Question

4. The length of a rectangle exceeds the width by 2 cm. If the diagonal is 10 cm long, find the width of the rectangle.​

Answer 1

Answer:

Let the breadth be x

so the length will be x+8

area of the rectangle=240cm²

∵area of rectangle=length*breadth

∴240=(x)(x+8)

240=x²+8x

x²+8x-240=0

x²+20x-12x-240=0

x(x+20)-12(x+20)=0

(x+20)(x-12)=0

so x=-20 or x=12

∵The dimensions cannot be negative

∴x=12

length=20cm

breadth=12cm

Answer 2

$\huge\underline{Given:-}$

➡.The length of a rectangle exceeds the width by 2 cm.

➡ the diagonal is 10 cm

$\huge\underline{To find:-}$

➡ width of the rectangle. = ?

$\huge\underline{SoLution:-}$

Let the length of the rectangle be x cm,

According to question :-

⭐ if length of a rectangle exceeds the width by 2 cm, the width be x + 2 cm

$\implies$ 10 = x² + (x + 2)²

$\implies$ 10 = x² + x² + 4x + 4

$\implies$ 10 = 2x² + 4x + 4

$\implies$ 2x² + 4x -6 = 0

Divide by '2' in above equation

$\implies$ x² + 2x - 3 = 0

$\implies$ x² +(3-1)x - 3 = 0

$\implies$ x² + 3x - x - 3 =0

$\implies$ x(x+3)-1(x+3)

$\implies$ (x-1)(x+3)

$\implies$ x = 1 & x = -3

⭐ Length can't be negative

so x = 1

Length of the rectangle = 1cm

and breadth be 1+3=4 cm

$\huge\underline{ThAnKs....}$