Please give me standard worksheet for math with solutions also of ch 13
Answers
Q.1: Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see in the figure below). Find how much he would spend on the tiles if the cost of the tiles is Rs.360 per dozen.
Solution:
Given,
Edge of the cubical tank (a) = 1.5 m = 150 cm
So, surface area of the tank = 5 × 150 × 150 cm2
The measure of side of a square tile = 25 cm
Area of each square tile = side × side = 25 × 25 cm2
Required number of tiles = (Surface area of the tank)/(area of each tile)
= (5 × 150 × 150)/(25 × 25)
= 180
Also, given that the cost of the tiles is Rs. 360 per dozen.
Thus, the cost of each tile = Rs. 360/12 = Rs. 30
Hence, the total cost of 180 tiles = 180 × Rs. 30 = Rs. 5400
Q.2: The paint in a certain container is sufficient to paint an area equal to 9.375 sq.m. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution:
Given,
Dimensions of the brick = 22.5 cm × 10 cm × 7.5 cm
Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm
Surface area of 1 brick = 2(lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm2
= 2(225 + 75 + 168.75) cm2
= 2 x 468.75 cm2
= 937.5 cm2
Area that can be painted by the container = 9.375 m2 (given)
= 9.375 × 10000 cm2
= 93750 cm2
Thus, the required number of bricks = (Area that can be painted by the container)/(Surface area of 1 brick)
= 93750/937.5
= 937500/9375
= 100
Q.3: The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs.7.50 per sq.m.
Solution:
Given,
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area of walls of the room = Lateral surface area of cuboid
= 2h(l + b)
= 2 × 3(5 + 4)
= 6 × 9
= 54 sq.m
Area of ceiling = Area of base of the cuboid
= lb
= 5 × 4
= 20 sq.m
Area to be white washed = (54 + 20) sq.m = 74 sq.m
Given that, the cost of white washing 1 sq.m = Rs. 7.50
Therefore, the total cost of white washing the walls and ceiling of the room = 74 × Rs. 7.50 = Rs. 555
Q.4: The curved surface area of a right circular cylinder of height 14 cm is 88 sq.cm. Find the diameter of the base of the cylinder.
Solution:
Let d be the diameter and r be the radius of a right circular cylinder.
Given,
Height of cylinder (h) = 14 cm
Curved surface area of right circular cylinder = 88 cm2
⇒ 2πrh = 88 cm2
⇒ πdh = 88 cm2 (since d = 2r)
⇒ 22/7 x d x 14 cm = 88 cm2
⇒ d = 2 cm
Hence, the diameter of the base of the cylinder is 2 cm.
Q.5: Curved surface area of a right circular cylinder is 4.4 sq.m. If the radius of the base of the cylinder is 0.7 m, find its height.
Solution:
Let h be the height of the cylinder.
Given,
Radius of the base of the cylinder (r) = 0.7 m
Curved surface area of cylinder = 4.4 m2
Thus,
2πrh = 4.4
2 × 3.14 × 0.7 × h = 4.4
4.4 × h = 4.4
h = 4.4/4.4
h = 1
Therefore, the height of the cylinder is 1 m.
Q.6: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
Given,
Length of the cylindrical pipe = h = 28 m
Diameter of the pipe = 5 cm
Now, the radius of piper (r) = 5/ 2 cm = 2.5 cm = 0.025 m
Total radiating surface in the system = Total surface area of the cylinder
= 2πr(h + r)
= 2 × (22/7) × 0.025 (28 + 0.025) m2
= (44 x 0.025 x 28.025)/7 m2
= 4.4 m2 (approx)
Q.7: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. (Take π = 3.14)
Solution:
Given
Height of a cone (h) = 16 cm
Radius of the base (r) = 12 cm
Now,
Slant height of cone (l) = √(r2 + h2)
= √(256 + 144)
= √400
= 20 cm
Curved surface area of cone = πrl
= 3.14 × 12 × 20 cm2
= 753.6 cm2
Total surface area = πrl + πr2
= (753.6 + 3.14 × 12 × 12) cm2
= (753.6 + 452.16) cm2
= 1205.76 cm2
Q.8: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution:
Given,
Diameter of the cone = 24 m
Radius of the cone (r) = 24/2 = 12 m
Slant height of the cone (l) = 21 m
Total surface area of a cone = πr(l + r)
= (22/7) × 12 × (21 + 12)
= (22/7) × 12 × 33
= 1244.57 m2
hope it helps you friend
Step-by-step explanation:
Finding out the surface area and the volume of different shapes is one of the fundamental studies of mathematics, and that's what students are going to learn from the important questions for class 9 maths chapter 13. In this chapter, we will showcase different methods to solve a problem related to surface area and volume of circle, square, rectangle, cylinder, pyramid, etc. in addition to this. The chapter starts with a brief introduction of the plane and solid figures which will help students understand the concepts such as finding out the area of a given square or finding out the volume of a given cone.