20
When the length of the rectangular plot is increased by four times its perimeter becomes 480 meters and area becomes 12800 sq.m. What is its original length(in meters)?
- A160
- B40
- C20
- DCannot be determined
Answer not set
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0
CCommunity
Solution
Let the length of the plot be $l$ meters and breadth = $b$ meters
New length = $4l$ meters
Perimeter = $2(4l+b)=480$
=> $4l+b=\frac{480}{2}=240$
=> $b=240-4l$ ---------(i)
Area = $(4l \times b)=12800$
=> $lb=\frac{12800}{4}=3200$
Substituting value of $b$ from equation (i)
=> $l(240-4l)=3200$
=> $240l-4l^2=3200$
=> $l^2-60l+800$
=> $l=\frac{-(-60) \pm \sqrt{(-60)^2-(4 \times 1 \times 800)}}{2}$
=> $l=\frac{60 \pm \sqrt{3600-3200}}{2} = \frac{60 \pm \sqrt{400}}{2}$
=> $l=\frac{60 \pm 20}{2}$
=> $l=\frac{60+20}{2},\frac{60-20}{2}$
=> $l=\frac{80}{2},\frac{40}{2}$
=> $l=40,20$ meters
=> Ans - (D)
Let the length of the plot be $l$ meters and breadth = $b$ meters
New length = $4l$ meters
Perimeter = $2(4l+b)=480$
=> $4l+b=\frac{480}{2}=240$
=> $b=240-4l$ ---------(i)
Area = $(4l \times b)=12800$
=> $lb=\frac{12800}{4}=3200$
Substituting value of $b$ from equation (i)
=> $l(240-4l)=3200$
=> $240l-4l^2=3200$
=> $l^2-60l+800$
=> $l=\frac{-(-60) \pm \sqrt{(-60)^2-(4 \times 1 \times 800)}}{2}$
=> $l=\frac{60 \pm \sqrt{3600-3200}}{2} = \frac{60 \pm \sqrt{400}}{2}$
=> $l=\frac{60 \pm 20}{2}$
=> $l=\frac{60+20}{2},\frac{60-20}{2}$
=> $l=\frac{80}{2},\frac{40}{2}$
=> $l=40,20$ meters
=> Ans - (D)
0
CCommunity
Ans - (D)