NCERT Solutions Class 6 Maths Chapter 6 Perimeter and Area

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 6 · M AT H S

NCERT Solutions

Chapter 6: Perimeter and Area

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

129 – 150 18 71 English

Solutions, notes, sample papers & more at 87 pages

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

CLASS 6 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 6: Perimeter and Area
Complete, step-by-step NCERT Solutions for Class 6 Maths Chapter 6 Perimeter and Area from the NCERT
textbook Ganita Prakash. Every Figure it Out question from pages 132 to 149 is solved, along with all the in-
text work — the Matha Pachchi running tracks, the Deep Dive race, straight and diagonal units, Split and
rejoin, the tangram, area by counting squares, area of a triangle, Charan's and Sharan's house plans and the
Area Maze puzzles.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 6) 129 – 150

SECTIONS QUESTIONS

18 71

MEDIUM

English

Figure it Out — Page 132
Section 6.1 Perimeter

Q1 Find the missing terms: a. Perimeter of a rectangle = 14 cm; breadth = 2 cm; length =
?. b. Perimeter of a square = 20 cm; side of a length = ?. c. Perimeter of a rectangle =
12 m; length = 3 m; breadth = ?.

Use the two formulas backwards.
(a) Perimeter = 14 cm, breadth = 2 cm

Perimeter = 2 × (length + breadth)

14 = 2 × (length + 2)

length + 2 = 14 ÷ 2 = 7

length = 7 − 2 = 5 cm

(b) Perimeter of a square = 20 cm

Page 1 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Perimeter = 4 × side

20 = 4 × side

side = 20 ÷ 4 = 5 cm

(c) Perimeter = 12 m, length = 3 m

12 = 2 × (3 + breadth)

3 + breadth = 12 ÷ 2 = 6

breadth = 6 − 3 = 3 m

Note: the answer key printed at the end of the book writes “3 cm” for part (c). The
measurements in the question are in metres, so the correct answer is 3 m.

Why halve first: a rectangle has two lengths and two breadths. Half the perimeter is
therefore exactly one length + one breadth, so dividing by 2 straight away is the
quickest first step.

Q2 A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the
wire is straightened and then bent to form a square, what will be the length of a
side of the square?

The same piece of wire is used, so the perimeter does not change.

Length of wire = perimeter of the rectangle

= 2 × (5 cm + 3 cm) = 2 × 8 cm = 16 cm

Side of the square = 16 ÷ 4 = 4 cm

Answer: each side of the square is 4 cm.

Page 2 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Did you notice? The wire covers the same distance, but the region it encloses
changes:
area of the rectangle = 5 × 3 = 15 sq cm, area of the square = 4 × 4 = 16 sq cm. Same
perimeter, more area — the square always encloses the most.

Q3 Find the length of the third side of a triangle having a perimeter of 55 cm and
having two sides of length 20 cm and 14 cm, respectively.

Perimeter of a triangle = sum of its three sides

55 = 20 + 14 + third side

55 = 34 + third side

third side = 55 − 34 = 21 cm

Answer: the third side is 21 cm.

Check it yourself: 20 + 14 + 21 = 55 ✔

Q4 What would be the cost of fencing a rectangular park whose length is 150 m and
breadth is 120 m, if the fence costs ₹40 per metre?

The fence runs all around the park, so we first need the perimeter.

Perimeter = 2 × (length + breadth)

= 2 × (150 m + 120 m)

= 2 × 270 m = 540 m

Now multiply by the rate:

Cost = 540 × ₹40 = ₹21,600

Answer: the fencing will cost ₹21,600.

Page 3 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
m.
Why perimeter and not area: fencing, lace, tape, rope and border designs run

as e
com so they need the area. l
along the boundary, so they always need the perimeter. Tiles, carpets, grass and paint
cover the.inside, a g
a s em
agl

co m
ag
A piece of string is 36 cm long. What will be the length of each side, if it is used to
.
Q5

e m
as
form: a. A square, b. A triangle with all sides of equal length, and c. A hexagon (a six

a g l
sided closed figure) with sides of equal length?

co m
In every case the whole string becomes the perimeter, so divide 36 cm by the number of equal
em.
m l as
.co g
sides.
m a
l a se
a g FIGURE NUMBER OF EQUAL SIDES WORKING LENGTH OF EACH SIDE

a s
a. Square 4
m
36 ÷ 4 9 cm

m .co agl
ase
b. Equilateral triangle 3 36 ÷ 3 12 cm

c. Regular hexagon 6 agl 36 ÷ 6 6 cm

co m
m .
e
The rule behind all three: for a regular polygon, perimeter = number of sides ×
m l as
.co g
length of a side. So length of a side = perimeter ÷ number of sides.
m a
l a se
ag
se m
com g l a
m . a
ase
agl

co m
m .
m ase
.co a g l
se m
g l a
a c
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g l as
a

co m
m .
m as e
.co


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Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q6 A farmer has a rectangular field having length 230 m and breadth 160 m. He wants
to fence it with 3 rounds of rope as shown. What is the total length of rope needed?

160 m

230 m
Redrawn sketch of the picture on page 132 — the rectangular field with a fence of three
rounds of rope tied all round on posts.

One round of rope goes once around the field — that is one perimeter.

Perimeter of the field = 2 × (230 m + 160 m)

= 2 × 390 m = 780 m

The farmer wants 3 rounds:

Total rope = 3 × 780 m = 2340 m

Answer: 2340 m of rope is needed (that is 2 km 340 m).

Figure it Out — Page 133

Page 5 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Matha Pachchi! — the two running tracks

MATH TALK

Q1 Find out the total distance Akshi has covered in 5 rounds.

Akshi runs on the outer track, which is a rectangle 70 m long and 40 m wide.

One round = 2 × (70 m + 40 m) = 2 × 110 m = 220 m

5 rounds = 5 × 220 m = 1100 m

Answer: Akshi covered 1100 m (1 km 100 m).

Q2 Find out the total distance Toshi has covered in 7 rounds. Who ran a longer
distance?

Toshi runs on the inner track, which is 60 m long and 30 m wide.

One round = 2 × (60 m + 30 m) = 2 × 90 m = 180 m

7 rounds = 7 × 180 m = 1260 m

Now compare the two totals:

Toshi 1260 m > Akshi 1100 m

1260 − 1100 = 160 m

Answer: Toshi ran the longer distance — 160 m more than Akshi, even though her track is
smaller. She simply ran two more rounds.

Math Talk: A smaller track does not mean a smaller distance. What matters is
perimeter × number of rounds.

Page 6 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q3 Think and mark the positions as directed— a. Mark ‘A’ at the point where Akshi will
be after she ran 250 m. b. Mark ‘B’ at the point where Akshi will be after she ran 500
m. c. Now, Akshi ran 1000 m. How many full rounds has she finished running around
her track? Mark her position as ‘C’. d. Mark ‘X’ at the point where Toshi will be after
she ran 250 m. e. Mark ‘Y’ at the point where Toshi will be after she ran 500 m. f.
Now, Toshi ran 1000 m. How many full rounds has she finished running around her
track? Mark her position as ‘Z’.

Both girls start at the bottom-right corner of their own track and run along the bottom side first,
then up the left side, then along the top, then down the right side. Each time, divide the
distance by one round and use the remainder to walk around the track.
Akshi — one round = 220 m

a. 250 = 220 + 30 → 1 full round, then 30 m more

A lies on the bottom (70 m) side, 30 m from her starting corner.

b. 500 = 2 × 220 + 60 → 2 full rounds, then 60 m more
B lies on the bottom side, 60 m from the start (10 m short of the corner).

c. 1000 = 4 × 220 + 120 → 4 full rounds, then 120 m more

120 = 70 (bottom) + 40 (left side) + 10 → C lies on the top side, 10 m from the top-left

corner.

Toshi — one round = 180 m

d. 250 = 180 + 70 → 70 = 60 (bottom) + 10

X lies on the left (30 m) side, 10 m above the bottom-left corner.

e. 500 = 2 × 180 + 140 → 140 = 60 + 30 + 50

Y lies on the top side, 50 m from the top-left corner.

f. 1000 = 5 × 180 + 100 → 5 full rounds, then 100 = 60 + 30 + 10

Z lies on the top side, 10 m from the top-left corner.

Page 7 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

70 m
C
Y
Z 60 m
40 m

30 m

X

start (Toshi)

B A start (Akshi)

Both girls run bottom → left side → top → right side
The two tracks with every marked position. A and B are on Akshi's bottom side; C, Y and Z are on the
top sides; X is on Toshi's left side.

Why division with remainder works: after each full round the runner is back at her
starting point. So only the remainder decides where she stops.

In-text Questions — Page 134

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Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

Deep Dive · Estimate and Verify
co m
e m.
m as
TRY THIS

.co a g l
a s em Dive: In races, usually there is a common finish line for all the runners. Here
gl
Deep
a
Q1
are two square running tracks with the inner track of 100 m each side and outer
track of 150 m each side. The common finishing line for both runners is shown by

. com
the flags in the figure which are in the center of one of the sides of the tracks. If the
ag
a s
total race is of 350 m, then we haveemto find out where the starting positions of the
a
two runners should be on thesegl two tracks so that they both have a common
finishing line after they run for 350 m. Mark the starting points of the runner on the
inner track as ‘A’ and the runner on the outer track as ‘B’.
co m
em.
m l as
m .co a g
l a se 150 m
a g
m a s
.co
m 100 m agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
Common Finishing Line
m .
o m l a se
m .c The two square tracks. The flags on the black line
ag common finishing line, at
mark the

l a se the middle of one side of each track; the red arrows show the direction of running.

ag
.c
s e m
m a
e m . co agl
g l as
Start at the flag and walk backwards 350 m along each track.
a
Inner track (side 100 m) — the flag is at the middle of a side, so it is 50 m from the corner.

co m
m .
m ase
.co


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Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

50 m (half a side) + 100 + 100 + 100 = 350 m

So A is the far corner of the inner square — the corner diagonally opposite the flag's side,
three full sides and one half-side away.
Outer track (side 150 m) — the flag is 75 m from the corner.

75 m (half a side) + 150 m (one full side) + 125 m = 350 m

So B lies on the opposite side of the outer square, 125 m from the corner — that is, only 25
m from the other corner.

outer track — side 150 m
B

inner track — side 100 m

A

common finishing line

Page 10 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Both runners finish at the red line in the middle of the bottom side. A on the inner track and B on the
outer track are each exactly 350 m away from it.

Why the starts are different: the tracks have different perimeters (400 m and 600
m). To make the race fair, the starting points are staggered so that each runner
covers exactly 350 m to the same finishing line — just as in a 200 m race on an
athletics track.

Q2 Estimate and Verify: Take a rough sheet of paper or a sheet of newspaper. Make a
few random shapes by cutting the paper in different ways. Estimate the total length
of the boundaries of each shape then use a scale or measuring tape to measure and
verify the perimeter for each shape.

This is an activity to do with a friend. Here is how to carry it out carefully:

1. Cut 4 or 5 odd shapes from an old newspaper — some with straight edges, some wavy.
2. First estimate. Look at the shape and guess its boundary length in centimetres. Write your
guess down before measuring.
3. Then measure. For straight edges use a scale and add the side lengths. For a curved edge,
lay a thread along the boundary, mark it, straighten the thread and measure it with a scale.
4. Make a table: shape → my estimate → measured perimeter → difference.

SHAPE ESTIMATE MEASURED PERIMETER DIFFERENCE

Triangle piece 20 cm 23 cm 3 cm

Wavy piece 30 cm 27 cm 3 cm

Try This: After 3 or 4 shapes your estimates will come much closer to the measured
values. Estimating is a skill that improves with practice — and it tells you at once if a
calculated answer is sensible.

Page 11 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q3 Akshi says that the perimeter of this triangle shape is 9 units. Toshi says it can’t be
9 units and the perimeter will be more than 9 units. What do you think?

The triangle drawn on the dot grid, page 134.

Toshi is right. The perimeter is more than 9 units.
The triangle is drawn on a dot grid. Two of its sides run along the grid (3 units + 3 units), but the
third side is a slanting (diagonal) line made of 3 diagonals of unit squares.

Perimeter = 3 straight units + 3 straight units + 3 diagonal units

= 6s + 3d units

Akshi counted every diagonal step as 1 unit, which is wrong: a diagonal of a unit square is
longer than its side.

Page 12 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Why a diagonal is longer: in any square, the diagonal joins two opposite corners,
and the straight path across is always longer than one side (it is the longest line you
can draw inside the square). A unit diagonal measures about 1.4 units, so

6 × 1 + 3 × 1.4 ≈ 10.2 units

which is clearly more than 9.

Tip: measure a red line and a blue line of the figure with your scale. The blue
(diagonal) line will come out longer every time.

In-text Questions — Page 135
Straight and diagonal units · Perimeter of a regular polygon

Q1 Write the perimeters of the figures below in terms of straight and diagonal units.

a b c d

The four figures drawn on the dot grid. Every side runs either along the grid (a straight
unit) or from corner to corner of one square (a diagonal unit).

Walk once round each figure and count the two kinds of steps separately: s for a straight (grid)
unit and d for a diagonal unit.

Page 13 of 87

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Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
em.
m l as
m .co a g
l a se
a g

com
e m . ag
g l as
a
a b c d

The four figures on the dot grid. Every side is either a straight unit (along the grid) or a diagonal unit
co m
e m.
(corner to corner of one square).

m l as
m .co a g
l a se
FIGURE STRAIGHT UNITS DIAGONAL UNITS PERIMETER

a g COUNTED COUNTED

m a s
.co agl
a. (the “F”) 3+1+1+3=8 1+1=2 8s + 2d

se m
l a
b. (the 1+2+1=4 6 4s + 6d
“heart”)
a g

com
c. (the “U”) 3 + 2 + 1 + 3 + 1 + 2 = 12 6 12s + 6d

m .
m ase
d. (the “N”) 4 + 1 + 3 + 1 + 4 + 1 + 3 + 1 = 18 3+3=6 18s + 6d

m . co agl
l a seWhy we do not just add s and d: they are two different lengths. Writing 8s + 2d
ag keeps the answer exact. If you want an approximate number, use d ≈ 1.4s — for

se m
com a
figure a that gives 8 + 2 × 1.4 ≈ 10.8 units.
. a g l
m
gl ase
a
What is a similarity between a square and an equilateral triangle?
Q2

co m
m .
m

o l a se
m .c are regular polygons — closed figures in which all sidesagare equal and all angles are
se equal.
Both

a
agl c
m .
m a s e
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g l as
a

com
m .
m ase
.co


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Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

EQUILATERAL TRIANGLE SQUARE

Number of sides 3 4

All sides equal? Yes Yes

All angles equal? Yes (60° each) Yes (90° each)

Perimeter 3 × side 4 × side

Because of this, in both figures the perimeter can be found by one multiplication instead of
adding the sides one by one.

Perimeter of a regular polygon = number of sides × length of a side

In-text Questions — Page 136
Regular polygons · Split and rejoin

TRY THIS

Q1 Find various objects from your surroundings that have regular shapes and find
their perimeters. Also, generalise your understanding for the perimeter of other
regular polygons.

Regular shapes are all around us. Measure one side with a scale, then multiply.

OBJECT REGULAR SHAPE ONE SIDE (TYPICAL) PERIMETER

Carrom board Square 74 cm 4 × 74 = 296 cm

₹1 coin face Circle-like (not a polygon) — measure with a thread

Bathroom floor tile Square 30 cm 4 × 30 = 120 cm

Stop-sign board Regular octagon 25 cm 8 × 25 = 200 cm

Honeycomb cell Regular hexagon 6 mm 6 × 6 = 36 mm

Set-square (60°) — — add the three sides

The generalisation:

Page 15 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Perimeter of a regular polygon = n × s

where n = number of sides and s = length of one side

Check it against the formulas you already know: n = 3 gives 3 × s (equilateral triangle), n = 4
gives 4 × s (square), n = 6 gives 6 × s (regular hexagon). ✔

Q2 Split and rejoin: A rectangular paper chit of dimension 6 cm × 4 cm is cut as shown
into two equal pieces. These two pieces are joined in different ways. For example,
the arrangement a. has a perimeter of 28 cm. Find out the length of the boundary
(i.e., the perimeter) of each of the other arrangements below.

4 cm

6 cm 6 cm

a.
6 cm 2 cm

b. c. d.

2 cm 2 cm 3 cm

2 cm

The 6 cm × 4 cm chit with the cut shown, arrangement a., and arrangements b., c. and
d., page 136.

The 6 cm × 4 cm chit is cut into two equal pieces, each 6 cm × 2 cm. Every arrangement uses
both pieces, so only the length of the joined edge changes.

Page 16 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Perimeter of one piece = 2 × (6 + 2) = 16 cm

Two separate pieces = 2 × 16 = 32 cm

When they are joined along a common edge of length L, that edge is hidden twice:

Perimeter = 32 − 2 × L

b. c. d. new

28 cm 28 cm 26 cm 22 cm

Arrangements b, c and d of the two 6 cm × 2 cm pieces, and (last) an arrangement whose perimeter is
22 cm.

ARRANGEMENT SHAPE FORMED LENGTH OF THE JOINED EDGE PERIMETER

a. (given) 12 cm × 2 cm rectangle 2 cm 32 − 4 = 28 cm

b. L-shape 2 cm 32 − 4 = 28 cm

c. T-shape 2 cm 32 − 4 = 28 cm

d. Z / step shape 3 cm 32 − 6 = 26 cm

Check b by walking round it: 8 + 6 + 2 + 4 + 6 + 2 = 28 cm ✔ Check d: 2 + 3 + 2 + 6 + 2 + 3 + 2 +
6 = 26 cm ✔

Why the area never changes: in every arrangement the same two pieces are used,
so the area stays 6 × 4 = 24 sq cm. Only the perimeter changes — a beautiful
example of “same area, different perimeter”.

Page 17 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q3 Arrange the two pieces to form a figure with a perimeter of 22 cm.

4 cm

6 cm 6 cm

a.
6 cm 2 cm

b. c. d.

2 cm 2 cm 3 cm

2 cm

The 6 cm × 4 cm chit with the cut shown, arrangement a., and arrangements b., c. and
d., page 136.

Use the rule found above and work backwards.

32 − 2 × L = 22

2 × L = 32 − 22 = 10

L = 5 cm

So the two pieces must touch along 5 cm. Place the two 6 cm × 2 cm pieces side by side with
their long edges together, but slide one of them 1 cm up. Then 5 cm of the 6 cm edges touch
(the last figure in the picture above).

Walking round: 2 + 1 + 2 + 6 + 2 + 1 + 2 + 6 = 22 cm ✔

Page 18 of 87

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Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
m.
Tip: if you join the long edges completely (6 cm), you get a 6 cm × 4 cm rectangle

as e
com the perimeter step by step. l
again, with perimeter 32 − 12 = 20 cm — the smallest possible. Sliding the pieces
. a g
em
apart increases
a s
agl

co m
ag
In-text Question — Page 137
m .
Section 6.2 Area
as e
a g l
Area of a square = ________ Area of a rectangle = _______
m
Q1

co
e m.
c o m g l as
Fill them . a
s e two boxes like this:

a gla
Area of a square = side × side
m a s
Area of a rectangle = length × width
m .co agl
l a se
a g
Where these formulas come from (grid paper): draw a rectangle 5 units long and

c o m
4 units wide on squared paper. It contains 4 rows of 5 unit squares, so counting
them gives 4 × 5 = 20 unit squares. You do not have to count one m .
s e by one — rows ×

.
squares
glaa rectangle whose
comin a row is the same as length × width. A square is ajust
m and width are equal, so its area is side × side.
ase
length

agl
se m
because we are counting squares..com
a
Remember the unit: area is always in square units — sq cm, sq m, sq km —

a g l
a s em
agl
Figure it Out — Page 138
co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


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Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Section 6.2 Area

Q1 The area of a rectangular garden 25 m long is 300 sq m. What is the width of the
garden?

Area = length × width

300 = 25 × width

width = 300 ÷ 25 = 12 m

Answer: the garden is 12 m wide.

Check it yourself: 25 × 12 = 300 ✔

Q2 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at
the rate of ₹8 per hundred sq m?

Step 1 — the area to be tiled.

Area = 500 m × 200 m = 1,00,000 sq m

Step 2 — how many “hundreds of sq m” is that?

1,00,000 ÷ 100 = 1000 hundreds

Step 3 — the cost.

Cost = 1000 × ₹8 = ₹8000

Answer: tiling will cost ₹8000.

Page 20 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Careful with the rate: the rate is per hundred square metres, not per square metre.
If you forget to divide by 100, the answer becomes ₹8,00,000 — a hundred times too
big.

Q3 A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree
requires 25 sq m, what is the maximum number of trees that can be planted in this
grove?

Area of the grove = 100 m × 50 m = 5000 sq m

Space needed by one tree = 25 sq m

Number of trees = 5000 ÷ 25 = 200

Answer: at most 200 coconut trees can be planted.

Picture it: 25 sq m is a square patch of 5 m × 5 m. Along the length you can fit 100 ÷
5 = 20 patches and along the width 50 ÷ 5 = 10 patches — that is 20 × 10 = 200
patches. ✔

Page 21 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q4 By splitting the following figures into rectangles, find their areas (all measures are
given in metres).

3 1
2 5
2
3
3
4 2
4
3
1 1
3 b.
a.
Figure (a) and figure (b), with the measurements printed in the book (all in metres).

Figure (a) — the staircase. Slice it into four horizontal strips (shown by the dashed lines) and
add.

Page 22 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

2
3 1 2

2 10 4
2
7
4
4
9 3

3
Figure (a) cut into four rectangles by horizontal lines. The red numbers are the areas of the strips.

Bottom strip 3 × 3 = 9

Next strip 7×1=7

Next strip 5 × 2 = 10

Top strip 2×1=2

Total = 9 + 7 + 10 + 2 = 28 sq m

Figure (b) — the arch. Here it is quicker to take the big rectangle and remove the notch.

Page 23 of 87

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Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
m.co 5 a g
l a se
a g

m .5com ag
as e
a g l
3 m
2 2 2 m .co
a se
. com ag l
m
ase
agl
s
1 m .co
m 1 agl
a
l a se
a g
Figure (b) as a 5 m × 3 m rectangle with a 3 m × 2 m piece cut out of the bottom — or as one bar plus

co m
.
two legs.

e m
m l as
.co
m rectangle = 5 × 3 = 15 sq m a g
a s eWhole
agl Notch removed = 3 × 2 = 6 sq m

se m
Area = 15 − 6 = 9 sq m
com g l a
m . a
gl ase
a
Splitting instead of subtracting gives the same answer: top bar 5 × 1 = 5, left leg 1 × 2 = 2, right
leg 1 × 2 = 2, and 5 + 2 + 2 = 9 sq m ✔

co m
m .
Two safe methods: (i) cut the figure into rectangles and add, or (ii) draw the
m as e
.co a g l
smallest rectangle around it and subtract the parts that are not in the figure. Choose

s e m whichever needs less work.
agla
.c
s e m
m a
Figure it Out — Page 139
e m . co agl
g l as
a

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Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

The tangram pieces

TRY THIS

Q1 Explore and figure out how many pieces have the same area.

B
E

A D

C

F
G

The seven tangram pieces A–G arranged into a square, page 139.

Place the pieces on top of one another. Taking the smallest triangle C as one unit of area, the
seven pieces measure:

Page 25 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

B
E

A D
C
F
G

The tangram square: A and B are the two big triangles, F the middle-sized triangle, C and E the two
small triangles, D the square and G the parallelogram.

Page 26 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

PIECE SHAPE AREA (TAKING C = 1)

A Large triangle 4

B Large triangle 4

C Small triangle 1

D Square 2

E Small triangle 1

F Middle-sized triangle 2

G Parallelogram 2

Answer — there are three groups of equal area:

A and B have the same area (4 units each).
C and E have the same area (1 unit each).
D, F and G all have the same area (2 units each), even though one is a square, one a triangle
and one a parallelogram.

Check the total: 4 + 4 + 1 + 2 + 1 + 2 + 2 = 16 units — and the seven pieces do make
one big square.

Page 27 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q2 How many times bigger is Shape D as compared to Shape C? What is the
relationship between Shapes C, D and E?

B
E

A D

C

F
G

The seven tangram pieces A–G arranged into a square, page 139.

Place C and E on top of the square D — together they cover it exactly.

Area of D = area of C + area of E

and area of C = area of E, so

Area of D = 2 × area of C

Answer: Shape D is twice as big as Shape C.

Page 28 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

The relationship: C and E are equal small triangles, and the two of them join along their
co m
longest sides to form the square D. So
e m.
m l as
m .co a g
l a se
D = C + E and C = E = half of D
a g

o m
Which shape has more area: Shapem
e
. c ag
s
D or F? Give reasons for your answer.
a
Q3

agl

co m
em.
m l as
m .co a g
l a se B
ag
E
m a s
m.co agl
l a se
A a g D
co m
m .
m C as e
.co a g l
se m
g l a
a F
G m
a se
. com a g l
m
ase
agl
The seven tangram pieces A–G arranged into a square, page 139.

co m
m .
as e
. com a g l
sem

a
agl
Neither — D and F have exactly the same area.

.c
s e m
m a
. co agl
Area of D = C + E = 1 + 1 = 2 units

e m
l as
Area of F = C + E = 1 + 1 = 2 units
a g

co m
m .
m ase
.co


a g l Page 29 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Reason: the square D can be covered exactly by the two small triangles C and E. The middle-
sized triangle F can also be covered exactly by the same two triangles — just place them side by
side along their short sides instead. Two shapes that are made from the same pieces must have
the same area.

The big idea: area does not depend on the shape. Cut a figure up and rearrange the
parts — the area stays the same, even though the figure now looks completely
different.

Q4 Which shape has more area: Shape F or G? Give reasons for your answer.

B
E

A D

C

F
G

The seven tangram pieces A–G arranged into a square, page 139.

Again, neither — F and G have equal areas (2 units each).

Page 30 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

F (middle triangle) = C + E = 2 units

G (parallelogram) = C + E = 2 units

Reason: both the triangle F and the parallelogram G can be exactly covered by the two small
triangles C and E. Cut the parallelogram G along its short diagonal and you get two small
triangles; put those two triangles together the other way and you get F.

Check it with the square: D, F and G are all 2 units, so together they make 6 units
— and 6 + 4 + 4 + 1 + 1 = 16, the whole square. ✔

Page 31 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q5 What is the area of Shape A as compared to Shape G? Is it twice as big? Four times
as big?

B
E

A D

C

F
G

The seven tangram pieces A–G arranged into a square, page 139.

Area of A = 4 units Area of G = 2 units
4÷2=2

Answer: Shape A is exactly twice as big as Shape G — not four times.
Shape A is four times as big as Shape C (the small triangle), because 4 ÷ 1 = 4. That is why it is
worth fixing one piece (C) as the unit and measuring everything against it.

Page 32 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

COMPARISON WORKING ANSWER

A compared with G 4÷2 2 times

A compared with C 4÷1 4 times

A compared with B 4÷4 equal

Page 33 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
m.
Can you now figure out the area of the big square formed with all seven pieces in
e
Q6

m l as
.co
terms of the area of Shape C?

a g
se m
g l a
a

co m
e m . ag
g l as
a B
E m
m .co
a se
. com a g l
m
ase
agl
A D s
m a
m .co agl
l a se
a g C
co m
m .
m l a se F
m .co G ag
ase
agl
se m
com g l a
m . a
ase
agl
The seven tangram pieces A–G arranged into a square, page 139.

com
m .
e

m l as
.co g
Just add up all seven pieces, measuring each in units of C.

m a
l a se
ag
c
A+B+C+D+E+F+G
m .
=4+4+1+2+1+2+2
m a s e
e m . co agl
as
= 16 × area of Shape C

a g l
Answer: the big square has an area equal to 16 small triangles (16 C).

co m
m .
m ase
.co


a g l Page 34 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Nice check: 16 is a square number, 16 = 4 × 4 — and indeed four small triangles C fit
along each half-diagonal of the big square.

Q7 Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle
in terms of the area of Shape C now? Give reasons for your answer.

B
E

A D

C

F
G

The seven tangram pieces A–G arranged into a square, page 139.

The rectangle also has an area of 16 × area of Shape C.

Page 35 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Reason: the rectangle is made from exactly the same seven pieces. Nothing was
added and nothing was thrown away, so the total region covered cannot change.
Rearranging pieces changes the shape, never the area.

Area of the square = 16 C

Area of the rectangle = 16 C

So they are equal

Try This: the seven pieces can also be arranged into a triangle, a trapezium and
dozens of animal and human figures. Every one of them has an area of 16 C.

Page 36 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q8 Are the perimeters of the square and the rectangle formed from these 7 pieces
different or the same? Give an explanation for your answer.

B
E

A D

C

F
G

The seven tangram pieces A–G arranged into a square, page 139.

The perimeters are different — the rectangle has the longer boundary, even though the two
figures have the same area.

Explanation: perimeter depends on how the pieces are placed, not only on how
much region they cover. When the pieces are packed into a square, the shape is as
“compact” as possible and the boundary is short. Stretching the same pieces into a
long rectangle pushes some edges outwards, so the boundary gets longer.

You can see the same effect with simple rectangles of area 16 sq units:

Page 37 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

RECTANGLE AREA PERIMETER

4 × 4 (a square) 16 sq units 16 units

8×2 16 sq units 20 units

16 × 1 16 sq units 34 units

Same area every time, but the more “stretched” the figure, the bigger the perimeter. Of all
figures with a given area, the square-like one has the least perimeter.

In-text Questions — Page 140
Estimating area by counting squares

Q1 Look at the figures below and guess which one of them has a larger area.

a. b.

The two closed shapes printed on page 140.

By eye it is almost impossible to say — and that is exactly the point of the question. Shape b has
sharp spikes that make it look bigger, while shape a is smooth and rounded.
The honest way to decide: trace each shape on transparent paper, place it on squared paper
and count the squares using the four rules:

1. one full square = 1 sq unit;
2. more than half a square = 1 sq unit;
3. exactly half a square = ½ sq unit;
4. less than half a square = ignore it.

Page 38 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

When this is done carefully, the smooth shape a turns out to cover slightly more squares than
co m
the spiky shape b — the spikes of b are thin, so they add very little region.
em.
m l as
.co a g
s em
Why our eyes get fooled: we tend to judge by how far a figure spreads out, not by
a
gl much region it actually fills. A long thin spike adds a lot of boundary (perimeter)
ahow
but very little area.

com
m . ag
l a se
Q2 ag figures.
Find the area of the following

co m
e m.
m l as
m .co a g
l a se
a g
m a s
m.co agl
bse
a
g l a c d
a
The same four figures on the dot grid, where four neighbouring dots enclose one square
unit.
co m
m .
m as e
.co a g l
a s em
gl

a Each figure is drawn on a dot grid where four neighbouring dots make one square unit. Break
each figure into rectangles, squares and half-squares (right triangles) and add.
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
a b c d
m .
m a s e
e m . co agl
as
The four figures again. Slanting sides always cut a unit square into two equal halves, each of area ½

a g l sq unit.

co m
m .
m ase
.co


a g l Page 39 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

FIGURE HOW IT SPLITS UP AREA

a. (the “F”) 2 full squares + 4 half-squares 4 sq units

b. (the “heart”) 6 full squares + 6 half-squares 9 sq units

c. (the “U”) 7 full squares + 6 half-squares 10 sq units

d. (the “N”) two 1 × 4 bars + a slanting bar 11 sq units

Answers: 4, 9, 10 and 11 square units.

Tip for the “N”: the two upright bars are 1 × 4 = 4 and 1 × 4 = 4 squares. The slanting
bar covers 3 more squares, giving 4 + 4 + 3 = 11 sq units.

Let’s Explore — Pages 141 & 142
Why squares? · Rectangles of a given area

MATH TALK

Q1 Why is area generally measured using squares? Draw a circle on a graph sheet with
diameter (breadth) of length 3. Count the squares and use them to estimate the
area of the circular region.

Counting the circle first. Draw a circle of diameter 3 units on graph paper. Using the four
counting rules you will find about

Area of the circular region ≈ 7 square units

(The exact value is about 7.1 square units, so counting squares comes very close.)
Now the main question — why squares and not circles?

Circles leave gaps. However you pack circles, small curved gaps are always left between
them, so they can never measure a region exactly. In the book's picture the same rectangle
holds 42 circles one way and 44 circles the other way — the answer changes with the
packing, which is useless for measurement.
Squares tile perfectly. Unit squares fit together with no gaps and no overlaps, and they
cover the whole region.
Squares match our length units. A square of side 1 cm gives 1 sq cm, so area and length
units stay linked.

Page 40 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Squares are easy to count in rows and columns, which is what gives length × width.

The key phrase: a good unit of area must tile the plane — cover it completely with no
gaps and no overlaps. Circles fail this test; squares pass it.

Q2 Try using different shapes (triangle and rectangle) to fill the given space (without
overlaps and gaps) and find out the merits associated with using a square shape to
find the area rather than another shape. List out the points that make a square the
best shape to use to measure area.

The same rectangle packed with circles in two different ways, page 141.

Triangles and rectangles do fill a space without gaps — so they could be used. Even so, the
square wins. Here is the comparison.

Page 41 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

UNIT FILLS SPACE WITHOUT PROBLEM WITH IT
SHAPE GAPS?

Circle No Always leaves gaps; the count changes with the
packing

Triangle Yes Must be flipped up and down alternately; hard to
count in rows

Rectangle Yes Has two different side lengths, so the unit must be
described twice

Square Yes No problem — the best choice

Points that make the square the best unit of area:

Squares fit together with no gaps and no overlaps.
All four sides are equal, so one number (the side) describes the unit completely.
They line up in neat rows and columns, which turns counting into a multiplication: rows ×
columns.
They can be cut into smaller squares (a 1 cm square = 100 squares of 1 mm), so we can
measure as finely as we like.
Half a square is easy to recognise, which helps when a shape cuts across a square.

Q3 Find the area (in square metres) of the floor outside of the corridor.

This one you measure in your own school. Here is how to do it and a sample calculation.

1. Take a measuring tape and measure the length and the breadth of the floor space outside
the corridor, in metres.
2. If the space is a rectangle, area = length × breadth.
3. If it bends around a corner, split it into rectangles, find each area and add — exactly like the
staircase figure on page 138.

Sample: length = 12 m, breadth = 3 m

Area = 12 m × 3 m = 36 sq m

Page 42 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Tip: if you have no tape, walk the length heel-to-toe and count steps. One ordinary
step of a Class 6 student is roughly 0.5 m, so 24 steps ≈ 12 m. Measure your own
step once and you will always have a “ruler” with you.

Q4 Find the area (in square metres) occupied by your school playground.

Measure your own playground — but here is the method with a worked example.

1. Measure the length and the breadth of the playground in metres (use a long tape, or count
steps and multiply).
2. Multiply: area = length × breadth.
3. For an L-shaped or irregular ground, split it into rectangles and add the areas.

Sample: a playground 60 m long and 40 m wide

Area = 60 m × 40 m = 2400 sq m

Perimeter (if you jog once round it) = 2 × (60 + 40) = 200 m

Did you know? A full-size cricket ground has a boundary about 65–70 m from the
pitch, and a standard football field is about 100 m × 64 m = 6400 sq m.

Q5 On a squared grid paper (1 square = 1 square unit), make as many rectangles as you
can whose lengths and widths are a whole number of units such that the area of
the rectangle is 24 square units. a. Which rectangle has the greatest perimeter? b.
Which rectangle has the least perimeter?

We need pairs of whole numbers whose product is 24 — that is, the factor pairs of 24.

Page 43 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
m.
RECTANGLE (LENGTH × WIDTH) AREA PERIMETER = 2 × (L + W)

m as e
1 × 24
.co
24 sq units
a g l
2 × 25 = 50 units

se m
g l
2 × 12a
a
24 sq units 2 × 14 = 28 units

3×8 24 sq units 2 × 11 = 22 units

. c24om ag
4×6
se m sq units 2 × 10 = 20 units

g l a
a. Greatest perimeter: the 1 × 24 a
rectangle, 50 units.

m
b. Least perimeter: the 4 × 6 rectangle, 20 units.
co
se m.
o m l a
Why: the long thin 1 × 24 strip has two very long sides, so its boundary is huge. The
gshape always has the
m .c a
se boundary for a given area.
4 × 6 rectangle is the one closest to a square, and a square-like

g l a
shortest
a
om a s
If you take a rectangle of areae32
. c
m sq cm, what will your answers be? Given any area, agl
a s
agl shape of the rectangle with the greatest perimeter as
Q6
is it possible to predict the
well as the least perimeter? Give examples and reasons for your answer.

co m
m .
as e

m l
.co
Area 32 sq cm — the whole-number rectangles are:
a g
a s em
agl RECTANGLE PERIMETER

se m
com l a
1 cm × 32 cm 2 × 33 = 66 cm

. a g
m
ase
2 cm × 16 cm

agl
2 × 18 = 36 cm

4 cm × 8 cm 2 × 12 = 24 cm

co m
Greatest perimeter: 1 cm × 32 cm (66 cm). Least perimeter: 4 cm × 8 cm (24 cm).
m .
m as e
.co
Yes — the shape can always be predicted:
a g l
se m The greatest perimeter comes from the longest, thinnest rectangle, that is 1 × (the area).
g l a
a The least perimeter comes from the rectangle whose sides are closest to each other — the
c
m .
e
most square-like one. If the area is a perfect square, that rectangle is a square (area 36 → 6 ×

m a s
co agl
6, perimeter 24).

m .
as e
a g l

co m
m .
m as e
.co


a g l Page 44 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Reason: for a fixed area, making one side longer forces the other to become shorter,
but the long side grows much faster than the short side shrinks. So the perimeter
shoots up as the rectangle gets thinner, and is smallest when the two sides are
balanced.

Math Talk: this is why a farmer fencing a fixed area of land prefers a square field —
it needs the least fencing and therefore costs the least.

In-text Questions — Page 142
Section 6.3 Area of a Triangle

Q1 Check! whether the two triangles overlap each other exactly. Do they have the
same area?

Yes — they overlap exactly, and so they have the same area.
Draw a rectangle, cut along one diagonal and place one triangle on top of the other (turn it
round if needed). Every corner matches and no part sticks out.

The two triangles are exactly the same size and shape.

So each one is half of the rectangle.

Try it with a long thin rectangle, a nearly square one, and a square. The result is always the
same. For a square, cutting along a diagonal gives two equal right-angled triangles.

Why the diagonal halves the rectangle: the diagonal splits the rectangle into two
triangles with the same base, the same height and the same right angles. Rotate
one by half a turn and it sits perfectly on the other.

Q2 Can you draw any inferences from this exercise? Please write it here.

Two clear inferences:

1. A diagonal divides a rectangle (or a square) into two triangles of equal area.

Page 45 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

2. Therefore the area of each such triangle = half the area of the rectangle, that is

Area of the triangle = ½ × length × width

A third, more general inference follows from this: two figures can look completely different yet
have exactly the same area, because area is about how much region is covered, not about shape.

Q3 Now, see the figures below. Is the area of the blue rectangle more or less than the
area of the yellow triangle? Or is it the same? Why?

The blue rectangle and the yellow triangle exactly as printed on page 142. The red lines
are the diagonal of the rectangle and the height of the triangle.

The two areas are the same.
Look carefully at the measurements in the figure. The yellow triangle stands on a base that is
twice as long as the blue rectangle, and both figures have the same height.

Let the rectangle be l long and h high → area = l × h

The triangle has base 2l and height h → area = ½ × 2l × h = l × h

So the two areas are equal.

Why: the dotted height line splits the yellow triangle into two right triangles. Each of
them is half of a small rectangle, and the two small rectangles together are the same
size as the blue rectangle counted twice. Halving and doubling cancel out — the
areas match.

Page 46 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q4 Can you see some relationship between the blue rectangle and the yellow triangle
and their areas? Write the relationship here.

The relationship is:

Area of a triangle = ½ × base × height

In words: a triangle covers half of the rectangle that has the same base and the same
height. Here the triangle's base is double, so its area comes out equal to the blue rectangle:

Triangle = ½ × (2 × rectangle's length) × height = rectangle's area

Check it on grid paper: draw a rectangle 6 units × 4 units (area 24). Now draw a
triangle with base 6 units and height 4 units — count its squares and you will get 12,
exactly half. ✔

In-text Questions — Pages 143 & 144
Areas of triangles on grid paper

Q1 Find the area of blue triangle BAD. __________

First read the grid. Rectangle ABCD is 5 units long and 4 units wide, and the blue triangle BAD
is one half of it, cut by the diagonal DB.

Page 47 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

D E C

A F B
Rectangle ABCD on grid paper. The blue triangle BAD is half the rectangle; the red triangle ABE stands
on the same base AB with the same height.

Area of rectangle ABCD = 5 × 4 = 20 sq units

Area of triangle BAD = ½ × 20 = 10 sq units

Answer: 10 square units.

Check by counting: the triangle covers 10 full squares plus 4 halves and leaves 8
halves out — 10 squares in all. ✔

Page 48 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
m.
Find the area of red triangle ABE. ___________
e
Q2

m l as
.co a g
a s em
a glthe line EF straight down from E to the base AB. It cuts the red triangle into two right
Drop
triangles, and the rectangle into two smaller rectangles.

co m
em
. ag
a s
AF = 3 units, FB = 2 units, height = 4 units

agl
Area of triangle AEF = ½ of rectangle AFED = ½ × (3 × 4) = 6 sq units
co m
em.
m as
Area of triangle BEF = ½ of rectangle BFEC = ½ × (2 × 4) = 4 sq units

.co a g l
a s em
a gl Area of triangle ABE = 6 + 4 = 10 sq units
m a s
.co agl
Answer: 10 square units — exactly the same as the blue triangle, although the two triangles
look completely different.
se m
g l a
a
Why they are equal: both triangles stand on the same base AB (5 units) and both
have the same height (4 units, the distance between the two horizontal lines). Any
co m
m .
m as e
triangle on this base with its top vertex anywhere on the line DC has area ½ × 5 × 4 =

.co
10 sq units.
a g l
se m
g l a
a
se m
com a
Area of rectangle ABCD = ________________
l
Q3

. a g
m
ase
agl

Area of rectangle ABCD = length × width
co m
m .
m
= AB × AD
as e
.=co5 units × 4 units = 20 square units a g l
se m
g l a
a c
And so, as the book states, the area of triangle BAD is half of this — 10 square units.
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 49 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Q4 Conclusion ____________________________________________________

Write a conclusion like this:

The area of a triangle is half the area of a rectangle with the same base and the same

height.

Area of a triangle = ½ × base × height

Two extra points that come out of this page:

A triangle does not have to be right-angled for the rule to work. Any triangle can be split by
its height into two right triangles, each of which is half of a rectangle (as with triangle ABE =
½ AFED + ½ BFEC = ½ ABCD).
Triangles that look very different can have the same area, as long as their base and height
are the same.

Figure it Out — Page 144

Page 50 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Areas by splitting into rectangles and triangles

Q1 Find the areas of the figures below by dividing them into rectangles and triangles.

a b c

d

e

Figures a to e on the square grid of unit squares (page 144).

Each figure is on a grid of unit squares. Cut it up with a few dashed lines, work out the pieces,
and add.
Figure a — a parallelogram. Cut off the top and bottom triangles and leave a rectangle in the
middle.

Page 51 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

2

4 × 5 = 20

2

Figure a = triangle (2) + rectangle (4 × 5 = 20) + triangle (2).

Page 52 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Rectangle 4 × 5 = 20 Two triangles ½ × 4 × 1 = 2 each

Area = 20 + 2 + 2 = 24 square units

Figure b — a trapezium. Same idea: a rectangle in the middle with a triangle above and below.

Page 53 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g

co m
e m . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

6

4 × 5 = 20

4
Page 55 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Figure b = triangle (6) + rectangle (4 × 5 = 20) + triangle (4).

Triangle above = ½ × 4 × 3 = 6 Rectangle = 4 × 5 = 20 Triangle below = ½ × 4 × 2 = 4

Area = 6 + 20 + 4 = 30 square units

Figure c — the kite-like figure. Here it is easier to draw the smallest rectangle around it and
subtract the four corner triangles.

Page 56 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

Page 57 of 87

Class 6 Maths Chapter 6 Perimeter and Area AglaSem · NCERT Solutions

3 3

72 − 24

15

3

Page 58 of 87

as e
Class 6 Maths Chapter 6 Perimeter and Area
a g l AglaSem · NCERT Solutions

c o m
.
m+ 3 = 24.
s e
Figure c sits inside a 6 × 12 rectangle. Subtract the four corner triangles: 3 + 3 + 15

. com a gla
a s em
gl
aSurrounding rectangle = 6 × 12 = 72

Corner triangles = 3 + 3 + 15 + 3 = 24
co m
Area = 72 − 24 = 48 square units
e m . ag
g l as
Figure d — the rectangle with a notch.
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g

co m
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