Question

The diagonals of a square with area 9 metre square divide the square into four nonoverlapping triangles what is the sum of the perimeter of the four triangles...pls don't post irrelevant answers..​

Answer 1

Area of square = 9m²

Area of a square = (side)²

9 = (side)²

side =  √9

side = 3m

The sides of the square = 3m

Diagonal =  √2 × a

= 4.24m

Perimeter of the one triangle = a + b + c

= 3 + 2.12 + 2.12

=7.24m

Perimeter of 4 triangles = 4×7.24

=28.9m

Answer 2

Given :

• The diagonals of a square with area 9 m² divide the square into four non overlapping triangles.

To find :

• Sum of the perimeter of the four triangles

Solution :

Area of a square is given by,

$\longmapsto\bold{Area \ of \ square = Side^2 } \\\\ \\ \longmapsto\sf 9 = Side^2 \\\\ \\ \longmapsto\displaystyle\sf Side = \sqrt{9} \\\\ \\ \longmapsto\bold{Side = 3 \ m}$

Now diagonal of a square is given by,

$\longmapsto\bold{Diagonal \ of \ square = \sqrt{2} \times Side} \\\\ \\ \longmapsto\sf Diagonal \ of \ square = \sqrt{2} \times 3 \\\\ \\ \longmapsto\bold{Diagonal \ of \ square = 4.24 \ m}$

Now from diagram we can see that the diagonals divide the triangle into 4 triangles which have 1 side of square and half the one diagonal and half the other diagonal as sides of Δ

Therefore,

$\longmapsto\sf Perimeter \ of \ \triangle = 3 + \dfrac{4.24}{2} + \dfrac{4.24}{2} \\\\ \\ \longmapsto\sf Perimeter \ of \ \triangle = 3 + 2.12 + 2.12 \\\\ \\ \longmapsto\bold{Perimeter \ of \ \triangle = 7.24 \ m}$

Now sum of perimeter of 4 triangles,

$\longmapsto\sf Perimeter \ of \ 4 \ triangles = (4 \times 7.24) \ m \\\\ \\ \longmapsto\bold{Perimeter \ of \ 4 \ triangles = 28.96 \ m \qquad [Required \ answer]}$

$\therefore\underline{\underline{\bold{Sum \ of \ perimeter \ of \ 4 \ triangles = 28.96 \ m}}}$