CBSE Class X

Question 4:

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboids are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the following figure).   [Use π = 22/7]

Answer:

Depth (h) of each conical depression = 1.4 cm

Radius (r) of each conical depression = 0.5 cm

Volume of wood = Volume of cuboid − 4 × Volume of cones

Question 3:

A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see the given figure).      [Use π = 22/7]

Answer:

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm

Length (h) of cylindrical part = 5 − 2 × Length of hemispherical part

= 5 − 2 × 1.4 = 2.2 cm

Volume of one gulab jamun = Vol. of cylindrical part + 2 × Vol. of hemispherical part

Volume of 45 gulab jamuns = 45 × 25.05 = 1,127.25 cm³

Volume of sugar syrup = 30% of volume

Question 2:

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)                [Use π = 22/7]

Answer:

From the figure, it can be observed that

Height (h₁) of each conical part = 2 cm

Height (h₂) of cylindrical part = 12 − 2 × Height of conical part

= 12 − 2 ×2 = 8 cm

Radius (r) of cylindrical part = Radius of conical part = 3/2 cm

Volume of air present in the model = Volume of cylinder + 2 × Volume of cones

Question 1:

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Answer:

Given that,

Height (h) of conical part = Radius(r) of conical part = 1 cm

Radius(r) of hemispherical part = Radius of conical part (r) = 1 cm

Volume of solid = Volume of conical part + Volume of hemispherical part

Question 9:

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in given figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.  [Use π = 22/7]

Answer:

Given that,

Radius (r) of cylindrical part = Radius (r) of hemispherical part = 3.5 cm

Height of cylindrical part (h) = 10 cm

Surface area of article = CSA of cylindrical part + 2 × CSA of hemispherical part

=2πrh + 2 × 2πr²

=2π × 3.5 × 10 + 2 × 2π ×3.5 × 3.5

= 70π + 49π

= 119π

= 17 × 22 = 374 cm²

Question 8:

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm². [Use π = 22/7]

Answer:

Given that,

Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm

Diameter of the cylindrical part = 1.4 cm

Therefore, radius (r) of the cylindrical part = 0.7 cm

Total surface area of the remaining solid will be

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

The total surface area of the remaining solid to the nearest cm² is 18 cm²

Question 7:

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m². (Note that the base of the tent will not be covered with canvas.)  [Use π = 22/7]

Answer:

Given that,

Height (h) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (l) of conical part = 2.8 m

Area of canvas used = CSA of conical part + CSA of cylindrical part

Cost of 1 m² canvas = Rs 500

Cost of 44 m² canvas = 44 × 500 = 22000

Therefore, it will cost Rs 22000 for making such a tent.

Question 6:

A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see the given figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. [Use π = 22/7]

Answer:

It can be observed that

Radius (r) of cylindrical part = Radius (r) of hemispherical part

= Diameter of the capsule/2 = 5/2

Length of cylindrical part (h) = Length of the entire capsule − 2 × r = 14 − 5 = 9 cm

Surface area of capsule = 2×CSA of hemispherical part + CSA of cylindrical part

Question 5:

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer:

Diameter of hemisphere = Edge of cube = l

Radius of hemisphere = 1/2

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part

Question 4:

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. [Use π = 22/7]

Answer:

From the figure, it can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm.

Radius (r) of hemispherical part = 7/2 = 3.5cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part

Question 3:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. [Use π = 22/7]

Answer:

It can be observed that the radius of the conical part and the hemispherical part is same (i.e., 3.5 cm).

Height of hemispherical part = Radius (r) = 3.5 = 7/2 cm

Height of conical part (h) = 15.5 −3.5 = 12 cm

Question 2:

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. [Use π = 22/7]

Answer:

It can be observed that radius (r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm).

Height of hemispherical part = Radius = 7 cm

Height of cylindrical part (h) = 13 −7 = 6 cm

Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part

= 2πrh + 2πr²

Question 1:

2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboids.

Answer:

Given that,

Volume of cubes = 64 cm³

(Edge) ³ = 64

Edge = 4 cm

If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm.

∴ Surface area of cuboids = 2(lb + bh + lh)

= 2(4 × 4 + 4 × 8 + 4 × 8)

= 2(16 + 32+ 32)

= 2(16 + 64)

= 2 × 80 = 160 cm²

Question 16:

Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.  [Use π = 22/7]

 

Answer:

 

The designed area is the common region between two sectors BAEC and DAFC.

Question 15:

In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. [Use π = 22/7]

 

Answer:

 

As ABC is a quadrant of the circle, ∠BAC will be of measure 90º.

In ΔABC,

BC² = AC² + AB²

= (14)² + (14)²

BC = 14√2

Question 14:

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If ∠AOB = 30°, find the area of the shaded region.  [Use π = 22/7]

Answer:

 

Area of the shaded region = Area of sector OAEB − Area of sector OCFD

Question 13:

In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [Use π = 3.14]

Answer:

 

In ΔOAB,

OB² = OA² + AB²

= (20)² + (20)²

OB = 20√2

Radius (r) of circle = 20√2 cm

Area of OABC = (Side)² = (20)² = 400 cm²

Area of shaded region = Area of quadrant OPBQ − Area of OABC

= (628 − 400) cm²

= 228 cm²

Question 12:

In the given figure, OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) Quadrant OACB             (ii) Shaded region          [Use π = 22/7]

 

Answer:

(i) Since OACB is a quadrant, it will subtend 90° angle at O.

Area of quadrant OACB = (90°/360°) ×  πr²

Question 11:

On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief.  [Use π = 22/7]

 

Answer:

 

From the figure, it can be observed that the side of the square is 42 cm.

Area of square = (Side)² = (42)² = 1764 cm²

Area of each circle = πr² = 22/7 × (7)² = 154 cm²

Area of 9 circles = 9 × 154 = 1386 cm²

Area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm²

Question 10:

The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (See the given figure). Find the area of shaded region. [Use π = 3.14 and √3 = 1.73205]

Answer:

Let the side of the equilateral triangle be a.

Area of equilateral triangle = 17320.5 cm²

Each sector is of measure 60°.