8级RD沙尔玛–第21章，测定II(长方体和立方体的体积和表面积)–练习21.3 |套装2

问题10.教室长11 m，宽8 m ，高5 m。找到地板和四面墙(包括门，窗等)的面积之和?

解决方案：

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

问题11：一个游泳池长20 m，宽15 m，深3 m。以Rs的比率找到修理地板和墙壁的成本。每平方米25?

解决方案：

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

问题12.房间地板的周长为30 m，高度为3 m。找出房间四面墙的面积吗?

解决方案：

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

问题13。显示长方体的地板和两个相邻墙的面积的乘积是其体积的平方吗?

解决方案：

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

问题14.房间的墙壁和天花板应抹灰。房间的长度，宽度和高度分别为4.5 m，3 m和350 cm。以Rs的比率查找抹灰的成本。每平方米8个?

解决方案：

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

问题15.长方体的总表面积为50 m ^2，而侧面表面积为30 m ² 。找出其基地的面积?

解决方案：

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

问题16.教室长7 m，宽6 m，高3.5 m。门窗面积为17 m ² 。^{以每m 2} 1.50卢比的价格粉刷墙壁的成本是多少?

解决方案：

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

问题17.学校的中央大厅长80 m，高8 m。它有10个大小为3m×1.5m的门和10个大小为1.5m×1m的窗口。^{如果以每m 2} 1.20卢比的价格粉刷大厅墙壁的成本为2385.60卢比，那么找到大厅的宽度了吗?

解决方案：

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