Question

The width of a field is 5m less than its length. If the area is 204 m^2, find the dimmensions of the field

Answer 1

Answer:-

$\red{\bigstar}$ Dimensions of the field are $\large\leadsto\boxed{\tt\green{17 \times 12 \: m^2}}$

• Given:-

• Width of field is 5m less than its length.

• Area of the field is 204 m².

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• To Find:-

• Dimensions of the field

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• Solution:-

Let the length of the field be 'x'.

Now, given that the width is 5m less than the length.

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Therefore,

The width of the field will be 'x-5'

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★ Figure:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(2,3.5){\sf\large x m}\put(-1.4,1.4){\sf\large x-5 m}\put(2,1.4){\large\bf 204 m^2 }\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

We know,

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{Area = Length \times Width}}}$

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➪ $\sf 204 = (x) \times (x-5)$

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➪ $\sf 204 = x^2 - 5x$

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➪ $\sf x^2 - 5x - 204 = 0$

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• Splitting the middle term:-

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➪ $\sf x^2 - 17x + 12x - 204 = 0$

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➪ $\sf x(x - 17) + 12(x - 17) = 0$

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➪ $\sf (x+12) (x-17) = 0$

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Now,

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✴ $\sf x + 12 = 0$

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➪ $\bf x = -12$

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✴ $\sf x - 17 = 0$

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➪ $\bf x = 17$

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✯ Length cannot be negative. Hence,

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★ $\large{\bf\pink{x = 17}}$

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Now,

Width will be x - 5

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➪ $\sf 17 - 5$

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★ $\large{\bf\pink{x = 12}}$

Hence,

• Length = 17m

• Width = 12m⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the dimensions of the the field are 17 × 12 m².

Answer 2

Answer:

$\large{\red{\bold{\underline{✫\: Given :-}}}}$

↦ The width of a field is 5 m less than its length.

↦ Area is 204 m²

$\large{\red{\bold{\underline{✫\: Find\: Out :-}}}}$

↦ What is the dimension of the field.

$\large{\red{\bold{\underline{✫\: Formula\: Required :-}}}}$

$\large{\boxed{\underline{\underline{\bf{✮\: A\: =\: L\: \times W\: ✮}}}}}$

【 where, A = Area, L = Length, W = Width 】

$\large{\red{\bold{\underline{✫\: Solution :-}}}}$

Let, the length be x m and the width will be (x - 5) m

According to the question, by substitute the values we get,

➳ 204 = x(x - 5)

➳ 204 = x² - 5x

➳ x² - 5x - 204 = 0

By doing middle term we get,

➳ x² - (17 - 12)x - 204 = 0

➳ x² - 17x + 12x - 204 = 0

➳ x(x - 17) + 12(x - 17) = 0

➳ x - 17 = 0 ; x + 12 = 0

➤ x = 17 ; x = - 12

We can't take length as negative so we take x = 17 as length.

Now, we have to find the width,

↬ (x - 5)

↬ (17 - 5)

➙ 12 m

Hence, width will be 12 m

Then, we get length = 17 m and breadth = 12 m

Hence, the dimension will be,

↪ Length × Width

↪ 17 m × 12 m

➠ 17 × 12 m²

∴ The dimensions of the field is 17 × 12 m² .