What is the area of a square (nearest whole number) with a diagonal one unit longer than the length of a side?

Question

The Answer Iz Not 1!!!!

Answer 1

let side length be x
diagonal = x + 1
(x + 1)² = x² + x²
x² + 2x + 1 = 2.x²
x² - 2x - 1 = 0
(x - 1).(x - 1) = 0
x = 1
Area = x² = 1 unit²

Answer 2

By Pythagoras, if one side of the square is x, and the diagonal is 1, we obtain a right angle triangle and can use Pythagoras' theorem.

x² + x² = (x + 1)²

2x² = (x + 1)²

x² = (1/2)(x + 1)²

Take the square root of both sides (we only take the positive solution since x cannot be negative),

x = (1/√(2)) (x + 1)

Solving for x,

x√(2) = x + 1
x√(2) - x = 1
x(√(2) - 1) = 1

x = 1/(√(2) - 1)

But we want to rationalize the denominator.

x = (√(2) + 1) / (2 - 1)
x = √(2) + 1

The area would be
A = x² = [√(2) + 1]² = 2 + 2√(2) + 1
A = 3 + 2√(2)

I can leave the rest for you to round.

Answer 3

Let the length of a side be x units; length of diagonal is therefore (x + 1) units.
Angle between one side and diagonal = 45 deg;
Sin 45 = x/ (x + 1); or 0.7071 = x/ (x + 1)
Solving for x gives x = 2.414
Area of square = x^2 = 5.82 = (approx whole number) = 6

Answer 4

Let the diagonal = x+1
So , the length is = x
With Pythagoras theorem we know that
d^2=l^2+l^2
(x+1)^2=x^2+x^2
(x+1)^2=2(x^2)
Then , we square root the equation and we"ll find that
x+1=x√2
x√2-x=1
x( √2 - 1) =1
x=1/( √2- 1)
=1/(1,4-1)
=1/(0,4)
=2,5
So the length is 2,5 length unit

Then , we can find the area
A=l^2
=x^2
=2,5^2
=6,25

The area is 6,25 square unit

Answer 5

in a square apply pythogarous theorom,

let x is length of side...
then,
(x+1)^2=x^2+x^2

x^2+1+2x=2x^2

it will b x^2-2x-1=0

x=2+sqrt8/2 [length canot b negative]

so...x=1+sqrt2

area of square=x^2

=(1+sqrt2)^2
=1+2+2sqrt2
=3+2*1.414
=5.828=6

units:cm^2 or m^2

Answer 6

Let the side of the square be a
Diagonal (b)= sqrt (2a^2)
b=a+1
Sqrt(2a^2) = a+1
squaring both sides
2a^2 = a^2 +1+2a
a^2-2a-1 =0
a=[2(+/-)sqrt(4+4)]/2
a=1(+/-)sqrt2
Area =(1-sqrt2)(1+sqrt2)
1-2
-1 say 1