ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions

ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions Ch-16. Step by Step Solutions of 16.2 Questions on Mensuration of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-16 Mensuration
Topics Solution of Exe-16.2 Questions
Academic Session 2024-2025

Solution of Exe-16.2 Questions on Mensuration

ML Aggarwal Class 9 ICSE Maths Solutions Ch-16.

Question 1.

(i) Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.

(ii) Find the area of quadrilateral whose diagonal are 18 cm and 12 cm and they intersect each other at right angle.

Answer :   

(i) Consider ABCD as a quadrilateral in which AC = 20 cm

We know that, BY = 9 cm and DY = 15 cm

Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.

Here

Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

We can write it as

= ½ × base × height + ½ × base × height

So we get

= (½ ×AC×BX) + (½ ×AC×DY)

= (½ × 20 × 9) + (½ × 20 × 15)

By further calculation

= (10×9) + (10×15)

= 90 + 150

= 240 cm2

(ii) Consider ABCD as a quadrilateral in which the diagonals AC and BD intersect each other at M at right angles

AC = 18 cm and BD = 12 cm

Find the area of a quadrilateral whose diagonals are of length 18 cm and 12 cm and they intersect each other at right angles.

We know that, Area of quadrilateral ABCD = ½ ×diagonal AC ×diagonal BD

= ½ ×18×12

By calculation

= 9×12

= 108 cm2

Question 2.  Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°

Answer :

ABCD is a quadrilateral field

AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°

Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°.

In triangle BAD

∠A = 900

Using the Pythagoras Theorem

BD2 = BA2 + AD2

BD2 = 402 + 92

By calculation

BD2 = 1600 + 81 = 1681

So ,

BD = 41

We know that, Area of quadrilateral ABCD = Area of △BAD + Area of △BDC

It can be written as

= ½ × base × height + Area of △BDC

= ½ × 40 × 9 + Area of △BDC

By calculation

= 180 m2 + Area of △BDC

Determining the area of △BDC

Consider a = BD = 41 m, b = CD = 15 m, c = BC = 28 m

S = Semi perimeter = (a + b + c)/2

= (41 + 15 + 28)/2

= 42 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 24

= 2×7×3×3

= 126 m2

So, the area of quadrilateral ABCD = 180 m2 + Area of △BDC

= 180 + 126

= 306 m2

Question 3. Find the area of the quadrilateral ABCD in which ∠BCA= 90°, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.

Find the area of the quadrilateral ABCD in which ∠BCA= 90°, AB = 13 cm and ACD is an equilateral triangle of side 12 cm
Answer :

ABCD is a quadrilateral in which ∠BCA = 90° and AB = 13 cm

ABCD is an equilateral triangle in which AC = CD = AD = 12 cm

In right angled △ABC

Using Pythagoras theorem,

AB2 = AC2 + BC2

132 = 122 + BC2

By calculation

BC2 = 132 – 122

⇒ BC2 = 169 – 144 = 25

BC = √25 = 5 cm

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

It can be written,

= (½ ×base× height) + (√3/4 × side2)

= (½ ×AC×BC) + (√3/4 × 122)

= (½ ×12×5) + (√3/4 × 12 × 12)

So we get

= (6 × 5) + (√3 × 3 × 12)

= 30 + 36√3

Substituting the value of √3

= 30 + (36 × 1.732)

= 30 + 62.28

= 92.28 cm2

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 4. Find the area of quadrilateral ABCD in which ∠B = 90°, AB = 6 cm, BC = 8 cm 13 and CD = AD = 13 cm.

Find the area of quadrilateral ABCD in which ∠B = 90°, AB = 6 cm, BC = 8 cm 13 and CD = AD = 13 cm

Answer :

ABCD is a quadrilateral in which ∠B = 90°, AB = 6 cm, BC = 8 cm and CD = AD = 13 cm

In △ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

AC2 = 62 + 82

By calculation

AC2 = 36 + 64 = 100

AC2 = 102

⇒ AC = 10 cm

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

It can be written as

= ½ × base × height + Area of △ACD

= ½ × AB × BC + Area of △ACD

= ½ × 6 × 8 + Area of △ACD

By calculation

= 24 cm2 + Area of △ACD …(1)

Finding the area of △ACD

Consider a = AC = 10 cm, b = CD = 13 cm, c = AD = 13 cm

We know that, S = Semi perimeter = (a + b + c)/2

Substituting the values

= (10 + 13 + 13)/ 2

= (10 + 26)/2

= 36/2

= 18 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 25

= 3 × 2 × 2 × 5

= 60 cm2

Using equation (1)

Area of quadrilateral ABCD = 24 cm2 + Area of △ACD

= 24 + 60

= 84 cm2

Question 5. The perimeter of a rectangular cardboard is 96 cm ; If its breadth is 18 cm, find the length and the area of the cardboard.

Answer :

Consider ABCD as a rectangle

Take length = l cm

Breadth = 18 cm

Perimeter = 96 cm

The perimeter of a rectangular cardboard is 96 cm; if its breadth is 18 cm, find the length and the area of the cardboard.

2 × (l + b) = 96 cm

2 × (l + 18) = 96 cm

By calculation

(l + 18) = 96/2

⇒ l + 18 = 48

So we get

⇒ l = 48 – 18 = 30 cm

Here,

Area of rectangular cardboard = l × b

= 30 × 18

= 540 cm2

Question 6.  The length of a rectangular hall is 5 m more than its breadth, If the area of the hall is 594 m2, find its perimeter.

The length of a rectangular hall is 5 m more than its breadth, If the area of the hall is 594 m2, find its perimeter

Answer :

Let ABCD be rectangular field.

Take Breadth = x m

Length = (x + 5) m

Area of rectangular field = l × b

594 = x (x + 5)

By calculation

594 = x2 + 5x

⇒ 0 = x2 + 5x – 594

⇒ x2 + 5x – 594 = 0

It can be written as

x2 + 27x – 22x – 594 = 0

Taking out the common terms

x (x + 27) – 22 (x + 27) = 0

So,

(x – 22) (x + 27) = 0

Here,

x – 22 = 0 or x + 27 = 0

We get,

x = 22 m or x = -27 which is not possible

Breadth = 22 m

Length = (x + 5) = 22 + 5 = 27 m

Perimeter = 2 (l + b)

= 2 (27 + 22)

By further calculation

= 2 × 49

= 98 m

Question 7.

(a) The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.
(b) In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.

(a) The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion. (b) In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.

Answer :

(a) Area of shaded portion = Area of rectangle ABCD + Area of rectangle PQRS – Area of square LMNO

ml class 9 Mensuration img 28

= (50×5) + (5×35) – (5×5)

By calculation

= 250 + 175 – 25

So ,

= 250 + 150

= 400 m2

(b) Area of shaded portion = Area of ABCD – 5 × Area of any small square

ml class 9 Mensuration img 29

It can be written as

= (l×b) – (5×side×side)

= (8×6) – (5×2×2)

By further calculation

= 48 – 20

= 28 cm2

Question 8.  A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all around. Find the area to be laid with grass.

Answer :

Consider ABCD as a plot

Length of plot = 20 m

Breadth of plot = 14 m

Take PQRS as the grassy plot

Here,

Length of grassy lawn = 20 – 2×2

By calculation

= 20 – 4

= 16 m

Breadth of grassy lawn = 14 – 2×2

By calculation

= 14 – 4

= 10 m

Area of grassy lawn = length × breadth

= 16 × 10

= 160 m2

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 9. The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.

(i) Find the length and the breadth of the lawn.
(ii) Hence, or otherwise, find the area of the flower – beds.
The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m

Answer :

BCDE is the lawn

(i) We know that

Length of lawn BCDE = BC

It can be written as

= AD – AB – CD

= 30 – 2 – 2

By calculation

= 30 – 4

= 26 m

Breadth of lawn BDCE = BE

It can be written as

= AG – GH

= 12 – 2

= 10 m

(ii) We know that, Area of flower beds = Area of rectangle ADFG – Area of lawn BCDE

It can be written as

= (AD×AG) – (BC×BE)

= (30×12) – (26×10)

By calculation

= 360 – 260

= 100 m2


Mensuration Exercise-16.2

ML Aggarwal Class 9 ICSE Maths Solutions

Page 364

Question 10. A foot path of uniform width runs all around the inside of a rectangular field 50 m long and 38m wide. If the area of the path is 492 m². Find its width.

Answer :

ABCD as a rectangular field having

Length = 50 m

Breadth = 38 m

A foot path of uniform width runs all around the inside of a rectangular field 50 m long and 38 m wide. If the area of the path is 492 m2, find its width.

We know that, Area of rectangular field ABCD = l × b

= 50 × 38

= 1900 m2

Let x m as the width of foot path all around the inside of a rectangular field

Length of rectangular field PQRS = (50 – x – x) = (50 – 2x) m

Breadth of rectangular field PQRS = (38 – x – x) = (38 – 2x) m

Here,

Area of foot path = Area of rectangular field ABCD – Area of rectangular field PQRS

492 = 1900 – (50 – 2x) (38 – 2x)

It can be written as

492 = 1900 – [50(38 – 2x) – 2x(38 – 2x)]

By calculation

492 = 1900 – (1900 – 100x – 76x + 4x2)

⇒ 492 = 1900 – 1900 + 100x + 76x – 4x2

On further simplification

492 = 176x – 4x2

Taking out 4 as common

492 = 4 (44x – x2)

⇒ 44x – x2 = 492/4 = 123

We get

x2 – 44x + 123 = 0

It can be written as

x2 – 41x – 3x + 123 = 0

Taking out the common terms

x (x – 41) – 3 (x – 41) = 0

⇒ (x – 3) (x – 41) = 0

Here

x – 3 = 0 or x – 41 = 0

So, x = 3 m or x = 41 m which is not possible

Hence, width is 3 m.

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 11. The cost of enclosing a rectangular garden with a fence all around at the rate of Rs. 15 per metre is Rs. 5400. If the length of the garden is 100 m And the area of the garden.

Answer :

ABCD as a rectangular garden

Length = 100 m

Take breadth = x m

Perimeter of the garden = 2 (l + b)

= 2 (100 + x)

= (200 + 2x) m

We know that, Cost of 1 m to enclosing a rectangular garden = Rs 15

So the cost of (200 + 2x) m to enclosing a rectangular garden = 15 (200 + 2x)

= 3000 + 30x

Given cost = Rs 5400

We get,

3000 + 30x = 5400

It can be written as

30x = 5400 – 3000

x = 2400/30 = 80 m

Breadth of garden = 80 m

So the area of rectangular field = l × b

= 100 × 80

= 8000 m2

Question 12. A rectangular floor which measures 15 m x 8 m is to be laid with tiles measuring 50 cm x 25 cm find the number of tiles required further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?

Answer :

ABCD as a rectangular field of measurement 15m × 8m

Length = 15 m

Breadth = 8 m

ml class 9 Mensuration img 31

Here the area = l × b = 15 × 8 = 120 m2

Measurement of tiles = 50 cm × 25 cm

Length = 50 cm = 50/100 = ½ m

Breadth = 25 cm = 25/100 = ¼ m

So, the area of one tile = ½ × ¼ = 1/8 m2

No. of required tiles = Area of rectangular field/Area of one tile

= 120/(1/8)

By calculation

= (120 × 8)/1

= 960 tiles

Length of carpet = 15 – 1 – 1

= 15 – 2

= 13 m

Breadth of carpet = 8 – 1 – 1

= 8 – 2

= 6 m

Area of carpet = l × b

= 13 × 6

= 78 m2

We know that, Area of floor which is uncovered by carpet = Area of floor – Area of carpet

= 120 – 78

= 42 m2

Fraction = Area of floor which is uncovered by carpet/ Area of floor

= 42/120

= 7/20

Question 13. The width of a rectangular room is 3/5 of its length x metres. If its perimeter isy metres, write an equation connecting x and y. Find the floor area of the room if its perimeter is 32 m.

Answer :

Length of rectangular room = x m

Width of rectangular room = 3/5 of its length

= 3x/5 m

Perimeter = y m

We know that, Perimeter = 2 (l + b)

y = 2 [(5x + 3x)/5]

By calculation

y = 2 × 8x/5

⇒ y = 16x/5

We get,

5y = 16x

⇒ 16x = 5y …(1)

Equation (1) is the required relation between x and y

Given perimeter = 32 m

So y = 32 m

Now substituting the value of y in equation (1)

16x = 5×32

By calculation

x = (5 × 32)/16

⇒ x = (5 × 2)/1

⇒ x = 10 m

Breadth = 3/5 × x

= 3/5 × 10

= 3×2

= 6 m

Here the floor area of the room = l × b

= 10×6

= 60 m2

Question 14. A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square meters, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Answer :

Consider ABCD as a rectangular garden

A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Length = 10 m

Breadth = 16 m

So the area of ABCD = l× b

= 10×16

= 160 m2

Consider x m as the width of the walk

Length of rectangular garden PQRS = 10 – x – x = (10 – 2x) m

Breadth of rectangular garden PQRS = 16 – x – x = (16 – 2x) m

Question 15. A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface area of the four walls.

Answer :

Length of rectangular room = 6 m

Breadth of rectangular room = 4.8 m

Height of rectangular room = 3.5 m

Here,

Inner surface area of four wall = 2 (l + b) × h

= 2(6 + 4.8)×3.5

By calculation

= 2×10.8×3.5

= 21.6×3.5

= 75.6 m2

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 16. A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.

Answer :

ml class 9 Mensuration img 33

Length of rectangular plot = 41 metres

Breadth of rectangular plot = 22.5 metres

Height of boundary wall = 2 metre

Here,

Boundary wall is built at a distance of 1.5 m

New length = 41 + 1.5 + 1.5

= 41 + 3

= 44 m

New breadth = 22.5 + 1.5 + 1.5

= 22.5 + 3

= 25.5 m

We know that, Inner surface area of the boundary wall = 2 (l + b) × h

= 2×(44 + 25.5)×2

By calculation

= 2 ×69.5× 2

= 2 ×139

= 278 m2

Question 17. 

(a) Find the perimeter and area of the figure
(i) given below in which all corners are right angled.
(b) Find the perimeter and area of the figure
(ii) given below in which all corners are right angles.
(c) Find the area and perimeter of the figure
(iii) given below in which all corners are right angles and all measurement in centimetres.
Question 17.   (Exercise-16.2 Mensuration ML Aggarwal ICSE Class-9 )
Answer :

(a) It is given that,

AB = 2m, BE = 4m, FE = 4m and FG = 1.5 m

So BD = 4 + 1.5 = 5.5 m

AC = BD = 5.5 m

CG = (4 + 2) = 6 m

We know that, Perimeter of figure (i) = AC + CG + GF + FE + EB + BA

= 5.5 + 6 + 1.5 + 4 + 4 + 2

= 23 m

Here,

Area of given figure = Area of ABEDC + Area of FEDG

It can be written as

= (length × breadth) + (length × breadth)

= (2 × 5.5) + (4 × 1.5)

= 11 + 6

= 17 m2

(b) It is given that

AB = CD = 3m

HI = AC = 7m

JE = BE = 5m

GF = DE = 2m

DG = EF = 8m

GH = JI = 2m

We know that,

CH = CD + DG + GH

= 3 + 8 + 2

= 13 m

Perimeter of the given figure = AB + AC + CH + HI + IJ + JF + FE + BE

= 3 + 7 + 13 + 7 + 2 + 5 + 8 + 5

= 50 m

Here,

Area of given figure = Area of first figure + Area of second figure + Area of third figure

= (7 × 3) + (8 × 2) + (7 × 2)

By calculation

= 21 + 16 + 14

= 51 m2

(c) It is given that

AB = 12 cm

AL = BC = 7 cm

JK = DE = 5 cm

HJ = GF = 3 cm

LK = HG = CD = 2 cm

Find the area and perimeter of the figure (iii) given below in which all corners are right angled and all measurement in centimetres.

We know that, Perimeter of given figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA

= 12 + 7 + 2 + 5 + 3 + 3 + 2 + 3 + 3 + 5 + 2 + 7

= 54 cm

Here

Area of given figure =Area of first part + Area of second part + Area of third part + Area of fourth part + Area of fifth part

= (7×2) + (2×3) + [(2+3)×2] + (2×3) + (7×2)

By calculation

= 14 + 6 + 10 + 6 + 14

= 50 cm2

Question 18. The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one third that of rectangle.

Answer :

Length of a rectangle = 12 cm

Breadth of a rectangle = 9 cm

So the area = l × b

= 12 × 9

= 108 cm2

The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one third that of rectangle.

Using the condition

Area of triangle ABC = 1/3 × area of rectangle

= 1/3 × 108

= 36 cm2

Consider h cm as the height of triangle ABC

Area of triangle ABC = ½ × base × height

36 = ½ × 9 × h

By further calculation

36 × 2 = 9 × h

⇒ h = (36 × 2)/9

So we get

h = 4 × 2

⇒ h = 8 cm

Hence, height of triangle ABC is 8 cm.

Question 19.  The area of a square plot is 484 mV Find the length of its one side and the length of its one diagonal.

The area of a square plot is 484 mV Find the length of its one side and the length of its one diagonal
Answer :

ABCD is a square plot having area = 484 m2

Sides of square are AB, BC, CD and AD

We know that, Area of square = side × side

484 = (side)2

So we get

Side = √484 = 22 m

⇒ AB = BC = 22 m

In triangle ABC

Using Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 222 + 222

⇒ AC2 = 484 + 484 = 968

By further calculation

AC = √968 = √(484×2)

⇒ AC = 22× √2

So we get

AC = 22 × 1.414 = 31.11 m

Hence, length of side is 22 m and length of diagonal is 31.11 m.

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 20. A square has the perimeter 56 m. Find its area and the length of one diagonal correct upto two decimal places.

Answer :

Consider ABCD as a square with side x m

Perimeter of square = 4 × side

56 = 4x

By calculation

x = 56/4 = 14 m

A square has the perimeter 56 m. Find its area and the length of one diagonal correct up to two decimal places.

In triangle ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

AC2 = 142 + 142

By calculation

AC2 = 196 + 196 = 392

So we get

AC = √392

⇒ AC = √(196×2)

⇒ AC = 14√2

Substituting the value of √2

AC = 14 × 1.414

⇒ AC = 19.80 m

Hence, the side of square is 14 m and diagonal is 19.80 m.


Mensuration Exercise-16.2

ML Aggarwal Class 9 ICSE Maths Solutions

Page 365

Question 21. A wire when bent in the form of an equilateral triangle encloses an area of 36√3 cm2. Find the area enclosed by the same wire when bent to form:

(i) a square, and
(ii) a rectangle whose length is 2 cm more than its width.

Answer :

Area of equilateral triangle = 36 √3 cm2

Consider x cm as the side of equilateral triangle

Area = √3/4 (side)2

36√3 = √3/4 ×(x)2

By calculation

x2 = (36√3 ×4)/√3

⇒ x2 = 36 × 4

So ,

x = √(36×4)

⇒ x = 6 × 2

⇒ x = 12 cm

Here,

Perimeter of equilateral triangle = 3 × side

= 3 × 12

= 36 cm

(i) We know that, Perimeter of equilateral triangle = Perimeter of square

It can be written as

36 = 4 × side

So we get

Side = 36/4 = 9 cm

Area of square = side × side

= 9 × 9

= 81 cm2

(ii) We know that, Perimeter of triangle = Perimeter of rectangle …(1)

According to the condition of rectangle

Length is 2 cm more than its width

Width of rectangle = x cm

Length of rectangle = (x + 2) cm

Perimeter of rectangle = 2 (l + b)

= 2 [(x+2) + x]

By further calculation

= 2 (2x + 2)

= 4x + 4

Using equation (1)

4x + 4 = Perimeter of triangle

⇒ 4x + 4 = 36

By calculation

4x = 36 – 4 = 32 cm

So we get

x = 32/4 = 8 cm

Here,

Length of rectangle = 8 + 2 = 10 cm

Breadth of rectangle = 8 cm

Area of rectangle = length × breadth

= 10×8

= 80 cm2

Question 22. Two adjacent sides of a parallelogram are: 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.

Answer :

ABCD as a parallelogram

Longer side AB = 15 cm

Shorter side = 10 cm

Distance between longer side DM = 8 cm

ml class 9 Mensuration img 38

Consider DN as the distance between the shorter side

Area of parallelogram ABCD = base × height

We can write it as

= AB×DM

= 15 × 8

= 120 cm2

If base is AD

Area of parallelogram = AD × DN

120 = 10 × DN

So ,

DN = 120/10 = 12 cm

Hence, the area of parallelogram is 120 cm2 and the distance between shorter side is 12 cm.

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 23. ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.

Answer :

It is given that

ABCD is a parallelogram

AB = 12 cm, BC = 10 cm and AC = 16 cm

ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.

Area of triangle ABC

BC = a = 10 cm

AC = b = 16 cm

AB = c = 12 cm

We know that

s = (a + b + c)/2

s = (10 + 16 + 12)/2

By further calculation

s = 38/2 = 19 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 40

3√399 cm2

Area of parallelogram = 2 × Area of triangle ABC

= 2 × 3√399

= 6 √399

So we get

= 6 × 19.96

= 119.8 cm2

Consider DM as the distance between the shorter lines

Base = AD = BC = 10 cm

Area of parallelogram = AD × DM

119.8 = 10 × DM

By calculation

DM = 119.8/10

⇒ DM = 11.98 cm

Hence, the distance between shorter lines is 11.98 cm.

Question 24. Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.

Answer :

It is given that

ABCD is a parallelogram

AC and BD are the diagonal which intersect at O

AB = 12 cm and DM = 6 cm

ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD

We know that

Area of parallelogram ABCD = AB×DM

= 12 × 6

= 72 cm2

Area of triangle AOD = ¼ × Area of parallelogram

= ¼ × 72

= 18 cm

Question 25. ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.

Answer :

ABCD is a parallelogram

 ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.

AB = 10 cm, AC = 12 cm

AO = CO = 12/2 = 6 cm

BD = 16 cm

BO = OD = 16/2 = 8 cm

In triangle AOB

a = 10 cm, b = AO = 6 cm, c = BO = 8 cm

s = (a + b + c)/2

s = (10 + 6 + 8)/2

By  calculation

s = 24/2 = 12 cm

ml class 9 Mensuration img 1
ml class 9 Mensuration img 43
So we get

= 12×2

= 24 cm2

We know that, Area of parallelogram ABCD = 4 × Area of triangle AOB

= 4×24

= 96 cm2

Question 26. The area of a parallelogram is p cm2 and its height is q cm. A second parallelogram has equal area but its base is ‘r’ cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.

Answer :

Area of a parallelogram = p cm2

Height of first parallelogram = q cm

We know that, Area of parallelogram = base × height

p = base × q

Base = p/q

Here

Base of second parallelogram = (p/q + r)

Taking LCM

= (p + qr)/ q cm

Area of second parallelogram = Area of first parallelogram = p cm2

It can be written,

Base × height = p cm2

[(p + qr)/q] × h = p

So we get

h = pq/(p + qr) cm

Hence, the height of second parallelogram is h = pq/(p + qr) cm.

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 27. What is the area of a rhombus whose diagonals are 12 cm and 16 cm ?

Answer :

It is given that

ABCD is a rhombus

BD = 12 cm and AC = 16 cm are diagonals

What is the area of a rhombus whose diagonals are 12 cm and 16 cm?

We know that

Area of rhombus ABCD = ½ × AC × BD

Substituting the values

= ½ × 16 × 12

By further calculation

= 8 × 12

= 96 cm2

Question 28. The area of a rhombus is 98 cm². If one of its diagonal is 14 cm, what is the length of the other diagonal?

Answer :

It is given that

Area of rhombus = 98 cm2

One of its diagonal = 14 cm

We know that, Area of rhombus = ½ ×product of diagonals

98 = ½ ×one diagonal ×other diagonal

⇒ 98 = ½ ×14 ×other diagonal

By calculation

Other diagonal = (98×2)/14

= 7 × 2

= 14 cm

Hence, the other diagonal is 14 cm.

Question 29. The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.

Answer :

It is given that

ABCD is a rhombus

Consider x cm as each side

What is the area of a rhombus whose diagonals are 12 cm and 16 cm?

Perimeter = 45 cm

AB + BC + CD + AD = 45 cm

x + x + x + x = 45

⇒ 4x = 45

By division

x = 45/4 cm

Height = 8 cm

Area of rhombus = base × height

= 45/4 ×8

= 45 ×2

= 90 cm2

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 30. PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.

Answer :

PQRS is a rhombus

PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.

PQ = 3 cm

Height = 2.5 cm

Consider PQ as the base of rhombus PQRS.

SM = 2.5 cm is the height of rhombus

We know that, Area of rhombus PQRS = base × height

= 3 ×2.5

= 7.5 cm2

Question 31. If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.

Answer :

ABCD as a rhombus with AC and BD as two diagonals

Here

AC = 8 cm and BD = 6 cm

AO = 4 cm and BO = 3 cm

If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.

In triangle ABC

Using Pythagoras theorem

AB2 = AO2 + BO 2

AB2 = 42 + 32

By calculation

AB2 = 16 + 9 = 25

So we get

AB = √25 = 5 cm

Side of rhombus ABCD = 5 cm

Here

Perimeter of rhombus = 4 × side

= 4 × 5

= 20 cm

Question 32. If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate

(i) the length of the other diagonal, and
(ii) the area of the rhombus.

Answer :

(i) It is given that

ABCD is a rhombus with side AB, BC, CD and AD

AB = BC = CD = AD = 5 cm

AC = 8 cm and AO = 4 cm

If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.

In triangle AOB

Using Pythagoras theorem

AB2 = AO2 + BO2

52 = 42 + BO2

By calculation

25 = 16 + BO2

⇒ BO2 = 25 – 16 = 9

So ,

BO = √9 = 3 cm

⇒ BD = 2 × BO = 2 × 3 = 6 cm

Length of other diagonal = 6 cm

(ii) Area of rhombus = ½ × product of diagonals

= ½ × 8 × 6

By calculation

= 4 × 6

= 24 cm2

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 33.

(a) The diagram (t) given below is a trapezium. Find the length of BC and the area of the trapezium Assume AB = 5 cm, AD = 4 cm, CD = 8 cm

(b) The diagram (ii) given below is a trapezium Find (i) AB (ii) area of trapezium ABCD.

(c) The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m², calculate its depth

a) The diagram (t) given below is a trapezium. Find the length of BC and the area of the trapezium Assume AB = 5 cm, AD = 4 cm, CD = 8 cm

Answer :

(a) It is given that

ABCD is a trapezium

AB = 5 cm, AD = 4 cm and CD = 8 cm

Construct BN perpendicular to CD

ml class 9 Mensuration img 47

Here,

BN = 4 cm

CN = CD – ND

⇒ CN = CD – AO

⇒ CN = 8 – 5 = 3 cm

In triangle BCN

Using Pythagoras theorem

BC2 = BN2 + CN2

BC2 = 42 + 32

By calculation

BC2 = 16 + 9 = 25

⇒ BC = √25 = 5 cm

Length of BC = 5 cm

Area of trapezium = ½ ×sum of parallel sides ×height

It can be written as

= ½ × (AB + CD) × AD

= ½ × (5 + 8) × 4

By calculation

= ½ × 13 × 4

So we get

= 13 × 2

= 26 cm2

Area of trapezium = 26 cm2

(b) From the figure (ii)

AD = 8 units

BC = 2 units

CD = 10 units

Construct CN perpendicular to AD

AN = 2 units

ml class 9 Mensuration img 48

We know that

DN = AD – DN

Substituting the values

= 8 – 6

= 2 units

In triangle CDN

Using Pythagoras theorem

CD2 = DN2 + NC2

Substituting the values

102 = 62 + NC2

By further calculation

NC2 = 102 – 62

⇒ NC2 = 100 – 36 = 64

So we get

NC = √64 = 8 units

From the figure NC = AB = 8 units

We know that

Area of trapezium = ½ × sum of parallel sides × height

It can be written as

= ½ ×(BC + AD)×AB

Substituting the values

= ½ ×(2 + 8)×8

By further calculation

= ½ ×10×8

= 5 × 8

= 40 sq. units

(c) Consider ABCD as the cross section of canal in the shape of trapezium.

AB = 6 m, DC = 8 m

Take AL as the depth of canal

ml class 9 Mensuration img 49

So the area of cross-section = 16.8 m2

It can be written as

½ × sum of parallel sides × depth = 16.8

⇒ ½ × (AB + DC) × AL = 16.8

Substituting the values

½ × (6 + 8) × AL = 16.8

By further calculation

½ × 14 × AL = 16.8

⇒ AL = (16.8 × 2)/14

So we get

AL = (16.8×1)/7

⇒ AL = 2.4 m


Mensuration Exercise-16.2

ML Aggarwal Class 9 ICSE Maths Solutions

Page 366

Question 34. The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.

Answer :

Consider ABCD as a trapezium in which AB || DC

Height CL = 12 cm

E and F are the mid-points of sides AD and BC

EF = 18 cm

The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.

We know that

EF = ½ (AB + DC) = 18 cm

Here

Area of trapezium ABCD = ½ (AB + DC) ×height

Substituting the values

= 18 × 12

= 216 cm2

Question 35. The area of a trapezium is 540 cm². If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.

Answer :

Area of trapezium = 540 cm2

Ratio of parallel sides = 7: 5

Consider 7x cm as one parallel side

Other parallel side = 5x cm

Distance between the parallel sides = height = 18 cm

We know that, Area of trapezium = ½ × sum of parallel sides × height

Substituting the values

540 = ½ × (7x + 5x) × 18

By calculation

540 = ½ × 12x × 18

⇒ 540 = 6x × 18

⇒ 540 = 108x

⇒ x = 540/108 = 5

Here

First parallel side = 7x = 7 × 5 = 35 cm

Second parallel side = 5x = 5 × 5 = 25 cm

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 36. The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm2, find the perimeter.

Answer :

ABCD is an isosceles trapezium

BC = AD

Height = 4 cm

Consider CD = 2x and AB = 3x

he parallel sides of an isosceles trapezium are in the ratio 2: 3. If its height is 4 cm and area is 60 cn2, find the perimeter.

Area of trapezium = ½ (sum of parallel sides)×height

Substituting the values

60 = ½ × (2x + 3x) × 4

By further calculation

60 = ½ × 5x × 4

⇒ 60 = 5x × 2

⇒ 60 = 10x

So we get

x = 60/10 = 6

Here,

CD = 2x = 2 × 6 = 12 cm

AB = 3x = 3 × 6 = 18 cm

AN = BM

We can write it as

AN = AB – BN

⇒ AN = AB – (MN + BM)

MN = CD

⇒ AN = AB – (CD + BM)

Similarly BM = AN

AN = AB – (CD + AN)

Substituting the values

AN = 18 – (12 + AN)

⇒ AN = 18 – 12 – AN

⇒ AN + AN = 6

⇒ 2AN = 6

By division

AN = 6/2 = 3

In triangle AND

Using Pythagoras theorem

AD2 = DN2 + AN2

Here DN = 4 cm

AD2 = 42 + 32

By further calculation

AD2 = 16 + 9 = 25

So we get

AD= √25 = 5 cm

Here AD = BC = 5 cm

Perimeter of trapezium = AB + BC + CD + AD

= 18 + 5 + 12 + 5

= 40 cm

Question 37. The area of a parallelogram is 98 cm². If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.

Answer :

Area of parallelogram = 98 cm2

Condition – If one altitude is half the corresponding base

Take base = x cm

Corresponding altitude = x/2 cm

Area of parallelogram = base × altitude

98 = x × x/2

⇒ 98 = x2/2

By cross multiplication

x2 = 98 × 2 = 196

So we get

x = √196 = 14 cm

Base = 14 cm

Here,

altitude = 14/2 = 7 cm

Question 38. The length of a rectangular garden is 12m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden

Answer :

The dimensions of rectangular garden are

Breadth = x m

Length = (x + 12) m

We know that

Area = l × b

Area = (x + 12) × x

⇒ Area = (x2 + 12x) m2

Perimeter = 2 (l + b)

= 2 [(x + 12) + x]

= 2 [ x + 12 + x]

= 2 [2x + 12]

= (4x + 24) m

The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Numerical value of area = 4 × numerical value of perimeter

2 + 12x = 4 × (4x + 24)

By calculation

x+ 12x = 16x + 96

⇒ x2 + 12x – 16x – 96 = 0

⇒ x2 – 4x – 96 = 0

It can be written as

x2 – 12x + 8x – 96 = 0

Taking out the common terms

x (x – 12) + 8 (x – 12) = 0

⇒ (x + 8) (x – 12) = 0

Here,

x + 8 = 0 or x – 12 = 0

So ,

x = -8 (not possible) or x = 12

Breadth of rectangular garden = 12 m

Length of rectangular garden = 12 + 12 = 24 m

Question 39. If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.

Answer :

It is given that

Perimeter of a rectangular plot = 68 m

Length of its diagonal = 26 m

ABCD is a rectangular plot of length x m and breath y m

 If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.

Perimeter = 2 (length + breadth)

68 = 2 (x + y)

By calculation

68/2 = x + y

⇒ 34 = x + y

So we get

x = 34 – y …(1)

In triangle ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

262 = x2 + y2

⇒ x2 + y2 = 676

Now substituting the value of x in equation (1)

(34 – y)2 + y2 = 676

⇒ 1156 + y2 – 68y + y2 = 676

By calculation

2y2 – 68y + 1156 – 676 = 0

⇒ 2y2 – 68y – 480 = 0

Taking 2 as common

2 (y2 – 34y – 240) = 0

⇒ y2 – 34y – 240 = 0

It can be written as

y2 – 24y – 10y – 240 = 0

Taking out the common terms

y (y – 24) – 10 (y – 24) = 0

⇒ (y – 10) (y – 24) = 0

Here,

y – 10 = 0 or y – 24 = 0

y = 10 m or y = 24 m

Now substituting the value of y in equation (1)

y = 10 m, x = 34 – 10 = 24 m

y = 24 m, x = 34 – 24 = 10 m

Area in both cases = xy

= 24×10 or 10×24

= 240 m2

Hence, the area of the rectangular block is 240 m2.

Question 40. A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than 2 side of a square. Find the perimeter of the square.

Answer :

Side of a square = x cm

Length of rectangle = (x + 12) cm

Breadth of rectangle = (x + 8) cm

We know that, Area of square = side × side = x×x = x2 cm2

Area of rectangle = l × b

= (x + 12) (x + 8) cm2

Based on the question

Area of rectangle = 2 ×area of square

(x + 12) (x + 8) = 2 × x2

It can be written as

x (x + 8) + 12 (x + 8) = 2x2

⇒ x2 + 8x + 12x + 96 = 2x2

⇒ x2 – 2x2 + 8x + 12x + 96 = 0

By calculation

-x2 + 20x + 96 = 0

⇒ -(x2 – 20x – 96) = 0

⇒ x2 – 20x – 96 = 0

We can write,

x2 – 24x + 4x – 96 = 0

⇒ x (x – 24) + 4 (x – 24) = 0

⇒ (x + 4) (x – 24) = 0

Here,

x + 4 = 0 or x – 24 = 0

⇒ x = -4 or x = 24 cm

Side of square = 24 cm

Perimeter of square = 4 × side

= 4 × 24

= 96 cm

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 41. The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.

Answer :

Perimeter of a square = 48 cm

Side = perimeter/4 = 48/4 = 12 cm

We know that, Area = side2 = 122 = 144 cm2

Area of rectangle = 144 – 4 = 140 cm2

Take breadth of rectangle = x cm

Length of rectangle = x + 4 cm

So the area = (x + 4) × x cm2

(x + 4) x = 140

By further calculation

x2 + 4x – 140 = 0

⇒ x2 + 14x – 10x – 140 = 0

Taking out the common terms

x (x + 14) – 10 (x + 14) = 0

⇒ (x + 14) (x – 10) = 0

Here,

x + 14 = 0 where x = – 14

x – 10 = 0 where x = 10

Breadth = 10 cm

Length = 10 + 4 = 14 cm

Perimeter = 2 (l + b)

= 2 (14 + 10)

= 2 × 24

= 48 cm

Question 42. In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangles; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.

Answer :

In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangle; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.

It is given that

ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm

HAD and BFC are equilateral triangle with each side = 8 cm

AEB and DCG are right angled isosceles triangles with hypotenuse = 10 cm

Consider AE = EB = x cm

In triangle ABE

AE2 + EB2 = AB2

Substituting the values

x2 + x2 = 102

⇒ 2x2 = 100

By further calculation

x2 = 100/2 = 50

⇒ x = √50 = √(25 × 2) = 5√2 cm

We know that

Area of triangle AEB = Area of triangle GCD

It can be written as

= ½ × x × x

= ½ x2 cm2

Substituting the value of x

= ½ × 50

= 25 cm2

Area of triangle HAD = Area of BFC

It can be written as

= √3/4 × 82

= √3/4 × 64

= 16√3 cm2

Area of shaded portion = Area of rectangle ABCD + 2 area of triangle AEB + 2 area of triangle BFC

Substituting the values

= (10×8) + (2×25) + (2×16√3)

By further calculation

= 80 + 50 + 32√3

So we get

= (130 + 32√3) cm2

Here

Perimeter of the figure = AE + EB + BF + FC + CD + GD + DH + HA

It can be written as

= 4AE + 4BF

Substituting the values

= (4×5√2) + (4×8)

So we get

= 20√2 + 32

= (32 + 20√2) cm

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 43.

(a) Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DGFG is an isosceles trapezium.
All measurements are in centmetres.

(b) Find the area enclosed by the figure (ii) given below. AH measurements are in centimetres.

(c) In the figure (iii) given below, from a 24. cm x 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.
where is an equilateral triangle and is an isosceles All measurements are in centmetres

Answer :

(a) It is given that

ABC is an equilateral triangle and DEFG is an isosceles trapezium

EF = GD = 5 cm

DE = 6 cm

GF = GB + BC + CF

= 3 + 6 + 3

= 12 cm

AB = AC = BC = 6 cm

Join BD and CE

In right triangle CEF

CE2 = EF2 – CF2

= 52 – 32

= 25 – 9

= 16

So,

CE = √16 = 4 cm

Area of triangle ABC = √3/4 × 62

By calculation

= √3/4 × 36

= 9√3 cm2

Area of trapezium DEFG = ½ (DE + GF) × CE

= ½ ×(6+12)× 4

By further calculation

= ½ ×18×4

= 36 cm2

So the area of figure = 9√3 + 36

= 9 × 1.732 + 36

= 15.59 + 36

= 51.59 cm2

(b) We know that, Length of rectangle = 2 + 2 + 2 + 2 = 8 cm

Width of rectangle = 2 cm

Area of rectangle = l × b

= 8 × 2

= 16 cm2

Here

Area of each trap = ½ (2+2) × (6–2)

By calculation

= ½ × 4 × 4

= 8 cm2

So the total area = area of rectangle + area of 2 trapezium

= 16 + 8 + 8

= 32 cm2

(c) We know that, Length of each rectangle = 24 cm

Width of each rectangle = 6 cm

Area of each rectangle = l × b

Substituting the values

= 24 × 6

= 144 cm2

Base of each parallelogram = 8 cm

Height of each parallelogram = 6 cm

So the area of each parallelogram = 8 × 6 = 48 cm2

Here,

Area of the M-shaped figure = (2×144) + (2×48)

So we get

= 288 + 96

= 384 cm2

Area of the square cardboard = 24 × 24 = 576 cm2

Area of the removing cardboard = 576 – 384 = 192 cm2


Mensuration Exercise-16.2

ML Aggarwal Class 9 ICSE Maths Solutions

Page 367

Question 44.

(a) The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.

(b) The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.

(c) Calculate the area of the pentagon ABCDE shown in fig. (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5cm,DX = 9cmand DX is perpendicular to EC and AB.
(a) The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section

Answer :

(a) From the figure (i)

AB = 1.8 m, CD = 0.6 m, DE = 1.2 m

EF = 0.3 m, AF = 2.4 m

Construct DE to meet AB in G

∠FEG = ∠GAF = 900

So, AGEF is a rectangle

ml class 9 Mensuration img 55

We know that, Area of given figure = Area of rectangle AGEF + Area of trapezium GBCD

It can be written as

= (l × b) + ½ (sum of parallel sides× height)

= (AF×AG) + ½ (GB+CD)×DG

= (2.4×0.3) + ½ [(AB – AG) + CD]×(DE + EG)

Here AG = FE and using EG = AF

= 0.72 + ½ [(1.8–0.3) + 0.6]×(1.2 + 2.4)

By calculation

= 0.72 + ½ [1.5+0.6]×3.6

= (0.72 + ½) × (2.1× 3.6)

So,

= 0.72 + (2.1×1.8)

= 0.72 + 3.78

= 4.5 m2

(b) ABCD is a pentagonal field

AX = 12 m, BX = 30 m, XZ = 15 m, CZ = 25 m,

DZ = 10 m, AD = 12 + 15 + 10 = 37 m, EY = 20 m

ml class 9 Mensuration img 56

We know that, Area of pentagonal field ABCDE = Area of triangle ABX + Area of trapezium BCZX + Area of triangle CDZ + Area of triangle AED

It can be written as

= (½ ×base × height) + [½ (sum of parallel sides)×height] + [½ × base × height] + [½ × base × height]

= [½ ×BX×AX] + [½ (BX + CZ)×XZ] + [½ ×CZ×DZ] + [½ ×AD×EY]

= [½×30×12] + [½ (30+25)×15] + [½×25×10] + [½ ×37×20]

By calculation

= [15×12] + [7.5×55 ]+ [25×5] + [37×10]

So ,

= 180 + 412.5 + 125 + 370

= 1087.5 m2

(c) It is given that ABCDE is a pentagon

AX = BX = 6 cm, EY = CY = 4 cm

DE = DC = 5 cm, DX = 9 cm

Construct DX perpendicular to EC and AB

Calculate the area of the pentagon ABCDE shown in fig (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5 cm, DX = 9 cm and DX is perpendicular to EC and AB.

In triangle DEY

Using Pythagoras Theorem

DE2 = DY2 + EY2

52 = DY2 + 42

⇒ 25 = DY2 + 16

⇒ DY2 = 25 – 16 = 9

So ,

DY = √9 = 3 cm

Here,

Area of pentagonal field ABCDE = Area of triangle DEY + Area of triangle DCY + Area of trapezium EYXA + Area of trapezium CYXB

It can be written as

= (½ ×base× height) + (½ ×base ×height) + [½ ×(sum of parallel sides) ×height] + [½ ×(sum of parallel sides) ×height]

= [½ ×EY ×DY] + [½ ×CY ×DY] + [½ ×(EY + AX) ×XY] + [½ ×(CY + BX) ×XY]

= [½ ×4 ×3] + [½ ×4 ×3] + [½ ×(4+6) ×(DX – DY)] + [½ (4+6) ×(DX – DY)]

By calculation

= (2×3) + (2×2) + [½× 10×(9 – 3)] + [½ ×10 ×(9 – 3)]

So we get

= 6 + 6 + (5×6) + (5×6)

= 6 + 6 + 30 + 30

= 72 cm2

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 45. If the length and the breadth of a room are increased by 1 metre the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the, area is decreased by 5 square metres. Find the perimeter of the room.

Answer :

Take length of room = x m

Breadth of room = y m

Here,

Area of room = l × b = xy m2

We know that, Length is increased by 1 m then new length = (x + 1) m

Breadth is increased by 1 m then new breadth = (y + 1) m

So the new area = new length × new breadth

= (x + 1) (y + 1) m2

Based on the question

xy = (x + 1) (y + 1) – 21

By further calculation

xy = x (y + 1) + 1 (y + 1) – 21

⇒ xy = xy + x + y + 1 – 21

So we get

0 = x + y + 1 – 21

⇒ 0 = x + y – 20

⇒ x + y – 20 = 0

⇒ x + y = 20 …(1)

Length is increased by 1 metre then new length = (x + 1) metre

Breadth is decreased by 1 metre than new breadth = (y – 1) metre

So the new area = new length × new breadth

= (x + 1) (y – 1) m2

Based on the question

xy = (x + 1) (y – 1) + 5

By calculation

xy = x (y – 1) + 1 (y – 1) + 5

⇒ xy = xy – x + y – 1 + 5

⇒ 0 = -x + y + 4

So we get

x – y = 4 …(2)

By adding equations (1) and (2)

2x = 24

⇒ x = 24/2 = 12 m

Now substituting the value of x in equation (1)

12 + y = 20

⇒ y = 20 – 12 = 8 m

Here

Length of room = 12 m

Breadth of room = 8 m

So the perimeter = 2 (l + b)

= 2 (12 + 8)

= 2 × 20

= 40 m

Question 46. A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Answer :

Sides of a triangle = 26 cm, 28 cm and 30 cm

We know that, s = (a + b + c)/2

Substituting the values

s = (26 + 28 + 30)/ 2

⇒ s = 84/2

⇒ s = 42 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 58

So we get

= 2 × 4 × 6 × 7

= 336 cm2

We know that, Base = 28 cm

Height = Area/base

= 336/28

= 12 cm

Question 47. A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that its perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of the rectangle.

Answer :

Area of rectangle = 105 cm2

Length of rectangle = x cm

We know that, Area = length × breadth

105 = x × breadth

⇒ x = 105/x cm

Perimeter of rectangle = 44 cm

So,

2 (l + b) = 44

⇒ 2 (x + 105/x) = 44

By calculation

(x2 + 105)/ x = 22

By cross multiplication

x2 + 105 = 22x

⇒ x2 – 22x + 105 = 0

x2 – 15x – 7x + 105 = 0

⇒ x(x – 15) – 7(x – 15) = 0

⇒ (x – 7) (x – 15) = 0

Here,

x – 7 = 0 or x – 15 = 0

⇒ x = 7 cm or x = 15 cm

If x = 7 cm,

Breadth = 105/7 = 15 cm

If x = 15 cm,

Breadth = 105/15 = 7 cm

Hence, the required dimensions of rectangle are 15 cm and 7 cm.

(ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions)

Question 48. The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the value of the length, breadth and the area to, write an equation in x. Solve the equation to calculate the length and breadth of the plot.

Answer :

Perimeter of a rectangular plot = 180 m

Area of a rectangular plot = 1800 m2

Take length of rectangle = x m

Here

Perimeter = 2 (length + breadth)

180 = 2 (x + breadth)

⇒ 180/2 = x + breadth

⇒ x + breadth = 90

⇒ Breadth = 90 – x m

Area of rectangle = l × b

1800 = x × (90 – x)

⇒ 90x – x2 = 1800

It can be written as

– (x2 – 90x) = 1800

⇒ x2 – 90x = – 1800

By calculation

x2 – 90x + 1800 = 0

⇒ x2 – 60x – 30x + 1800 = 0

⇒ x (x – 60) – 30 (x – 60) = 0

⇒ (x – 30) (x – 60) = 0

Here,

x – 30 = 0 or x – 60 = 0

⇒ x = 30 or x = 60

If x = 30 m

Breadth = 90 – 30 = 60 m

If x = 60 m

Breadth = 90 – 60 = 30 m

Hence, the required length of rectangle is 60 m and the breadth of rectangle is 30 m.

 

—  : End of ML Aggarwal Mensuration Exe-16.2 Class 9 ICSE Maths Solutions :–

Return to :-   ML Aggarawal Maths Solutions for ICSE  Class-9

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