the diagram shows two paths drawn inside a rectangular feild 50 m long and 35 m wide . The width of each path is 5 meters. Find the area of shaded portion
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295
Solution:-
Area of the shaded portion :
There are two rectangles and a square in this shaded portion. I have enclosed a diagram so that you can understand easily.
Length and breadth of rectangle ABCD are 50 m and 5 m respectively.
Area of rectangle ABCD = L × B
= 50 × 5
= 250 sq m
Length and breadth of the rectangle EFGH are 35 m and 5 m respectively.
Area of rectangle EFGH = 35 × 5
= 175 sq m
Length of each side of the square IJKL is 5 m.
Area of square = 5 × 5
= 25 sq m
Now, Area of the shaded portion = (Area of rectangle ABCD + Area of rectangle EFGH) - Area of square IJKL
= (250 sq m + 175 sq m) - 25 sq m
= 425 sq m - 25 sq m
= 400 sq m
So, the area of the shaded portion is 400 sq m.
Answer.
Area of the shaded portion :
There are two rectangles and a square in this shaded portion. I have enclosed a diagram so that you can understand easily.
Length and breadth of rectangle ABCD are 50 m and 5 m respectively.
Area of rectangle ABCD = L × B
= 50 × 5
= 250 sq m
Length and breadth of the rectangle EFGH are 35 m and 5 m respectively.
Area of rectangle EFGH = 35 × 5
= 175 sq m
Length of each side of the square IJKL is 5 m.
Area of square = 5 × 5
= 25 sq m
Now, Area of the shaded portion = (Area of rectangle ABCD + Area of rectangle EFGH) - Area of square IJKL
= (250 sq m + 175 sq m) - 25 sq m
= 425 sq m - 25 sq m
= 400 sq m
So, the area of the shaded portion is 400 sq m.
Answer.
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Answer:
Step-by-step explanation:let the area of rectangle ABCD=35*5=175+
Area of rectangle EFGH=50*5=250 &- AREA of square MNOP=5*5=25=400cm2
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