RD Sharma solution class 8 chapter 21 Mensuration II(Volume and Surface Area of Cuboid and a Cube) Exercise 21.3

Exercise 21.3

Page-21.22

Question 1:

Find the surface area of a cuboid whose
(i) length = 10 cm, breadth = 12 cm, height = 14 cm
(ii) length = 6 dm, breadth = 8 dm, height = 10 dm
(iii) length = 2 m, breadth = 4 m, height = 5 m
(iv) length = 3.2 m, breadth = 30 dm, height = 250 cm.

Answer 1:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Dimension} \mathrm{of} \mathrm{the} \mathrm{cuboid}: \\ \mathrm{Length}=10 \mathrm{cm} \\ \mathrm{Breadth}=12 \mathrm{cm} \\ \mathrm{Height}=14 \mathrm{cm} \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (10\times 12+12\times 14+10\times 14) \\ =2\times (120+168+140) \\ =856 {\mathrm{cm}}^2 \\ \left(\mathrm{ii}\right) \\ \mathrm{Dimensions} \mathrm{of} \mathrm{the} \mathrm{cuboid}: \\ \mathrm{Length}=6 \mathrm{dm} \\ \mathrm{Breadth}=8 \mathrm{dm} \\ \mathrm{Height}=10 \mathrm{dm} \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (6\times 8+8\times 10+6\times 10) \\ =2\times (48+80+60) \\ =376 {\mathrm{dm}}^2 \\ \left(\mathrm{iii}\right) \\ \mathrm{Dimensions} \mathrm{of} \mathrm{the} \mathrm{cuboid}: \\ \mathrm{Length}=2 \mathrm{m} \\ \mathrm{Breadth}=4 \mathrm{m} \\ \mathrm{Height}=5 \mathrm{m} \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (2\times 4+4\times 5+2\times 5) \\ =2\times (8+20+10) \\ =76 {\mathrm{m}}^2 \\ \left(\mathrm{iv}\right) \\ \mathrm{Dimensions} \mathrm{of} \mathrm{the} \mathrm{cuboid}: \\ \mathrm{Length}=3.2\mathrm{m} \\ =3.2\times 10 \mathrm{dm} (1 \mathrm{m}=10 \mathrm{dm}) \\ =32 \mathrm{dm} \\ \mathrm{Breadth}=30 \mathrm{dm} \\ \mathrm{Height}=250 \mathrm{cm} \\ =250\times \frac{1}{10}\mathrm{dm} (10\mathrm{cm} = 1 \mathrm{dm}) \\ =25 \mathrm{dm} \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (32\times 30+30\times 25+32\times 25) \\ =2\times (960+750+800) \\ =5020 {\mathrm{dm}}^2 \end{gathered}$$

Question 2:

Find the surface area of a cube whose edge is
(i) 1.2 m
(ii) 27 cm
(iii) 3 cm
(iv) 6 m
(v) 2.1 m

Answer 2:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Edge} \mathrm{of} \mathrm{the} \mathrm{a} \mathrm{cube}=1.2 \mathrm{m} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (1.2{)}^2=6\times 1.44=8.64 {\mathrm{m}}^2\mathrm{.} \\ \left(\mathrm{ii}\right) \\ \mathrm{Edge} \mathrm{of} \mathrm{the} \mathrm{a} \mathrm{cube}=27 \mathrm{cm} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (27{)}^2=6\times 729=4374 {\mathrm{cm}}^2 \\ \left(\mathrm{iii}\right) \\ \mathrm{Edge} \mathrm{of} \mathrm{the} \mathrm{a} \mathrm{cube}=3 \mathrm{cm} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (3{)}^2=6\times 9=54 {\mathrm{cm}}^2 \\ \left(\mathrm{iv}\right) \\ \mathrm{Edge} \mathrm{of} \mathrm{the} \mathrm{a} \mathrm{cube}=6 \mathrm{m} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (6{)}^2=6\times 36=216 {\mathrm{m}}^2 \\ \left(\mathrm{v}\right) \\ \mathrm{Edge} \mathrm{of} \mathrm{the} \mathrm{a} \mathrm{cube}=2.1 \mathrm{m} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (2.1{)}^2=6\times 4.41=26.46 {\mathrm{m}}^2 \end{gathered}$$

Question 3:

A cuboidal box is 5 cm by 5 cm by 4 cm. Find its surface area.

Answer 3:

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{box} \mathrm{are} 5 \mathrm{cm}\times 5 \mathrm{cm}\times 4 \mathrm{cm}. \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{box}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (5\times 5+5\times 4+5\times 4) \\ =2\times (25+20+20) \\ =130 {\mathrm{cm}}^2 \end{gathered}$$

Question 4:

Find the surface area of a cube whose volume is
(i) 343 m³
(ii) 216 dm³

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{cube}=343 {\mathrm{m}}^3 \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3 \\ \Rightarrow (\mathrm{side}{)}^3=343 \\ \mathrm{i}.\mathrm{e}., \mathrm{side} =\sqrt[3]{343}=7 \mathrm{m} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (7{)}^2=294 {\mathrm{m}}^2 \\ \left(\mathrm{ii}\right) \\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{cube}=216 {\mathrm{dm}}^3 \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{volume} \mathrm{of} \mathrm{a} \mathrm{cube}=(\mathrm{side}{)}^3 \\ \Rightarrow (\mathrm{side}{)}^3=216 \\ \mathrm{i}.\mathrm{e}., \mathrm{side}=\sqrt[3]{216}=6 \mathrm{dm} \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cube}=6\times (\mathrm{side}{)}^2=6\times (6{)}^2=216 {\mathrm{dm}}^2 \end{gathered}$$

Question 5:

Find the volume of a cube whose surface area is
(i) 96 cm²
(ii) 150 m²

Answer 5:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{cube}=96 {\mathrm{cm}}^2 \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{cube}=6\times (\mathrm{side}{)}^2 \\ \Rightarrow 6\times (\mathrm{side}{)}^2=96 \\ \Rightarrow (\mathrm{side}{)}^2\mathrm{=}\frac{96}{6}=16 \\ \mathrm{i}.\mathrm{e}., \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cube}=\sqrt{16}=4 \mathrm{cm} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=(4{)}^3=64 {\mathrm{cm}}^3 \\ (\mathrm{ii}) \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{cube}=150 {\mathrm{m}}^2 \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{cube}=6\times (\mathrm{side}{)}^2 \\ \Rightarrow 6\times (\mathrm{side}{)}^2=150 \\ \Rightarrow (\mathrm{side}{)}^2\mathrm{=}\frac{150}{6}=25 \\ \mathrm{i}.\mathrm{e}., \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cube}=\sqrt{25}=5 \mathrm{m} \\ \therefore \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cube}=(\mathrm{side}{)}^3=(5{)}^3=125 {\mathrm{m}}^3 \end{gathered}$$

Question 6:

The dimensions of a cuboid are in the ratio 5 : 3 : 1 and its total surface area is 414 m². Find the dimensions.

Answer 6:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{cuboid} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 5:3:1. \\ \mathrm{Suppose} \mathrm{that} \mathrm{its} \mathrm{sides} \mathrm{are} x \mathrm{multiple} \mathrm{of} \mathrm{each} \mathrm{other}, \mathrm{then} \mathrm{we} \mathrm{have}: \\ \mathrm{Length}=5x \mathrm{m} \\ \mathrm{Breadth}=3x \mathrm{m} \\ \mathrm{Height}=x \mathrm{m} \\ \mathrm{Also}, \mathrm{total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=414 {\mathrm{m}}^2 \\ \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ \Rightarrow 414=2\times (5x\times 3x+3x\times 1x+5x\times x) \\ \Rightarrow 414=2\times (15x^2+3x^2+5x^2) \\ \Rightarrow 414=2\times \left(23x^2\right) \\ \Rightarrow 2\times (23\times x^2)=414 \\ \Rightarrow (23\times x^2)=\frac{414}{2}=207 \\ \Rightarrow x^2\mathrm{=}\frac{207}{23}=9 \\ \Rightarrow x=\sqrt{9}=3 \\ \mathrm{Therefore}, \mathrm{we} \mathrm{have} \mathrm{the} \mathrm{following}: \\ \mathrm{Lenght} \mathrm{of} \mathrm{the} \mathrm{cuboid}=5\times \mathrm{x}=5\times 3=15 \mathrm{m} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{cuboid}=3\times \mathrm{x}=3\times 3=9 \mathrm{m} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{cuboid}=\mathrm{x}=1\times 3=3 \mathrm{m} \end{gathered}$$

Question 7:

Find the area of the cardboard required to make a closed box of length 25 cm, 0.5 m and height 15 cm.

Answer 7:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{box}=25 \mathrm{cm} \\ \mathrm{Width} \mathrm{of} \mathrm{the} \mathrm{box}=0.5 \mathrm{m} \\ =0.5\times 100 \mathrm{cm} (\because 1 \mathrm{m}= 100 \mathrm{cm}) \\ =50 \mathrm{cm} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{box}=15 \mathrm{cm} \\ \therefore \mathrm{Surface} \mathrm{are}a \mathrm{of} \mathrm{the} \mathrm{box}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (25\times 50+50\times 15+25\times 15) \\ =2\times (1250+750+375) \\ =4750 {\mathrm{cm}}^2 \end{gathered}$$

Question 8:

Find the surface area of a wooden box whose shape is of a cube, and if the edge of the box is 12 cm.

Answer 8:

$$\begin{gathered} \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{cubical} \mathrm{wooden} \mathrm{box} \mathrm{is} 12 \mathrm{cm}. \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cubical} \mathrm{box}=6\times (\mathrm{side}{)}^2=6\times (12{)}^2=864 {\mathrm{cm}}^2 \end{gathered}$$

Question 9:

The dimensions of an oil tin are 26 cm × 26 cm × 45 cm. Find the area of the tin sheet required for making 20 such tins. If 1 square metre of the tin sheet costs Rs 10, find the cost of tin sheet used for these 20 tins.

Answer 9:

$$\begin{gathered} \mathrm{Dimensions} \mathrm{of} \mathrm{the} \mathrm{oil} \mathrm{tin} \mathrm{are} 26 \mathrm{cm}\times 26 \mathrm{cm}\times 45 \mathrm{cm}. \\ \mathrm{So}, \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{tin} \mathrm{sheet} \mathrm{required} \mathrm{to} \mathrm{make} \mathrm{one} \mathrm{tin}=2\times (\mathrm{length}\times \mathrm{breadth}+\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (26\times 26+26\times 45+26\times 45) \\ =2\times (676+1170+1170)=6032 {\mathrm{cm}}^2 \\ \mathrm{Now}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{tin} \mathrm{sheet} \mathrm{required} \mathrm{to} \mathrm{make} 20 \mathrm{such} \mathrm{tin}s=20\times \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{one} \mathrm{tin} \\ =20\times 6032 \\ =120640 {\mathrm{cm}}^2 \\ \mathrm{It} \mathrm{can} \mathrm{be} \mathrm{observed} \mathrm{that} 120640 {\mathrm{cm}}^2=120640\times 1\mathrm{cm}\times 1\mathrm{cm} \\ =120640\times \frac{1}{100}\mathrm{m}\times \frac{1}{100}\mathrm{m} (\because 100 \mathrm{cm}=1 \mathrm{m}) \\ =12.0640 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{cost} \mathrm{of} 1 {\mathrm{m}}^2 \mathrm{of} \mathrm{tin} \mathrm{sheet}=\mathrm{Rs} 10 \\ \therefore \mathrm{The} \mathrm{cost} \mathrm{of} 12.0640 {\mathrm{m}}^2 \mathrm{of} \mathrm{tin} \mathrm{sheet}=12.0640\times 10=\mathrm{Rs} 120.6 \end{gathered}$$

Question 10:

A cloassroom is 11 m long, 8 m wide and 5 m high. Find the sum of the areas of its floor and the four walls (including doors, windows, etc.)

Answer 10:

$$\begin{gathered} \mathrm{Lenght} \mathrm{of} \mathrm{the} \mathrm{classroom}=11 \mathrm{m} \\ \mathrm{Width}=8 \mathrm{m} \\ \mathrm{Height}=5 \mathrm{m} \\ \mathrm{We} \mathrm{have} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{its} \mathrm{floor} \mathrm{and} \mathrm{the} \mathrm{four} \mathrm{walls} (\mathrm{i}.\mathrm{e}., \mathrm{like} \mathrm{an} \mathrm{open} \mathrm{box}). \\ \therefore \mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{and} \mathrm{the} \mathrm{four} \mathrm{walls}=(\mathrm{length}\times \mathrm{width})+2\times (\mathrm{width}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =(11\times 8)+2\times (8\times 5+11\times 5) \\ =88+2\times (40+55) \\ =88+190 \\ =278 {\mathrm{m}}^2 \end{gathered}$$

Question 11:

A swimming pool is 20 m long 15 m wide and 3 m deep. Find the cost of repairing the floor and wall at the rate of Rs 25 per square metre.

Answer 11:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{swimming} \mathrm{pool}=20 \mathrm{m} \\ \mathrm{Breadth}=15 \mathrm{m} \\ \mathrm{Height}=3 \mathrm{m} \\ \mathrm{Now}, \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{and} \mathrm{all} \mathrm{four} \mathrm{walls} \mathrm{of} \mathrm{the} \mathrm{pool}=(\mathrm{length}\times \mathrm{breadth})+2\times (\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =(20\times 15)+2\times (15\times 3+20\times 3) \\ =300+2\times (45+60) \\ =300+210 \\ =510 {\mathrm{m}}^2 \\ \mathrm{The} \mathrm{cost} \mathrm{of} \mathrm{repairing} \mathrm{the} \mathrm{floor} \mathrm{and} \mathrm{the} \mathrm{walls} \mathrm{is} \mathrm{Rs} 25/{\mathrm{m}}^2. \\ \therefore \mathrm{The} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{repairing} 510 {\mathrm{m}}^2 \mathrm{area}=510\times 25=\mathrm{Rs} 12750 \end{gathered}$$

Question 12:

The perimeter of a floor of a room is 30 m and its height is 3 m. Find the area of four walls of the room.

Answer 12:

$$\begin{gathered} \mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{of} \mathrm{the} \mathrm{room}=30 \mathrm{m} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{oom}=3 \mathrm{m} \\ \mathrm{Perimeter} \mathrm{of} \mathrm{a} \mathrm{rectangle}=2\times (\mathrm{length}+\mathrm{breadth})=30 \mathrm{m} \\ \mathrm{So}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{four} \mathrm{walls}=2\times (\mathrm{length}\times \mathrm{height}+\mathrm{breadth}\times \mathrm{height}) \\ =2\times (\mathrm{length}+\mathrm{breadth})\times \mathrm{height} \\ =30\times 3=90 {\mathrm{m}}^2 \end{gathered}$$

Question 13:

Show that the product of the areas of the floor and two adjacent walls of a cuboid is the square of its volume.

Answer 13:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{length}, \mathrm{breadth} \mathrm{and} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cuboidal} \mathrm{floor} \mathrm{are} l \mathrm{cm}, b \mathrm{cm} \mathrm{and} h \mathrm{cm}, \mathrm{respectively}. \\ \mathrm{Then}, \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{floor}=l\times b {\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{the} \mathrm{wall}=b\times h {\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{its} \mathrm{adjacent} \mathrm{wall}=l\times h {\mathrm{cm}}^2 \\ \mathrm{Now}, \mathrm{product} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{and} \mathrm{the} \mathrm{two} \mathrm{adjacent} \mathrm{walls}=(l\times b)\times (b\times h)\times (l\times h)=l^2\times b^2\times h^2=(l\times b\times h{)}^2 \\ \mathrm{Also}, \mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{cuboid}=l\times b\times h {\mathrm{cm}}^2 \\ \therefore \mathrm{Product} \mathrm{of} \mathrm{the} \mathrm{areas} \mathrm{of} \mathrm{the} \mathrm{floor} \mathrm{and} \mathrm{the} \mathrm{two} \mathrm{adjacent} \mathrm{walls}=(l\times b\times h{)}^2=(\mathrm{volume}{)}^2 \end{gathered}$$

Question 14:

The walls and ceiling of a room are to be plastered. The length, breadth and height of the room are 4.5 m, 3 m and 350 cm, respectively. Find the cost of plastering at the rate of Rs 8 per square metre.

Answer 14:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{a} \mathrm{room}=4.5 \mathrm{m} \\ \mathrm{Breadth}=3 \mathrm{m} \\ \mathrm{Height}=350 \mathrm{cm} \\ =\frac{350}{100}\mathrm{m} (\because 1 \mathrm{m}= 100 \mathrm{cm} ) \\ =3.5 \mathrm{m} \\ \mathrm{Since} \mathrm{only} \mathrm{the} \mathrm{walls} \mathrm{and} \mathrm{the} \mathrm{ceiling} \mathrm{of} \mathrm{the} \mathrm{room} \mathrm{are} \mathrm{to} \mathrm{be} \mathrm{plastered}, \mathrm{we} \mathrm{have}: \\ \mathrm{So}, \mathrm{total} \mathrm{area} \mathrm{to} \mathrm{be} \mathrm{plastered}=\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{ceiling}+\mathrm{area} \mathrm{of} \mathrm{the} \mathrm{walls} \\ =(\mathrm{length}\times \mathrm{breadth})+2\times (\mathrm{length}\times \mathrm{height}+\mathrm{breadth}\times \mathrm{height}) \\ =(4.5\times 3)+2\times (4.5\times 3.5+3\times 3.5) \\ =13.5+2\times (15.75+10.5) \\ =13.5+2\times (26.25) \\ =66 {\mathrm{m}}^2 \\ \mathrm{Again}, \mathrm{cost} \mathrm{of} \mathrm{plastering} \mathrm{an} \mathrm{area} \mathrm{of} 1 {\mathrm{m}}^2=\mathrm{Rs} 8 \\ \therefore \mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{plastering} \mathrm{an} \mathrm{area} \mathrm{of} 66 {\mathrm{m}}^2=66\times 8=\mathrm{Rs} 528 \end{gathered}$$

Page-21.23

Question 15:

A cuboid has total surface area of 50 m² and lateral surface area is 30 m². Find the area of its base.

Answer 15:

$$\begin{gathered} \mathrm{Total} \mathrm{sufrace} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=50 {\mathrm{m}}^2 \\ \mathrm{Its} \mathrm{lateral} \mathrm{surface} \mathrm{area}=30 {\mathrm{m}}^2 \\ \mathrm{Now}, \mathrm{total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cuboid}=2\times (\mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{base})+(\mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} 4 \mathrm{walls}) \\ \Rightarrow 50=2\times (\mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{base})+\left(30\right) \\ \Rightarrow 2\times (\mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{base})=50-30=20 \\ \therefore \mathrm{Surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{base}=\frac{20}{2}=10 {\mathrm{m}}^2 \end{gathered}$$

Question 16:

A classroom is 7 m long, 6 m broad and 3.5 m high. Doors and windows occupy an area of 17 m². What is the cost of white-washing the walls at the rate of Rs 1.50 per m².

Answer 16:

$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{classroom}=7\mathrm{m} \\ \mathrm{Breadth} \mathrm{of} \mathrm{the} \mathrm{classroom}=6 \mathrm{m} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{classroom}=3.5 \mathrm{m} \\ \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{classroom} \mathrm{to} \mathrm{be} \mathrm{whitewashed}=\mathrm{areas} \mathrm{of} \mathrm{the} 4 \mathrm{walls} \\ =2\times (\mathrm{breadth}\times \mathrm{height}+\mathrm{length}\times \mathrm{height}) \\ =2\times (6\times 3.5+7\times 3.5) \\ =2\times (21+24.5) \\ =91 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{the} \mathrm{doors} \mathrm{and} \mathrm{windows} \mathrm{occupy} 17 {\mathrm{m}}^2. \\ \mathrm{So}, \mathrm{the} \mathrm{remaining} \mathrm{area} \mathrm{to} \mathrm{be} \mathrm{whitewashed}=91-17=74 {\mathrm{m}}^2 \\ \mathrm{Given} \mathrm{that} \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{whitewashing} 1 {\mathrm{m}}^2 \mathrm{of} \mathrm{wall}=\mathrm{Rs} 1.50 \\ \therefore \mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{whitewashing} 74 {\mathrm{m}}^2 \mathrm{of} \mathrm{area}=74\times 1.50=\mathrm{Rs} 111 \end{gathered}$$

Question 17:

The central hall of a school is 80 m long and 8 m high. It has 10 doors each of size 3 m × 1.5 m and 10 windows each of size 1.5 m × 1 m. If the cost of white-washing the walls of the hall at the rate of Rs 1.20 per m² is Rs 2385.60, fidn the breadth of the hall.

Answer 17:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{hall} \mathrm{is} b \mathrm{m}. \\ \mathrm{Lenght} \mathrm{of} \mathrm{the} \mathrm{hall}=80 \mathrm{m} \\ \mathrm{Height} \mathrm{of} \mathrm{the} \mathrm{hall}=8 \mathrm{m} \\ \mathrm{Total} \mathrm{surface} \mathrm{area} \mathrm{of} 4 \mathrm{walls} \mathrm{including} \mathrm{doors} \mathrm{and} \mathrm{windows}=2\times (\mathrm{length}\times \mathrm{height}+\mathrm{breadth}\times \mathrm{height}) \\ =2\times (80\times 8+\mathrm{b}\times 8) \\ =2\times (640+8\mathrm{b}) \\ =1280+16\mathrm{b} {\mathrm{m}}^2 \\ \mathrm{The} \mathrm{walls} \mathrm{have} 10 \mathrm{doors} \mathrm{each} \mathrm{of} \mathrm{dimensions} 3 \mathrm{m}\times 1.5 \mathrm{m}. \\ \mathrm{i}.\mathrm{e}., \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{door}=3\times 1.5=4.5 {\mathrm{m}}^2 \\ \therefore \mathrm{Area} \mathrm{of} 10 \mathrm{doors}=10\times 4.5=45 {\mathrm{m}}^2 \\ \mathrm{Also}, \mathrm{there} \mathrm{are} 10 \mathrm{windows} \mathrm{each} \mathrm{of} \mathrm{dimensions} 1.5 \mathrm{m}\times 1 \mathrm{m}. \\ \mathrm{i}.\mathrm{e}., \mathrm{area} \mathrm{of} \mathrm{one} \mathrm{window}=1.5\times 1=1.5 {\mathrm{m}}^2 \\ \therefore \mathrm{Area} \mathrm{of} 10 \mathrm{windows}=10\times 1.5=15 {\mathrm{m}}^2 \\ \mathrm{Thus}, \mathrm{total} \mathrm{area} \mathrm{to} \mathrm{be} \mathrm{whitwashed}=(\mathrm{total} \mathrm{area} \mathrm{of} 4 \mathrm{walls})-(\mathrm{areas} \mathrm{of} 10 \mathrm{doors}+\mathrm{areas} \mathrm{of} 10 \mathrm{windows}) \\ =(1280+16b)-(45+15) \\ =1280+16b-60 \\ =1220+16b {\mathrm{m}}^2 \\ \mathrm{It} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{cost} \mathrm{of} \mathrm{whitewashing} 1 {\mathrm{m}}^2 \mathrm{of} \mathrm{area}=\mathrm{Rs} 1.20 \\ \therefore \mathrm{Total} \mathrm{cost} \mathrm{of} \mathrm{whitewashing} \mathrm{the} \mathrm{walls}=(1220+16b)\times 1.20 \\ =1220\times 1.20+16b\times 1.20 \\ =1464+19.2b \\ \mathrm{Since} \mathrm{the} \mathrm{total} \mathrm{cost} \mathrm{of} \mathrm{whitewashing} \mathrm{the} \mathrm{walls} \mathrm{is} \mathrm{Rs} 2385.60, \mathrm{we} \mathrm{have}: \\ 1464+19.2b=2385.60 \\ \Rightarrow 19.2b=2385.60-1464 \\ \Rightarrow 19.2b=921.60 \\ \Rightarrow b=\frac{921.60}{19.2}=48 \mathrm{m} \\ \therefore \mathrm{The} \mathrm{breadth} \mathrm{of} \mathrm{the} \mathrm{central} \mathrm{hall} \mathrm{is} 48 \mathrm{m}. \end{gathered}$$