pls send me some extra questions of surface area and volumes class 10 for practice...pls send photos (5mark question of this chapter pls.....)
Answers
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Answer:
.1: Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 13.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = 22/7)
Important questions class 10 maths chapter 13.1
Solution:
TSA of the toy = CSA of hemisphere + CSA of cone
Curved surface area of the hemisphere = 1/ 2 (4πr2) = 2π r2 = 2(22/7)× (3.5/2) × (3.5/2) cm2
Height of the cone = Height of the top – Radius of the hemispherical part
= (5 – 3.5/2) cm = 3.25 cm
Sant height of the cone (l) =r2+h2−−−−−−√=(3.52)2+(3.25)2−−−−−−−−−−−−√=3.7 cm (approx.)
Therefore, CSA of cone = πrl = (22/7) × (3.5/2) × 3.7 cm2
Hence, the surface area of the top = [2(22/7)× (3.5/2) × (3.5/2) + (22/7) × (3.5/2) × 3.7] cm2
= (22/7) × (3.5/2) (3.5+3.7) cm2
= (11/2) × (3.5 + 3.7) cm2
= 39.6 cm2 (approx.)
Q. 2: Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end as shown in the figure. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = 22/7)
Important Question class 10 maths chapter 13.2
Solution:
Let h be the height of the cylinder and r be the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of the hemisphere
= 2πrh + 2πr2
= 2π r(h + r)
= 2 (22/7) × 30 × (145 + 30) cm2
= 33000 cm2 = 3.3 m2
Q. 3: 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
The diagram will obtain as below:
Important Question class 10 maths chapter 13.3
Given,
The Volume (V) of each cube is = 64 cm3
This implies that a3 = 64 cm3
∴ The side of the cube i.e. a = 4 cm
Also, the breadth and length of the resulting cuboid will be 4 cm each while its height will be 8 cm.
So, the surface area of the cuboid (TSA) = 2(lb + bh + lh)
Now, by putting the values we get,
= 2(8×4 + 4×4 + 4×8) cm2
= (2 × 80) cm2
Hence, TSA of the cuboid = 160 cm2
Q. 4: 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.)
Solution:
It is known that a tent is a combination of a cone and a cylinder as shown below.
Important question class 10 maths chapter 13.4
From the question, we know that
The diameter = D = 4 m
l = 2.8 m (slant height)
The radius of the cylinder is equal to the radius of the cylinder
So, r = 4/2 = 2 m
Also, we know the height of the cylinder (h) is 2.1 m
So, the required surface area of the given tent = surface area of the cone (the top) + surface area of the cylinder(the base)
= πrl + 2πrh
= πr (l+2h)
Now, substituting the values and solving it we get the value as 44 m2
∴ The cost of the canvas at the rate of Rs. 500 per m2 for the tent will be
= Surface area × cost/ m2
= 44 × 500
So, Rs. 22000 will be the total cost of the canvas.
Q. 5: A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference between the volumes of the cylinder and the toy. (Take π = 3.14)
Solution:
Important question class 10 maths chapter 10.5
Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere as shown in the above figure.
The radius BO of the hemisphere (as well as of the cone) =( ½) × 4 cm = 2 cm.
So, volume of the toy = (⅔) πr3 + (⅓) πr2h
= (⅔) × 3.14 × 23 + (⅓)× 3.14 × 22 × 2
= 25.12 cm3
Now, let the right circular cylinder EFGH circumscribe the given solid.
The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is
EH = AO + OP = (2 + 2) cm = 4 cm
So, the volume required = volume of the right circular cylinder – volume of the toy
= (3.14 × 22 × 4 – 25.12) cm3
= 25.12 cm3
Hence, the required difference of the two volumes = 25.12 cm3
Q. 6: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.do it yourself