Question

find the area of rectangle whose length is 4 x + 3 units and breadth is 2 x minus 3 units​

Answer 1

✬ Area = 8x² – 6x – 9 ✬

Step-by-step explanation:

Given:

• Measure of length of rectangle is (4x + 3).
• Measure of breadth of rectangle is (2x – 3).

To Find:

• What will the area of rectangle ?

Solution: As we know that area of rectangle is given by

★ Ar. of Rectangle = Length • Breadth ★

$\implies{\rm }$ Area = (4x + 3)(2x – 3)

$\implies{\rm }$ 4x(2x – 3) + 3(2x – 3)

$\implies{\rm }$ 8x² – 12x + 6x – 9

$\implies{\rm }$ 8x² – 6x – 9

Hence, area of given rectangle will be 8x² – 6x – 9.

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• Perimeter of rectangle = 2(Length + Breadth)

• Opposite sides of rectangle are equal and also parallel to each other.

• Each angle of rectangle is of 90°.

Answer 2

✴ $\mathcal{\huge{\underline{\underline{\pink{Answer:-}}}}}$

➡The area of rectangle is 0 units.

⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐⭐

$\mathcal{\LARGE{\fbox{\red{Given:-}}}}$

• Length of rectangle = (4x + 3)......(1)

• Breadth of rectangle = (2x – 3)......(2)

$\mathcal{\LARGE{\fbox{\green{To\:Find:-}}}}$

▶ The Area of Rectange.

$\mathcal{\LARGE{\fbox{\purple{Solution:-}}}}$

✰$\mathcal{\fbox{\blue{Ar.\:of\:Rectangle\:=\:Length\:×\:Breadth }}}$

⟿ Area of Δ= (4x + 3)(2x – 3)

⟿ 4x(2x – 3) + 3(2x – 3)

⟿ 8x² – 12x + 6x – 9

⟿ ( 8x² – 6x – 9 )

Using, Factorisation of middle term.

( 8x² – 6x – 9 )

➙ 8x² – 12x + 6x – 9

➙ 4x(2x – 3) + 3(2x – 3)

Taking (2x – 3) common & 4x with +3 ,

➙ (4x + 3) (2x – 3)

➙ (4x + 3 = 0) ( 2x – 3 = 0)

➙ x = –3/4 & x = 3/2

Taking, the + ve value of x. So, x = 3/2

From 1 & 2 finding, length & breadth of the rectangle.

• Length of rectangle = (4x + 3) = (4×3/2+3)= (2×3+3)= (6 + 3) = 9 units

• Breadth of rectangle = (2x – 3) = (2×3/2-3)= ( 3-3 ) = 0 units

➣ So, the area of Δ is l × b = 9 × 0 = 0 units .

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