The outer length and breadth of a pointing is $150 c m \times 120 c m$ . If there is a margin of $8 c m$ width and around the painting. Find the area of the actual painting (without margin).

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Solution

The dimensions of the rectangular painting ,

Length $= L = 150 c m$

Breadth $= B = 120 c m$

According to the problem,

The width of the marigin around the painting $= w = 8 c m$

Now , the diminsions of the actual painting are

Length of the actual painting $= l = L - 2 w$

$l = 150 - 2 \times 8$

$l = 150 - 16$

$l = 134 c m$

Breadth of the actual painting $= b = B - 2 w$

$b = 120 - 2 \times 8$

$b = 104 c m$

Therefore,

Area of the actual painting $= l \times b$

$= 134 c m \times 104 c m$

$= 13936 s q u a r e c m$

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The outer length and breadth of a pointing is 150 cm×120 cm. If there is a margin of 8 cm width and around the painting. Find the area of the actual painting (without margin).

The outer length and breadth of a pointing is $150 c m \times 120 c m$ . If there is a margin of $8 c m$ width and around the painting. Find the area of the actual painting (without margin).