Mathematics Solutions Solutions for Class 8 Math Chapter 1 Rational And Irrational Numbers are provided here with simple step-by-step explanations. These solutions for Rational And Irrational Numbers are extremely popular among Class 8 students for Math Rational And Irrational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

#### Page No 108:

#### Question 1:

Find the volume of a box if its length, breadth and height are 20 cm, 10.5 cm and 8 cm respectively.

#### Answer:

$\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{box}=\mathrm{length}\times \mathrm{breadth}\times \mathrm{height}\phantom{\rule{0ex}{0ex}}=20\times 10.5\times 8\phantom{\rule{0ex}{0ex}}=1680{\mathrm{cm}}^{2}$

#### Page No 108:

#### Question 2:

A cuboid shape soap bar has volume 150 cc. Find its thickness if its length is 10 cm and breadth is 5 cm.

#### Answer:

Volume of soap bar = 150 cc

$lbh=150\phantom{\rule{0ex}{0ex}}\Rightarrow 10\times 5\times h=150\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{150}{10\times 5}=3\mathrm{cm}$

#### Page No 108:

#### Question 3:

How many bricks of length 25 cm, breadth 15 cm and height 10 cm are required to build a wall of length 6 m, height 2.5 m and breadth 0.5 m?

#### Answer:

Let the number of bricks required be *n. *

Volume of each brick = $25\times 15\times 10=3750{\mathrm{cm}}^{3}$

Volume of the wall = $6\times 2.5\times 0.5=7.5{\mathrm{m}}^{3}$

Number of bricks required(*n*) = $\frac{\mathrm{Vol}\mathrm{of}\mathrm{wall}}{\mathrm{Vol}\mathrm{of}\mathrm{brick}}=\frac{7.5\times 1000000}{3750}=2000$

Thus, 2000 bricks are required to make the wall.

#### Page No 109:

#### Question 4:

For rain water harvesting a tank of length 10 m, breadth 6 m and depth 3 m is built. What is the capacity of the tank? How many litre of water can it hold?

#### Answer:

length = 10 m

breadth = 6 m

depth = 3 m

Volume of the tank = $10\times 6\times 3=180{\mathrm{m}}^{3}$

$1{\mathrm{m}}^{3}=1000\mathrm{L}\phantom{\rule{0ex}{0ex}}\Rightarrow 180{\mathrm{m}}^{3}=180000\mathrm{L}$

#### Page No 110:

#### Question 1:

In each example given below, radius of base of a cylinder and its height are given. Then find the curved surface area and total surface area.

(1) *r* = 7 cm, *h* = 10 cm

(2) *r* = 1.4 cm, *h* = 2.1 cm

(3) *r* = 2.5 cm, *h* = 7 cm

(4) *r* = 70 cm, *h* = 1.4 cm

(5) *r* = 4.2 cm, *h* = 14 cm

#### Answer:

We know that

Curved surface area of a cylinder = $2\mathrm{\pi}rh$

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$

(1) *r* = 7 cm, *h* = 10 cm

Curved surface area of a cylinder = $2\mathrm{\pi}rh$ = $2\mathrm{\pi}\times 7\times 10=140\mathrm{\pi}=140\times \frac{22}{7}=440$ sq cm

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$ = $2\times \frac{22}{7}\times 7\left(10+7\right)=748$ sq cm

(2) *r* = 1.4 cm, *h* = 2.1 cm

Curved surface area of a cylinder = $2\mathrm{\pi}rh$ = $2\times \frac{22}{7}\times 1.4\times 2.1=18.48$ sq cm

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$ = $2\times \frac{22}{7}\times 1.4\left(2.1+1.4\right)=30.8$ sq cm

(3) *r* = 2.5 cm, *h* = 7 cm

Curved surface area of a cylinder = $2\mathrm{\pi}rh$ = $2\times \frac{22}{7}\times 2.5\times 7=110$ sq cm

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$ = $2\times \frac{22}{7}\times 2.5\left(2.5+7\right)=149.28$ sq cm

(4) *r* = 70 cm, *h* = 1.4 cm

Curved surface area of a cylinder = $2\mathrm{\pi}rh$ = $2\times \frac{22}{7}\times 70\times 1.4=616$ sq cm

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$ = $2\times \frac{22}{7}\times 70\left(70+1.4\right)=31416$ sq cm

(5) *r* = 4.2 cm, *h* = 14 cm

Curved surface area of a cylinder = $2\mathrm{\pi}rh$ = $2\times \frac{22}{7}\times 4.2\times 14=369.6$ sq cm

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$ = $2\times \frac{22}{7}\times 4.2\left(4.2+14\right)=480.48$ sq cm

#### Page No 110:

#### Question 2:

Find the total surface area of a closed cylindrical drum if its diameter is 50 cm and height is 45 cm. (π = 3.14)

#### Answer:

Diameter = 50 cm

Radius = 25 cm

Height = 45 cm

Total surface area = $2\mathrm{\pi}r\left(h+r\right)$

$=2\times 3.14\times 25\left(25+45\right)\phantom{\rule{0ex}{0ex}}=10990\mathrm{sq}\mathrm{cm}$

Thus, the total surface area of the closed cylinderical drum is 10990 sq cm.

#### Page No 111:

#### Question 3:

Find the area of base and radius of a cylinder if its curved surface area is 660 sqcm and height is 21 cm.

#### Answer:

Curves surface area = 660 sq cm

Height = 21 cm

Curved surface area = $2\mathrm{\pi}rh=2\times \frac{22}{7}\times r\times 21=660\phantom{\rule{0ex}{0ex}}$

$\Rightarrow 132r=660\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{660}{132}=5\mathrm{cm}$

Area of base = $\mathrm{\pi}{r}^{2}=\frac{22}{7}\times {5}^{2}=78.57\mathrm{sq}\mathrm{cm}$

#### Page No 111:

#### Question 4:

Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.

#### Answer:

Diameter = 28 cm

Radius = 14 cm

Height = 20 cm

Curved surface area of the cylinder + area of the base = area of the sheet

$\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{sheet}=2\mathrm{\pi}rh+\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{sheet}=\mathrm{\pi}r\left(2h+r\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{sheet}=\frac{22}{7}\times 14\left(2\times 20+14\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{sheet}=2376$

Thus, area of sheet required = 2376 cm^{2 }

Area of sheet required to make a lid of height 2 cm will be

$\mathrm{Area}\mathrm{of}\mathrm{sheet}=2\mathrm{\pi}rh+\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}r\left(2h+r\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 14\left(2\times 2+14\right)\phantom{\rule{0ex}{0ex}}=44\times 18\phantom{\rule{0ex}{0ex}}=792$

Thus, area of sheet used to make the lid is 792 sq cm.

#### Page No 112:

#### Question 1:

Find the volume of the cylinder if height (*h*) and radius of the base (*r*) are as given below.

(1) *r* = 10.5 cm, *h* =8 cm

(2) *r* = 2.5 m, *h* = 7 m

(3) *r* = 4.2 cm, *h* = 5 cm

(4) *r* = 5.6 cm, *h* = 5 cm

#### Answer:

Volume of the cylinder = $\mathrm{\pi}{r}^{2}h$

(1) *r* = 10.5 cm, *h* =8 cm

Volume = $\mathrm{\pi}{r}^{2}h$ = $\frac{22}{7}\times 10.5\times 10.5\times 8=2772$ cubic cm

(2) *r* = 2.5 m, *h* = 7 m

Volume = $\mathrm{\pi}{r}^{2}h$ = $\frac{22}{7}\times 2.5\times 2.5\times 7=137.5$ cubic cm

(3) *r* = 4.2 cm, *h* = 5 cm

Volume = $\mathrm{\pi}{r}^{2}h$ = $\frac{22}{7}\times 4.2\times 4.2\times 5=277.2$ cubic cm

(4) *r* = 5.6 cm, *h* = 5 cm

Volume = $\mathrm{\pi}{r}^{2}h$ = $\frac{22}{7}\times 5.6\times 5.6\times 5=492.8$ cubic cm

#### Page No 112:

#### Question 2:

How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?

#### Answer:

Total length = 90 cm

Diameter = 1.4 cm

Radius = 0.7 cm

Total iron required = Volume of cylinder

= $\mathrm{\pi}{r}^{2}h=\frac{22}{7}\times 0.7\times 0.7\times 90=138.6$ cm^{3 }

Thus, 138.6 cm^{3 }of iron is required.

#### Page No 112:

#### Question 3:

How much water will a tank hold if the interior diameter of the tank is 1.6 m and its depth is 0.7 m?

#### Answer:

Diameter = 1.6 m

Radius = 0.8 m

Depth = 0.7 m

Volume of the tank = ${\mathrm{\pi r}}^{2}\mathrm{h}=\frac{22}{7}\times {\left(0.8\right)}^{2}\times 0.7=1.408$ m^{3}.

$1{\mathrm{m}}^{3}=1000\mathrm{L}\phantom{\rule{0ex}{0ex}}1.408{\mathrm{m}}^{3}=1.408\times 1000=1408\mathrm{L}$

Thus, the tank will hold 1408 L.

#### Page No 112:

#### Question 4:

Find the volume of the cylinder if the circumference of the cylinder is 132 cm and height is 25 cm.

#### Answer:

Circumference = 132 cm

$\Rightarrow 2\mathrm{\pi}r=132\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times \frac{22}{7}r=132\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{132\times 7}{2\times 22}=21\mathrm{cm}$

Height = 25 cm

Volume = $\mathrm{\pi}{r}^{2}h=\frac{22}{7}\times 21\times 21\times 25=34650$ cm^{3}

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