Question

Edit: I figured it out, it's 14+7(sqrt sign) 2

Answer 1

Answer:

$Perimeter = 14 + 7\sqrt{2}$

Step-by-step explanation:

Given:

Area of the square = 49 in²

Required

Determine the perimeter of the one of the congruent triangles

First, we'll determine the length of the square;

$Area = Length * Length$

Substitute 49 for Area

$49 = Length * Length$

$49 = Length^2$

Take Square root of both sides

$7 = Length$

$Length = 7$

When the square is divided into two equal triangles through the diameter;

2 sides of the square remains and the diagonal of the square forms the hypotenuse of the triangle;

Calculating the diagonal, we have;

$Hypotenuse^2 = Length^2 + Length^2$ -- Pythagoras Theorem

$Hypotenuse^2 = 7^2 + 7^2$

$Hypotenuse^2 = 2(7^2)$

Take square root of both sides

$Hypotenuse = \sqrt{2} * \sqrt{7^2}$

$Hypotenuse = \sqrt{2} * 7$

$Hypotenuse = 7\sqrt{2}$

The perimeter of one of the triangles is the sum of the 2 Lengths and the Hypotenuse

$Perimeter = Length + Length + Hypotenuse$

$Perimeter = 7 + 7 + 7\sqrt{2}$

$Perimeter = 14 + 7\sqrt{2}$