Question

The perimeter of the rectangle is 80 cm. The area of the rectangle is acm2 . (i) show that x 2 40x + a = 0.

Answer 1

GIVEN :

The perimeter of the rectangle is 80 cm

The area of the rectangle is $a cm^2$

TO SHOW :

i) $x^2- 40x + a = 0$

SOLUTION :

From the given we have that the perimeter of the rectangle is 80 cm.

Let x be the length of the rectangle

Let y be the width of the rectangle

The formula for perimeter of rectangle is :

$Perimeter=2(l+w)$ units

∴ Perimeter=80=2(x+y)

80=2(x+y)

Rewritting the above equation we get

2(x+y)=80

$x+y=\frac{80}{2}$

∴ x+y=40

⇒ y=40-x

The area of the rectangle is $a cm^2$

The formula for area of rectangle is :

Area=lw square units

Area=a=xy square units

a=x(40-x) (∵ y=40-x )

a=x(40)+x(-x)

$a=40x-x^2$

Rewritting the above equation

$-x^2+40x=a$

$-x^2+40x-a=0$

$x^2-40x+a=0$

Hence $x^2-40x+a=0$ is showed.

Answer 2

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