Question

Area of a redangular plot is 528 m² . The lenghts of plot is one more than twice of its breadth We need to find length and breadth of plot .​

Answer 1

Given:

• Area of a rectangular plot is 528 m².
• The lenght of plot is one more than twice of its breadth

To Find:

• Length and breadth

Solution:

• Let the Breadth be x m
• and Length be 2x + 1 m

We know that,

• Area of Rectangle = L × B

Where,

• L = Length
• B = Breadth

➠ 528 = x × (2x + 1)

➠ 528 = 2x² + x

➠ 2x² + x - 528 = 0

Now,

➠ 2x² + x - 528 = 0

➠ 2x² - 32x + 33x - 528 = 0

➠ x (2x + 33) - 16 (2x + 33) = 0

➠ (x - 16)(2x + 33) = 0

➠ x - 16 = 0, 2x + 33 =0

➠ x = 16 and x = -33/2

Length and breadth of the rectangle can't negative.

• Breadth of Rectangle is 16 m

• Length of Rectangle = 2x + 1 = ( 2 × 16 + 1) = 33 m

Answer 2

Given: The Length of a rectangular plot is one more than twice it's Breadth. & Area of the plot is 528 m².

Need to find: The Length & Breadth of plot?

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❍ Let's say, that the Breadth of plot be x m. Then, Length of the plot would be (2x + 1) m respectively.

⠀

» We'll use Formula of Area of a rectangle to find out the required dimensions. To Calculate the Area of rectangle formula is Given by :

⠀

$\quad\star\:\underline{\boxed{\pmb{\frak{Area_{\:(rectangle)} = Length \times Breadth}}}}$

⠀$\underline{\bf{\dag} \:\mathfrak{Substituting\: Values\;in\; above\;formula\: :}}$⠀⠀⠀⠀

$:\implies\sf x \times \Big(2x + 1\Big) = 528\\\\\\:\implies\sf 2x^2 + x = 528\\\\\\:\implies\sf 2x^2 + x - 528 = 0\\\\\\:\implies\sf 2x^2 - 32x + 33x - 528 = 0\\\\\\:\implies\sf 2x\Big(x - 16\Big) + 33\Big(x - 16\Big) = 0\\\\\\:\implies\sf \Big(x - 16\Big) \Big(2x + 33\Big) = 0\\\\\\:\implies\underline{\boxed{\pmb{\frak{\purple{x = 16\;or\;x = - \dfrac{33}{2}}}}}}\;\bigstar$

⠀

• Ignoring –ve value of x because we know that side can't be negative, Taking x = 16.

⠀

Therefore,

⠀

• Breadth of plot, x = 16 m.
• Length of plot, (2x + 1) = [2(16) + 1] = (32 + 1) = 33 m.

⠀

$\therefore{\underline{\textsf{Hence, Length and Breadth of plot are \textbf{16 m} \sf{\&} \textbf{33 m} respectively}.}}$

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