Класс 8 Р. Д. Шарма - Глава 21 Измерение II (Объемы и площади поверхности кубоида и куба) - Упражнение 21.3 | Комплект 2
Вопрос 10. Класс имеет длину 11 м, ширину 8 м и высоту 5 м. Найдите сумму площадей пола и четырех стен (включая двери, окна и т. Д.)?
Решение:
Given, the Length of classroom = 11 m
The Breadth of classroom = 8 m
The Height of the classroom = 5 m
Area of floor = length × breadth = 11 × 8 = 88 m2
And, the Area of four walls (including doors, windows) = 2 (lh + bh)
= 2 (11 × 5 + 8 × 5)
= 190 m2
The total area of four walls and floor = Area of floor + Area of four walls (including doors, windows) = 88 + 190 = 278 m2
Hence, the sum of the areas of its floor and the four walls is 278 m2
Вопрос 11. Бассейн имеет длину 20 м, ширину 15 м и глубину 3 м. Найдите стоимость ремонта пола и стен из расчета рупий. 25 за квадратный метр?
Решение:
Given, the Length of the swimming pool = 20 m
The Breadth of the swimming pool = 15 m
The Height of the swimming pool = 3 m
Area of the swimming pool floor = length × breadth = 20 × 15 = 300 m2
Area of the swimming pool walls = 2 (lh + bh)
= 2 (20 × 3 + 15 × 3)
= 210 m2
The total area of swimming pool = Area of floor + Area of four walls = 300 + 210 = 510 m2
Since, the cost of repairing 1 m2 swimming pool area = Rs 25
So, the cost of repairing 1 m2 swimming pool area = Rs 25 × 510 = Rs 12750
Hence, the cost of repairing the floor and wall of the swimming pool is Rs 12750
Вопрос 12. Периметр этажа помещения 30 м, высота 3 м. Найдите площадь четырех стен комнаты?
Решение:
Given, the perimeter of a room floor = 30m
It means 2(l + b) = 30
So, l + b = 15 m
The height of the room = 3 m
Now, the Area of the room walls = 2 (lh + bh)
= 2h (l + b) = 2 × 3 × 15 = 90 m22
Hence, the area of the four walls of the room is 90 m2
Вопрос 13. Показать, что произведение площадей пола и двух смежных стен кубоида равно квадрату его объема?
Решение:
Let l be the length of the cuboid
Let b be the breadth of cuboid
Let h be the height of the cuboid
So, the Area of floor = length X breadth = l × b
And, the product of the area of two adjacent walls of cuboid = (l × h) × (b × h) = lbh2
And, finally the product of the areas of the floor and two adjacent walls of a cuboid = lb X lbh2 = (l × b × h)2
{We know that Volume of cuboid = l × b × h} = (Volume of the cuboid)2
Hence, proved that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume
Вопрос 14. Оштукатурить стены и потолок помещения. Длина, ширина и высота помещения составляют 4,5 м, 3 м и 350 см соответственно. Найдите стоимость штукатурки из расчета Rs. 8 за квадратный метр?
Решение:
Given, the Length of room = 4.5 m
The Breadth of room = 3 m
The Height of room = 350 cm = 3.5 m
Area of the room ceiling = length × breadth = 4.5 × 3 = 13.5 m2
Area of the room walls = 2 (lh + bh)
= 2 (4.5 × 3.5 + 3 × 3.5)
= 52.5 m2
So, the Sum of the area of ceiling and area of four walls = 13.5 + 52.5 = 66m2
Also, the Cost of plastering room 1 m2 area = Rs 8
So, the Cost of plastering room 66 m2 area = Rs 8 × 66 = Rs 528
Hence, the cost of plastering is Rs 528
Вопрос 15. Кубовид имеет общую площадь 50 м 2 и площадь боковой поверхности 30 м 2 . Найдите площадь его базы?
Решение:
Given, the Total surface area of cuboid = 50 m2
The Lateral surface area of cuboid = 30 m2
As we know the lateral surface area of cuboid = Area of 4 walls of cuboid
And, the Total surface area of a cuboid = 2 (Area of Base) + Area of 4 walls
50 = 2 (Area of Base) + 30
So, Area of Base = 10 m2
Hence, the area of the base of cuboid = 10 m2
Вопрос 16. Класс имеет длину 7 м, ширину 6 м и высоту 3,5 м. Двери и окна занимают площадь 17 м 2 . Сколько стоит побелка стен из расчета 1,50 рупий за м 2 ?
Решение:
Given, the Length of classroom = 7 m
The Breadth of classroom = 6 m
The height of the classroom = 3.5 m
The area occupied by doors and windows = 17 m2
So, area of 4 walls (including doors & windows) = 2(lh + bh) = 2(7 × 3.5 + 6 × 3.5) = 91m2
Since, whitewashing can’t be done on doors and windows,
So, the area of 4 walls excluding doors & windows = 91 – 17 = 74 m2
Also, the cost of whitewashing 1 m2 classroom = Rs 1.50
So, the cost of whitewashing 74 m2 classroom = Rs 1.50 × 74 = Rs 111
Hence, the cost of whitewashing the walls is Rs 111
Вопрос 17. Центральный зал школы 80 м в длину и 8 м в высоту. Он имеет 10 дверей размером 3 × 1,5 м и 10 окон размером 1,5 × 1 м. Если стоимость побелки стен зала из расчета 1,20 рупий за м 2 составляет 2385,60 рупий, найдите ширину зала?
Решение:
Given, the Length of central hall = 80 m
The height of the central hall = 8 m
And let b be the Breadth of the hall
For Doors,
Length of door = 3 m
Width of door = 1.5 m
So, the Area of Door = l × b = 3 × 1.5 = 4.5 m2
And, the Area of 10 doors = 4.5 × 10 = 45 m2
For Windows,
Length of window = 1.5 m
Breadth of window = 1 m
So, the Area of window = l × b = 1.5 × 1 = 1.5 m2
And, the Area of 10 windows = 1.5 × 10 = 15 m2
So, the total area occupied by doors & windows = 45 + 15 = 60 m2
The Area of 4 walls (including doors & windows) = 2(lh + bh) = 2(80 × 8 + b × 8) = 2(640 + 8b) m2
Since, whitewashing can’t be done on doors and windows,
So, area of 4 walls excluding doors & windows = 2(640+8b) – 60 = (1220 + 16b) m2
Also, the cost of whitewashing 1 m2 central hall = Rs 1.20
So, the cost of whitewashing (1220 + 16b) m2 central hall = (1220 + 16b) × 1.20
Total whitewashing cost = Rs 2385.60
It means,
2385.60 = (1220 + 16b) × 1.20
b = 48 m
Hence, the breadth of central hall is 48 m