9类RD Sharma解决方案-第18章，长方体和立方体的表面积和体积–练习18.2 |套装1

问题1.长方体水箱长6 m，宽5 m ，深4.5 m。它可以容纳多少升水?

解决方案：

Given, the length of water tank = 6 m

The breadth of water tank = 5 m

The height of water tank = 4.5 m

The quantity of water that tank can hold = Volume of Cuboid

= length × breadth × height

= 6 × 5 × 4.5 = 135 m³

As we know 1 m³ = 1000 liters

So, 135 m³ = 135 × 1000 = 135000 liters

Hence. the water tank can hold 1,35,000 liters of water.

问题2：一个长10 m，宽8 m的长方体容器。容纳380立方米液体必须达到多高?

解决方案：

Given, the length of the cuboidal vessel = 10 m

The breadth of cuboidal vessel = 8 m

The volume of cuboidal vessel = 380 cubic meters

Volume of Cuboid = length × breadth × height

380 = 10 × 8 × height

So, height is 4.75 m

Hence, the vessel should be 4.75 m high in order to hold 380 cubic meters of a liquid

问题3.找到以8卢比/ m ³的速率开凿长8m，宽6m，深3m的长方体坑的成本。

解决方案：

Given, the length of the cuboidal pit = 8 m

The breadth of cuboidal pit = 6 m

The height of cuboidal pit = 3 m

We know Volume of Cuboid = length × breadth × height

= 8 × 6 × 3 = 144 m³

Also, the cost of digging 1 m³ = Rs 30

So, cost of digging 144 m³ = Rs 30 × 144

= Rs 4320

Hence, the cost of digging the cuboidal pit is Rs 4320

问题4.如果V是长方体a，b，c和S的体积是其表面积，则证明1 / V = 2 / S(1 / a + 1 / b + 1 / c)

解决方案：

Let a be the length of cuboid, b be the breadth of cuboid and c be the height

Surface area of cuboid = 2(a × b + b × c + c × a)

S = 2(a × b + b × c + c × a)

Volume of cuboid = a × b × c

V = a × b × c

We need to prove 1/V = 2/S(1/a + 1/b + 1/c)

Taking L.H.S, 2/S(1/a + 1/b + 1/c)

= 2/aS + 2/bS + 2/aS = 2(bc+ca+ab)/S × abc

= 2(bc+ca+ab)/(ab + bc + ac) × abc

= 2/abc = 2/V

So, L.H.S = R.H.S

Hence proved

问题5.长方体的三个相邻面的面积分别是x，y和z。如果体积为V，则证明V ² = xyz。

解决方案：

Let a be the length of cuboid, b be the breadth of cuboid and c be the height

Given, the areas of three adjacent faces of a cuboid are x, y and z

So, x = ab

y = bc

z = ac

We know, Volume of cuboid = a × b × c

Multiplying x, y and z we get,

xyz = ab X bc X ac

xyz = (abc)²

So, xyz = (V)²

Hence, proved

问题6.如果长方体的三个相邻面的面积8 cm ²时，18厘米²和25平方厘米^。找到长方体的体积。

解决方案：

Let x, y, and z be the area of adjacent faces of cuboid

So, x = 8 cm²

y = 18 cm²

z = 25 cm²

We know that, product of area of adjacent faces of cuboid will be equal to square of volume of cuboid

It means, xyz = V²

So, V² = 8 × 18 × 25 = 3600

V = 60 cm³

Hence, the volume of cuboid is 60 cm³

问题7.一个房间的宽度是其高度的两倍，是其长度的一半，房间的体积为512立方米。查找其尺寸。

解决方案：

Given, volume of room is 512 cu.dm

Let l be the length of room, and h be the height of room

Also, b = 2 × h and b = l/2

So, we can say h = b/2 and l = 2 × b

Volume of cuboid = l × b × h

So, 512 = 2b × b × b/2

So, b = 8 cm

l = 2b = 16 cm

h = b/2 = 4 cm

Hence, the length, breadth and height of room are 16cm, 8 cm and 4 cm respectively

问题8：一条3 m深，40 m宽的河流以每小时2 km的速度流动。一分钟内有多少水会掉入海中?

解决方案：

Given, the depth of the river is 3 m

The width of the river is 40 m

The water flowing in the river at the rate of 2 km/hour = 100/3 m/minute

So, we can derive the volume of water flowing in 1 min = 100/3 × 40 × 3

= 4000 m³

Or 4000000 liters

Hence, 4000000 liters of water will fall into the sea in 1 minute

问题9.一条30 dm宽，12 dm深的运河中的水每小时以100 km的速度流动。如果需要8厘米的静水，将在30分钟内灌溉多少面积?

解决方案：

Given, the width of canal = 30 dm = 3 m

The depth of canal = 12 dm = 1.2 m

And, the length of the canal will be given as the distance travelled with the speed of 100 km per hour in 30 minutes

So, length = 100 × 30/60 = 50000 m

We can say, the volume of water for irrigation = l × b × h

= 50000 × 3 × 1.2 m³

As we know the water in the field will form a cuboid, whose area will be equal to the area of field and height will be 8/100 m

So, we can conclude Area of field × 8/100 = 50000 × 3 × 1.2

Area of field = 2250000 m³

Hence, the desired area is 2250000 m³

问题10.将三个分别具有6cm，8cm和10cm边缘的金属立方体熔化在一起，并形成一个立方体。找到新立方体的体积，表面积和对角线。

解决方案：

Let a be the edge of the new cube

So, the volume of cube = a³

Also, we know that sum of volumes of given three cubes will be equal to the volume of the new cube 

So, 6³ + 8³ + 10³ = a³ 

a = 12

So, the volume of the new cube = a³ = 12³

= 1728 cm³

Also, the surface area of the new cube = 6 × (a)²

= 6 × 12² = 864 cm²

And, the diagonal of the new cube = √3a = √3 × 12 

= 12√3 cm

Hence, the volume, surface area and diagonals of the new cube is 1728 cm³, 864 cm² and 12√3 cm

问题11.两个立方体，每个立方体的体积为512 cm ³ ，首尾相连。找到所得长方体的表面积。

解决方案：

Let a be the edge of the cube

Given the volume of the cube is 512 cm³

So, a³ = 512 cm³

a = 8 cm

So we can say the edge of each cube is 8 cm

Now the dimensions of the resulting cuboid will be,

Length is 8 + 8 = 16 cm

Breadth is 8 cm

Height is 8 cm

So, the surface area of the cuboid = 2 (l × b + b × h + h × l)

= 2 (16 × 8 + 8 × 8 + 8 × 12)

= 640 cm²

Hence, the surface area of the resulting cube is 640 cm²

问题12.通过锤击将半立方米的金箔延展至1公顷的面积。找到金片的厚度。

解决方案：

Given, the volume of the gold-sheet is 1/2 m³

Also, the area of the gold-sheet is 1 hectare = 10000 m²

So the thickness of the gold-sheet is given by the Volume of the gold-sheet/ Area of the gold-sheet

= 0.5/10000 = 1/20000 m = 1/200 cm

Hence, the thickness of the gold-sheet is 1/200 cm

问题13：将一个边缘为12 cm的金属立方体熔化，并形成三个较小的立方体。如果两个较小的多维数据集的边缘分别为6厘米和8厘米，请找到第三个较小的多维数据集的边缘。

解决方案：

Let a be the edge of the third cube

Also, we know that sum of volumes of given three cubes will be equal to the volume of the new cube 

So, 6³ + 8³ + a³ = 12³ 

216 + 512 + a³ = 1728

a = 10 cm

Hence, the edge of the third cube is 10 cm.