Question

HIUJIU
EXAMPLE 2
The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more
than twice its breadth. Formulate the quadratic equation to determine the length and breadth of the
[NCERT]
TON Let the breadth of the plot be x metres.
plot.
Can U Solve This For Me please​

Answer 1

Answer

Area =528m²

Let Breadth be b

Then length l=2b+1

So (b)(2b+1)=528

⇒2b² +b−528=0

⇒2b² −32b+33b−528 =0

⇒2b(b−16)+33(b−16) =0

⇒(2b+33)(b−16) =0

So

b=16

l=33

hope it helps

Answer 2

Appropriate Question :

The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. Formulate the quadratic equation to determine the length and breadth of the rectangular plot.

Given :

• Area of the rectangular plot = 528 m²
• Length of plot is one more than twice its breadth.

To find :

• Length of the rectangle
• Breadth of the rectangle

Concept :

To find the dimensions of the rectangular plot.. Firstly, we will assume the breadth of the rectangular plot as any variable, let it be x and the length of the rectangular plot as one more than twice its breadth. That is, 2x + 1. Then by using the formula of area of rectangle calculate the value of x. Substitute the value of x in the dimensions of the rectangular plot, the resultant value will be our required answer.

➳ Formula of Area of rectangle :-

$\boxed{ \pmb{ \rm{Area \: of \: rectangle = l \times b}}}$

where,

• L stands for length of the rectangle
• B stands for breadth of the rectangle

Solution :

Let,

• Breadth of the rectangular plot = x meter
• Length of the rectangular plot = 2x + 1 meter

Using formula,

★ Area of rectangle = l × b

Substituting the given values,

$\\ \twoheadrightarrow \quad\sf Area \: of \: rectangle = l \times b$

$\\ \twoheadrightarrow \quad\sf 528 = 2x + 1 \times x$

$\\ \twoheadrightarrow \quad\sf 528 = 2x^{2} + 1x$

$\\ \twoheadrightarrow \quad\sf 0 = 2x^{2} + 1x - 528$

$\\ \twoheadrightarrow \quad\sf 0 = 2x^{2} + 33x - 32x - 528$

$\\ \twoheadrightarrow \quad\sf 0 = x(2x + 33)- 16(2x - 33)$

$\\ \twoheadrightarrow \quad\sf 0 = x(2x + 33)- 16(2x + 33)$

$\\ \twoheadrightarrow \quad\sf 0 = (x - 16)(2x + 33)$

$\\ \twoheadrightarrow \quad\sf 0 = x - 16$

$\\ \twoheadrightarrow \quad\sf 16 = x$

$\\ \twoheadrightarrow \quad\sf 0 = 2x + 33$

$\\ \twoheadrightarrow \quad\sf - 33= 2x$

$\\ \twoheadrightarrow \quad\sf \dfrac{- 33}{2}= x$

$\\ \to \quad\tt The \: value \: of \: x = 16 \: or \: \: \dfrac{-33}{2}$

As we know that the dimension of any rectangular field cannot be in negative. So, -33/2 will be rejected.

Therefore, the value of x = 16.

Substitute the value of x in the dimensions of the rectangular plot :-

DIMENSIONS OF THE RECTANGULAR PLOT :-

→ Length of the rectangular = 2x + 1 = 2(16) + 1 = 32 + 1 = 33 m

→ Breadth of the rectangular plot = x = 16 m

Therefore,

• Length and breadth of the rectangular plot are 33 m and 16 m respectively.

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VERIFICATION :-

To verify the dimensions of the rectangular plot find the area of the plot. If the area will be equal to 528 m² as mentioned in the question, then the values are right.

→ Area of rectangular plot = l × b

→ Area = 33 × 16

→ Area = 528

Area of the rectangular plot = 528 m².

HENCE, VERIFIED.