Book Name |
: Ganit Prakash |

Subject |
: Mathematics |

Class |
: 10 (Madhyamik) |

Publisher |
: Prof Nabanita Chatterjee |

Chapter Name |
: Rectangular Parallelopiped Or Cuboid (4th Chapter) |

Table of Contents

Toggle**Application 1. **

**I am observing that the length, breadth and height of the rectangular parallelopiped box, brought by my brother, are 40 cm, 25 cm and 15 cm respectively. **

**Solution : **

**Total surface area of a rectangular box**

= 2 (l × b+l × h+b × h)

= 2 (40 × 25 + 40 × 15+25 × 15) sqcm.

= 2 (1000+600+375) sqcm.

= 2 × 1975 sqcm.

= 3950 sqcm.

**Application 2. **

**The length, breadth and height of a rectangular parallelopiped box are 15 cm, 12 cm and 20 cm respectively. Let me write its total surface area. [Let me do it myself]**

**Solution : **

The total Surface area of a rectangular parallelopiped

= 2 (15 × 12+15 × 20+12 × 20) sqcm.

= 2 (180+300+240) sqcm = 2 × (720) sqcm

= 1440 sq cm.

**Application 3. **

**By measuring, I see that the length of one side of the cube, brought by Rajia is 27 cm.**

**Solution : **

Length of one side of a cube = 27 cm.

∴ The total surface area of the cube = 6 × (27)^{2} sq cm.

= 4374 sq cm.

**Application 4. **

**Let us write by calculating the quantity of coloured papers required to cover the whole surface of a cube whose length of one side is 12 cm. [Let me write it myself].**

**Solution : **

Total area of the colour paper required for the cube = 6 × (12)^{2} sqcm .

= 6 × 144 sqcm.

= 884 sqcm. Ans.

**Application 5. **

**The length, breadth and height of our living room are 7 m, 5 m and 4 m respectively. The shape of the room is ____ [cube/rectangular parallelopiped]**

**Let us calculate the total area to colour four walls of the room.**

**Solution : **

The length, breadth & height of the room arc 7 m, 5 m, and 4 m respectively.

∴ The room is a rectangular parallelopiped

**Calculation of Area**

The area of the 4 walls of the room

= 2 × (l+b) × h = 2(7+5) × 4 sqcm.

= 96 sq cm

**Application 6. **

**Let us write the length of one side of a cube whose total surface area is 150 sq cm.**

**Solution : **

Let one side of a cube = a cm.

The total surface area of the cube = 6a^{2}

or, 6a^{2 }= 150

or, a^{2 }= 150/6

or, a^{2 }= 25

or, a = √25 = 5

∴ Length of the cube = 5 cm

**Application 7. **

**Let us calculate the length of one side of a cube whose total surface area is 486 sq.m. [Let me do it myself]**

**Solution : **

Let one side of a cube = a m.

∴ 6a^{2} = 486

or, a^{2 }= 81

or, a = √81 =9

One side of the cube = 9 m

**Application 9. **

**A cuboidal room has its length, breadth and height as a, b and c units respectively and if a + b + c = 25 units, ab + bc + ca = 240.5 units, then let us write the length of the longest rod that can be kept in this room.**

**Solution : **

\mathrm{ (a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a) }

or, \mathrm{ (25)^2=a^2+b^2+c^2+2.240 .5 }

or, \mathrm{ 625=a^2+b^2+c^2+481 }

\therefore \mathrm{ a^2+b^2+c^2=625-481=144 }

\therefore The diagonal of the room \mathrm{ =\sqrt{a^2+b^2+c^2} }

\mathrm{ =\sqrt{144} } 12 m

**Application 10. **

**I have made a rectangular parallelopiped by joining two cubes side by side made by Mita, whose edge is 8 cm in length. Let us calculate the total surface area and the length of the diagonal of the rectangular parallelopiped which is made in this way. [Let me do it myself]**

**Hints: The length of the rectangular parallelopiped which is made = (8+8) cm = 16 cm, breadth = 8 cm, height = 8 cm.**

**Solution : **

If two cubes are placed side by side. The new length of the parallelopiped

= (8+8) = 16 cm

breadth = 8 cm

height = 8 cm.

\therefore Total area of the rectangular parallelopiped

= 2(16 × 8 + 16 × 8+8 × 8) sqcm.

= 2(128+128+64) sqcm = 2 × 320=640 sqcm

∴ Total area of the rectangular parallelopiped

= 2(16 × 8+16 × 8+8 × 8) sqcm.

= 2(128+128+64) sqcm = 2 × 320 = 640 sqcm.

\therefore Diagonal of the rectangular parallelopiped

=\sqrt{(16)^2+(8)^2+(8)^2}=\sqrt{256+64+64}=\sqrt{384} \\

= 8 \sqrt{6} cm

If the height of a rectangular parallelopiped is increased, its volume will be increased.

**Application 11. **

**By measuring the cuboidal at box, I am observing that the length is 32 cm, breadth is 21 cm, and height is 15 cm.**

**Solution : **

Volume of the sand = 32 × 21 × 15 cu cm

= 10080 cu cm

**Application 12. **

**The length of one side of a cube is 5 cm, its volume is ____ c.c = ____ cc [Let me do it myself]**

**Solution : **

Length of one side of a cube = 5 cm.

∴ Volume of the cube = (5)^{3} cu cm = 125 cu cm

**Application 15. **

**If the length, breadth and volume of a cuboidal room are 8 m, 6 m and 192 cubic m respectively, then let us calculate the height of the room and the area of four walls of the room.**

**Solution : **

Length (l) = 8 m; Breadth (b) = 6 m; Volume (V) = 192 cu cm

According to the problem,

8 × 6 × h = 192

h =\frac{192}{8 × 6} = 4

∴ Area of the 4 walls of the room = 2 × (8+6) × 4

= 2 × 14 × 4 sq m = 112 sq m

**Application 17. **

**The length and breadth of cuboidal water land in neighbouring villages are 18 m and 11 m respectively. In this water land, water is being irrigated from a nearby pond with a pump. If the pump can irrigate 39,600 lit. water per hour then for how much time is required to raise the height of the water level by 3.5 dcm of the water land – let us write by calculating it. [ 1 lit. -1 cubic dcm.]**

**Solution : **

The pump will run =\frac{180 × 110 × 3.5}{39600}hr.

=\frac{180 × 110 × 35}{39600 × 10} × 60 min

= 35 × 3 = 105 min

= 1 hr 45 min.

**Application 18. **

**If the pump can fill 37,400 liters of water in 1 hour, then what time will be required to raise the height of water 17 dcm of a cuboidal water land whose length is 18 m and breadth is 11 m? Let us write by calculating it. [Let me do myself]**

**Solution : **

The length, Breadth and height of the tanks are 18 m, 11 m and 17 dcm respectively

The volume of the tank = (180 × 110 × 17) cu dcm.

= 180 × 110 × 17 litre.

∴ The pump will run =\frac{180 × 110 × 17}{37400} hours = 9 hours.

**Application 19.**

** From a wooden log with a length of 5 dcm, and a breadth of 5 dcm. and the thickness of 3 dcm, 40 planks are of 2 dcm length and 2 dcm breadth are cloven. For cleaving wood has been destroyed. But still, now 108 cubic dcm of wood remains in the log. What is the thickness of each plank that is cloven, let us calculate and write it.**

**Solution : **

The total volume of the log =40 dcm × 5 dcm × 3 dcm.

= (40 × 5 × 3) cu dcm.

= 600 dcm

Total volume of wood wasted = 600 × \frac{2}{10} = 12cu dcm.

Let the thickness of each plate = x dcm.

\therefore Volume of each plate = 20 × 2 × x cu dcm.

Volume of 40 plate = 40 ×(20 × 2 × x) = 1600 × cu.dcm.

According to the problem,

1600 x + 108 + 12 = 600

1600 x = 600 – 120 = 480

∴ x = \frac{480}{1600}= 0.3 dcm .

∴ The thickness of each plate = 0.3 dcm.

## Let Us Work Out – 4

**Question 1**

**Let us write the names of 4 cuboids and 4 cube-shaped solid things in our environment.**

**Solution : **

(a) Name of 4 rectangular parallelopipeds:

Brick; Matchbox; Mathematics book; Geometry box.

(b) Name of 4 cubes :

Dice of ludo; Ice cube; Rubix cube; Square suitcase.

**Question 2**

**Let us write the names of surfaces, edges and vertices of the adjoining cuboidal figure.**

**Solution : **

Name of the surfaces : ABCD; EFGH; ADEF; BCGH; ABEH; CDFG.

Name of the edges: AB, BC, CD, DA, EF, FG, EH, GH, AE, BH, CG & FG.

Vertexes are A, B, C, D, E, F, G & H.

**Question 3**

**The length, breadth and height of a cuboidal room are 5 m, 4 m, and 3 m respectively. Let us write the length of the longest rod which can be kept in that room.**

**Solution : **

Length of the rod = Diagonal of the room

=\sqrt{1^2+b^2+h^2} \\

=\sqrt{5^2+4^2+3^2}=\sqrt{25+16+9}=\sqrt{50}=5 \sqrt{2} m

**Question 4**

**The area of one surface of a cube is 64 sq m let us calculate the volume of the cube**

**Solution : **

The surface area of the cube = 64 sqm.

∴ Length of one side of the cube =\sqrt{64} = 8 m

∴ Volume of the cube = (8 m)^{3} = 512 cu m.

**Question 5**

**In our Bokultala village, a canal is cut whose breadth is 2 m and depth is 8 dcm. If the total quantity of soil extracted is 240 cubic metres, then let us calculate the total surface area of the cube.**

**Solution : **

Let the length of the canal = x m

Breadth = 2 m.

Depth = 8 dcm = 0.8 m,

According to the problem,

x×2×(0.8) = 240

∴ x = \frac{240}{2 × 0.8} =\frac{240 × 10}{2 × 8} = 150 m

Total Surface Area = 2 (l x b + b x h + l x h)

= 2 (150 × 2 + 2 × 0.8 + 150 × 0.8)

= 843.2 m^{2}

**Question 6**

**If the length of the diagonal of a cube is 4 \sqrt{3}cm, then let us calculate the total surface area of the cube.**

**Solution : **

Diagonal of a cube = 4 \sqrt{3}cm.

As diagonal of a cube =\sqrt{3} × side.

Length of one side of the cube =\frac{4 \sqrt{3}}{\sqrt{3}}=4 cm

\therefore Total surface area of a cube = 6 × (side)^{2}

= 6 × (4)^{2} = 16 × 6 sq cm

= 96 sq cm

∴ 96 sq cm

**Question 7**

**The sum of the length of the edges of a cube is 60 cm, let us out the volume of the cube.**

**Solution : **

There are 12 edges of a cube & all the edges are equal.

∴ 12 × length one edge = 60 cm

∴ Length of one edge =\frac{60}{12} cm = 5 cm.

The volume of a cube = (one edge)^{3 }= (5)^{3} cu cm. = 125 cu cm.

**Question 8**

**If the sum of areas of 6 surfaces of a cube is 216 sq cm then let us calculate the volume of the cube.**

**Solution : **

Total area of 6 surfaces of cube = 6 × (side )^{2}

or, 6 × (side )^{2}= 216 sq cm.

or, (side )^{2 }=\frac{216}{6}=36 sq cm.

or, Length of one side =\sqrt{36}= 6 cm

or, Volume of a cube = (side)^{3 }= (6 cm)^{3 }= 216 cu cm.

**Question 9**

**The volume of a rectangular parallelopiped is 432 sq cm. If it is converted into two cubes of equal volumes, then let us calculate the length of each edge of each cube.**

**Solution : **

The volume of a rectangular parallelepiped is 432 cu cm

If it is bisected into two equal cubes

Volume of each cube =\frac{432}{2} = 216 cu cm

(Length of each edge)^{3 }= 216

Length of each edge =\sqrt[3]{216}= 6 cm

**Question 10**

**Each side of a cube is decreased by 50%. Let us calculate the ratio of the volumes of the original cube and the changed cube.**

**Solution : **

Let one side of a cube = x cm.

∴ Its volume = x^{3} cu.cm.

If the length is reduced by 50% then the new length of the cube =\frac{x}{2} cm.

∴ Volume of the new cube =(\frac{x}{2} cm)^3=\frac{x^3}{8}

∴ The ratio of the volume of the original cube & volume of the new cube

=x^3: \frac{x^3}{8}=1: \frac{1}{8}= 8 : 1

**Question 11**

**If the ratio of length, breadth and height of a cuboidal box is 3 : 2 : 1 and its volume is 384 cc then let us calculate the total surface area of the box.**

**Solution : **

Let the length, breadth & height of the rectangular box be 3x cm, 2x cm & x cm respectively.

According to the problem,

3x . 2x . x = 384

or, 6x^{3 }= 384

x^{3}=\frac{384}{6}= 64 = (4)^{3}

\therefore x = 4

Length = 3 × 4 = 12 cm.

Breadth = 2 × 4 = 8 cm.

& Height = 1 × 4 = 4 cm.

The total area of 6 surfaces of the rectangular parallelopiped

= 2(l × b + l × h + b × h)

= 2(12 × 8 + 12 × 4 + 8 × 4) sqcm.

= 2(96 + 48 + 32) sqcm.

= 2(96 + 48 + 32) sqcm = 2 × 176 sq cm = 352 sq cm

**Question 12**

**The inner length, breadth and height of a box of tea are 7.5 dcm, 6 dcm and 5.4 dcm respectively; if the weight of the box filled with tea is 52 kg 350 gm, but in empty state, its weight is 3.75 kg then let us write by calculating, the weight of 1 cubic dcm. tea.**

**Solution : **

Inside length = 7.5 dcm.

Breadth = 6 dcm.

and Height = 5.4 dcm.

The inside volume of the box = (7.5 × 6 × 5.4) cu. dcm. = 243 cu dcm

The weight of tea = (52.350-3.750) kg

= 48.600 kg

∴ Weight of 243 cu. dcm tea = 48.60 kg

∴ Weight of 1 cu.dcm tea =\frac{48.60}{243} kg = 0.2 kg = 200 gm

∴ Weight of 1 cu dcm tea = 200 gm

**Question 13**

**The length, breadth and weight of a brass plate with squared base are x cm, 1 mm and 4725 gm respectively, if the weight of 1 cubic dcm of brass is 8.4 gm, then let us write by calculating the value of x.**

**Solution : **

Length of the square plate = x cm.

Breadth of the square plate = x cm

Height of the square plate = 1 mm = 0.1 cm.

Volume of the square plate = 0.1 x^{2} cu. cm

Wt. of 1 cu.cm brass plate = 8.4 gm.

Wt. of 0.1 x^{2} cu.cm brass plate = 0.1 x^{2} × 8.4 gm.

According to the problem, 0.1 x^{2} × 8.4 = 4725

∴ x^{2 }= \frac{4725}{0.1 × 8.4} = 225 × 25

∴ x = \sqrt{225 × 25} = 15 × 5 = 75 cm .

∴ Value of x = 75 cm

**Question 14**

**The height of Chandamri Road is to be raised. So, 30 cuboidal holes with equal depth and of equal measure are dug out on both sides of the road and with this soil the road is elevated. If the length and breadth of each hole are 14m and 8m respectively and if the total quantity of soil required to make the road be 2520 cubic metres then let us calculate the depth of each hole.**

**Solution : **

Let the depth of each hole = x m.

Volume of each hole = 14 × 8 × x cum.

Volume of 30 holes = 30 × 14 × 8 × x cum.

According to the problem,

30 × 14 × 8 × x = 2520

∴ x =\frac{2520}{30 × 14 × 8}=\frac{6}{8}=\frac{3}{4} m

∴ Depth of each hole =\frac{3}{4} m=0.75 m.

**Question 15**

**If 64 water filled buckets of equal measure are taken out from a cubical water led tank, then \frac{1}{3} \mathrm{rd} of water remain in the tank. If the length of one edge of tank is 2m, then let us calculate and write the quantity of water that can be hold in each bucket?**

**Solution : **

Length of each side of the tank = 1.2 m

Volume of the tank = (1.2 m)^{3} m^{3 }= 1.728 m^{3} = 1728 cu.dcm. = 1728 litre.

In 64 buckets water contained = (1-\frac{1}{3})=\frac{2}{3} part of the tank

=\frac{2}{3} × 1728 litre = 2 × 576 litre.

∴ 1 bucket contains =\frac{2 × 576}{64}= 18 litres

**Question 16**

**If the length, breadth and height of one packet of one gross match box are 2.8 dcm, 1.5 dcm and 0.9 dcm respectively, then let us calculate the volume of one match box [one gross = 12 dozen]. But if the length and breadth of one match box be 5 cm and 3.5 cm, then let us calculate the height of it.**

**Solution : **

Volume of 1 gross (12 × 12) match boxes = (2.8 × 1.5 × 0.9) cu. dcm.

Volume of 1 match box =\frac{2.8 × 1.5 × 0.9}{12 × 12} cu.dcm.

=\frac{3.780}{12 × 12} \text { cu.dcm }= 0.02625 cu dcm

= 0.02625 × 100 cu cm

= 26.25 cu cm.

Let the height of a match box = h cm.

∴ Volume = 5 × 3.5 × h cu.cm.

∴ 17.5 h = 26.25

∴ h =\frac{26.25}{17.5}=1.5 cm .

∴ Height of each match box = 1.5 cm.

**Question 17**

**Half of a cuboidal water tank with length of 2.1 m and breadth of 1.5 m is filled with water. If 630 litres water is poured more into the tank, then let us calculate and write the depth that will be increased by.**

**Solution : **

Length of the tank = 2.1 m =21 dcm.

Breadth of the tank = 1.5 m =15 dcm

Let the height of increased water = x dcm.

According to the problem,

21 × 15 × x = 630.

∴ x=\frac{630}{21 × 15}=2

∴ Height of the water level increased by 2 dcm.

**Question 18**

**The length and breadth of a rectangular field of the village are 20 m and 15 m respectively. For construction of pillars in the 4 corners of that field 4 cubic holes having length of 4 m are dug out and the soils removed are dispersed on the remain. ing land. Let us calculate and write the height of the surfaces of the field that is increased by.**

**Solution : **

Area of the field = 20 m × 15 m = 300 sqm.

Area of the land for 4 pillars = 4 × (4)^{2} Sqm. = 64 sqm.

∴ Remaining part of the land = (300-64) Sqm. = 236 sqm.

Volume of ear’ removed =4 × (4)^{3} cum. = 256 cum.

Let increased height of the remaining field x m.

∴ 236 × x = 256

x =\frac{256}{236}=\frac{64}{59}=1 \frac{5}{59} m.

**Question 19**

**For elevating 6.5 dcm of a low land with length of 48 m and breadth of 31.5 m, it is decided that the soil will be collected by scooping a hole in a nearby land with length of 27 m and breadth of 18.2 m, let us calculate the depth of the hole in metre. **

**Solution : **

Let the depth of the hole = x m.

Volume of earth removed from the hole = 27 × 18.2 × x cum.

According to the problem,

27 × 18.2 × x = 48 × 31.5 × 0.65

∴ x=\frac{48 × 31.5 × 0.65}{27 × 18.2} m

=\frac{48 × 315 × 65}{27 × 182 × 100} = 2 m .

∴ Depth of the hole = 2 m.

**Question 20**

**There were 800 lit, 725 lit and 575 lit kerosene oil in three kerosine oil drums the house. The oil of these three drums is poured into a cuboidal pot and for this, the depth of oil in drums becomes 7 dcms. If the ratio of the length and breadth of the cuboidal pot is 4 : 3, then let us write by calculating the length and breadth of the pot. If the depth of the cuboidal pot would be 5 dcm, then let us calculate whether 162 lit oil can be kept or not in that pot.**

**Solution : **

Let the length & the breadth of the rectangular tank are 4x dcm & 3x dc respectively.

Total volume of soil = (800+725+575) litre.

\therefore Volume of the tank = (4x × 3x × 7) cu.dcm.

According to the problem,

4x × 3x × 7 = 2100

or, 84x^{2 }= 2100

∴ x^{2 }=\frac{2100}{84}= 25

∴ x = \sqrt{25} = 5

Length & breadth of the tank = (4 × 5) dcm and (3 × 5) dcm = 20 dcm & 15 dcm.

**2nd part :**

If the height tank = 5 dcm.

∴ Its volume will be = (20 × 15 × 5) cu.dcm. = 1500 cu.dcm = 1500 litre.

But total volume of oil = 2100 litre.

∴ It is not possible to keep 1620 litre oil in the 2nd tank.

**Question 21**

**The daily requirements of water of three families in our three-storyed flat are 1200 lit, 1050 lit and 950 lit respectively. After fulfilling these requirements in order to put up a tank again and to deposit to store 25% of the required water, only a land having the length of 2.5m and breadth of 1.6m has be produced. Let us calculate the depth of the tank in metre that should be made.**

**If the breadth of the land would be more by 4 dcm, then let us calculate the depth of the tank to be made.**

**Solution : **

Total volume of water required for 3 families = (1200+1050+950) litre = 3200 litre.

Now to keep 25% of more water, i.e., 3200 × \frac{25}{100}= 800 litre.

Total volume of water = (3200+800) = 4000 litre.

Length of the tank = 2.5 m = 25 dcm.

Breadth of the tank = 1.6 m = 16 dcm.

Let the height of the tank = x dcm.

\therefore 25 × 16 × x = 4000

\therefore x =\frac{4000}{25 × 16}= 10 dcm.= 1 m.

**2nd part :**

If the breadth of tank = (16+4) = 20 dcm.

\therefore Volume = 25 × 20 × h = 4000 [h = New height of the tank]

\therefore h =\frac{4000}{25 × 20}= 8 dcm

\therefore Height of the tank = 8 dcm. Ans.

**Question 22**

**The weight of a wooden box made of wooden planks with the thickness 5 cm along with its covering is 115.5kg. But the weight of the box filled with rice is 880.5 kg. The length and breadth of inner side of the box are 12 dcm and 8.5 dcm respectively and the weight of 1 cubic dcm rice is 1.5kg. Let us write the inner height of the box after calculation. Let us calculate the total expenditure ot colour the outside of the box, if the rate is Rs. 1.50 per sq.dcm.**

**Solution : **

Weight of rice inside the box = (880.5-115.5) kg. = 765 kg.

Volume of the box =\frac{765}{1.5}= 510 cu. dcm.

Let the height of the box = h dcm.

According to the problem,

12 × 8.5 × h = 510

h =\frac{510}{12 × 8.5} = 5 dcm.

Outside length of the box = (12+2 × 0.5) = 13 dcm.

Outside breadth of the box = (8.5+2 × 0.5) = 9.5 dcm.

Outside height of the box =(5+2 × 0.5)=6 dcm.

Total outside surface area of the box

= 2(13 × 9.5+13 × 6+9.5 × 6) sq. dcm.

= 2(123.5+78+57) = 2 × (258.5) = 517 sq. dcm.

∴ Total cost for painting the outside surface area

= 517 × Rs. 1.50 = Rs. 775.50

**Question 23**

**The depth of a cuboidal pond with length of 20 m and breadth of 18.5 m is 3.2 m, let us write by calculating the time required to irrigate whole water of the pond with a pump having the capacity to irrigate 160 kilo lit water per hour. If that quantity of water is poured on a paddy field with a ridge having the length of 59.2 m and breadth of 40 m, then what is the depth of water in that land – let us write by calculating it. [ 1 cubic metre =1 kilo litre]**

**Solution : **

Volume of the water in the rectangular tank

= (20 × 18.5 × 3.2) cu m. = 1184 cu m.

= 1184 kilolitre.

Time required to remove the water from the tank by pump =\frac{1184}{160}=\frac{37}{5} \mathrm{hr}=7 \mathrm{hr} 24 m.

**2nd part :**

Let the height of water in the paddy field = h m.

Volume of water in the paddy field will be 59.2 m × 40 m × h m =2368 h cu m

∴ 2368 h = 1184

∴ h =\frac{1184}{2368}= 0.5 m

∴ Height of water level = 0.5 m

## M.C.Q.

**Question 1**

**The inner volume of a cuboidal box is 440 cc. and the area of inner base is 88 sq cm, the inner height of the box is**

**(a) 4 cm**

**(b) 5 cm**

**(c) 3 cm**

**(d) 6 cm**

**Solution : **

(b) 5 cm

**Explanation**

Inner volume of the rectangular box = 440 cu cm and inner area of the bottom of the box = 88 sq cm.

Height of the box =\frac{440}{88} = 5 cm

**Question 2**

** The length, breadth and height of a cuboidal hole are 40 m, 12 m and 16 respectively. The number of planks having the height of 5 m the breadth of 4 m an the thickness of 2 m, can be kept in that hole is**

**(a) 190**

**(b) 192**

**(c) 184**

**(d) 180**

**Solution : **

(b) 192

**Explanation:**

The length, breadth & height of a rectangular hole are 40 m, 12 m & 16 m respectively volume = (40 × 12 × 16) m^{3}.

And the length, breadth & thickness of a rectangular wooden plank are 5 m, 4 m & 2 respectively.

∴ Its volume = (5 × 4 × 2) m^{3}

∴ No. of planks =\frac{40 × 12 × 16}{5 × 4 × 2}= 12 × 16 = 192

**Question 3**

**The surface area of a cube is 256 sq m, the volume of the cube is**

**(a) 64 cubic metre**

**(b) 216 cubic metre**

**(c) 256 cubic metre**

**(d) 512 cubic metre**

**[Hints: The number of surfaces is 6]**

**Solution : **

(d) 512 cubic metre

**Explanation**

Side faces of a cube = 256sqm

∴ Area of one surface =\frac{256}{4} = 64sqm.

[∴ Each side of the cube =\sqrt{64} = 8m

∴ Volume of the cube = (8)^{3 }= 512 m^{3 }

**Question 4**

**The ratio of the volumes of two cubes is 1 : 27, the ratio of total surface areas of two cubes is**

**(a) 1 : 3**

**(b) 1 : 8**

**(c) 1 : 9**

**(d) 1 : 18**

**Solution : **

1 : 9

**Exlanation**

As the ratio of volume of two cubes are 1 : 27.

∴ The ratio of each side of the cubes

=3 \sqrt{1}: 3 \sqrt{27}=1: 3

The ratio of total surface areas of the two cubes = 6(1)^{2}: 6(3)^{2 }= 1^{2 }: 3^{2 }= 1 : 9

**Question 5**

**If total surface area of a cube is s sq. unit and the length of the diagonal is d unit, then the relation between s and d is**

**(a) s = 6 d ^{2}**

**(b) 3s = 7d**

**(c) s ^{3 }= d^{2}**

**(d) d ^{2 }=\frac{s}{2}**

**Solution : **

d^{2}

**Explanation**

Total surface area of a cube = s sq. unit

Diagonal of a cube = d unit

Let one side of cube x unit.

\therefore Total surface area (s) = 6x^{2} sq unit

Diagonal (d)=\sqrt{3 x} unit

∴ s = 2d^{2}

∴ \frac{s}{2}= d^{2}

## Let us write True or False against the following statements :

**Question 1**

**If the length of each edge of a cube is twice of that 1st cube then the volume of this cube is 4 times more than that of the 1st cube.**

**Solution : **

False

**Question 2**

**In rainy season, the height of rainfall in 2 hectre land is 5cm, the volume of rain water is 1000 cubic metre.**

**Solution : **

True

## Let us fill in the blanks :

**Question 1**

**The number of diagonals of a cuboid is ____**

**Solution : **

No. of diagonals = 16

**Explanation**

A cube is a three-dimensional shape, that has six square faces of equal dimensions. It has 12 edges and 8 vertices. The primary diagonals of the cube are the straight lines that pass through the centre of the cube and join the opposite vertices. The diagonals of the faces of the cube are the straight lines that join the opposite vertices on each face.

Hence,

- Number of primary diagonals of cube = 4
- Number of diagonals on the faces of cube = 12
- Total diagonals of the cube = 12 + 4 = 16

**Question 2**

**The length of the diagonal on surface of a cube = __________________ x the length of one edge.**

**Solution : **

\sqrt{2}.

**Question 3**

**If the length, breadth and height of a rectangular parallelopiped are equal, then the special name of this solid is _________________.**

**Solution : **

If the length, breadth & height of a rectangular parallelopiped, it is called cube.

## S.A.

**Question 1**

**If the number of surfaces of a cuboid is x, the number of edges is y, the number of vertices is z and the number of diagonals is p, let us write the value of x – y + x + p.**

**Solution : **

No. of surfaces (x) = 6.

No. of edges (y) = 12

No. of vertices (z) = 8

No. of diagonals (p) = 4

\therefore x-y+x+p = 6-12+8+4 = 6

**Question 2**

**The lengths of the dimensions of two cuboids are 4, 6, 4 units and 8, (2h – 1), 2 units respectively. If the volumes of two cuboids are equal, then let us write the value of h.**

**Solution : **

4 × 6 × 4=8 × (2h – 1) × 2

\therefore 2h – 1=\frac{4 × 6 × 4}{8 × 2} = 6

\therefore h =\frac{6+1}{2}=\frac{7}{2} = 3.

**Question 3**

**If each edge of a cube is increased by 50%, then how much the total surface area of the cube will be increased in percent – let us write by calculating it.**

**Solution : **

Let one side of a cube = x units. \therefore Total surface area of the cube = 6x^{2} sq. unit.

Now, if length increases by 50%,

\therefore New length of the cube \frac{150}{100} x=\frac{3 x}{2} unit

& its total surface area = 6 × (\frac{3 x}{2})^{2}=\frac{27 x^{2}}{2}

Increased percentage of area \frac{(\frac{27 x^{2}}{2}-6 x^{2})}{6 x^{2}} × 100% = 125%.

**Question 4**

**The lenghts of the edges of three solid cubes are 3 cm, 4 cm and 5 cm respectively, a new solid is made by melting these three solid cubes. Let us write the length of one edge of the new cube.**

**Solution : **

Let one edge of the new cube = a unit.

\therefore a^{3} = 3^{3 }+ 4^{3 }+ 5^{3} = 27+64+

\therefore a = 3\sqrt{216} = 6

\therefore One edge of the new cube = 6 unit.

**Question 5**

**The lengths of two adjacent walls of a room are 12 m and 8 m respectively. If the height of the room is 4 m, then let us write by calculating, the area of the floor of the room.**

**Solution : **

Area of the floor of the room = 12 m × 8 m = 96 sqm.