Question

ABCD is a square of area 4 with diagonals AC and BD, dividing square into 4 congruent triangles.
Figure looks like four non-over lapping triangles. Then the sum of the perimeters of the triangles is
a. 8(2 + √2)
b. 8(1+ √2)
c. 4(1 + √2)
d. 4(2 + √2)

Answer 1

Total perimeter is four times the per of one triangle

Answer 2

Answer:

d. 4(2+√2)

Step-by-step explanation:

Given:

Area of square = 4

To find: Perimeter of all triangles

Solution

Area of square = side² = 4 => side = 2

Hence AB = BC = CD = DA =2

Let the diagonals of the square intersect at O.

The perimeter of the triangle = Sum of three sides of the triangle.

Perimeter of all four triangles = Perimeter of the square + Sum of lengths of diagonals

(Refer attachment)

Applying Pythagoras theorem, which states

hypotenuse² = perpendicular² + base²

to Δ ABC, we have

AC² = AB² + BC²

AC² = 2²+ 2² = 4 + 4 = 8

AC = 2√2

Similarly, BD =2√2

Now, perimeter of the 4 triangles = 4(2) + 2√2 + 2√2 = 8 + 4√2  = 4(2+√2)

∴ The sum of perimeters of the triangles is d. 4(2+√2)