Question

Edit: I figured it out, it's 14+7(sqrt sign) 2
A square piece of paper is folded once so that one pair of opposite corners coincide. When the paper is unfolded, two congruent triangles have been formed. Given that the area of the original square is $49$ square inches, what is the number of inches in the perimeter of one of these triangles? Express your answer in simplest radical form.

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

$Perimeter = Length + Length + Hypotenuse$

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

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