Question

If the length of a rectangle is a two-digit number with identical digits and the width is 1/10 the length and the perimeter is 2 times the area of the rectangle, what is the the length and the width

Answer 1

If the length of a rectangle is a two-digit number with identical digits and the width is 1/10 the length and the perimeter is 2 times the area of the rectangle, what is the the length and the width

Solution:

Let the length of rectangle=x

Width of rectangle=x/10

Perimeter is 2(Length+Width)

= 2(x+x/10)

Area of Rectangle= Length* Width=x*x/10

As, Perimeter=2(Area)

So,2(x+x/10)=2(x*x/10)

Multiplying the equation with 10, we get,

2(10x+x)=2x²

Adding Like terms, 10x+x=11x

2(11x)=2x^2

22x=2x²

2x²-22x=0

2x(x-11)=0

By Zero Product property, either x=0

or, x-11=0

or, x=11

So, Width=x/10=11/10=1.1

Checking:

So, Perimeter=2(Length +Width)=2(11+1.1)=2*(12.1)=24.2

Area=Length*Width=11*1.1=12.1

Hence, Perimeter= 2 Area

As,24.2=2*12.1=24.2

So, Perimeter=2 Area

So, Answer:Length of Rectangle=11 units

Width of Rectangle=1.1 units

Answer 2

The length is 11 and the width is 1.1