Mensuration Made Simple : With Word Problems For Class 7th,8th,9th,10th [4]

Q.1) IN a bakery store, 250 cubical boxes are required. The length of the edge box
measures 20 cm. How many cardboard sheets of size 50 cm*50 cm will be
required to make these boxes ?

SOLUTION : Cubical boxes
Given : No. of cubical boxes = 250
edge of each box = a = 20 cm.
Step 1.) Total surface area = TSA= ?
TSA= 6a²
TSA= 6 * 20²
TSA= 6* 400
= 2400 cm²
Step 2) TSA of 250 boxes = 2400 * 250 cm²

Step 3) Rectangular cardboard
Length = L= 50 cm Breadth = B = 50 cm
Area of the cardboard = ?
A = L * B
= 50 * 50
= 2500 cm²

Step4) No. of cardboard sheets required =
= Area of cubical boxes / Area of cardboard
= 2400 *250 / 2500
= 240 sheets.
Ans: 240 cardboard sheets are required.


Q2) A cylindrical roller is used to polish a rectangular table. The length of the roller
is 0.75m and the diameter is 0.21m. If the roller rolls over the table 20 times completely
find the area of the table?
SOLUTION : Cylindrical roller
Given : Length = height = h = 0.75m
Diameter = d = 0.21m radius = r = 0.21/2m
No. of rounds = 20 times
Area of the table =?
Step 1) Area covered by roller in one round = 2*pie *r*h
= 2*22/7*0.21/2*0.75
= 0.495m²
Step 2) Area of rectangular table = Area covered by the roller in one round *No. of rounds
= 0.495 * 20
= 9.9m²



Q.3) The sum of the area of floor and ceiling of a room is equal to the area of four walls. If the
room is 24m long and 16m wide, find the height of the room?

SOLUTION : Room
Given: Length = L = 24m Breadth = b = 16m Height =h=?
The sum of the area of floor and ceiling = Area of four walls
Step 1) Area of the floor = L*B Area of the ceiling = L*B
Area of the four walls = Lateral surface area = LSA= 2( b*h + h*l)

According to the problem,
Area of the floor +Area of the ceiling = Area of four walls
L*B + L*B = 2(B*H +H*L)
2(L*B) = 2 H(B + L)
24*16 = H ( 24 +16)
24*16 = H (40)
24*16/40 = H
96/10 = H
9.6m = H
Ans : The height of the room is 9.6m.


Q.4) Find the cost of plastering the inner surface of a well, with diameter 3.5m and depth 10m
at the rate of Rs. 45 per m²?
SOLUTION : Well
Given : diameter = d = 3.5m radius =r = 3.5/2 m
depth = h = 10 m
Rate of plastering = Rs. 45 per m²
Step1) Area of the inner surface of the well = Lateral surface area =?
LSA = 2*pie*r*h
= 2*22/7*3.5/2*10
=110m²
STEP 2) Rate of plastering perm² = Rs.45
Cost of plastering 110m² = ?
= 110*45
=Rs.4950
Ans : The cost of plastering the inner surface of the well is Rs. 4950.