Question

ABCD is a square in which AB =3X +5 and BC =4x-7.Find the value of x and the perimeter of the square

Answer 1

$\large\underline{\sf{Given- }}$

• ABCD is a square.

and

• AB = 3x + 5

• BC = 4x - 7

$\large\underline{\sf{To\:Find - }}$

• Value of x

• Perimeter of square

$\large\underline{\sf{Solution-}}$

Given that,

• ↝ ABCD is a square

and

• ↝ AB = 3x + 5

and

• ↝ BC = 4x - 7

We know that,

• ↝ In square, all the sides are equal.

$\rm :\implies\:AB = BC$

$\rm :\longmapsto\:3x + 5 = 4x - 7$

$\rm :\longmapsto\:4x - 3x = 7 + 5$

$\rm :\implies\: \boxed{ \bf \: x \: = \: 12}$

So,

↝ Side of square,

• AB = 3x + 5 = 3 × 12 + 5 = 36 + 5 = 41 units

Now,

We know that,

$\boxed{ \bf \: Perimeter_{(square)} \: = \: 4 \: \times \: side}$

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times 41$

$\rm :\longmapsto\:Perimeter_{(square)} = 164 \: units$

Additional Information :-

Properties of a Square

The properties of a square are listed below:

• All four interior angles are equal to 90°

• All four sides of the square are equal.

• The opposite sides of the square are parallel to each other

• The diagonals of the square bisect each other at 90°

• The diagonals of the square are equal.

• The square has 4 vertices and 4 sides

• The diagonal of the square divide it into two congruent triangles.

• The length of diagonal is greater than the side of the square.

$\underline{ \boxed{ \bf \: Area_{(square)} = {(side)}^{2} = \dfrac{1}{2} {(diagonal)}^{2}}}$

$\underline{ \boxed{ \bf \: diagonal_{(square)} = \sqrt{2} \times side_{(square)}}}$