Question
Given the area of rectangle is A=25a2−35a+69. The length is given as (5a−3).Find the width of the rectangle.
Answer: Option B
:
B
Let ybe the width.
Given: Area=25a2−35a+69
We know that, area of a rectangle= Length×breadth
Hence,y×(5a−3)=25a2−35a+69
y=(25a2−35a+69)÷(5a−3)
By long division method,
5a−35a−23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a2−35a+6925a2−12a−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−23a+69−23a+69+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0
∴y=5a−23Width of the rectangle =5a−23
Was this answer helpful ?
:
B
Let ybe the width.
Given: Area=25a2−35a+69
We know that, area of a rectangle= Length×breadth
Hence,y×(5a−3)=25a2−35a+69
y=(25a2−35a+69)÷(5a−3)
By long division method,
5a−35a−23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a2−35a+6925a2−12a−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−23a+69−23a+69+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0
∴y=5a−23Width of the rectangle =5a−23
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution