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# R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 14 – Areas of Triangles and Quadrilaterals

## Chapter 14 – Areas of Triangles and Quadrilaterals Exercise Ex. 14

Question 1

Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.

Solution 1
Question 2

The base of the isosceles triangle measures 80 cm and its area is 360cm2.Find the perimeter of the triangle.

Solution 2

Question 3

The perimeter of an isosceles
triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle.

HINT Ratio of sides = 3 : 3 : 2.

* Back answer incorrect

Solution 3

It
is given that the ratio of equal side to its base is 3 :
2.

Ratio of sides
of isosceles triangle = 3 : 3 : 2

i.e.
a : b : c = 3 : 3 : 2

a = 3x, b = 3x
and c = 2x

Given,
perimeter = 32 cm

3x + 3x + 2x =
32

8x = 32

x = 4

So,
the sides of the triangle are

a
= 3x = 3(4) = 12 cm

b
= 3x = 3(4) = 12 cm

c
= 2x = 2(4) = 8 cm

Question 4

The perimeter of a triangle is 50
cm. One side of the triangle is 4 cm longer than the smallest side and the
third side is 6 cm less than twice the smallest side. Find the area of the triangle.

Solution 4

Let
the three sides of a triangle be a, b and c respectively such that c is the
smallest side.

Then,
we have

a
= c + 4

And,
b = 2c – 6

Given,
perimeter = 50 cm

a + b + c = 50

(c + 4) + (2c
– 6) + c = 50

4c – 2 = 50

4c = 52

c = 13

So,
the sides of the triangle are

a
= c + 4 = 13 + 4 = 17 cm

b
= 2c – 6 = 2(13) – 6 = 20 cm

c
= 13 cm

Question 5

The triangular side walls of a
flyover have been used for advertisements. The sides of the walls are 13 m,
14 m, 15 m. The advertisements yield an earning of
Rs.2000 per m2 a year. A company hired one of its
walls for 6 months. How much rent did it pay?

Solution 5

Three
sides of a wall are 13 m, 14 m and 15 m respectively.

i.e.

a
= 13 m, b = 14 m and c = 15 m

Rent
for a year = Rs. 2000/m2

Rent for 6
months = Rs. 1000/m2

Thus,
total rent paid for 6 months = Rs. (1000
× 84) = Rs. 84,000

Question 6

The perimeter of the isosceles triangle is 42 cm and its base is times with each of the equal sides. Find (i) the length of the equal side of the triangle (ii) the area of the triangle (iii)the height of the triangle. (Given, )

Solution 6

Question 7

If the area of an equilateral triangle is cm2, find its perimeter.

Solution 7
Question 8

If the area of an equilateral triangle is cm2, find its height.

Solution 8
Question 9

Each side of the equilateral
triangle measures 8 cm. Find (i) the area of the triangle, correct to 2
places of decimal and (ii) the height of the triangle, correct to 2 places of
decimal. Take =1.732.

Solution 9

(i) Area of an equilateral
triangle=

Where a is the side of the
equilateral triangle

Question 10

The height of an equilateral triangle measures 9 cm. Find its area, correct to 2 places of decimal. Take =1.732.

Solution 10

Question 11

The base of the right -angled triangle measures 48 cm and its hypotenuse measures 50 cm; find the area of the triangle.

Solution 11
Question 12

The base of the triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height.

Solution 12
Question 13

Find the area of the shaded region
in the figure given below.

Solution 13

In
right triangle ADB, by Pythagoras theorem,

AB2
= AD2 + BD2 = 122 + 162 = 144 +
256 = 400

AB = 20 cm

For
ΔABC,

Thus,
area of shaded region

=
Area of
ΔABC – Area of ΔABD

=
(480 – 96) cm2

=
384 cm2

Question 14

The sides of a quadrilateral ABCD
taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle
between the first two sides is a right angle. Find its area. (Given, )

Solution 14

Let
ABCD be the given quadrilateral such that
∠ABC = 90° and AB = 6 cm,
BC = 8 cm, CD = 12 cm and AD = 14 cm.

In
ΔABC, by
Pythagoras theorem,

AC2
= AB2 + BC2 = 62 + 82 = 36 + 64 =
100

AC = 10 cm

In
ΔACD, AC = 10
cm, CD = 12 cm and AD = 14 cm

Let
a = 10 cm, b = 12 cm and c = 14 cm

Thus,
area of quadrilateral ABCD

=
A(
ΔABC) + A(ΔACD)

=
(24 + 58.8) cm2

=
82.8 cm2

Question 15

Find the perimeter and area of a
quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15 cm, DA = 17 CM and
∠ABD
= 90°.

Solution 15

In
ΔABD, by
Pythagoras theorem,

AB2
= AD2 – BD2 = 172 – 152 = 289 –
225 = 64

AB = 8 cm

Perimeter of quadrilateral ABCD =
AB + BC + CD + AD

= 8 + 12 + 9 + 17

= 46 cm

In
ΔBCD, BC = 12
cm, CD = 9 cm and BD = 15 cm

Let
a = 12 cm, b = 9 cm and c = 15 cm

Thus,
area of quadrilateral ABCD

=
A(
ΔABD) + A(ΔBCD)

=
(60 + 54) cm2

=
114 cm2

Question 16

Find the perimeter and area of the
quadrilateral ABCD in which AB = 21 cm,
∠BAC
= 90°,
AC = 20 cm, CD = 42 cm and AD = 34 cm.

Solution 16

In
ΔBAC, by
Pythagoras theorem,

BC2
= AC2 + AB2 = 202 + 212 = 400 +
441 = 841

BC = 29 cm

Perimeter of quadrilateral ABCD =
AB + BC + CD + AD

= 21 + 29 + 42 + 34

= 126 cm

In
ΔACD, AC = 20
cm, CD = 42 cm and AD = 34 cm

Let
a = 20 cm, b = 42 cm and c = 34 cm

Thus,
area of quadrilateral ABCD

=
A(
ΔABC) + A(ΔACD)

=
(210 + 336) cm2

=
546 cm2

Question 17

Find the area of the quadrilateral ABCD in which BCD is an equilateral triangle, each of whose side is 26 cm, AD=24 cm and.Also, find the perimeter of the quadrilateral [Given =1.73]

Solution 17

Perimeter of quad. ABCD = AB + BC + CD + DA = 10 + 26 + 26 + 24 = 86 cm

Question 18

Find the area of a parallelogram ABCD in which AB= 28 cm , BC=26 cm and diagonal AC=30 cm.

Solution 18

Question 19

Find the area parallelogram ABCD in which AB=14 cm, BC=10 cm and AC= 16 cm. [Given =1.73]

Solution 19

Question 20

In the given figure ABCD is a quadrilateral in which diagonal BD=64 cm, AL BD and CM BD such that AL= 16.8 cm and CM=13.2 cm. Calculate the area of quadrilateral ABCD.

Solution 20

Question 21

The area of a trapezium is 475 cm2
and its height is 19 cm. Find the lengths of its two parallel sides if one
side is 4 cm greater than the other.

Solution 21

Let
the smaller parallel side of trapezium = x cm

Then,
larger parallel side = (x + 4) cm

Thus,
the lengths of two parallel sides are 23 cm and 27 cm respectively.

Question 22

In the given figure, a ΔABC
has been given in which AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC, a
parallelogram DBCE of the same area as that of ΔABC
is constructed. Find the height DL of the parallelogram.

Solution 22

In ΔABC, AB = 7.5
cm, BC = 7 cm and AC = 6.5 cm

Let a = 7.5 cm, b = 7 cm
and c = 6.5 cm

Question 23

Find the area of triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side.

Solution 23

Question 24

A field is in the shape of a
trapezium having parallel sides 90 m and 30 m. These sides meet the third
side at right angles. The length of the fourth side is 100 m. If it costs
Rs.5 to plough 1 m2 of the field, find the total
cost of ploughing the field.

Solution 24

Construction:
Draw BT
⊥ CD

In
ΔBTC, by
Pythagoras theorem,

BT2
= BC2 – CT2 = 1002 – 602 = 10000
– 3600 = 6400

BT = 80 m

AD = BT = 80 m

Cost
of ploughing 1 m2 field = Rs. 5

Cost of ploughing
4800 m2 field = Rs. (5 × 4800) = Rs. 24,000

Question 25

A rectangular plot is given for
constructing a house, having a measurement of 40 m long and 15 m in the
front. According to the laws, a minimum of 3-m-wide space should be left in
the front and back each and 2 m wide space on each of the sides. Find the largest
area where house can be constructed.

Solution 25

Length
of rectangular plot = 40 m

Width
of rectangular plot = 15 m

Keeping
3 m wide space in the front and back,

length
of rectangular plot = 40 – 3 – 3 = 34 m

Keeping
2 m wide space on both the sides,

width
of rectangular plot = 15 – 2 – 2 = 11 m

Thus,
largest area where house can be constructed

=
34 m
× 11 m

=
374 m2

Question 26

A rhombus -shaped sheet with
perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate
of
Rs.5 per cm2.
Find the cost of painting.

Solution 26

Let
ABCD be the rhombus-shaped sheet.

Perimeter
= 40 cm

4 × Side = 40 cm

Side = 10 cm

AB = BC = CD =
AD = 10 cm

Let
diagonal AC = 12 cm

Since
diagonals of a rhombus bisect each other at right angles,

AO
= OC = 6 cm

In
right
ΔAOD, by
Pythagoras theorem,

OD2
= AD2 – AO2 = 102 – 62 = 100 – 36
= 64

OD = 8 cm

BD = 2 × OD =
2 × 8 = 16 cm

Now,

Cost of painting = Rs. 5/cm2

Cost of
painting rhombus on both sides = Rs. 5 × (96 + 96)

=
Rs. 5
× 192

=
Rs. 960

Question 27

The difference between the semiperimeter and sides of a ΔABC
are 8 cm, 7 cm, and 5 cm respectively. Find the area of the triangle.

Solution 27

Let
the sides of a triangle be a, b, c respectively and ‘s’
be its semi-perimeter.

Then,
we have

s
– a = 8 cm

s
– b = 7 cm

s
– c = 5 cm

Now,
(s – a) + (s – b) + (s – c) = 8 + 7 + 5

3s – (a + b +
c) = 20

3s – 2s = 20

s = 20

Thus,
we have

a
= s – 8 = 20 – 8 = 12 cm

b
= s – 7 = 20 – 7 = 13 cm

c
= s – 5 = 20 – 5 = 15 cm

Question 28

A floral design on a floor is made up of 16 tiles , each triangular in shape having sides 16 cm, 12 cm and 20 cm. Find the cost of polishing the tiles at Re 1 per sq cm.

Solution 28
Question 29

An umbrella is made by stitching by 12 triangular pieces of cloth, each measuring ((50 cmx20 cm x50 cm).Find the area of the cloth used in it.

Solution 29
Question 30

In the given figure, ABCD is a
square with diagonal 44 cm. How much paper of each shade is needed to make a
kite given in the figure?

Solution 30

In ΔAEF, AE = 20
cm, EF = 14 cm and AF = 20 cm

Let a = 20 cm, b = 14 cm
and c = 20 cm

Question 31

A rectangular lawn, 75 m by 60 m,
has two roads, each road 4 m wide, running through the middle of the lawn,
one parallel to length and the other parallel to breadth, as shown in the
figure. Find the cost of gravelling the roads at
Rs.50 per m2.

Solution 31

For
road ABCD, i.e. for rectangle ABCD,

Length
= 75 m

= 4 m

Area
of road ABCD = Length
× Breadth = 75 m × 4m = 300 m2

For
road PQRS, i.e. for rectangle PQRS,

Length
= 60 m

= 4 m

Area
of road PQRS = Length
× Breadth = 60 m × 4 m = 240 m2

For
road EFGH, i.e. for square EFGH,

Side
= 4 m

Area
of road EFGH = (Side)2 = (4)2 = 16 m2

Total
area of road for gravelling

=
Area of road ABCD + Area of road PQRS – Area of road EFGH

=
300 + 240 – 16

=
524 m2

Cost
of gravelling the road = Rs. 50 per m2

Cost of
gravelling 524 m2 road = Rs. (50 × 524) = Rs. 26,200

Question 32

The shape of the cross section of
canal is a trapezium. If the canal is 10 m wide at the top, 6 m wide at the
bottom and the area of its cross section is 640 m2, find the depth
of the canal.

Solution 32

Area of cross section =
Area of trapezium = 640 m2

Length of top + Length
of bottom

= sum of parallel sides

= 10 m + 6 m

= 16 m

Thus, the depth of the canal is 80 m.

Question 33

Find the area of a trapezium whose
parallel sides are 11 m and 25 m long, and the nonparallel sides are 15 m and
13 m long.

Solution 33

From
C, draw CE
∥ DA.

Clearly,
∥ EC and AE ∥ DC such that
AD = 13 m and D = 11 m.

AE
= DC = 11 m and EC = AD = 13 m

BE = AB – AE =
25 – 11 = 14 m

Thus,
in
ΔBCE, we have

BC
= 15 m, CE = 13 m and BE = 14 m

Let
a = 15 m, b = 13 m and c = 14 m

Question 34

Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side.

Solution 34
Question 35

The difference between the lengths
of the parallel sides of a trapezium is 8 cm, the perpendicular distance
between these sides is 24 cm and the area of the trapezium is 312 cm2.
Find the length of each of the parallel sides.

Solution 35

Let
the smaller parallel side = x cm

Then,
longer parallel side = (x + 8) cm

Height
= 24 cm

Area
of trapezium = 312 cm2

Thus,
the lengths of parallel sides are 9 cm and 17 cm respectively.

Question 36

A parallelogram and a rhombus are
equal in area. The diagonals of the rhombus measure 120 m and 44 m. If one of
the sides of the parallelogram measures 66 m, find its corresponding
altitude.

Solution 36

Area
of parallelogram = Area of rhombus

Question 37

A parallelogram and a square have
the same area. If the sides of the square measures 40 m and altitude of the
parallelogram measures 25 m, find the length of the corresponding base of the
parallelogram.

Solution 37

Area
of parallelogram = Area of square

Question 38

Find the area of a rhombus one
side of which measures 20 cm and one of whose diagonals is 24 cm.

Solution 38

Let
ABCD be a rhombus and let diagonals AC and BD intersect each other at point
O.

We
know that diagonals of a rhombus bisect each other at right angles.

Thus,
in right-angled
ΔAOD, by Pythagoras theorem,

OD2
= AD2 – OA2 = 202 – 122 = 400 –
144 = 256

OD = 16 cm

BD = 2(OD) =
2(16) = 32 cm

Question 39

The area of a rhombus is 480 cm2,
and one of its diagonals measures 48 cm. Find (i) the length of the other
diagonal, (ii) the length of each of its sides, and (iii) its perimeter.

Solution 39

(i) Area of a rhombus = 480 cm2

(ii) Let diagonal AC = 48 cm and diagonal BD =
20 cm

We
know that diagonals of a rhombus bisect each other at right angles.

Thus,
in right-angled
ΔAOD, by Pythagoras theorem,

= OA2 + OD2 = 242 + 102 = 576 +
100 = 676

AD = 26 cm

AD = BC = CD =
AD = 26 cm

Thus,
the length of each side of rhombus is 26 cm.

(iii) Perimeter of a rhombus = 4 × side = 4 × 26 = 104 cm

Question 40

Find the area of a triangular field whose sides are 91m, 98 m, 105m in length. Find the height corresponding to the longest side.

Solution 40
Question 41

The sides of a triangle are in the ratio 5: 12:13 and its perimeter is 150 m. Find the area of the triangle.

Solution 41
Question 42

The perimeter of a triangular
field is 540 m and its sides are in the ratio 25 :
17 : 12. Find the area of the field. Also, find the cost of ploughing the field at
Rs. 5 per m2.

Solution 42

It
is given that the sides a, b, c of the triangle are in the ratio 25 : 17 :
12,

i.e.
a : b : c = 25 : 17 : 12

a = 25x, b =
17x and c = 12x

Given,
perimeter = 540 m

25x + 17x +
12x = 540

54x = 540

x = 10

So,
the sides of the triangle are

a
= 25x = 25(10) = 250 m

b
= 17x = 17(10) = 170 m

c
= 12x = 12(10) = 120 m

Cost
of ploughing the field = Rs.
5/m2

Cost of ploughing 9000 m2 = Rs.
(5 × 9000) = Rs. 45,000

Question 43

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

Solution 43
Question 44

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measure is 20 cm.

Solution 44

## Chapter 14 – Areas of Triangles and Quadrilaterals Exercise MCQ

Question 1

In a ∆ABC it is given
that base = 12 cm and height = 5 cm. Its area is

(a) 60 cm2

(b) 30 cm2

(d) 45 cm2

Solution 1

Question 2

The sides of a triangle
are in the ratio 5:12:13 and its perimeter is 150 cm. The area of the
triangle is

(a) 375 cm2

(b) 750 cm2

(c) 250 cm2

(d) 500 cm2

Solution 2

Question 3

The lengths of the three
sides of a triangle are 30 cm, 24 cm and 18 cm respectively. The length of
the altitude of the triangle corresponding to the smallest side is

(a) 24 cm

(b) 18 cm

(c) 30 cm

(d) 12 cm

Solution 3

Question 4

The base of an isosceles
triangle is 16 cm and its area is 48 cm2. The perimeter of the
triangle is

(a) 41 cm

(b) 36 cm

(c) 48 cm

(d) 324 cm

Solution 4

Question 5

Solution 5

Question 6

Each of the equal sides
of an isosceles triangle is 13 cm and its base is 24 cm. The area of the
triangle is

(a)156 cm2

(b)78 cm2

(c) 60 cm2

(d) 120 cm2

Solution 6

Question 7

The base of a right
triangle is 48 cm and its hypotenuse is 50 cm long. The area of the triangle
is

(a) 168 cm2

(b) 252 cm2

(c) 336 cm2

(d) 504 cm2

Solution 7

Question 8

Solution 8

Question 9

The lengths of three
sides of a triangle are 20 cm, 16 cm and 12 cm. The area of the triangle is

(a) 96 cm2

(b) 120 cm2

(c) 144 cm2

(d) 160 cm2

Solution 9

Question 10

Each side of an
equilateral triangle measures 8 cm. The area of the triangle is

Solution 10

Question 11

The base of an isosceles
triangle is 8 cm long and each of its equal sides measures 6 cm. The area of
the triangle is

Solution 11

Question 12

The base of an isosceles
triangle is 6 cm and each of its equal sides is 5 cm. The height of triangle
is

Solution 12

Question 13

Each of the two equal
sides of an isosceles right triangle is 10 cm long. Its area is

Solution 13

Question 14

Each side of an
equilateral triangle is 10 cm long. The height of the triangle is

Solution 14

Question 15

The height of an
equilateral triangle is 6 cm. Its area is

Solution 15

Question 16

The lengths of the three
sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of
the triangle is

(a) 480 m2

(b) 320m2

(c) 384 m2

(d) 360m2

Solution 16

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