problems unless specified otherwise). Thus:
Cost = LSA × Rate,
15000 = 250 h × 10.
Simplify to find h: 15000 = 2500 h,
h = 15000 / 2500 = 6 m.

The surface area of a sphere is the total area covering its curved surface.
It equals four times the area of a circle with the same radius.
So, the formula is 4πr², where r is the radius.
This was derived using experimental wrapping with string.

The volume of a sphere is derived by experimentally finding how much water it displaces when submerged.
The formula is (4/3)πr³, where r is the radius. This is supported by repeated experiments and accepted as true at this stage of learning.

A hemisphere is half of a sphere, so its curved surface area is half the total surface area of a sphere.
Since a sphere has surface area 4πr², the curved surface area of a hemisphere becomes 2πr², excluding the base circular face.

The total surface area of a hemisphere includes its curved surface area and the circular base.
The curved surface area is 2πr² and the base area is πr².
Therefore, total surface area = 2πr² + πr² = 3πr².
This includes all external faces.

A hemisphere is half of a sphere.
The volume of a sphere is (4/3)πr³.
So, volume of a hemisphere is (1/2) × (4/3)πr³ = (2/3)πr³.
This formula is applied for bowl-shaped objects or half-spherical tanks in practical problems.

Each cube has an edge length of 12 cm. When two such cubes are joined, they form a new cuboid. The dimensions of the new cuboid are as follows:
– Length (l) = 12 + 12 = 24 cm (since the two cubes are joined along their edges),
– Breadth (b) = 12 cm (same as the edge of one cube),
– Height (h) = 12 cm (same as the edge of one cube).
The surface area of a cuboid is given by:
Surface Area = 2(lb + bh + lh).
Substitute the values of l = 24 cm, b = 12 cm and h = 12 cm:
Surface Area = 2[(24 × 12) + (12 × 12) + (24 × 12)].
Calculate each term:
24 × 12 = 288,
12 × 12 = 144,
24 × 12 = 288.
Add these values:
288 + 144 + 288 = 720.
Surface Area = 2 × 720 = 1440 cm².

The curved surface area (CSA) of a cylinder is given by the formula:
CSA = 2πrh,
where:
r is the radius of the circular base,
h is the height of the cylinder.
From the problem, the CSA is 88 cm² and the height (h) is 14 cm. Substituting these values into the formula: 88 = 2πr(14).
88 = 28πr.
Divide both sides by 28π to isolate r:
r = 88 / (28π).
Using π ≈ 22/7 for calculation:
r = 88 / (28 × 22/7),
r = 88 / (4 × 22),
r = 88 / 88,
r = 1 cm.
The diameter (d) = 2r = 2 × 1 = 2 cm.

In a cone, the curved surface is calculated using the slant height l not the vertical height h.
The formula is πrl where r is the radius and l is the slant height.
This does not include the base which is only added in total surface area.

A right circular cone must have a circular base and the line joining vertex to base center must be perpendicular to the base.
If the base is not circular, it is not a right circular cone.
Only cones with circular bases and right angles qualify.

An ice cream cone resembles a cone because it has a pointed vertex and a circular base. It tapers from the wide circular base to a sharp tip and has a curved surface that wraps around. 

By filling a cone and emptying it into a cylinder with the same base and height, we see it takes three full cones to fill one cylinder.
This proves the volume of a cone is one-third that of a cylinder with the same base and height.

A cube is a prism because it is formed by stacking congruent square faces along a height.
Prisms are solids with identical cross-sections throughout their height. A cone or hemisphere has curved surfaces and does not maintain a uniform cross-section along height.

The inner curved surface area (CSA) of a cylindrical pipe is given by the formula:
CSA = 2πrh,
where:
r is the inner radius of the circular base,
h is the height (or length) of the cylinder.
The length (h) of the pipe is 77 cm,
The inner diameter of the cross-section is 4 cm, so the inner radius (r) is:
r = Diameter / 2 = 4 / 2 = 2 cm.
Substitute the values of r = 2 cm and h = 77 cm into the formula:
CSA = 2πrh.
Using π ≈ 22/7 for calculation:
CSA = 2 × (22/7) × 2 × 77.
CSA = 2 × (22/7) × 154,
CSA = 2 × 22 × 22,
CSA = 968 cm².

The roller is a cylinder, and the area it covers in one complete revolution is equal to its curved surface area (CSA). The formula for the CSA of a cylinder is:
CSA = 2πrh,
Where: r is the radius of the circular base,
h is the length (or height) of the cylinder.
The diameter of the roller is 84 cm, so the radius (r) = Diameter / 2 = 84 / 2 = 42 cm = 0.42 m (converted to meters),
The length (h) of the roller is 120 cm = 1.2 m (converted to meters).
Substitute the values of r = 0.42 m and h = 1.2 m into the formula:
CSA = 2πrh.
Using π ≈ 22/7 for calculation:
CSA = 2 × (22/7) × 0.42 × 1.2.
CSA = 2 × (22/7) × 0.504,
CSA = 2 × 22 × 0.072,
CSA = 3.168 m².
This is the area covered by the roller in one complete revolution. Since the roller takes 500 revolutions to level the playground, the total area of the playground is:
Total Area = CSA × Number of Revolutions,
Total Area = 3.168 × 500,
Total Area = 1584 m².

Using the Pythagoras theorem,
l = √(r² + h²)
= √(7² + 24²)
= √(49 + 576)
= √625 = 25 m.
This is used to compute the slant height which is essential in calculating curved surface area of the cone using πrl. 

A cone consists of a circular base and one curved surface that tapers to a point called the vertex. The only surface that wraps around is curved. There are no flat vertical faces or edges, making it different from polyhedra like cubes.

When a sphere is covered with string and the string is used to fill circles of the same radius, it covers exactly four circles. So, the surface area of a sphere is four times the area of a circle, which equals 4πr².

A sphere is a perfectly round three-dimensional object with only one curved surface.
It has no edges or vertices and does not have any flat surfaces.
This makes it unique compared to other solid figures which may have multiple surfaces or edges.

The total surface area of a cone combines the curved surface area and the base area.
The curved surface area is πrl and the base area is πr².
Therefore, the total surface area becomes πr(l + r) by combining both expressions using factorization.

The inner curved surface area (CSA) of a circular well (cylinder) is given by the formula:
CSA = 2πrh,
Where, r is the radius of the circular base,
h is the depth (or height) of the well.
The inner diameter of the well is 3.5 m, so the radius (r) is:
r = Diameter / 2 = 3.5 / 2 = 1.75 m,
The depth (h) of the well is 10 m.
Substitute the values of r = 1.75 m and h = 10 m into the formula:
CSA = 2πrh.
Using π ≈ 22/7 for calculation:
CSA = 2 × (22/7) × 1.75 × 10.
Simplify step by step:
CSA = 2 × (22/7) × 17.5,
CSA = 2 × 22 × 2.5,
CSA = 110 m².

The curved surface area (CSA) of a right circular cone is given by the formula:
CSA = πrl, where:
r is the radius of the base,
l is the slant height of the cone.
The radius (r) of the base is 7 cm,
The slant height (l) is 10 cm.
Substitute the values of r = 7 cm and l = 10 cm into the formula:
CSA = πrl.
Using π ≈ 22/7 for calculation:
CSA = (22/7) × 7 × 10.
Simplify step by step:
CSA = 22 × 10,
CSA = 220 cm².

Volume = (1/3)πr²h
= (1/3) × (22/7) × (3.5)² × 12
= (1/3) × (22/7) × 12.25 × 12
= 158.6 cm³ approximately.
This calculation uses radius squared and height in the formula for the volume of a cone.

Volume measures the three-dimensional space occupied by a solid and is expressed in cubic units. The unit cm³ (cubic centimetre) represents volume. Units like cm or cm² are for linear and area measurements and cm⁻³ is not applicable for volume.

A bowl often resembles a hemisphere because it is half of a sphere and has one curved surface with a flat circular base.
Bowls used in kitchens and tanks in practical problems are typical examples of hemispherical shapes.

Surface area = 4πr² = 616.
So, r² = 616 ÷ (4 × 22 ÷ 7)
= 616 ÷ (88 ÷ 7)
= 49.
Therefore, r = √49 = 7 cm.
This reverse application of the formula helps in finding radius when area is known.

The volume of a cone is given by (1/3)πr²h which depends on both the base radius and the vertical height.
The slant height does not affect volume directly.
Volume increases with increasing r or h and changes with the square of radius.

The slant height (l) of a cone can be calculated using the Pythagorean theorem, as the slant height forms the hypotenuse of a right triangle where:
– The height (h) of the cone is one leg,
– The radius (r) of the base is the other leg.
The formula for the slant height l = √(r² + h²).
The height (h) of the cone is 16 cm,
The radius (r) of the base is 12 cm.
Substitute the values of r = 12 cm and h = 16 cm into the formula:
l = √(12² + 16²).
Simplify step by step:
l = √(144 + 256),
l = √400,
l = 20 cm.

The height (h) of a cone can be calculated using the Pythagorean theorem, as the slant height (l), radius (r), and height form a right triangle. The relationship is given by:
l² = r² + h², where:
l is the slant height,
r is the radius of the base,
h is the height of the cone.
The slant height (l) is 26 cm,
The diameter of the base is 20 cm, so the radius r = Diameter / 2 = 20 / 2 = 10 cm.
Substitute the values of l = 26 cm and r = 10 cm into the formula:
l² = r² + h².
Rearrange to solve for h²: h² = l² – r².
Substitute the values: h² = 26² – 10²,
h² = 676 – 100,
h² = 576.
Take the square root of both sides:
h = √576,
h = 24 cm.