Question

eoc 1: the perimeter of the square below is 36. what is the length of the diagonal, x?
image of a square with diagonal x
a. 6
b. 9
c. 6√2
d. 9√2

eoc 3: what is the value of x for the triangle below?
image of a right triangle with a 45° angle, one leg 12, hypotenuse x
a. √2
b. 12√2
c. 12√3
d. 24

Explanation:

EOC 1

Step1: Find side length of square

Perimeter of square \( P = 4s \), where \( s \) is side length. Given \( P = 36 \), so \( 4s = 36 \), then \( s = \frac{36}{4} = 9 \).

Step2: Use Pythagorean theorem for diagonal

In square, diagonal \( x \) forms right triangle with two sides. By Pythagorean theorem, \( x^2 = s^2 + s^2 = 9^2 + 9^2 = 81 + 81 = 162 \). Then \( x = \sqrt{162} = \sqrt{81\times2} = 9\sqrt{2} \).

Step1: Identify triangle type

The triangle is a right - isosceles triangle (one angle \( 45^\circ \), one right angle), so the legs are equal. Let the leg length be \( l \), and we know one leg is 12, so the other leg is also 12.

Step2: Find hypotenuse \( x \)

For a right - isosceles triangle, hypotenuse \( x=\sqrt{l^{2}+l^{2}} \) (by Pythagorean theorem) or using the ratio for \( 45 - 45 - 90 \) triangle (\( \text{hypotenuse}=\text{leg}\times\sqrt{2} \)). Here \( l = 12 \), so \( x = 12\sqrt{2} \)? Wait, no, wait. Wait, the triangle: the right - angled side is 12, and the angle is \( 45^\circ \), so the other leg is also 12. Then hypotenuse \( x=\sqrt{12^{2}+12^{2}}=\sqrt{144 + 144}=\sqrt{288}=\sqrt{144\times2}=12\sqrt{2} \)? But wait, looking at the options, option B is \( 12\sqrt{2} \). Wait, maybe I misread the triangle. Wait, the triangle: the vertical side is 12, the angle at the bottom is \( 45^\circ \), and it's a right triangle. So in a \( 45 - 45 - 90 \) triangle, legs are equal, hypotenuse is leg\( \times\sqrt{2} \). If one leg is 12, hypotenuse \( x = 12\sqrt{2} \).

Answer:

D. \( 9\sqrt{2} \)

EOC 3