Answers
Answer:
★ Concept :-
Here concept of volume is used. In first question firstly we will find volume of each brick and volume of wall, then we will divide volume of wall by volume of brick to get number of bricks used, then we will easily find cost of bricks. In second question firstly we will find volume of earth taken out, then we will compare it with volume of earth spread on rectangular plot to get height of platform.
Let's do it !!
★ Formula used :-
★ Solution :-
⠀⠀⠀⠀⠀⠀☯ Question ①
Given,
~ Dimensions of brick ::
»» Length of brick (L) = 10 cm
»» Breadth of brick (B) = 7.5 cm
»» Height of brick (H) = 2.5 cm
~ Dimensions of wall ::
»» Length of wall (L) = 30 m = 3000 cm
»» Breadth of wall (B) = 0.25 m = 25 cm
»» Height of wall (H) = 2 m = 200 cm
~ Finding volume of brick ::
~ Finding volume of wall ::
~ Finding number of bricks ::
~ Finding cost of bricks ::
⠀⠀⠀⠀⠀⠀⠀☯ Question ②
Given,
~ Dimensions of well ::
»» Depth (height) of well (h) = 25 m
»» Diameter of well (d) = 5.6 m
»» Radius of well (r) = 2.8 m
~ Dimensions of plot ::
»» Length of plot (L) = 22 m
»» Breadth of plot (B) = 16 m
~ Finding volume of earth taken out ::
Now, we know that earth taken out is spread on rectangular plot to make a platform. Therefore,
__________________________
★ More to know :-
↠ TSA of cube = 6a²
↠ CSA of cube = 4a²
↠ Volume of cube = a³
↠ TSA of cuboid = 2(lb + bh + hl)
↠ CSA of cuboid = 2(l + b)h
↠ Volume of cuboid = l × b × h
↠ TSA of cylinder = 2πr(r + h)
↠ CSA of cylinder = 2πrh
↠ Volume of cylinder = πr²h
↠ TSA of cone = πr(l + r)
↠ CSA of cone = πrl
↠ Volume of cone = 1/3πr²h
↠ SA of sphere = 4πr²
↠ Volume of sphere = 4/3πr³
↠ TSA of hemisphere = 3πr²
↠ CSA of hemisphere = 2πr²
↠ Volume of hemisphere = 2/3πr³
Answer:
We use possessive adjectives to express who owns (or 'possesses') something. A possessive adjective is used in front of a noun (a thing). For example: My computer.