Question

if the side of square is 1/2 [x+1] units and diagonal is 3-x/ 2 units then the length of the side square would be

Answer 1

Step-by-step explanation:

The extremities of a diagonal of a square are given by (1,−2,3) and (2,−3,5).

The length of this diagonal will be given by

(1

2

+1

2

+2

2

)

=

6

Length of a side ×

2

= Length of diagonal

So, the length of the side of the square will be given by =

3

Answer 2

Question:

If the side of the square is 1/2 × (x + 1) units and the diagonal is (3 - x)/√2 units, then the length of the side of the square would be?

Answer:

$\bigstar{\bold{Side\:of\:the\:square=1\:unit}}$

Step-by-step explanation:

$\Large{\underline{\rm{Given:}}}$

• Side of the square = 1/2 × (x + 1) units

• Diagonal of the square = (3 - x)/√2 units

$\Large{\underline{\rm{To\:Find:}}}$

• The length of side of the square

$\Large{\underline{\rm{Solution:}}}$

➙ Here we are given the length and diagonal of the square in terms of x.

➙ Now we know that,

    $\boxed{\rm{Diagonal\:of\:a\:square=\sqrt{2}\:a}}$

    where a is a side of the square

➙ Hence substituting the value for side we get,

    $\sf{\dfrac{3-x}{\sqrt{2} } =\sqrt{2} \times \dfrac{1}{2} (x + 1)}$

➙ Simpifying,

    $\sf{\dfrac{3-x }{\sqrt{2} } =\sqrt{2}\times \dfrac{1}{\sqrt{2}\times \sqrt{2} }(x+1)}$

➙ Cancelling √2 on both numerator and denominator,

    $\sf{\dfrac{3-x }{\sqrt{2} } = \dfrac{1}{\sqrt{2} }(x+1)}$

➙ Now cancelling √2 on both sides of the equation,

     3 - x = 1 × (x + 1)

     3 - x = x + 1

     3 - 1 = x + x

     2x = 2

     x = 2/2

     x = 1

➙ Hence the value of x is 1 .

➙ Now,

    Side of the square = 1/2 × (x + 1)

    Substitute value of x

    Side of the square = 1/2 × (1 + 1)

    Side of the square = 1/2 × 2

    Side of the square = 1 unit

    $\boxed{\bold{Side\:of\:the\:square=1\:unit}}$

$\Large{\underline{\rm{Notes:}}}$

➙ The diagonal of a square = √2 a

➙ The perimeter of a square = 4a

➙ The area of a square = a²

    where a is a side of the square.