Question

a triangle has side lengths of 3, 4, and 6. is it a right triangle? explain.
yes, $3^2 + 4^2 = 6^2$
no, $3^2 + 4^2 = 6^2$
no, $3^2 + 4^2 \
eq 6^2$
yes, $3^2 + 4^2 \
eq 6^2$
question 17
5 pts
to complete a tile pattern in her bathroom, lina has to cut square tiles diagonally as shown below. what is the perimeter of the square?
image of a square with a diagonal labeled 12, one side labeled x, a 45° angle at the bottom left, and right angles at the corners

Explanation:

First Question (Triangle Side Lengths)

Step1: Recall Pythagorean theorem

For a right triangle, \(a^2 + b^2 = c^2\) (where \(c\) is hypotenuse, longest side). Here, sides are 3, 4, 6 (longest is 6).

Step2: Calculate \(3^2 + 4^2\) and \(6^2\)

\(3^2 + 4^2 = 9 + 16 = 25\), \(6^2 = 36\). So \(3^2 + 4^2
eq 6^2\), so not a right triangle.

Step1: Identify triangle type

The triangle formed by the diagonal of the square is a 45 - 45 - 90 triangle (since square has right angles and diagonal splits it, so two 45° angles). In a 45 - 45 - 90 triangle, legs are equal, and hypotenuse \(h = x\sqrt{2}\) (where \(x\) is leg length, also side of square).

Step2: Solve for \(x\)

Given hypotenuse \(h = 12\), so \(x\sqrt{2}=12\), then \(x=\frac{12}{\sqrt{2}} = 6\sqrt{2}\) (rationalize: \(\frac{12\sqrt{2}}{2}=6\sqrt{2}\)).

Step3: Calculate perimeter of square

Perimeter of square \(P = 4x\), so \(P = 4\times6\sqrt{2}=24\sqrt{2}\)? Wait, no, wait. Wait, maybe I made a mistake. Wait, in a square, diagonal \(d = x\sqrt{2}\), so if diagonal is 12, then \(x=\frac{12}{\sqrt{2}}=6\sqrt{2}\), perimeter is \(4x = 24\sqrt{2}\approx33.94\). But maybe another approach: Wait, the triangle is isosceles right triangle, so legs are equal, so \(x^2 + x^2=12^2\) (Pythagorean theorem). So \(2x^2 = 144\), \(x^2 = 72\), \(x = \sqrt{72}=6\sqrt{2}\), same as before. Then perimeter is \(4\times6\sqrt{2}=24\sqrt{2}\). But maybe the problem expects a different approach? Wait, no, let's check again. Wait, the square's side is \(x\), diagonal is 12. So by Pythagoras, \(x^2 + x^2 = 12^2\) => \(2x^2 = 144\) => \(x^2 = 72\) => \(x = 6\sqrt{2}\). Then perimeter is \(4x = 24\sqrt{2}\approx33.94\). But maybe the problem has a typo or I misread. Wait, maybe the diagonal is 12, and we can also think that in a square, diagonal \(d = s\sqrt{2}\), so \(s=\frac{d}{\sqrt{2}}\), perimeter \(P = 4s=\frac{4d}{\sqrt{2}} = 2d\sqrt{2}\). Wait, no, \(4\times\frac{d}{\sqrt{2}}=\frac{4d\sqrt{2}}{2}=2d\sqrt{2}\). If \(d = 12\), then \(P = 2\times12\times\sqrt{2}=24\sqrt{2}\approx33.94\). But maybe the problem is expecting an integer? Wait, maybe I made a mistake in the triangle type. Wait, the angle is 45°, so it's a 45 - 45 - 90 triangle, so legs are equal, hypotenuse is leg\(\times\sqrt{2}\). So leg is \(12/\sqrt{2}=6\sqrt{2}\), perimeter is 4 times that, so \(24\sqrt{2}\). But let's check with Pythagoras again. \(x^2 + x^2 = 12^2\) => \(2x^2 = 144\) => \(x^2 = 72\) => \(x = 6\sqrt{2}\), correct. So perimeter is \(4\times6\sqrt{2}=24\sqrt{2}\).

Answer:

C. No, \(3^2 + 4^2
eq 6^2\)

Second Question (Square Perimeter)