0 votes

$\text{ABCD}$ is a square of area $4$ with diagonals $\text{AC}$ and $\text{BD}$, dividing square into $4$ congruent triangles. Figure looks like four non-over lapping triangles. Then the sum of the perimeters of the triangles is

- $8\left ( 2+\sqrt{2} \right )$
- $8\left ( 1+\sqrt{2} \right )$
- $4\left ( 1+\sqrt{2} \right )$
- $4\left ( 2+\sqrt{2} \right )$

0 votes

We can draw the diagram.

DIAGRAM

The triangle $\triangle \text{ABC}$ is a right angle triangle, so we can apply the Pythagorean theorem.

$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$

$\Rightarrow \text{(AC)}^{2} = \text{(AB)}^{2} + \text{(BC)}^{2}$

$\Rightarrow \text{(AC)}^{2} = \text{(2)}^{2} + \text{(2)}^{2}$

$\Rightarrow \text{(AC)}^{2} = 4 +4$

$\Rightarrow \text{(AC)}^{2} = 8$

$\Rightarrow \boxed{\text{AC} = \sqrt{8}}$

Similarly $\boxed{\text{BD} = \sqrt{8}}$

$\therefore$ The sum of the perimeters of the triangles = sum of all sides of square + $2 \times $ sum of diagonal lengths

$= 4 \times 2 + 2 \times \sqrt{8}$

$= 8 + 2 \times \sqrt{4 \times 2} = 8 + 4 \sqrt{2} = 4(2 + \sqrt{2}).$

Correct Answer $: \text{D}$

DIAGRAM

The triangle $\triangle \text{ABC}$ is a right angle triangle, so we can apply the Pythagorean theorem.

$\text{(Hypotenuse)}^{2} = \text{(Perpendicular)}^{2} + \text{(Base)}^{2}$

$\Rightarrow \text{(AC)}^{2} = \text{(AB)}^{2} + \text{(BC)}^{2}$

$\Rightarrow \text{(AC)}^{2} = \text{(2)}^{2} + \text{(2)}^{2}$

$\Rightarrow \text{(AC)}^{2} = 4 +4$

$\Rightarrow \text{(AC)}^{2} = 8$

$\Rightarrow \boxed{\text{AC} = \sqrt{8}}$

Similarly $\boxed{\text{BD} = \sqrt{8}}$

$\therefore$ The sum of the perimeters of the triangles = sum of all sides of square + $2 \times $ sum of diagonal lengths

$= 4 \times 2 + 2 \times \sqrt{8}$

$= 8 + 2 \times \sqrt{4 \times 2} = 8 + 4 \sqrt{2} = 4(2 + \sqrt{2}).$

Correct Answer $: \text{D}$