# 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

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

The base of the isosceles triangle measures 80 cm and its area is 360cm^{2}.Find the perimeter of the triangle.

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

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

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.

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

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 m^{2} a year. A company hired one of its

walls for 6 months. How much rent did it pay?

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/m^{2}

⇒ Rent for 6

months = Rs. 1000/m^{2}

Thus,

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

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, )

If the area of an equilateral triangle is cm^{2}, find its perimeter.

If the area of an equilateral triangle is cm^{2}, find its height.

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.

(i) Area of an equilateral

triangle=

Where a is the side of the

equilateral triangle

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

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

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.

Find the area of the shaded region

in the figure given below.

In

right triangle ADB, by Pythagoras theorem,

AB^{2}

= AD^{2} + BD^{2} = 12^{2} + 16^{2} = 144 +

256 = 400

⇒ AB = 20 cm

For

ΔABC,

Thus,

area of shaded region

=

Area of ΔABC – Area of ΔABD

=

(480 – 96) cm^{2}

=

384 cm^{2}

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, )

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,

AC^{2}

= AB^{2} + BC^{2} = 6^{2} + 8^{2} = 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) cm^{2}

=

82.8 cm^{2}

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°.

In

ΔABD, by

Pythagoras theorem,

AB^{2}

= AD^{2} – BD^{2} = 17^{2} – 15^{2} = 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) cm^{2}

=

114 cm^{2}

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.

In

ΔBAC, by

Pythagoras theorem,

BC^{2}

= AC^{2} + AB^{2} = 20^{2} + 21^{2} = 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) cm^{2}

=

546 cm^{2}

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]

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

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

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

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.

The area of a trapezium is 475 cm^{2}

and its height is 19 cm. Find the lengths of its two parallel sides if one

side is 4 cm greater than the other.

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.

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.

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

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.

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 m^{2} of the field, find the total

cost of ploughing the field.

Construction:

Draw BT ⊥ CD

In

ΔBTC, by

Pythagoras theorem,

BT^{2}

= BC^{2} – CT^{2} = 100^{2} – 60^{2} = 10000

– 3600 = 6400

⇒ BT = 80 m

⇒ AD = BT = 80 m

Cost

of ploughing 1 m^{2} field = Rs. 5

⇒ Cost of ploughing

4800 m^{2} field = Rs. (5 × 4800) = Rs. 24,000

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.

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 m^{2}

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 cm^{2}.

Find the cost of painting.

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,

OD^{2}

= AD^{2} – AO^{2} = 10^{2} – 6^{2} = 100 – 36

= 64

⇒ OD = 8 cm

⇒ BD = 2 × OD =

2 × 8 = 16 cm

Now,

Cost of painting = Rs. 5/cm^{2}

∴ Cost of

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

=

Rs. 5 × 192

=

Rs. 960

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.

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

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.

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.

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?

In ΔAEF, AE = 20

cm, EF = 14 cm and AF = 20 cm

Let a = 20 cm, b = 14 cm

and c = 20 cm

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 m^{2}.

For

road ABCD, i.e. for rectangle ABCD,

Length

= 75 m

Breadth

= 4 m

Area

of road ABCD = Length × Breadth = 75 m × 4m = 300 m^{2}

For

road PQRS, i.e. for rectangle PQRS,

Length

= 60 m

Breadth

= 4 m

Area

of road PQRS = Length × Breadth = 60 m × 4 m = 240 m^{2}

For

road EFGH, i.e. for square EFGH,

Side

= 4 m

Area

of road EFGH = (Side)^{2} = (4)^{2} = 16 m^{2}

Total

area of road for gravelling

=

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

=

300 + 240 – 16

=

524 m^{2}

Cost

of gravelling the road = Rs. 50 per m^{2}

∴ Cost of

gravelling 524 m^{2} road = Rs. (50 × 524) = Rs. 26,200

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 m^{2}, find the depth

of the canal.

Area of cross section =

Area of trapezium = 640 m^{2}

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.

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.

From

C, draw CE ∥ DA.

Clearly,

ADCE is a parallelogram having AD ∥ 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

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.

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 cm^{2}.

Find the length of each of the parallel sides.

Let

the smaller parallel side = x cm

Then,

longer parallel side = (x + 8) cm

Height

= 24 cm

Area

of trapezium = 312 cm^{2}

Thus,

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

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.

Area

of parallelogram = Area of rhombus

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.

Area

of parallelogram = Area of square

Find the area of a rhombus one

side of which measures 20 cm and one of whose diagonals is 24 cm.

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,

OD^{2}

= AD^{2} – OA^{2} = 20^{2} – 12^{2} = 400 –

144 = 256

⇒ OD = 16 cm

⇒ BD = 2(OD) =

2(16) = 32 cm

The area of a rhombus is 480 cm^{2},

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.

(i) Area of a rhombus = 480 cm^{2}

(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,

AD^{2}

= OA^{2} + OD^{2} = 24^{2} + 10^{2} = 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

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

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

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 m^{2}.

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/m^{2}

⇒ Cost of ploughing 9000 m^{2} = Rs.

(5 × 9000) = Rs. 45,000

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.

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

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

In a ∆ABC it is given

that base = 12 cm and height = 5 cm. Its area is

(a) 60 cm^{2}

(b) 30 cm^{2}

(d) 45 cm^{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 cm^{2}

(b) 750 cm^{2}

(c) 250 cm^{2}

(d) 500 cm^{2}

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

The base of an isosceles

triangle is 16 cm and its area is 48 cm^{2}. The perimeter of the

triangle is

(a) 41 cm

(b) 36 cm

(c) 48 cm

(d) 324 cm

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 cm^{2}

(b)78 cm^{2}

(c) 60 cm^{2}

(d) 120 cm^{2}

The base of a right

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

is

(a) 168 cm^{2}

(b) 252 cm^{2}

(c) 336 cm^{2}

(d) 504 cm^{2}

The lengths of three

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

(a) 96 cm^{2}

(b) 120 cm^{2}

(c) 144 cm^{2}

(d) 160 cm^{2}

Each side of an

equilateral triangle measures 8 cm. The area of the triangle is

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

The base of an isosceles

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

is

Each of the two equal

sides of an isosceles right triangle is 10 cm long. Its area is

Each side of an

equilateral triangle is 10 cm long. The height of the triangle is

The height of an

equilateral triangle is 6 cm. Its area is

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 m^{2}

(b) 320m^{2}

(c) 384 m^{2}

(d) 360m^{2}