Question

a rectangle is 4 times as long as it is wide. if the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches . what were the dimensions of the original rectangle?

Answer 1

Given :

• A triangle is 4 times as long as it's wide.
• If the length is increased by 4 inch and the width is decreased by 1 inch, the area will be 60 square inches.

To find :

• Dimensions of original rectangle.

Solution :

Consider,

• Original length of the rectangle = x inches
• Original width of the rectangle = y inches.

According to the 1st condition :-

• A triangle is 4 times as long as it's wide.

$\implies\sf{x=4y..................(1)}$

According to the 2nd condition :-

• If the length is increased by 4 inch and the width is decreased by 1 inch, the area will be 60 square inches.

Therefore ,

Length = (x+4) inches

Width = (y-1) inches

Formula Used :-

${\boxed{\bold{Area\:of\: rectangle=Length\times\: Width}}}$

$\implies\sf{(x+4)(y-1)=60}$

$\implies\sf{(4y+4)(y-1)=60\:[Put\:x=4y\: from\:eq(1)]}$

$\implies\sf{4y^2-4y+4y-4=60}$

$\implies\sf{4y^2-4=60}$

$\implies\sf{4(y^2-1)=60}$

$\implies\sf{y^2-1=15}$

$\implies\sf{y^2=15+1}$

$\implies\sf{y=\sqrt{16}}$

$\implies\sf{y=4}$

• Width = 4 inches

Now put y = 4 in eq(1) for getting the value of x.

$\implies\sf{x=4y}$

$\implies\sf{x=4\times\:4}$

$\implies\sf{x=16}$

Therefore, the length of the rectangle is 16 inches and the width of the rectangle is 4 inches.

Answer 2

$\bigstar$ Given :

• A rectangle is 4 times as long as it is wide

• If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches

$\bigstar$ To Find :

• The dimensions of the original rectangle

$\bigstar$ Solution :

• Let the width of the rectangle be x
• So the length of the rectangle = 4 x

A/c , "  If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches "

⇒ Length = 4 x + 4 inches

⇒ Width = x - 1 inches

So ,

Since , Length * Width = Area of Rectangle

⇒ ( 4 x + 4 ) ( x - 1 ) = 60 inch²

⇒ 4 ( x + 1 ) ( x - 1 ) = 60

⇒ x² - 1 = 15

⇒ x² = 16

⇒ x = ± 4

But lengths can't be negative . So , x = 4

⇒  $\bold{\bf{\red{Width = x = 4 \;inches}}}$

⇒  $\bold{\bf{\blue{Length = 4 x = 4 * 4 = 16 \;inches }}}$