Assignment no 5 pdf

Assignment no: 5

Class: Nine
Lesson: Mensuration

Short Questions :
1. Find the area and perimeter of the given plane figure.

Solution:

Length (l) = 7.5cm 5cm

Breadth (b) = 5cm

Now, 7.5cm

Area of given rectangle = l * b

= 7.5cm * 5cm

= 37.5cm2

Also,

Perimeter = 2(l + b)

= 2 (7.5 + 5)cm

= 2 * 12.5cm

= 25cm

2. A rectangular room is 8m long and 6m wide. What length of carpet of 3.5 m wide is needed
to carpet room?
Solution:

Length of carpet (l1) = ?

Breadth of carpet (b1) = 3.5m

Length of room (l) = 8m

Breadth of room (b) = 6m

Area of carpet (A1) = Area of room (A)

i.e. l1 * b1 = l * b

or, l1 = !∗#
#$

= %&∗'&
(.*&

= 13.71m

∴ 13.71m long carpet is required.

3. What length of wall paper of breadth 50cm is reqiured to cover a wall of 5m long and 3m
high?
Solution:

Length of wall paper (l1) = ?

Breadth of wall paper (b1) = 50cm = 0.5m

Length of wall (l) =5m

Breadth of wall (b) =3m

While carpeting a room,

Area of wall paper (A1) = Area of wall (A)

i.e. l1 * b1 = l*b

or, l1 = l*b / b1

= 5m∗3m / 0.5m
= 15m / 0.5m

= 30m
∴ 30m long wall paper is required.
4. Find the number of stone to pave 12m long and 8m wide courtyard if the length of the stone
is 2m and width is 1.6m?
Solution:
Length of courtyard (l) = 12m
Breadth of courtyard(b) = 8m
Area of courtyard (A) = l * b

= 12m * 8m
= 96m2
Again,
Length of stone (l1) = 2m
Breadth of stone (b1) = 1.6m
Area of stone (a) = l1 * b1
= 2m * 1.6m
= 3.2m2
Now,
Required number of stone(N) = Area of courtyard / Area of stone

= 96m2 / 2m2
= 30
Hence, 30 stones are needed to pave the courtyard.

5. The path having area 75m2 and width 2.5m is surrounded inside a garden. Find the length of
the path.
Solution:

Area of square garden (A) = 75m2

Breadth (b) = 2.5m

Length (l) = ?

Now,

Area of square = l * b

or, 75m2 = l * 2.5m

or, l = 75m2 / 2.5m
∴ l = 30m

Hence, the length of the path is 30m.

6. Calculate the volume and total surface area of the cylinder whose radius is 4cm and height

is 6cm.

Volume=π r²h = π × 4² × 6 = 96 π 6cm
=301.5928947 cm³
=302 cm³ (to 3 significant figures)

Area of curved surface=2π rh = 2 × π × 4 × 6
=48π
=150.7964474 cm²

Area of each end=π r² = π × 4²
=16π
=50.26548246 cm²

Total surface area=150.7964474 + (2 × 50.26548246)
=251.3274123 cm²
=251 cm² (to 3 significant figures)

7. The cost of carpeting a room of 7m long, 6m wide at the rate of Rs.200/m is Rs.2400. Find
the breadth of the carpet.
Here,

Length of the room (l) = 7m
Breadth of room (b0 = 6m
Cost of carpeting the room (C) = Rs.200/m
Total cost (T= Rs.2600
∴ Length of carpet (l1) = T/C
= Rs.2400/Rs.200 = 12m
Breadth of carpet (b1) = ?
When we carpet the room,
Area of carpet = Area of room
i.e. l1*b1 = l*b
or, 12m*b1 = 7m*6m
or, b1 = 42m2/12m
or, b1 = 3.5 m
∴ The breadth of the carpet is 3.5m.
8. Find the number of stones of size 40cm long and 30 cm wide to cover a courtyard of length
30 m and breadth is 15 m?
Here,
Length of stone (l1) = 40 cm = 0.4m
Breadth of stone (b1) = 30 cm = 0.3m
∴ Area of each stone (a) = l1*b1
= 0.4m * 0.3m = 0.12m2

And, length of courtyard (l) = 30m
breadth of courtayard (b) = 15m
∴ Area of courtyard (A) = l*b
= 30 m * 15 m
= 450 m2
Now,
Required number of stones (N) = Area of courtyard(A) / Area of stone(a)
= 450m2 / 0.12m2
= 3750
Hence, 3750 stones of given dimension are required to cover the courtyard.

Long Questions
1. Find the length of a carpet of width 50cm to carpet a hall 8m long and 6m wide. Also, find
the cost of carpeting at the rate of Rs 450/metre.
Solution:
Length of carpet(l1) = ?
Breadth of carpet (b1) = 50cm = 0.5m
Length of room (l) = 8m
Breadth of room (b) = 6m
While carpeting a room,
Area of carpet (A1) = Area of room (A)
i.e. l1 * b1 = l*b

or, l1 = l*b / b1

= 8m∗6m / 0.5m
= 48m2 /0.5m
= 96m

∴ 96m long carpet is required.

Again,

Length of carpet (l1) = 96m

Cost of carpet/metre (C) = Rs 450/m

Toal cast (T) = l1 * C

= 96 * Rs 450

Rs 43,200

Hence, it costs Rs 43,200 in order to carpet the room.

2. Find the perimeter of the given window. Also, find its area, (π =22 / 7)
Solution:

The given window frame consists of a rectangle and a semi-circle.

For semi-circle,

diameter (d) = 280cm

radius (r) = 1 / 2d 150cm 150cm
=1/2 * 280cm
= 140cm

Semicircle (c) =1/2πd 280cm

=,- ∗ -- ∗ 280
.

= 440cm
Now, perimeter of the window frame = l + b + b + semicircle
= 280cm + 150cm + 150cm + 440cm
= 1,020cm
Now, Area of rectangular part (A1) = l * b
= 280cm * 150cm
= 42,000cm2
Area of semi circle (A2) =1/2πr2
=1/2*22/7*(140cm)2
= 30,800cm2
∴ Area of window frame (A) = A1 + A2
= 42,000cm2 + 30,800cm2
= 72,800cm2
Hence,
Perimeter of window frame (P) = 1020cm
Area ofwindow frame (A) = 72,800cm2 = 7.28m2
3. Find the area of the shaded region.

Solution:

From the figure,
Length of rectangle (l) = 50m
Breadth (b) = 30m
Width of the path (d) = 5m
Now,
Area of EFGH = l * d
= 50m * 5m
= 250m2
Again,
Area of WXYZ = b * d
= 30m * 5m
= 150m2
Also,
Area of crossing path ABCD = d2
= (5m)2
= 25m2
Hence,
Area of shaded region = Area of EFGH + Area of WXYZ- Area of crossing path ABCD
= 250m2 + 150m2- 25m2
= 375m2

Alternatively,
Area of crossing paths/shaded region = d(l + b - d)
= 5m (50m - 30 - 5m)
= 5m * 15m
= 375m2
4. A room is 8m long, 6m wide and 4m height. It has door of 2m × 1.5m and two windows of
size 2m × 3m. Find the Area of 4 walls Area of 4 walls and ceiling Area of 4 walls excluding
door and windows Cost of painting 4 walls and ceiling at Rs 80/m2
Solution:
Given,
Length of wall (l) = 8m
Breadth of wall (b) = 6m
Height of wall (h) = 4m
Length of door (l1) = 2m
Height of door (h1) = 1.5m
Length of window (l2) = 2m
Height of window (h2) = 3m
Cost of painting walls (C) = Rs 80/m2
Now,
Area of 4 walls = 2h(l +b)
= 2 × 4m (8m + 6m)

= 8m × 14m
= 112m2
Area of 4 walls and ceiling =2h(l +b) +lb
= 2 × 4m (8m + 6m) + 8m × 6m
= 8m × 14m + 48m2
= 112m2 + 48m2
= 160m2
Area of 4 walls excluding door and windows (A) =2h(l +b) -l1h1- 2( l2h2)
= 2 × 4m (8m + 6m) -2m × 1.5m - 2(2m × 3m)
=8m × 14m - 3m2 - 2 × 6m2
= 112m2 - 3m2 - 12m2
= 97m2
Cost of painting 4 walls and ceiling = C * A
= Rs 80/m2 × 97m2
= Rs 7760
Hence,
Area of 4 walls = 112m2
Area of 4 walls and ceiling = 160m2
Area of 4 walls excluding door and windows = 97m2
Cost of painting 4 walls and ceiling = Rs 7760

5. Find the length of carpet of width 50cm to carpet a hall 8m long and 6 m wide. Also find the
cost of carpeting at the rate of Rs.400/m.
Here,
Length of carpet (l1) = ?
Breadth of carpet (b1) = 50cm = 0.5m
Length of room (l) = 8m
Breadth of room (b) = 6m
While carpeting a room,
Area of carpet (A1) = Area of room (A)
i.e. l1*b1= l*b
or, l1 = l*b / b1
= 8m * 6m / 0.5 m
= 96m
∴ 96m long carpet is required.
Again,
Length of carpet (l1) = 96m
cost of carpet/ metre (C) = Rs.400/m
Total cost (T) = l1*C
= 96 * Rs.400
= Rs.38400
Hence, it costs Rs.38400 in order to carpet the room.

6. A hall is 30 m long and 20 wide. The area of four wall is 400 m2. Find the height of the wall.
Also, find a) The area of four wall and ceiling. b) The area of the four wall, ceiling and floor
(T.S.A). c) The area of floor and ceiling. d) Find the cost of painting four wall and ceiling at the
rate of Rs.200/m2
Hence,
length (l) = 40m
breadth (b) = 30 m
area of 4 walls (A) = 600m2
height (h) =?
Area of 4 walls (A) = 400m2
i.e. 2h(l + b) = 400 m2
or, 2h( 30m + 20m) = 400m2
or, 2h * 50m = 400m2
or, h = 4m
∴ height of the wall (h) = 4m
a) Now, Area of 4 walls and ceiling
= 2h(l + b) + lb
= 2*4m( 30m + 20m) + 30m*20m
= 400m2 + 600m2
= 1000m2
b) Area of 4 walls, ceiling and floor (T.S.A)

= 2h (l + b) + 2lb
= 2*4m (30m + 20m ) + 2*30m*20m
= 400m2 + 1200m2
= 1600m2
c) Area of floor and ceiling
= 2 lb
= 2*30m*20m
=1200m2
d) Area of 4 walls and ceiling (A) = 1000m2
cost of painting the wall (C) = Rs.200/m2
Total cost (T) = C*A
= Rs.200/m2*1000m2
= Rs.200000
Total cost of painting four wall and ceiling is Rs.200000.
7. A room is 5m long, 4m wide and 3m high. Find the following: a) Area of floor of a room b)
Area of 4 walls of a room c) Area of ceiling d) Area of 4 wall and ceiling e) Area of 4 wall,
ceiling and floor (TSA of a room)
Here,
length (l) = 5m
breadth (b) = 4 m
height (h) = 3m

Now,
a) Area of floor (A)
= l*b
= 5m * 4m
= 20m2
b) Area of 4 walls (A)
= 2h (l + b)
= 2*3m (5m + 4m)
= 54m2
c) Area of ceiling (A)
=l*b
= 5m * 4m
= 20m2
d) Area of 4 wall and ceiling (A)
= 2h (l + b) + lb
= 2*3m (5m + 4m) + 5m * 4m
= 54 m2+ 20m2
= 74 m2
e) Area of 4 wall, ceiling and floor (TSA)
= 2h (l + b) + lb + lb
= 2*3m(5m + 4m) + 5m * 4m + 5m * 4m
=54m2 + 20m2+ 20m2
=94m2

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