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snow_tiger [21]
4 years ago
8

Do you length of a rectangle is five times the width. If the length and width are both increased by 2, the new area would be 85.

What is the original area?
Mathematics
1 answer:
puteri [66]4 years ago
8 0

Answer:

The original of the given rectangle = 45 sq units.

Step-by-step explanation:

Here, let us assume the actual width of the rectangle = a

So, the actual length of the rectangle  = 5 x ( width) = 5 (a)  = 5 a

Now, the new width w'  = ( a + 2)

and the new length l' =  ( 5 a + 2)

AREA OF THE RECTANGLE = LENGTH x WIDTH

So, the area of the new rectangle =  (a + 2)(5 a +2)

Also, new area  = 85      ⇒(a + 2)(5 a +2)  = 85

\implies 5a^2 + 10 a + 2a + 4 = 85\\\implies 5a^2 + 12a - 81 = 0\\\implies 5a^2 - 15 a + 27 a - 81 = 0\\\implies 5a(a -3) + 27(a -3) = 0\\\implies (5a +27)(a-3)= 0

⇒ ( 5a +27) =0 or (a-3) = 0

⇒ a =  -27/5 or a = 3

But, <u>a  = Width of a rectangle , so a CANNOT be Negative</u>

⇒ a ≠ -27/5 and a =  3

So, the actual width of the rectangle = a  =  3

The length of the rectangle = 5 a  = 5 (3) = 15

The original area = Original L x Original W = 3 x 15 = 45 sq units.

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