NCERT Solutions Class 8 Maths Chapter 14 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 8 · M AT H S

NCERT Solutions

Chapter 14: Area

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 148 – 171 23 98 English

Solutions, notes, sample papers & more at 83 pages

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

CLASS 8 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 14: Area
The last chapter of Class 8 Maths — Ganita Prakash Part II, Chapter 7. It builds every area formula you need
from one idea: cutting a figure up and moving the pieces around never changes its area. Triangles,
parallelograms, rhombuses and trapeziums all fall out of that single principle, with the ancient Śulba-Sūtras
supplying several of the dissections.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 8) Part II, 148 – 171

SECTIONS QUESTIONS

23 98

MEDIUM

English

In-text Questions — Page 148
Section 7.1 Rectangle and Squares

MATH TALK

Q1 How many different ways can you divide a square into 4 parts of equal area?

Infinitely many. Start with any one division into 4 equal parts — say the two lines through the
centre that cut the square into 4 small squares — and then reshape the pieces without changing
their areas.
Take the boundary between two neighbouring parts and push it into one part. That part loses
some area and its neighbour gains exactly the same area. So push the boundary in at one place
and push it out by the same amount somewhere else along the same boundary:

Page 1 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

4 equal squares still 4 equal parts
Every bump pushed out of one part is a dent of the same size pushed into its neighbour, so all four
areas stay equal.

Why it happens: the four parts together always make up the whole square. If a
piece of area x is moved from one part to the next, and an equal piece of area x is
moved back, both parts end with the area they began with. Since the size of the
bump can be chosen in infinitely many ways, there are infinitely many such divisions.

Q2 Try to think of different creative ways to divide a square into 4 parts of equal area.
[Math Talk]

Here are four genuinely different families to try:

Four strips. Cut the square into 4 equal strips with three parallel lines. Each strip is side ×
(side ÷ 4) = one quarter of the square.
Four triangles from the centre. Join the centre to the four corners. Each triangle has base =
side and height = half the side, so its area is ½ × s × (s/2) = s²/4.
Any four lines through the centre, at 45° apart? Not quite — but any two perpendicular
lines through the centre do work, whatever their slope. The two lines cut the square into 4
pieces that map onto each other under a quarter-turn about the centre, so they are
congruent.
Bumpy pieces. Start with any of the above and trade equal bumps and dents across a
boundary, as in Q1.

Page 2 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Check it yourself: for the four triangles from the centre, add the four areas: 4 × s²/4
= s². They do fill the square exactly.

In-text Question — Page 149
Section 7.1 Rectangle and Squares

Q1 Which of these rectangles requires more rangoli powder to be coloured, if the
colouring is done evenly?

The 7 cm × 4 cm rectangle.

7 cm × 4 cm rectangle: 7 × 4 = 28 unit squares

8 cm × 3 cm rectangle: 8 × 3 = 24 unit squares
28 > 24, so the first rectangle needs more powder.

Why it happens: if the colouring is even, the powder used is proportional to the
region's area, not to how long or how wide it looks. Counting the non-overlapping 1
cm squares that pack into each rectangle is exactly what measuring area means —
and 4 rows of 7 squares is 28 squares, while 3 rows of 8 squares is only 24.

Notice: both rectangles have almost the same perimeter (22 cm and 22 cm). Equal
perimeters, unequal areas — the very point made on the next page.

In-text Questions — Page 150
Why Can't Perimeter be a Measure of Area?

MATH TALK

Q1 What is the area of each triangle in this rectangle?

14 cm² each.

Page 3 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
e m.
Area of the rectangle = 7 × 4 = 28 cm²
m l as
.co
The diagonal cuts it into two congruent triangles
m a g
l a se
g
Area of each triangle = ½ × 7 × 4 = 14 cm²
a

. com
Why it happens: a diagonal of a rectangle splits it into two triangles that fit exactly
ag
on each other (turn one through halfeamturn about the centre of the rectangle).
g l as and the two areas add up to 28 cm², so each
a
Congruent figures have equal areas,
must be 14 cm².

co m
se m.
o m l a
g for area? Couldn't
m .c do we count the number of unit squares to assign measures
a
se we have just used the perimeter of a region, i.e., the length of its boundary as a
Q2 Why

l a
ag measure of its area?

m a s
m .co agl
l a se
g
No. Perimeter measures the boundary; area measures the region enclosed. They are different
a
quantities, and one does not determine the other.

m
A unit square is the natural yardstick for a region because copies of it can be packed to fill up a

. co
m
region without gaps or overlaps. Counting how many fit tells us exactly "how much surface"

m as e
l
there is.
.co a g
m it happens: perimeter is a length (measured in cm), area is a two-dimensional
a s eWhy
agl measure (measured in cm²). A rangoli of perimeter 22 cm might use 28 cm² of

se m
a
powder or 24 cm² of powder — the boundary length simply does not decide the
answer.
. com a g l
m
ase
agl

co m
If two regions have the same perimeter, can't we conclude that they have the same
.
Q3

em
area? Or, if one region has a larger perimeter than another region, can't we

m l as
.co g
conclude that it also has a larger area?

emANSWER a
a s
agl
.c
e m
No to both.

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

co m
m .
m ase
.co


a g l Page 4 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

RECTANGLE PERIMETER AREA

7 cm × 4 cm 2(7 + 4) = 22 cm 28 cm²

8 cm × 3 cm 2(8 + 3) = 22 cm 24 cm²

10 cm × 1 cm 2(10 + 1) = 22 cm 10 cm²

Same perimeter, three different areas. So equal perimeters do not force equal areas.

Why it happens: with the perimeter fixed at 22 cm, the two sidelengths must add to
11 cm, but the product can be anything from just above 0 up to 5.5 × 5.5 = 30.25 cm².
Fixing a sum does not fix a product.

Q4 Find two rectangles that are examples of such regions. If needed, use a grid paper
(given at the end of the book) for this.

We need Region 1 with the larger perimeter but the smaller area.

REGION RECTANGLE PERIMETER AREA

Region 1 12 cm × 1 cm 26 cm 12 cm²

Region 2 4 cm × 4 cm 16 cm 16 cm²

Perimeter of Region 1 = 26 cm > 16 cm = Perimeter of Region 2

Area of Region 1 = 12 cm² < 16 cm² = Area of Region 2 ✓

Why it happens: a long thin rectangle spends almost all of its boundary on the two
long sides while enclosing very little. Stretching a rectangle out increases its
perimeter but squeezes its area down.

Try This: on grid paper draw a 20 × 1 rectangle and a 5 × 5 square. The thin one has
perimeter 42 units and area 20 squares; the square has perimeter 20 units and area
25 squares.

Page 5 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q5 Also give an example of two regions of other shapes, where the region with the
larger perimeter has the smaller area! This property should be visually clear in your
example. [Math Talk]

Take a square of side 6 cm and cut deep, thin notches into it, like the teeth of a comb.

Region 2: 6 × 6 Region 1: the comb
Every extra slit adds a lot of boundary but removes area.

Region 2 (the plain square): perimeter 24 cm, area 36 cm².
Region 1 (the comb): each slit is 5 cm deep and very thin. Cutting it away removes only a
sliver of area, but adds about 5 + 5 = 10 cm of new boundary. With 6 slits the perimeter
grows past 80 cm while the area drops below 36 cm².

Why it happens: a slit of depth d and width w takes away area dw — which is tiny
when w is tiny — but it adds boundary of length about 2d, which does not shrink at
all as w shrinks. So perimeter can be made as large as we like while area only goes
down.

Figure it Out — Pages 150 – 152

Page 6 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Section 7.1 Rectangle and Squares

MATH TALK

Q1 Identify the missing sidelengths.

(i)
4 in
28 in²
21 in²
3 in 7 in

35 in²
14 in²

2 in

? in

(i) Four rectangles pinned round one point. The four given lengths are 4 in, 7 in, 3 in and
2 in; the side marked “? in” is the one to find.

Page 7 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

(ii) ? ?

4m 29 m² 11 m²

?

Area = 50 m²
(ii) The thin outline is the whole 4 m high strip; the thick outline is the rectangle of area
50 m². Three sidelengths are marked “?”.

(i) ? = 2 in. Each rectangle hands you a length that the next one needs.

Page 8 of 83

ase
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
e m.
m l as
m .co a g
l a se
a g 4 in
28 in²
4×7 . com ag
a s em 21 in²
ag l
3 in 7×3
7 in co m
em.
c o m 35 in² g l as
m . a
l a se 7×5 14 in²
a g 2×7
m a s
.co agl
2 in
a s em
agl ? = 2 in

co m
m .
e
Four rectangles round one point. Each one shares a full side with the next, so the chain of deductions

m l as
.co g
never breaks.

m a
l a se
ag The right edge of the 28 in² rectangle = 4 in (given) + 3 in (the height of the 21 in² rectangle)
se m
com g l a
.
= 7 in
m a
ase
agl
So its other side = 28 ÷ 7 = 4 in

Top of the 35 in² rectangle = 3 in (overhang) + 4 in = 7 in
co m
m .
m
So its height = 35 ÷ 7 = 5 in
as e
.co a g l
se m
g l a
a Left edge of the 14 in² rectangle = 5 in + 2 in (the stub below) = 7 in
c
m .
So ? = 14 ÷ 7 = 2 in
m a s e
e m . co agl
g l as
(The 21 in² rectangle starts the chain: its width is the given 7 in, so its height is 21 ÷ 7 = 3 in.)
a
(ii) The top strip has height 4 m throughout, so divide each area by 4.

co m
m .
m ase
.co


a g l Page 9 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Width of the 29 m² part = 29 ÷ 4 = 7.25 m

Width of the 11 m² part = 11 ÷ 4 = 2.75 m

The bold rectangle has area 50 m², and its top part is 29 m²,

so its lower part = 50 − 29 = 21 m²

Its width is the same 7.25 m, so

height of the lower part = 21 ÷ 7.25 = 84/29 = 2.90 m (approx.)

Check it yourself: the whole top strip is 7.25 + 2.75 = 10 m long and 4 m high, giving
40 m² = 29 + 11. ✓

Why it happens: whenever two rectangles sit side by side sharing a full edge, that
edge is a common factor of both areas. Dividing an area by the shared side is what
recovers the unknown side.

Page 10 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q2 The figure shows a path (the shaded portion) laid around a rectangular park EFGH.

D C

H G

E F

A B
The shaded (khaki) border is the path; the green rectangle EFGH is the park, inside the
outer rectangle ABCD.

(i) What measurements do you need to find the area of the path? Once you identify
the lengths to be measured, assign possible values of your choice to these
measurements and find the area of the path. Give a formula for the area. An
example of a formula — Area of a rectangle = length × width. [Hint: There is a
relation between the areas of EFGH, the path, and ABCD.] (ii) If the width of the path
along each side is given, can you find its area? If not, what other measurements do
you need? Assign values of your choice to these measurements and find the area of
the path. Give a formula for the area using these measurements. [Hint: Break the
path into rectangles.] (iii) Does the area of the path change when the outer
rectangle is moved while keeping the inner rectangular park EFGH inside it, as
shown?

Page 11 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

D C D C D C
H G
H G
H G

E F
E F
E F
A B A B A B

The same park EFGH sitting in three different positions inside the outer rectangle ABCD.

(i) Four measurements are enough: the two sidelengths of the outer rectangle ABCD and the
two sidelengths of the park EFGH.

Area of ABCD = Area of EFGH + Area of the path

so Area of the path = Area of ABCD − Area of EFGH

Area of the path = (AB × BC) − (EF × FG)

Taking AB = 20 m, BC = 15 m, EF = 16 m, FG = 11 m:

Area of the path = (20 × 15) − (16 × 11) = 300 − 176 = 124 m²

(ii) The four widths alone are not enough — you also need the length and width of the park.
With the park l by b, and the path of width p on the left, q on the right, t on top and d at the
bottom, break the path into eight rectangles: four strips and four corners.

top and bottom strips: l(t + d)

left and right strips: b(p + q)

the four corner rectangles: pt + qt + pd + qd = (p + q)(t + d)

Area of the path = l(t + d) + b(p + q) + (p + q)(t + d)

With l = 16 m, b = 11 m and every width 2 m (so p = q = t = d = 2):

= 16 × 4 + 11 × 4 + 4 × 4 = 64 + 44 + 16 = 124 m²

Page 12 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Same answer as (i) — as it must be, since the outer rectangle is then 20 m by 15 m.
(iii) No, it does not change.

Why it happens: however the outer rectangle is shifted, its size is unchanged and
the park's size is unchanged. The path is exactly "outer minus inner", so its area
stays Area(ABCD) − Area(EFGH). What changes is only the shape of the path — wide
on one side, narrow on the other. Area is preserved; the shape is not.

Page 13 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
The figure shows a plot with sides 14m and 12m, and with a crosspath.
e
Q3

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

com
e m . ag
g l as
a

co m
e m.
om m g l as
m.c12 a
l a se
a g
m a s
m .co agl
l a se
a g

co m
m .
14 m
as e
. com a g l
m
ase
agl
A 14 m × 12 m plot with a crosspath — one strip running across and one running down.
The width of the strips is not marked.

se m
com g l a
. a
What other measurements do you need to find the area of the crosspath? Once you
m
ase
identify the lengths to be measured, assign some possible values of your choice and

a gl
find the area of the path. Give a formula for the area based on the measurements
you choose. [Math Talk]

co m
m .
ase

m l
.co
You need the widths of the two strips — nothing else.
a g
a s emLet the plot be 14 m long and 12 m wide. Let the strip that runs across the length have width a,
agl and the strip that runs across the width have width b.
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 14 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Horizontal strip = 14 × a

Vertical strip = 12 × b

But the little square where they cross has been counted twice, and it measures a × b.

Area of the crosspath = 14a + 12b − ab

Choosing a = 2 m and b = 1 m:

= 14 × 2 + 12 × 1 − 2 × 1

= 28 + 12 − 2 = 38 m²

Check it yourself: the crosspath leaves four rectangles of grass. Along the 14 m side
the path takes away 1 m, leaving 13 m; along the 12 m side it takes away 2 m,
leaving 10 m. So the grass measures 13 × 10 = 130 m², and 130 + 38 = 168 m² = 14 ×
12. ✓

Why it happens: the crossing square belongs to both strips. Adding the two strip
areas counts it twice, so it must be subtracted once. This is the same "add, then
remove the overlap" idea used whenever two regions meet.

Page 15 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q4 Find the area of the spiral tube shown in the figure. The tube has the same width
throughout.

20

1

15

5
10 20
1

15 5

10

20
The spiral tube. It has the same width (1 unit) all the way along; the outer square is 20 ×
20.

[Hint: There are different ways of finding the area. Here is one method.]

Page 16 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

5
?

5

The hint: a bent tube of width 1 with outer arms 5 and 5, and beside it a straight tube of
the same width whose length is to be found.

What should be the length of the straight tube if it is to have the same area as the
bent tube on the left?

Area of the spiral = 112 sq. units.
First settle the hint. The bent tube is an L whose two outer arms are 5 and 5, and whose width is
1.

Area of the L = 5 × 1 + 5 × 1 − 1 × 1 = 9 sq. units

(the corner square lies in both arms, so it is subtracted once)

A straight tube of width 1 and length ℓ has area ℓ × 1.

So ℓ = 9 units.

The same idea straightens the whole spiral. Every bend costs one unit square. Add all nine
outer arms, then subtract 1 for each of the 8 bends.

ARM (MEASURED ALONG THE 1 2 3 4 5 6 7 8 9 TOTAL
OUTER EDGE)

Length 20 20 20 15 15 10 10 5 5 120

Page 17 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Straightened length = 120 − 8 = 112 units

Width = 1 unit

Area = 112 × 1 = 112 sq. units

Why it happens: unrolling the tube is a dissection — cut it at the bends and lay the
pieces end to end. Nothing is added or thrown away, so the area is unchanged. Each
bend is a square of side 1 shared by the two arms meeting there, which is why it is
counted once instead of twice.

Check it yourself: the inner edge of the spiral measures 19 + 18 + 18 + 13 + 13 + 8 +
8 + 3 + 4 = 104 units. The average of the outer and inner edges is (120 + 104) ÷ 2 =
112 — the length of the middle line of the tube, and the same answer again.

Page 18 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
In this figure, if the sidelength of the square is doubled, what is the increase in the
e
Q5

m l as
.co
areas of the regions 1, 2 and 3? Give reasons.

a g
se m
g l a
a

co m
e
.
m2 ag
g l as
a

co m
em.
m l as
m .co a g
l a se 1
a g
m a s
m .co 3 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
A square with the diagonal from the top-right corner to the bottom-left corner, and a
second cut from the top-left corner to the centre. The two ticks show that the centre is
the midpoint of the diagonal.

co m
m .
m as e
.co a g l
emEvery region becomes 4 times as big, so each increases by 3 times its original area.

a s
agl In the figure the diagonal from the bottom-left corner to the top-right corner splits the square
.c
s e m
m a
in half. Region 3 is that lower-right half. The segment from the top-left corner to the centre of

m . co
the square then splits the upper half into regions 1 and 2. Let the sidelength be a.
e agl
g l as
a

com
m .
m ase
.co


a g l Page 19 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

REGION SIDE A SIDE 2A INCREASE

1 a²/4 a² 3a²/4

2 a²/4 a² 3a²/4

3 a²/2 2a² 3a²/2

Region 3 = half the square = a²/2 → (2a)²/2 = 4a²/2 = 2a²

The centre is the midpoint of the diagonal, so the segment from the top-left corner is a

median of the upper triangle

Region 1 = Region 2 = ½ × a²/2 = a²/4 → a²

Why it happens: doubling the sidelength doubles every length in the figure, because
all the cuts are described by the corners and the centre. A region that was l by w
becomes 2l by 2w, so its area is multiplied by 2 × 2 = 4. Areas scale by the square of
the length factor, so "twice as long" means "four times the area" — never twice.

Tip: the same rule answers many questions at once. Triple the sidelength and every
area is multiplied by 9; halve it and every area is divided by 4.

Page 20 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q6 Divide a square into 4 parts by drawing two perpendicular lines inside the square as
shown in the figure.

Two perpendicular cuts inside the square, each running from one side to the opposite
side, dividing it into four pieces.

Rearrange the pieces to get a larger square, with a hole inside. You can try this
activity by constructing the square using cardboard, thick chart paper, or similar
materials. [Math Talk]

Draw the two cuts as in the figure: one line from a point on the left side to a point on the right
side, and the other perpendicular to it, from a point on the top side to a point on the bottom
side. Cut out the four pieces.

Page 21 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Now slide the four pieces outwards, keeping each one's slant direction, until each cut edge of
one piece lies against the matching cut edge of the next. They close up into a bigger square
with a square hole in the middle.

hole

square, side s
bigger square, side L
The four pieces are only moved, never stretched — so the hole is exactly the extra area of the bigger
square.

Area of the four pieces = area of the original square = s² (unchanged)

Area of the big square = area of the pieces + area of the hole

so area of the hole = L² − s²

The side L of the new square is the length of each cut, and the side of the hole is the offset d —
the distance between the two points where one cut meets the pair of opposite sides. So L² = s² +
d², exactly the Baudhāyana–Pythagoras relation.

Why it happens: a dissection can never create or destroy area — that is the one
thing which stays invariant when pieces are rearranged. The puzzle only looks
paradoxical because the outline has grown; the growth is precisely accounted for by
the hole. If you make the two cuts pass exactly through the centre, all four pieces are
congruent and the hole is a neat square in the middle.

Try This: make the two cuts nearly parallel to the sides. The offset d becomes tiny,
and so does the hole. Make them steep and the hole grows.

Page 22 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

In-text Questions — Page 153
Triangles

Q1 In the given figure, which triangle has a greater area: ∆XDC or ∆YDC, if both the
rectangles are identical?

A X B A Y B

D C D C

Two identical rectangles ABCD. In the first, X sits near A on the side AB; in the second, Y
sits at the middle of AB. Both triangles stand on the same side DC.

Neither — the two triangles have exactly the same area, and each is half of the rectangle.
X and Y both lie on the side AB, and AB ‖ DC. So the perpendicular distance from X to DC and
the perpendicular distance from Y to DC are both equal to the width of the rectangle.

Area (∆XDC) = ½ × DC × (distance from AB to DC)

Area (∆YDC) = ½ × DC × (the same distance)

So Area (∆XDC) = Area (∆YDC) = ½ × area of the rectangle

Why it happens: sliding the apex along a line parallel to the base changes the shape
of a triangle but not its height, and the base has not moved either. Area depends
only on base and height, so it does not change at all.

Page 23 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
In the given figure, which triangle has a greater area: ∆XDC or ∆YBC, if both the
e
Q2

m l as
.co
rectangles are identical?

a g
se m
g l a
a A X B A B

co m
em . ag
g l as Y
a
m
.co
D C D C

se m
com g l a
Two identical rectangles ABCD. ∆XDC stands on DC with X on AB; ∆YBC stands on BC with

m . a
ase
Y on AD.

agl
om a s
c agl
.
m of its rectangle, even though they sit on different
s e
They are equal again. Each triangle is half
a
agl
sides.

co m
.
∆XDC: base DC, apex X on AB, height = AD

e m
m l as
.co
Area (∆XDC) = ½ × DC × AD = ½ × area of the rectangle
a g
a s em
agl ∆YBC: base BC, apex Y on AD, height = AB
se m
com a
Area (∆YBC) = ½ × BC × AB = ½ × area of the rectangle
. a g l
e m
g l asboth halves are the same. So Area (∆XDC) = Area (∆YBC).
Since the rectangles are identical,
a
m
Why it happens: in a rectangle, opposite sides are parallel and equal. Whichever
. co
m
side you choose as the base, the opposite side is exactly one "height" away — so a

o m l a se
ag
triangle with its base on one side and its apex anywhere on the opposite side always

m .cfills half the rectangle. The choice of side makes no difference.
l a se
ag
.c
s e m
m a
. co agl
Q3 Find the area of ∆ XDC.

e m
g l as
ANSWER a
m
From Fig. 7.1 the enclosing rectangle ABCD has DC = 5 and AD = 4, and X lies on AB.

. co
e m
m l as
.co a g
Page 24 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

base = DC = 5

height = distance from X to DC = AD = 4

Area (∆XDC) = ½ × 5 × 4 = 10 sq. units

Tip: you do not need to know where X sits on AB. Only its distance from DC matters,
and that is fixed at 4.

Q4 To find the area of a triangle, what measurements do we need?

Just two: one side (the base) and the height to that side — that is, the perpendicular distance
from the opposite vertex to the line of that base.

Area of a triangle = ½ × base × height

Why it happens: the triangle is exactly half of the rectangle built on that base with
that height (Fig. 7.1). The rectangle needs only its two sidelengths, so the triangle
needs only the matching two measurements. The other two sides of the triangle,
and its angles, do not enter the calculation at all.

Tip: a triangle has three sides, so it has three base–height pairs. All three give the
same area — a fact used again and again in this chapter.

Q5 How do we get the outer rectangle from the given triangle?

Draw the line l through the apex A parallel to the base BC. Then drop perpendiculars to l from B
and from C.

Page 25 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Step 1: draw l ‖ BC through A

Step 2: at B draw BE ⊥ BC, meeting l at E

Step 3: at C draw CD ⊥ BC, meeting l at D
BCDE is the required rectangle, and ∆ABC lies inside it with the same base BC.

Why it happens: BE and CD are both perpendicular to BC, so they are parallel to
each other and equal in length (both equal the distance between the parallel lines l
and BC). With ED ‖ BC as well, BCDE has four right angles — it is a rectangle whose
base is BC and whose height is the height of the triangle.

In-text Questions — Page 154
Triangles

Q1 BCDE is a rectangle (how?)

Because it has been built to have four right angles.

BE ⊥ BC and CD ⊥ BC by construction, so ∠EBC = ∠DCB = 90°.
BE and CD are both perpendicular to BC, so BE ‖ CD.
ED lies along the line l, and l ‖ BC by construction.

A quadrilateral with both pairs of opposite sides parallel and one angle 90° has all four angles
90°. So BCDE is a rectangle, with BC as its base and BE as its height.

Q2 BXAE is a rectangle (how?), so the height of the rectangle is the same as the height
of the triangle.

X is the foot of the perpendicular from A to BC, so ∠AXB = 90°. Also ∠XBE = 90° (BE ⊥ BC), and EA
lies along l which is parallel to BX.

Page 26 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

∠BXA = 90°, ∠XBE = 90°, EA ‖ XB, AX ‖ EB

So BXAE has four right angles → it is a rectangle

Opposite sides of a rectangle are equal, so AX = BE

AX is the height of the triangle and BE is the height of the rectangle. They are equal — which is
exactly why the rectangle's height can be used in the triangle's area formula.

Q3 Will this formula hold for the kind of triangle, around which we cannot draw a
rectangle with BC as the base?

Yes, it still holds. For an obtuse triangle the foot of the perpendicular from A falls outside BC, at
D on the line CB extended. Then ∆ABC is the difference of two right-angled triangles, each of
which does sit inside a rectangle.

Area (∆ABC) = Area (∆ADC) − Area (∆ADB)

= ½ × h × DC − ½ × h × DB

= ½ × h × (DC − DB)

= ½ × h × BC

Why it happens: ∆ADC and ∆ADB share the same height h = AD, so their areas differ
only through their bases DC and DB. Removing the smaller from the larger removes
the overlap exactly and leaves ∆ABC, whose base is DC − DB = BC. The formula ½ ×
base × height therefore covers acute, right-angled and obtuse triangles alike.

In-text Questions — Page 155
Some Applications of the Area Formula

Q1 Find BY.

In the figure, AX ⊥ BC with AX = 5, BC = 3, and AC = 4, while BY ⊥ AC.
The trick is to compute the same area twice, once with each base–height pair.

Page 27 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Using base BC: Area (∆ABC) = ½ × AX × BC = ½ × 5 × 3 = 15/2 sq. units

Using base AC: Area (∆ABC) = ½ × BY × AC = ½ × 4 × BY = 2 BY

So 2 BY = 15/2

BY = 15/4 = 3.75 units

Why it happens: a triangle has one area but three base–height pairs, and all three
must give the same number. Writing that equality down turns a known area into an
unknown altitude. This "two ways, one area" move is used again in Q2 and Q3 of the
next exercise.

Q2 What is Area (∆ABC)?

Use the base and height that are already marked in the figure.

Area (∆ABC) = ½ × AX × BC

=½×5×3

= 15/2 = 7.5 sq. units

Tip: AX is the perpendicular from A to BC, so BC and AX are a matching base–height
pair. Never multiply a base by a slanted side.

Page 28 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
Are the 4 triangles obtained by drawing the diagonals of a rectangle (regions 1 – 4
e
Q3

m l as
.co
in the figure) of equal areas?

a g
se m
g l a
a A B
m
co2
em . ag
g l as
a
1 3 .com
O em
m a s
. co a gl
m
ase
agl 4

m a s
agl
D C
.co
a s em
agl ABCD meet
The two diagonals of rectangle at O and cut it into the four triangles marked
1, 2, 3 and 4.

co m
m .
m as e
.co a g l
a s em— all four have equal areas, even though they are not all congruent.
gl
Yes
a Let the diagonals meet at O. Take two adjacent triangles, say 1 and 2, and choose OD and OB as
their bases. Both bases lie on the same straight line BD, and both triangles have the same apex
se m
com g l a
.
A, so they have the same height — the perpendicular from A to BD.
m a
ase
agl
The diagonals of a rectangle bisect each other, so OB = OD

Same base length, same height → Area(1) = Area(2)
co m
Repeating around the point O: Area(2) = Area(3), Area(3) = Area(4) m.
o m l a se
c
.Area(1) = Area(2) = Area(3) = Area(4) = ¼ × area of the a g
se m rectangle

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

co m
m .
m ase
.co


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Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Why it happens: triangles 1 and 3 are congruent to each other, and so are 2 and 4
— but a triangle from the first pair need not be congruent to one from the second.
Equal area is a weaker demand than congruence, and the bisected diagonals supply
exactly that. This gives the general statement: in a triangle, the line joining a
vertex to the midpoint of the opposite side divides it into two triangles of equal
areas.

In-text Questions — Page 156
Triangles between Parallel Lines with a Common Base

Q1 (i) Which of these triangles has the maximum area, and which has the minimum
area?

None of them — every one of these triangles has exactly the same area. There is no
maximum and no minimum.
All the triangles share the base BC, and every third vertex lies on the line l, which is parallel to
BC. The distance between two parallel lines is the same everywhere, so every one of these
triangles has the same height.

height = distance between l and BC = d (the same for every position of the apex)

Area = ½ × BC × d, whatever the apex

All the areas are equal.

Why it happens: the apex sliding along l stretches the triangle sideways. Its two
slanted sides get longer, but the base and the height — the only two numbers in the
formula — never change. This is the single most useful fact in the chapter: triangles
on the same base and between the same parallels are equal in area.

Q2 (ii) Which of these triangles has the maximum perimeter, and which has the
minimum perimeter?

There is no maximum perimeter, but there is a definite minimum.

Page 30 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

No maximum: push the apex A further and further along l and the two slanted sides AB and
AC grow without limit, so the perimeter grows without limit.
Minimum: BC is common to every triangle, so we only have to make AB + AC as small as
possible. Reflect C in the line l to get C′. Then AC = AC′, so AB + AC = AB + AC′, which is a path
from B to C′ through A. The shortest such path is the straight segment BC′, so choose A
where BC′ meets l.

For the minimum: A = the point where BC′ cuts l, where C′ is the reflection of C in l

Why it happens: reflecting in l is a mirror move — it preserves every length, so AC
and AC′ are equal for every position of A. That turns "make a bent path as short as
possible" into "make a path from B to C′ as short as possible", and a straight line is
the shortest path between two points.

In-text Questions — Page 157
Triangles between Parallel Lines with a Common Base

MATH TALK

Q1 What can we say about the lengths of AB and its reflection AB´?

They are equal: AB = AB′.

X is the point where BB′ crosses l, and the mirror sends B to B′ with BX = B′X

∠AXB = ∠AXB′ = 90°, and AX is common

So ∆AXB ≅ ∆AXB′ (SAS)

Hence AB = AB′, and in the same way AC = AC′

So the bent path B → A → C and the bent path B → A → C′ have exactly the same length.

Why it happens: a reflection is a rigid motion — it flips the plane over without
stretching it. Every length in the figure survives the flip unchanged, which is why the
mirror can be used to replace an awkward path by an equal one that is easier to
shorten.

Page 31 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q2 Analyse whether A lies on the perpendicular bisector of BC. [Math Talk]

Yes, it does — so the triangle of least perimeter is the isosceles one.
Set BC along a line and let l be parallel to it at distance d. Reflecting C in l puts C′ directly above C
at height 2d, while B is at height 0.

The segment BC′ rises from height 0 to height 2d

It crosses l at exactly half that rise, i.e. at the midpoint of BC′

That crossing point sits horizontally halfway between B and C

So A is directly above the midpoint of BC → AB = AC

Why it happens: because l ‖ BC, the mirror image C′ is exactly as far above l as C is
below it. The straight line BC′ therefore meets l at its own midpoint, and that point
lies on the perpendicular bisector of BC. Intuition and proof agree here — but notice
that the proof, not the picture, is what settles it.

Figure it Out — Pages 157 – 159

Page 32 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Triangles

MATH TALK TRY THIS

Q1 Find the areas of the following triangles:

N
E
A
N
5 4 cm
cm
3 cm
3.2 cm

B E 4 cm C D F A 3 cm T

(i) (ii) (iii)

(i) ∆ABC with AE ⊥ BC, AE = 3 cm and BC = 4 cm. (ii) ∆DEF with DN ⊥ EF, DN = 3.2 cm and EF
= 5 cm. (iii) ∆NAT, right-angled at A, with AN = 4 cm and AT = 3 cm.

In each case pick the matching base–height pair and halve the product.

TRIANGLE BASE HEIGHT AREA

(i) ∆ABC BC = 4 cm AE = 3 cm ½ × 4 × 3 = 6 cm²

(ii) ∆DEF EF = 5 cm DN = 3.2 cm ½ × 5 × 3.2 = 8 cm²

(iii) ∆NAT AT = 3 cm AN = 4 cm ½ × 3 × 4 = 6 cm²

(i) 6 cm² (ii) 8 cm² (iii) 6 cm²

Why it happens: in (i) the foot E lies inside BC, in (ii) the height DN falls on the side
EF, and in (iii) the two perpendicular sides are themselves a base–height pair — no
extra line is needed. The formula does not care which case it is; it only needs a side
and the perpendicular distance from the opposite vertex to that side's line.

Tip: in (i) the whole base BC is 4 cm. Where the foot E sits along it makes no
difference to the area.

Page 33 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
Find the length of the altitude BY.
e
Q2

m l as
.co a g
a s em
a gl figure AX ⊥ BC (with X outside the segment), AX = 4 units, BC = 6 units, AC = 8 units, and
In the
BY ⊥ AC.

. com ag
Using base BC: Area (∆ABC) = ½ × BCe×mAX = ½ × 6 × 4 = 12 sq. units
g l as
a
Using base AC: Area (∆ABC) = ½ × AC × BY = ½ × 8 × BY = 4 BY
co m
em.
com g l as
m . a
e
So 4 BY = 12
s
a glaBY = 3 units
m a s
m .co agl
se
Why it happens: the triangle is obtuse at B, so the foot X of the altitude from A

g l a
lands outside BC. That does not matter — as shown on page 154, ½ × base × height
a
is still the area. Once the area is known, the second base–height pair gives the

m
second altitude immediately.

. co
e m
m l as
.co a g
a s em Find the area of ∆SUB, given that it is isosceles, SE is perpendicular to UB, and the
gl
Q3

a area of ∆SEB is 24 sq. units.

se m
com g l a
. a

m
ase
48 sq. units.

agl
∆SUB is isosceles with SU = SB, and SE is the perpendicular from the apex S to the base UB. In an
isosceles triangle that perpendicular also bisects the base, so UE = EB.

co m
m .
as e
com l
∆SEU and ∆SEB have equal bases (UE = EB) and the same height SE
.So a g
se m Area (∆SEU) = Area (∆SEB) = 24 sq. units
g l a
a c
m .
Area (∆SUB) = 24 + 24 = 48 sq. units
m a s e
e m . co agl
g l as
a

co m
m .
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.co


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Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Why it happens: ∆SEU ≅ ∆SEB by RHS — right angle at E, hypotenuses SU = SB, and
SE common — so UE = EB and the two halves are congruent. Congruent pieces have
equal areas, so the altitude from the apex of an isosceles triangle splits it into two
equal halves.

Q4 [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.

Take the rectangle ABCD with DC as base and height h.

1. Extend DC beyond C to a point E so that CE = DC. Now DE = 2 × DC.
2. Mark any point P on the line AB (the side opposite DC) — the vertex A itself will do.
3. Join PD and PE. Then ∆PDE is the required triangle.

base DE = 2 × DC, height of ∆PDE = distance from AB to DC = h

Area (∆PDE) = ½ × (2 × DC) × h = DC × h

= Area of rectangle ABCD

Why it happens: the triangle keeps the rectangle's height but is given twice its base,
and the ½ in the triangle formula cancels the doubling exactly. P may be anywhere
on the line AB because the apex may slide along a line parallel to the base without
changing the area.

Q5 [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

Take ∆ABC with base BC and height h, and cut it at half the height.

1. Mark M, the midpoint of AB, and N, the midpoint of AC. Join MN — this line is parallel to BC
and is at height h/2.
2. Cut along MN. The top piece is the small triangle ∆AMN.
3. Rotate ∆AMN through half a turn about M. The vertex A lands on B, and the piece fills the
gap on the left. Do the same on the right with the other half — or simply cut ∆AMN along the
altitude and swing the two halves out to the sides.

Page 35 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

The result is a rectangle of base BC and height h/2

Area = BC × h/2 = ½ × BC × h = Area (∆ABC) ✓

Why it happens: M is the midpoint of AB, so turning ∆AMN about M carries A onto B
and MN onto a segment of the same line — nothing is stretched, so no area is
created or lost. This is a dissection, and dissection is exactly the operation that
preserves area.

Try This: cut a paper triangle along the line joining the midpoints of two sides and
fold the top down. It lands flat on the base and the figure becomes a rectangle of
half the height.

Page 36 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q6 ABCD, BCEF, and BFGH are identical squares.

H G

A B F

D C E

ABCD, BCEF and BFGH are identical squares. DH is drawn from D to H; the red region and
the blue region are the two parts it makes.

(i) If the area of the red region is 49 sq. units, then what is the area of the blue
region? (ii) In another version of this figure, if the total area enclosed by the blue
and red regions is 180 sq. units, then what is the area of each square? [Math Talk]

Let each square have side s. Place D at the corner, so that D, C, B, H run as in the figure: H is
directly above B, and B is directly above C.

Page 37 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

H
G

A B
F

blue red = ∆DCH

D C E

The line DH cuts off the blue triangle; C, B and H lie on one vertical line, which makes the red region a
single triangle.

Red region. C, B and H are collinear, so the red region is the triangle DCH with base CH = CB +
BH = 2s and height DC = s.

Area (red) = ½ × 2s × s = s² = the area of one square

Blue region. DH rises 2s across a horizontal run of s, so at half the run it has risen half the way.
It therefore crosses AB at its midpoint P, with AP = s/2.

Area (blue) = Area (∆ADP) = ½ × AP × AD = ½ × (s/2) × s = s²/4

(i) Red = 49, so s² = 49.

Area (blue) = 49 ÷ 4 = 12.25 sq. units

(ii) Blue + red = 180.

Page 38 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
e m.
s²/4 + s² = 180
m l as
(5/4) s² = 180
m .co a g
l a se
g
s² = 180 × 4/5 = 144 sq. units — the area of each square
a

. com
Why it happens: the red triangle's area equals exactly one square, however big the
ag
a s em long and its height is one side long — and
squares are, because its base is two sides
the ½ halves the product back. g
aOncel both regions are expressed as multiples of s², a
single given number fixes s².

co m
se m.
o m l a
c and N are the midpoints of XY and XZ, what fraction ofagthe area of ∆XYZ is the
If .M
Q7
m
se area of ∆XMN? [Hint: Join NY] [Try This]
l a
ag
m a s
.co agl

One quarter.
se m
g l a
a
Join NY, as the hint suggests, and use the midpoint fact twice.

In ∆XYZ, N is the midpoint of XZ, so YN is a median
co m
m .
m as e
l
Area (∆XNY) = ½ × Area (∆XYZ)
.co a g
a s em
agl In ∆XNY, M is the midpoint of XY, so NM is a median
se m
com a
Area (∆XMN) = ½ × Area (∆XNY) = ½ × ½ × Area (∆XYZ)
. a g l
m
ase
agl
Area (∆XMN) = ¼ × Area (∆XYZ)

co m
m .
e
Why it happens: a median cuts a triangle into two triangles with equal bases and
m l as
.co g
the same height, so it halves the area every time. Halving twice gives a quarter.

em a
s
Notice the answer does not depend on the shape of ∆XYZ at all.
l a
ag c
m .
Did you know? MN is parallel to YZ and half as long. Since ∆XMN has half the base
m a s e
. co agl
and half the height of ∆XYZ, its area is ½ × ½ = ¼ of it — the scaling rule of Q5 on
e m
as
page 152, seen again.

a g l

co m
m .
m ase
.co


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Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q8 Gopal needs to carry water from the river to his water tank. He starts from his
house. What is the shortest path he can take from his house to the river and then to
the water tank? Roughly recreate the map in your notebook and trace the shortest
path.

River

Water tank

House

The map: the river is the strip between the two lines; the water tank and the house are
both on the same side of it.

[Math Talk]

Reflect the water tank in the river, join the house to that image by a straight line, and fetch the
water where that line meets the river.

1. Let H be the house, T the water tank, and let the river be the line r. (Both H and T are on the
same side of r.)
2. Draw T′, the mirror image of T in r.
3. Join HT′ by a straight line. Let it cut r at P.
4. The shortest journey is H → P → T.

For any point Q on the river, QT = QT′ (mirror)

So HQ + QT = HQ + QT′, a path from H to T′ through Q
The shortest path from H to T′ is the straight segment HT′

That segment meets the river at P, so HP + PT is the least possible

Page 40 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Why it happens: this is the same mirror argument used on page 157 for the triangle
of least perimeter. The reflection does not change any distance on the far side of the
river, but it straightens a bent path into a single line — and between two points the
straight line is the shortest.

Did you know? Light does exactly this. A ray reflecting off a mirror takes the
shortest path, which is why the angle of incidence equals the angle of reflection at P.

In-text Questions — Page 159
Area of any Polygon

Q1 How do we find the area of this quadrilateral? What measurements do we need for
this?

Split it into two triangles with a diagonal, and add their areas.
For quadrilateral ABCD, join BD. Now ABCD = ∆ABD + ∆CBD, and both triangles stand on the
same base BD.

Measure: the diagonal BD, and the two heights — the perpendicular from A to BD and the

perpendicular from C to BD.

Area (ABCD) = ½ × BD × h₁ + ½ × BD × h₂

= ½ × BD × (h₁ + h₂)

Why it happens: a diagonal separates the quadrilateral into two pieces that do not
overlap and together make the whole, so the areas simply add. Because both
triangles share the diagonal as base, the two heights can be added first, which saves
work.

Page 41 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q2 How do we find the area of this pentagon?

The same way — cut it into triangles.
Pick one vertex of the pentagon and join it to the two non-neighbouring vertices. That gives
three triangles. Measure each triangle's base and height, and add the three areas.

A pentagon (5 sides) → 3 triangles

In general, an n-sided polygon → n − 2 triangles

Tip: for an awkward polygon it is often easier to enclose it in a rectangle and
subtract the corner triangles, rather than to add up many small pieces.

Q3 Can any polygon be divided into triangles?

Yes. Every polygon can be cut into triangles by drawing diagonals inside it.

Why it happens: a polygon has straight sides, and any region bounded by straight
lines can be sliced up by more straight lines until only three-sided pieces are left. For
a convex polygon this is easy — join one vertex to all the others. For a polygon that
caves inwards you may have to choose the diagonals more carefully, but a set that
works always exists.

This is why the triangle formula is the key to the whole chapter:

Know ½ × base × height → know the area of any polygon

Figure it Out — Page 160

Page 42 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Area of any Polygon

MATH TALK

Q1 Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm,
BM is perpendicular to AC, and DN is perpendicular to AC.

The diagonal AC splits ABCD into ∆ABC and ∆ACD, and both stand on AC.

Area (∆ABC) = ½ × AC × BM = ½ × 22 × 3 = 33 cm²

Area (∆ACD) = ½ × AC × DN = ½ × 22 × 3 = 33 cm²

Area (ABCD) = 33 + 33 = 66 cm²

Or in one step:

Area (ABCD) = ½ × AC × (BM + DN) = ½ × 22 × 6 = 66 cm²

Why it happens: B and D lie on opposite sides of AC, so the two triangles do not
overlap and their areas add. Both use the same base AC, so the two heights can be
collected together — 22 × 6 ÷ 2 is quicker than doing two separate multiplications.

Q2 Find the area of the shaded region given that ABCD is a rectangle.

ABCD is 18 cm by 10 cm. E lies on AB with AE = 10 cm and EB = 8 cm; F lies on AD with AF = 6 cm
and FD = 4 cm. The segments FE and EC cut off two triangles at the corners A and B, and the
shaded region is what remains — the quadrilateral DFEC.

Page 43 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
e m.
Area of rectangle ABCD = 18 × 10 = 180 cm²
m l as
m .co a g
l a se
g
Corner at A: Area (∆AFE) = ½ × AE × AF = ½ × 10 × 6 = 30 cm²
aCorner at B: Area (∆EBC) = ½ × EB × BC = ½ × 8 × 10 = 40 cm²

co m
e m . ag
Shaded area = 180 − 30 − 40 = 110 cm²
g l as
a

. c om
Why it happens: each corner triangle is right-angled, so its two perpendicular sides
msplit the four-
are already a base–height pair. Subtracting is far quicker than tryingeto
s
. com region into triangles, because the two pieces being
sided shaded
a glaremoved are the
a s em ones.
gl
simple
a
m a s
agl
Check it yourself: split DFEC instead — join FC. Then ∆FDC = ½ × 18 × 4 = 36 cm² and

m .co
∆FEC has base FC and is harder to handle. Subtracting from the rectangle is the

l a se
g
better route.
a

c o m
.
m hexagon? [Math
s e
What measurements would you need to find the area of a regular

om gla
Q3

. cTalk] a
a s em
ag l ANSWER
Two are enough: the sidelength and the distance from the centre to a side. In fact the
se m
com g l a
. a
sidelength alone determines the hexagon completely.
m
ase
Join the centre O to all six vertices. The six triangles are congruent, each with base s (a side) and

agl
height h (the perpendicular from O to that side).

co m
Area = 6 × (½ × s × h) = 3sh
m .
o m l a se
m .ca regular hexagon those six triangles are equilateral, so haisgfixed once s is known:
se
For
a
agl c
h = √(s² − (s/2)²) = (√3/2) s (Baudhāyana–Pythagoras)
m .
m a s e
. co agl
Area = 3s × (√3/2)s = (3√3/2) s²
e m
g l as
a

com
m .
m ase
.co


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Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Why it happens: "regular" means all sides equal and all angles equal, so the shape
is pinned down by a single number. Measuring one side is enough — everything
else, including h, follows from it.

Q4 What fraction of the total area of the rectangle is the area of the blue region? [Math
Talk]

Exactly one half — and it does not matter where the meeting point sits.
The blue region is two triangles that meet at a point P inside the rectangle. One has the whole
top side as its base; the other has the whole bottom side as its base.

h₁
H
P
h₂

base = w
Both blue triangles span the full width. Their heights add up to the height of the rectangle.

Page 45 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Let the rectangle be w wide and H high, and let P be at heights h₁ and h₂ from the two sides,

so h₁ + h₂ = H

Blue area = ½ × w × h₁ + ½ × w × h₂

= ½ × w × (h₁ + h₂)

=½×w×H

= ½ × area of the rectangle

Why it happens: both triangles use the full width as their base, so their areas
depend only on their heights — and those two heights must together make up the
height of the rectangle, wherever P is placed. Slide P anywhere inside and one
triangle grows by exactly as much as the other shrinks. The fraction stays ½.

Q5 Give a method to obtain a quadrilateral whose area is half that of a given
quadrilateral. [Math Talk]

Take the given quadrilateral ABCD.

1. Draw the diagonal AC and mark M, the midpoint of AC.
2. Join BM and DM.
3. ABMD is the required quadrilateral.

In ∆ABC, BM is a median (M is the midpoint of AC), so

Area (∆ABM) = ½ × Area (∆ABC)

In ∆ACD, DM is a median, so

Area (∆AMD) = ½ × Area (∆ACD)

Adding: Area (ABMD) = ½ [Area (∆ABC) + Area (∆ACD)] = ½ × Area (ABCD)

Page 46 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Why it happens: the diagonal splits the quadrilateral into two triangles, and one
median halves each of them. Halving both halves of a whole halves the whole.

Another way: join the midpoints of the four sides in order. The quadrilateral you get
always has exactly half the area of the original — each of the four corner triangles
cut off is a quarter of the triangle it sits in, and the four together come to half.

In-text Questions — Page 161
Parallelogram

Q1 Give a method to convert a parallelogram into a rectangle of equal area. You can try
this using a cut-out of a parallelogram.

One cut and one slide.

1. In parallelogram ABCD, drop a perpendicular from A to the side DC. Call the foot X, so AX ⊥
DC. AX is a height of the parallelogram.
2. Cut along AX. This separates ∆AXD from the trapezium ABCX.
3. Slide ∆AXD across to the right-hand end and fit it against BC. The result is a rectangle.

Area of the rectangle = Area of the parallelogram, because nothing was added or removed

— only moved

Why it happens: the piece that is cut off on the left is exactly the piece that is
missing on the right. Extending XC and dropping BY ⊥ XC produces ∆BYC, and ∆AXD
≅ ∆BYC by RHS, so ∆AXD fits over ∆BYC perfectly. This process of cutting a figure and
rearranging the pieces into a different figure of the same area is called dissection.

Page 47 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q2 Can ∆AXD and ABCX fit together, as shown in the figure, to get a rectangle?

A A B

D X X C Y

The two pieces of the parallelogram: ∆AXD on the left, and the trapezium ABCX on the
right with ∆BYC (dashed, shaded) shown in the place where ∆AXD would fit.

Yes. The test is to find the triangle that would complete ABCX into a rectangle, and then check
whether ∆AXD is congruent to it.

Extend XC to the right, and drop BY ⊥ XC.

ABYX now has four right angles → a rectangle,

and ∆BYC is exactly the missing corner piece.

BY = AX (opposite sides of rectangle ABYX)

∠BYC = ∠AXD = 90°

BC = AD (opposite sides of parallelogram ABCD)

So by RHS, ∆BYC ≅ ∆AXD

Since the two triangles are congruent, ∆AXD can be laid exactly over the gap ∆BYC. So yes — the
two pieces fit together into rectangle ABYX.

Page 48 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
We need another right angle to get a rectangle (what about the fourth angle?)
e
Q3

m l as
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a
s em
a l angle looks after itself.
Thegfourth

co m
. ag
The four angles of a quadrilateral add up to 360°
e m
g l as
If three of them are 90°, then the fourth = 360° − (90° + 90° + 90°) = 90°
a

co m
So only three right angles need to be arranged; the fourth is forced. That is why constructing BY

m.
⊥ XC is enough to finish the rectangle.

m as e
.co a g l
a s em
a glQ4 Is ∆AXD congruent to it?
m a s
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m
ase criterion.
≅ ∆BYC by the RHS congruency
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Yes — ∆AXD

m
R — right angle: ∠AXD = ∠BYC = 90°
. co
e m
m AX = BY (opposite sides of the rectangle ABYX) glas
H — hypotenuse: AD = BC (opposite sides of the parallelogram)

. c o a
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S— side:

a s e
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So ∆AXD ≅ ∆BYC, and hence Area (∆AXD) = Area (∆BYC)
se m
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m . a
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asfigures can be laid one on top of the other, so they must
g
Why it happens: congruentl
a is what makes the dissection honest — the piece that
have the same area. That

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leaves the left-hand end is the exact size of the hole it fills at the right-hand end.
. co
em
m l as
.co a g
a s em Questions — Page 162
gl
In-text
a c
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m a s e
e m . co agl
g l as
a

com
m .
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.co


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Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Parallelogram

Q1 How do we find the area of a parallelogram by dissecting it into a rectangle?

The dissection turns parallelogram ABCD into rectangle ABYX, so their areas are equal — and a
rectangle's area we already know.

Area of parallelogram ABCD = Area of rectangle ABYX

= AX × XY

where AX is the height of the parallelogram

Once XY is shown to equal DC (see the next question), this becomes

Area of a parallelogram = base × height

Q2 Is there a relation between XY and DC?

Yes: XY = DC.

From the dissection, DX = CY (they are matching sides of the congruent triangles ∆AXD

and ∆BYC)

Add the common part XC to both:

DX + XC = CY + XC

DC = XY

Why it happens: the triangle that leaves the left end takes the piece DX with it and
puts it down as CY at the right end. So the base of the rectangle is the base of the
parallelogram slid along — the same length, only shifted. That is why the base of the
parallelogram may be used directly in the formula.

Page 50 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q3 Can the area of the parallelogram be determined by taking another side as the base
and its corresponding height?

Yes. Any side may be used as the base, provided its own height is used with it.

Using DC as base with height AX: Area = DC × AX

Using AD as base with height CZ: Area = AD × CZ

Both give the same area, so DC × AX = AD × CZ

Why it happens: the parallelogram has only one area, so every correct base–height
pair must give the same number. A long side goes with a short height and a short
side goes with a long height; the product is fixed. This is exactly the relation used to
find an unknown height from a known one.

Q4 Can the parallelogram be cut along CZ and rearranged to form a rectangle?

Yes. Take AD as the base and let CZ be the perpendicular from C to AD (extended if necessary).
Cutting along CZ and sliding the piece across gives a rectangle with base AD and height CZ, by
exactly the same congruence argument as before — the triangle cut off at one end is congruent
to the gap at the other.

Area of the parallelogram = AD × CZ

So the choice of side is free. Whichever side you can measure, together with the height to it, will
do.

Figure it Out — Pages 162 – 164

Page 51 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Parallelogram

Q1 Observe the parallelograms in the figure below.

(a) (b) (c) (d)

(e) (f) (g)

Seven parallelograms (a) to (g). Each is drawn on a base of the same length and between
the same pair of parallel lines, so all have the same height — only the lean changes.

(i) What can we say about the areas of all these parallelograms? (ii) What can we say
about their perimeters? Which figure appears to have the maximum perimeter, and
which has the minimum perimeter?

(i) All seven have the same area. Every one of them is drawn on a base of the same length,
between the same pair of parallel lines — so every one has the same base and the same height.

Area = base × height, and both are the same for (a) to (g)

So Area (a) = Area (b) = … = Area (g)

(ii) Their perimeters are all different. The base stays fixed, but the two slanting sides get
longer as the parallelogram leans over further.

Minimum perimeter: (a) — the one that leans least, so its slant sides are closest to being
upright.
Maximum perimeter: (g) — the one that leans the most, so its slant sides are the longest.

Why it happens: the slant side is the hypotenuse of a right triangle whose vertical
leg is the fixed height and whose horizontal leg is the lean. The more the
parallelogram leans, the longer that hypotenuse — while the height, and therefore
the area, does not budge. Equal areas with unequal perimeters, once again.

Page 52 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q2 Find the areas of the following parallelograms:

4 cm
3 cm

7 cm 5 cm
(i) (ii)

5 cm
2 cm
4.4 cm
4.8 cm

(iii) (iv)

(i) base 7 cm, height 4 cm. (ii) base 5 cm, height 3 cm (the diagonal is not needed). (iii)
base 5 cm — the slanting right side — with height 4.8 cm drawn to it. (iv) base 2 cm —
again the slanting right side — with height 4.4 cm.

Each time, multiply a side by its own height.

PARALLELOGRAM BASE HEIGHT TO THAT BASE AREA

(i) 7 cm 4 cm 7 × 4 = 28 cm²

(ii) 5 cm 3 cm 5 × 3 = 15 cm²

(iii) 5 cm 4.8 cm 5 × 4.8 = 24 cm²

(iv) 2 cm 4.4 cm 2 × 4.4 = 8.8 cm²

Page 53 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

co m
m.
Tip: in (ii) the diagonal drawn across the figure is not needed at all. In (iii) and (iv) the

m l a se
height is drawn to the slanting side, so that slanting side is the base — do not reach
o
.c a g
m
for the horizontal side instead.

l a se
g
aWhy it happens: the formula base × height is only valid when the height is

o m
gives a number that means nothing. m.c ag
measured perpendicular to the chosen base. Pairing a base with the wrong height

l a se
ag

co m
m.
Q3 Find QN.

o m l a se
ANSWER .c a g
m
seis a parallelogram with SR = 12 cm, QM ⊥ SR with QM = 6 cm, PS = 7.6 cm, and QN ⊥ PS.
l a
ag
PQRS

m a s
.co agl
Using SR as base: Area (PQRS) = SR × QM = 12 × 6 = 72 cm²

se m
g l a
a
Using PS as base: Area (PQRS) = PS × QN = 7.6 × QN

com
m .
m as e
l
So 7.6 × QN = 72
. c o a g
s e m = 72 ÷ 7.6 = 720/76 = 180/19
QN
a
agl QN ≈ 9.47 cm

se m
com g l a
m . a
e
Why it happens: the parallelogram has one area, so both base–height pairs must
as than SR (12 cm), so its height QN must be
give 72 cm². PS (7.6 cm) is lshorter
a g
correspondingly longer than 6 cm — and indeed 9.47 > 6.

co m
m .
Check it yourself: 7.6 × 9.47 = 71.97 ≈ 72 ✓
m as e
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se m
g l a
a c
Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm.
m .
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Q4

m a s
co agl
Which has the greater area? [Hint: Imagine constructing them on the same base.]

m .
ase
a g l
The rectangle has the greater area.

co m
m .
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.co


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Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Rectangle: area = 5 × 4 = 20 cm²

Parallelogram on the same base of 5 cm:

its height is the perpendicular distance from the opposite side,

which is less than the slanting side of 4 cm

So area = 5 × height < 5 × 4 = 20 cm²

Why it happens: the 4 cm side of the parallelogram is a slant, not a height. A
perpendicular is always the shortest segment from a point to a line, so the height is
strictly shorter than the 4 cm side — unless the parallelogram is standing upright, in
which case it is the rectangle. The more it leans, the smaller its area, even though its
four sides never change length.

Did you know? Push the top of a rectangular gate sideways and it becomes a
parallelogram with the same four sides but less area. That is why a diagonal brace is
nailed onto a gate — to stop it collapsing.

Q5 Give a method to obtain a rectangle whose area is twice that of a given triangle.
What are the different methods that you can think of?

Given ∆ABC with base BC and height h, its area is ½ × BC × h. We want a rectangle of area BC × h.
Method 1 — the enclosing rectangle. Draw the line l through A parallel to BC, then drop BE ⊥
BC and CD ⊥ BC meeting l at E and D. The rectangle BCDE has base BC and height h, so its area
is BC × h = twice the triangle.
Method 2 — two copies. Make a second copy of ∆ABC and rotate it through a half-turn about
the midpoint of AC. The two copies join into a parallelogram of base BC and height h. Now
dissect that parallelogram into a rectangle (cut along a height and slide).
Method 3 — choose your own shape. Any rectangle whose two sides multiply to BC × h will do,
for instance one of base 2 × BC and height h/2.

Area (∆ABC) = ½ × BC × h

Area (rectangle) = BC × h = 2 × Area (∆ABC) ✓

Page 55 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Why it happens: the triangle formula already contains the factor ½. Removing that
factor — by keeping the base and height but building a rectangle instead — is
precisely doubling the area.

Q6 [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given
triangle.

Cut the triangle at half its height and fold the top down.

1. In ∆ABC with base BC, mark M and N, the midpoints of AB and AC. Join MN — it is parallel to
BC and lies at height h/2.
2. Cut along MN.
3. Drop the small triangle ∆AMN down onto the base strip: turn it half a turn about M to fill the
left gap, and cut-and-turn the other part about N to fill the right gap.

The pieces close up into a rectangle with base BC and height h/2

Area = BC × h/2 = ½ × BC × h = Area (∆ABC) ✓

Why it happens: a half-turn about the midpoint of a side is a rigid motion, so no
area is gained or lost. The trapezium that is left after the cut is short of exactly the
two corner triangles that the top piece supplies. This is a dissection: the triangle and
the rectangle are made of the very same pieces.

Q7 [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection
in a simpler way. Can you find out how to do it? [Hint: Show that triangles ∆ADB and
∆ADC can be made into halves of a rectangle. Figure out how they should be
assembled to get a rectangle. Use cut-outs if necessary.]

Cut along the axis of symmetry, then fit the two right triangles together along their slant sides.

1. In the isosceles ∆ABC (AB = AC), let AD be the perpendicular from A to BC. Then D is the
midpoint of BC, and ∆ADB ≅ ∆ADC by RHS.
2. Cut along AD. You now hold two right triangles, each with legs AD (the height) and DB = DC
(half the base).

Page 56 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

3. Keep ∆ADC where it is. Turn ∆ADB over and lay its slant side AB along the slant side AC, so
that the two right angles land at opposite corners.

The result is a rectangle with sides AD and DC

Area = AD × DC = AD × (½ BC) = ½ × BC × AD = Area (∆ABC) ✓

Why it happens: a rectangle is cut by a diagonal into two right triangles with the
same legs. Here the two legs are AD and DC, and both of our pieces have exactly
those legs — so each is a "half rectangle". Placing them slant-against-slant, with the
right angles diagonally opposite, restores the rectangle they came from. Since only
cutting and moving are involved, the area is unchanged.

Try This: cut a paper isosceles triangle along its height and try to make the rectangle
without turning a piece over. You will find you must flip one piece — the two halves
are mirror images of each other.

Q8 [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by
dissection.

Run Q7 backwards.

1. Take rectangle PQRS with base PQ = p and height QR = q.
2. Cut along the diagonal PR into two right triangles.
3. Keep one of them. Turn the other over and join it to the first along the side of length q, so
that the two right angles sit side by side and their bases form one straight line.

The result is an isosceles triangle with base 2p and height q

Area = ½ × 2p × q = pq = Area of the rectangle ✓

Why it happens: the two right angles at the join add to 180°, so the two bases really
do line up into a single straight base. The two slanting sides are the two halves of
the diagonal PR, which are equal — which is exactly what makes the triangle
isosceles.

Page 57 of 83

Class 8 Maths Chapter 14 Area AglaSem · NCERT Solutions

Q9 Which has greater area — an equilateral triangle or a square of the same sidelength
as the triangle? Which has greater area — two identical equilateral triangles
together or a square of the same sidelength as the triangle? Give reasons.

The square is greater in both cases.
Let the common sidelength be s.

Square: area = s²

Equilateral triangle: its height h is the perpendicular from a vertex to the opposite side.

A perpendicular is the shortest distance from a point to a line, so h < s.

Area = ½ × s × h < ½ × s × s = s²/2

One triangle < s²/2 < s² → the square is larger

Two triangles < 2 × s²/2 = s² → the square is still larger

The exact values confirm it. By the Baudhāyana–Pythagoras theorem,

h = √(s² − (s/2)²) = (√3/2)s ≈ 0.866 s

Area of one equilateral triangle = (√3/4)s² ≈ 0.433 s²

Two of them ≈ 0.866 s², and the square is 1.000 s²

Why it happens: the square has two of its sides meeting at a right angle, so one
side is the full height for the other. In the equilateral triangle the sides lean at 60°, so
the height falls short of the side — and that shortfall is what costs the triangle its
area. Notice that two triangles come surprisingly close to the square (about 87%)
without ever reaching it.

In-text Question — Page 164

Page 58 of 83

as e
Class 8 Maths Chapter 14 Area
a g l AglaSem · NCERT Solutions

Rhombus
co m
e m.
m l as
Q1
.co a g
Try working this out! [Transform a rhombus into a rectangle of the same area by
m
l a se
dissection, by the method that occurs in one of the Śulba-Sūtras.]

a g

com
. ag
A rhombus is a parallelogram, so base × height already works. But its extra properties give a

se m
a
neater dissection.

ag l
1. In rhombus ABCD, draw both diagonals. They meet at O, and — this is the key — they are
perpendicular bisectors of each other.

co m
m.
2. The diagonal BD cuts the rhombus into ∆ABD and ∆CBD. Since AB = AD and CB = CD, both

m
are isosceles triangles on the base BD.
as e
.co a g l
m of symmetry (AO for the first, CO for the second) and fit the two right-triangle halves
3. Convert each isosceles triangle into a rectangle by the dissection of Q7 (page 164): cut along

a s eaxis
gl together along their slant sides.
its

a
4. Join the two rectangles side by side. They fit, because both have the same height, ½BD.
m a s
m .co agl
Result: a single rectangle WXYZ with
l a se
XW = AO + OC = AC and WZ = ½ BD a g
Area = AC × ½BD = ½ × AC × BD
co m
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o m l a se
g angles, which is
.c it happens: the two diagonals of a rhombus meet ataright
m
ase
Why

agl
exactly what makes ∆ABD and ∆CBD isosceles with BD as base — and an isosceles

m
triangle is the easiest shape of all to turn into a rectangle. Every step is a cut and a

a se
com l
move, so the area never changes.
. a g
m
ase
agl
In-text Questions — Page 165
co m
Rhombus
m .
m as e
c o
. What are the sidelengths of the rectangle WXYZ? a g l
s e mQ1
a
agl c
m .
e

m a s
. co agl
From the dissection:

e m
g l as
a

com
m .
m ase
.co


a g l Page 59 of 83