•• Mensuration

Learn and Remember

1. Perimeter: It is the distance along the line forming a closedfigure when you go round the figure once.Perimeter of Regular ShapesPerimeter of square = 4 x side.Perimeter of rectangle = 2 x (length + breadth)Equilateral triangle = 3 x side

2. Area: The amount of a surface enclosed by a closed figure iscalled its area.Area of Regular FigureSquare = side x sideRectangle = length x breadthCircle = 1t(radius)2To calculate the area of a figure using a squared paper, thefollowing conventions are adopted:

(a) Ignore position of the area that are less than half a square.(b) If more than half a square is in a region count it as one square.(c) If exactly halfthe square is in a region count it as one square.

TEXTBOOK QUESTIONS SOLVEDEXERCISE 10.1Q1. Find the perimeter of each of the following figures:

~v~

15 em

15em

o,/ 13

5em 40 em

(b) (c)(a)

126 MATHEMATiCS-VI

Oc,~"tC'/);I

% ~< ::t

4em

.~0.5em '?'SC

1t r::::;------:-4-e-m-~~. ~'&

em+ 4 em ----+ )fc,~

0.5 em 'l-~.X'.(e)(d)

1 em 4 em.--; w E

3emo o E3 CO) o

(\J

2em 3em

3em 2emE w Ec o(\J 3 o

3em CO)

II'-----

1 emEov

Eov

1 em4em

1 em(f)

Sol. (a)

~c,~<C'/);I

(")

/ ,35em

Perimeter = 4 em + 2 em + 1 em + 5 em = 12 em.

(b)

40 emPerimeter = 23 em + 35 em + 40 em + 35 em = 133 em.

15 em

(c)

§.:::

15 emPerimeter = 15 em + 15 em + 15 em + 15 em = 60 em.

MENSURATION 127

(d) Oc,~"tC'/);I

~ -A.U o< ::t

4emPerimeter = 4 em + 4 em + 4 em + 4 em + 4 em = 20 em.

(e) .~05em '?'sr.--~--'-~ C'/);It 4 em ----+ '&

1 em+ 4 em ----+ )fc,~

0.5 em 'l-~.X'Perimeter = 1 em + 4 em + 0.5 em + 2.5 em + 2.5 em + 0.5 em + 4 em

= 15 em.

(j)1 em 4 em~

i w E3em

t o c E3 CO) o

(\J

2em 3em

3em 2emE co Eo o(\J 3 o

3em CO)

~

1 emEov

5v

1 em4em

1 emPerimeter = 4 em + 1 em + 3 em + 2 em + 3 em + 4 em + 1 em + 3 em + 2 em

+ 3 em + 4 em + 1 em + 3 em + 2 em + 3 em + 4em + 1 em + 3 em+2em+3em

= 52 em.Q2. The lid of a rectangular box of sides 40 em by 10 em is sealed all

round with tape. What is the length of the tape required?

40 em

II I~40 em

Thtallength of tape required is equal to the perimeter of rectangle.Perimeter ofreetangular box = 2 x (length + breadth) = 2 x (40em + 10 em)

= 2 x (50 em) = 100 em = 1 m.Therefore the total length of tape required is 100 em or 1 m.

Sol.

128 MATHEMATIcs-VI

Q3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is theperimeter of the table-top?

Sol. Length of table top = 2 m 25 em = 2.25 mBreadth of table top = 1 m 50 em = 1.50 mPerimeter of table top = 2 x (length + breadth)

2 x (2.25 m + 1.50 m)2 x (3.75 m) = 7.50 m.

Thus perimeter of table top is 7.5 m.Q4. What is the length of the wooden strip required to frame a

photograph of length and breadth 32 cm and 21 em,respectively?

Sol. Length of wooden strip is equal to the perimeter of the photograph.Perimeter of photograph = 2 x (length + breadth)

= 2 x (32 em + 21 em)= 2 x (53 em) = 106 em.

Thus the length of the wooden strip required is equal to 106 em.Q5. A rectangular piece of land measures 0.7 km by 0.5 km. Each

side is to be fenced with 4 rows of wires. What is the length ofthe wire needed?

Sol. Sinee the 4 rows of wires are needed. Therefore the total length ofwires is equal to 4 times the perimeter of reetangle.Perimeter of field = 2 x (length + breadth)

= 2 x (0.7 km + 0.5 km) = 2 x (1.2 km)= 2.4 km = 2.4 x 1000 m = 2400 m

Thus the length of wire = 4 x 2400 m = 9600 m or 9.6 km.Q6. Find the perimeter of each of the following shapes.

(a) A triangle of sides 3 cm, 4 cm and 5 em,(b) An equilateral triangle of side 9 cm.(c) An isosceles triangle with equal sides 8 cm each and third

side6cm ••Sol. (a) A.s/>«:r>.

B 5cm C

Perimeter of MBC = AB + BC + AC= 3 em + 5 em + 4 em = 12 em.

A

_§1\09,,~LiB s cm C

(b)

MENSURATION 129

Perimeter of MBC = AB + BC + AC= 9 em + 9 em + 9 em = 27 em.

Q7.

_§AO'o..~B 6cm C

Perimeter of MBC = AB + BC + AC= 8 em + 6 em + 8 em = 22 em.

Find the perimeter of a triangle with sides measuring 10 cm, 14cmandl5cm.Perimeter of triangle = Sum of length of sides

= 10 em + 14 em + 15 em = 39 em.Find the perimeter of a regular hexagon with each sidemeasuring 8 m.Perimeter of hexagon = 6 x length of one side

=6 x8m=48m.Thus the perimeter of hexagon is 48 m.Find the side of the square whose perimeter is 20 m.Perimeter of square = 4 x side:. 4 x side = 20 m

(c)

SoL

Q8.

Sol.

Q9.Sol.

side20

= - m e fi m.4

Thus side of square is 5 m.QI0. The perimeter of a regular pentagon is 100 em. How long is its

each side?Sol. Perimeter of regular pentagon = 5 x side.

Therefore 5 x side = 100 em

side = 100 em = 20 em5

Thus side of regular pentagon is 20 em.QU. A piece of string is 30 cm long. What will be the length of each

side if the string is used to form.(a) a square? (b) an equilateral triangle?(c) a regular hexagon?

Sol. Length of string is equal to the perimeter of eaeh figure.(a) For a square :

Perimeter of square = 4 x sideThus 4 x side = 30 em

130 MATHEMATICS-VI

'd 30 5S1 e = - cm = 7. cm4

Thus length of each side of square is 7.5 em.An equilateral trianglePerimeter of equilateral triangle = 3 x sideThus 3 x side = 30 em

30- = 10cm3

Thus the length of each side of equilateral!! is 10 cm.(c) A regular hexagon

Perimeter of hexagon6 x side

(b)

side

6 x side30 cm

id 30 5S1 e = - = cm6

Thus the length of each side is 5 cm.Q12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of

the triangle is 36 cm. What is the third side?Sol. Let the length of third side = x

The length of other two sides are = 12 cm, 14 cmTherefore 12 em + 14 cm + x = 36 cm

26 cm + x = 36 emx = 36 em - 26 emx = 10 em

Thus the length of third side is 10 cm.Q13. Find the cost of fencing a square park of side 250 m at the rate of

~ 20 per metre.Sol. Side of square = 250 m

Perimeter of square = 4 x side= 4 x 250 = 1000 m

Cost of fencing = ~ 20 per metre.Therefore total cost offencing = ~ 20 x 1000 = ~ 20,000.

Q14. Find the cost of fencing a rectangular park oflength 175 m andbreadth 125 m at the rate of ~ 12 per metre.

Sol. Length of rectangular park = 175 mBreadth of rectangular park = 125 m

Perimeter of park = 2 x (length + breadth)= 2(175 + 125) m= 2(300) m = 600 m

Cost of fencing park = ~ 12 per metreTotal cost of fencing = ~ 12 x 600 = ~ 7200.Thus the total cost of fencing the park is ~ 7200.

Q15. Sweety runs around a square park of side 75 m. Bulbul runsaround a rectangular park with length 60 m and breadth 45 m.Who covers less distance?

MENSURATION

Sol. Distance covered by Sweety is equal to the perimeter of square park.Perimeter of squared park = 4 x side

= 4 x 75 = 300 m.Thus distance cover by Sweety is 300 m.Distance covered by Bulbul is equal to the perimeter of rectangularpark.Perimeter of rectangular park = 2 x (length + breadth)

= 2(60 + 45) m= 2(105) m = 210 m

Thus distance cover by Bulbul is 210.

Thus Bulbul covers less distance.Q16. What is the perimeter of each of the following figures? What do

you infer from the answers.

25cmEDI\)o c.nII) 0C\I 3

40cm

25cm

(a)

~I I;40cm

(b)

30cm "n0Ar>.40cm

EooC\I

I\)oo3

30cm(c) (d)

Sol. (a) Square of 25 cm side:Perimeter of square = 4 x side

= 4 x 25 = 100 cm.(b) Rectangle of 40 cm by 10 cm

Perimeter of rectangle = 2 x (length + breadth)= 2(40 + 10)= 2(50) = 100 em.

(c) Rectangle of 30 cm by 20 cmPerimeter of rectangle = 2 x (length + breadth)

= 2(30 + 20) cm = 100 em.(d) Isosceles a with equal sides 30 cm and other side 40 cm.

Perimeter of!! = (2 x 30 + 40) em= (60 + 40) em = 100 em

Thus all the figures has same perimeter.

131

132 MAn-tEMAncs-VI

Q17. Avneet buys 9 square paving slabs, each with a side im. He

lays them in the form of a square.

(a) What is the perimeter of his arrangement.(b) Shari does not like his arrangement. She gets him to lay

them out like a cross. What is the perimeter of herarrangement.

(c) Which has greater perimeter?(d) Avneet wonders, ifthere is a way of getting an even greater

perimeter. Can you find a way of doing this? (The pavingslabs must meet along complete edges i.e., they cannot bebroken.)

Sol. (a) 6 m(b) 10 m(c) Second arrangement has greater perimeter.(d) Yes, if all the square are arranged in row, the perimeter be 10 cm.

EXERCISE 10.2Q1. Find the areas of the following figures by counting square:... ':' ...~...:....~...~... ':' ...~...~....~...~....~...~....~...~... ': ...~....~...~....~... :'.."' ...

· . . . . . . . . . . . . . . . . . . . .

...[] ....[]... ·······~·············rn····LtBd···············--·'" . . .'" . . ,'" . . .

....... ; ;......... . .; ; - -;-_ .. ----;--- -'-, ; -; ... .. ....,. ... .. ....,. .·.. ....~... ~.. ..... . '.... ' . - ..; ~. .. . . '. ... . :. ... .... .- . :.... . :....... ,..... .."... '",.. " '. . .... . ., , , , , , . . . . , ,., .,.· . , . . . , . , . . .., .· . . . . . . . . . . ... .• ••• ' ••••••••• I •••• ' •••• ~ •••••••••••••• I •••• ' •••• I •••• ' •••• l •••• '. • • • • • • • ••••••· . . . . . . . . . . . . . . . . . . . .· . . . . . . . . . . . . . . . . . . . .-.:.z=2: : : ... -:-- .. ~: :-.--~ .. .: :----t±R: .. .: .. --: : .. --;.---~:.- .. : : : : .... :-: .. :---.· . .. ... .. .. .. ..· . " '" .. .. .. ..

:::r-·j"·::::f::C:"--j··-::::'::::: ---.: j··-··:::C:. ~·-·LT·-~i.· . . , . . . ,. "... ..,

·...:....~...:....:....~...~....~...!....:.... ....:....:....:....:.... ;.... ....:....:...'.'.... . . . . . . . . , , , , . ,, . . . . . . , . , , , , , ,

, , . . ". . . . . . . . . . ,••• I·· •• •••••••••• •••••••••• I •••• ' •••••••• i •••••••• i ••••••••• ' ••••••••••••••••••• ' ••••••• ,i ••••••••• I •••

MENSURATION 133

•... - : ... A: .... : .•• ~ .... ~ ..• : ... - ~ ••. : ..•. ~ ..•

, . . . . , . . .---·~----~---·----:--·-:---tED·----· .--.-~-.-.--- .. - .., . . , . . , . .. . . . . . . , ,- _. -:-- - -. - - -,. _.. - - - -: .. - -. - -! _.. -:-- - - . - - -;-. - -;..- -/-;-. --

: : : :: :: . r-,, . . ., ".... ... ..., , , I , I I , ,

\ ., . . , , . . . .

~....:.... ~... ~.... :.... ~... ~.... !... ~.... ; ... :(. .. ;....:....

:---~----:----/1- --~---;-..-;.---;.---~..-:-

~................ : : .

~... ... - .... . .Sol. (a) Full filled square = 9

Area covered by full squares = (9 x 1) sq. units= 9 sq. units

Thus the area of given fig. is 9 sq. units.(b) Full filled square = 5

Area covered by full square = (5 x 1) sq. units= 5 sq. units

Thus the area of given fig. is 5 sq. units.(c) Full filled square = 2

Half filled square = 4Area covered by full square = (2 x 1) sq. units = 2 sq. units

Area covered by half square = j x ~ = 2 sq. units

Thtal area = 2 sq. unit~ + 2 sq. units = 4 sq. units.(d) Full filled square = 8

Area covered by full square = (8 x 1) sq. units= 8 sq. units

Thus the area of given fig. is 8 sq. units.(e) Full filled square =- 10

Area covered by full square = (10 x 1) sq. units= 10 sq. units

Thus the area of given fig. is 10 sq. units.

134 MATHEMATICS-VI.--------------------(f) Full filled square = 2

Half filled square = 4Area covered by full square = (2 x 1) sq. units = 2 sq. units

Area covered by half square = (t x ~) sq. units = 2 sq. units

Total area covered = 2 sq. units + 2 sq. units = 4 sq. units.(g) Full filled square = 4

Half filled square = 4Area covered by full square = (4 x 1) sq. units = 4 sq. units

Area covered by half square = (t x ~) sq. units = 2 sq. units

Total area covered = 4 sq. units + 2 sq. units = 6 sq. units.(h) Full filled square = 5

Area covered by full square = (5 x 1) sq. units = 5 sq. unitsThus the area of given fig. is 5 sq. units.

(i) Full filled square = 9Area covered by full square = (9 x 1) sq. units = 9 sq. unitsThus the area of given fig. is 9 sq. units.

(j) Full filled square = 2Half filled square = 4

.Area of full square = (2 x 1) sq. units = 2 sq. units

Area of half filled square = (t x ~) sq. units = 2 sq. units

Total area = 2 sq. units + 2 sq. units = 4 sq. units(k) Full filled square = 4

Half filled square = 2Area offull square = (4 x 1) sq. units = 4 sq. units

Area of half square = (2 x ~ J sq. units = 1 sq. unit

Total area = 4 sq. units + 1 sq. unit = 5 sq. units(l) Full square = 3

Half square filled = 10Area of full square = (3 x 1) sq. units = 3 sq. units

Area of half sq. units = (10 x ~ J sq. units = 5 sq. units

'Ictal area = 3 sq. units + 5 sq. units = 8 sq. units.

MENSURATlON ' 135

(m) Full square = 7Half square = 14Area of full square = (7 x 1) sq. units = 7 sq. units

Area of half sq.' units = (14X~J sq. units = 7 sq. units

Total area = 7 + 7 = 14 sq. units.(n) Full square = 10

Half square = 16Area of full square = (10 x 1) sq. units = 10 sq. units

Area of half square = (16 x ~ J = 8 sq. units

Total area = 10 + 8 = 18 sq. units.

EXERCISE 10.3Q1. Find the areas of the rectangles whose sides are:

(a) 3cmand4cm (b) 12mand21m(c) 2kmand3km (d) 2mand70cm

SoL (a) Area ofrectangle (b) Area ofrectangle= length x breadth = length x breadth= 3 cm x 4 em = 12 cm2. = 12 m x 21 m = 252 mi.

(c) Area of rectangle (d) Area of rectangle= length x breadth = length x breadth= 2 km x 3 km = 2 m x 70 em=6km2. =2mxO.7m=1.4m2.

Q2. Find the areas of the squares whose sides are:(a) 10 cm (b) 14 cm (c) 5 m

Sol. (a) Area of square (b) Area of square= side x side = side x side= 10 em x 10 cm = 100 cm2. = 14 em x 14 em = 196 cm2.

(c) Area of square= side x side= 5 m x 5 m = 25 m2.

Q3. The length and breadth of three rectangles are as given below:(a)'9mand6m (b) 17mand3m (c) 4mand14mWhich one has the largest area and which one has the smallest?

Sol. (a) Area of first rectangle = length x breadth

136 MATIiEMATlcs-VI

= 9 m x 6 m = 54 m2(b) Area of second rectangle = 3 m x 17 m = 51 m2.(c) Area of third rectangle = 4 m x 14 m = 56 m2.

Thus the third rectangle has largest area i.e., 56 m2 and secondrectangle has smallest area i.e., 51 m2.

Q4. The area of a rectangular garden 50 m long is 300 m2, find thewidth of the garden.

SoL Length of rectangle = 50 mArea of rectangle = 300 m2Now, Area of rectangle = length x breadth

= Area of Rectangle = BreadthLength

pOOBreadth = ¢ = 6 m

Thus the breadth of the garden is of 6 m.Q5. What is the cost of tilling a rectangular plot of land 500 m long

and 200 m wide at the rate on 8 per hundred sq. m?SoL Length of land = 500 m

Breadth of land = 200 mArea of land = 500 m x 200 m = 1,00,000 m2Cost of tilling 1 sq. m ofland = ~ 8Thus cost of tilling 1,00,000 m2 of land

8xl000,OO' = ~ 8000.= 1-J

Q6. A table-top measures 2 m by 1 m 50 em, What is its area insquare metres?

Sol. Length of table = 2 m

Breadth of table = 1 m 50 cm = 1.50 mArea oftable = length x breadth

= 2 m x 1.50 m = 3 m2.Q7. A room is 4 m long and 3 m 50 cm wide. How many square

metres of carpet is needed to cover the floor of the room?Sol. Length of room = 4 m

Breadth of room = 3 m 50 em = 3.50 mArea of carpet = length x breadth

= 4 x 3.50 = 14 m2.

MENSURATION 137

Q8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m islaid on the floor. Find the area of the floor that is not carpeted.

Sol. Length of floor = 5 mBreadth of floor = 4 mArea of floor = (5 x 4) m2 = 20 m2Side of carpet = 3 mArea of carpet = side x side

=3m x 3m =9m2

Floor that remain uncarpeted = 20 m2_ 9 m2 = 11 m2.Q9. Five square flower beds each of sides 1 m are dug on a piece

of land 5 m long and 4 m wide. What is the area of the remainingpart ofthe land?

Sol. Side of square bed = 1 mArea of square bed = 1 xl = 1 m2Area of 5 square beds = 5 x 1 m2

=5m2

Length ofland = 5 m, breadth ofland = 4 mArea of land = length x breadth

= (5 x 4) m2=20m2

Area of remaining part = Area of land - Area of flower beds= 20 m2_5 m2

= 15m2•

QI0. By splitting the following figures into rectangles, find their areas(The measures are given in centimetres).

5

33

42

41 13L-_

1L--

3

(a)

SoL (a) Area ofHKLM

= 3 x 3 = 9 cm2

(b)

138 ------_._-------------------_._._--MATHEMATICS-VI

A B~

: 14IIIII

Area ofIJGH

= 1 x 2 = 2 cm2 F "I

AreaofFEDG 2

= 3 x 3 = 9 cm2 I 2 J

Area ofABCD i~f-----~1 M

= 2 x 4 = 8cm2 4

~ 31 13Total area = 9 + 2 + 9 + 8 = 28 cm2.

(b) Area of rectangle ABCD K L

= 3 xl = 3 cm2 < 7

Area ofBDEF A B 5

= 3 xl = 3 cm2 ~

o 2 C4

F G

Area ofFGHI

= 3 xl = 3 cm2

.- 2Total area = 3 + 3 + 3 = 9 CD;l • 01 ·C I 1 H

QU. Split the following shapes into rectangles and find their areas.(The measures are given in·centimetres.)

o23

72

7 77 7

7 77 7

7

3 E

2

5

2212 10 7

8

1 12

10 7

4 4

(a) (b) (c)

Sol. (a) Area ofrectangle ABCD A2B= 2 x 10 = 20 cm2

Area of rectangle DEFG 121 110= 10 x 2 = 20 cm2

Area of total fig. DrJC 8 I~= 20 cm2 + 20 cm2 = 40 cm2. E 10 F

~

MENSURATION 139

(b) There are five squares eachof side 7 cmArea of one square

= 7 x 7 = 49 cm2

Area of five square= (5 x 49) cm2 = 245 cm2.

7

7

7

5 BA(c) Area of rectangle ABCD

= 5 xl =5 cm2

Area of rectangle EFGH=4 xl =4cm2

Thtal area= 5 cm2 + 4 cm2 = 9 cm2. G 1 H

Q12. How many tiles whose length and breadth are 12 cm and 5 emrespectively will be needed to fit in a rectangular region whoselength and breadth are, respectively:(a) 100 cm and 144 cm. (b) 70 em and 36 em,

Sol. (a) Area of region = 100 em x 144 em = 14400 cm2

Area of tile = 5 em x 12 cm = 60 cm2

1L..-

o 21..J

2 CE F

44

Area of region _Number of tiles = Area of one tile -

214400 em = 240

260cm

Thus 240 tiles are required.(b) Area of region = 70 cm x 36 em = 2520 cm2

Area of tile = 5 cm x 12 em = 60 em2

Area of regionNumber of tiles = Area of one tile

2520 cm2

= 42 tiles260cm

Thus 42 tiles are required.DO

7