a closed rectangular box is made of wood of 1.5cm thickness. the exterior length and breadth are respectively 78cm and 19cm, and the capacity of the box is 15 cubic decimeter. calculate the exterior height of the box.
Answers
Given,
The thickness of the wood of the box is 1.5 centimetres.
The exterior length and breadth are 78 centimetres and 19 centimetres, respectively.
Capacity of the box = 15 dm³
To find,
The exterior height of the box.
Solution,
The thickness of wood = 1.5 cm = 0.15 dm
Exterior length of the box = 78 cm = 7.8 dm
Exterior width of the box = 19 cm = 1.9 dm
Interior length of the box = 7.8 - (0.15×2) = 7.5 dm
Interior width of the box = 1.9 - (0.15×2) = 1.6 dm
Let,the exterior height of the box = x dm
Interior height of the box = x - (0.15×2) = (x - 0.3) dm
So,the internal volume = 7.5 × 1.6 × (x-0.3) = 12 × (x-0.3) = 12x - 3.6 dm³
Now,if we compare the value of internal volume that we have calculated and the value of the internal volume that is given in the question,we will get the following mathematical equation.
12x - 3.6 = 15
12x = 15+3.6
12x = 18.6
x = 1.55
Exterior height = 1.55 dm = 15.5 cm
Hence,the exterior height of the box is 15.5 centimetres.
Answer:
15.5
Step-by-step explanation:
Given, the external dimensions of a closed wooden box are 78 cm, 19 cm.
Let the external height be h cm respectively.
As the thickness is 1.5 cm , the internal length, breadth and height of the closed box are 78−2(1.5)=75 cm, 19−2(1.5)=16 cm
and h−2(1.5)=(h−3) cm respectively.
Given, capacity of the box =15cudm=15000cucm
As the box is cuboidal, its volume = length × breadth × height
⇒(75×16×(h−3))=15000
⇒(h−3)=12.5 cm
⇒h=15.5 cm