the lenght of a rectangle is twice its width.If the width is reduced by 1cm and the lenght is also reduced by 2cm,the area will be 15cm(square).Find the dimensions of the original rectangle.
SIMPLIFY
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To find the dimensions of the original rectangle, we can start by setting up equations based on the given information.
Let's denote the width of the original rectangle as "w" cm and the length as "L" cm.
We are given that the length of the rectangle is twice its width, so we have L = 2w.
We are also given that if the width is reduced by 1 cm and the length is reduced by 2 cm, the area will be 15 cm^2.
The area of a rectangle is given by the formula: A = length × width. Using the original dimensions, the area is L × w.
So, we have the equation L × w = 15.
Now, let's substitute the value of L from the first equation into the second equation:
(2w) × w = 15.
Simplifying this equation, we get: 2w^2 = 15.
Next, we can rearrange it to isolate "w":
w^2 = 15/2.
w^2 = 7.5.
Taking the square root of both sides, we have:
w = √7.5.
w ≈ 2.74 cm (rounded to two decimal places).
Now that we have the value of "w," we can substitute it back into the first equation to find the length, L:
L = 2w.
L = 2 × 2.74.
L ≈ 5.48 cm (rounded to two decimal places).
Therefore, the dimensions of the original rectangle are approximately 5.48 cm (length) and 2.74 cm (width).