The area of rectangle whose length is five more than twice of breadth is 75sq.cm find the length
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ok quite difficult ask one more time
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Let the width of the rectangle = x units
Length = (2 x + 5) units
According to the question,
Area = x(2x + 5)
=> 75 = 2x2+5x2x2+5x
=> 2x2+5x−75=02x2+5x−75=0
=> 2x2+15x−10x−75=02x2+15x−10x−75=0
=> x(2x+15)-5(2x+15)=0
=> (2x+15)(x-5)=0
=> x = 5 and −152−152
Width cannot be negative.
Width = 5 units
Length=2x+5 =2×5+52×5+5=15 units
Perimeter of the rectangle
= 2(15 + 5) = 40 units
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