Question

1. higher order thinking alejandro is making a frame for a rectangular photo. the diagonal of the photo is 12 inches. he has 24 inches of framing material. does he have enough framing material to make the frame? explain.

Explanation:

Step1: Define variables for rectangle

Let length = $l$, width = $w$, diagonal $d=12$. By Pythagoras: $l^2 + w^2 = 12^2 = 144$.

Step2: Minimize perimeter of rectangle

For fixed diagonal, the rectangle with minimum perimeter is a square. For square: $l=w$, so $2l^2=144 \implies l^2=72 \implies l=\sqrt{72}=6\sqrt{2}$.

Step3: Calculate minimum perimeter

Perimeter of square: $P=4l=4\times6\sqrt{2}=24\sqrt{2}$.

Step4: Compare to available material

$24\sqrt{2} \approx 24\times1.414=33.936$ inches.

Answer:

No, he does not have enough framing material. The minimum perimeter of a rectangle with a 12-inch diagonal (a square) is approximately 33.94 inches, which is greater than the 24 inches of material he has.