CalculatorUser

bruh moderation

leg 1 = 3

leg 2 = x

hypotenuse = 2x - 3

set up equation using pythagorean theorem

3² + x² = (2x-3)²

now solve it

you should get an integer answer for x

Set up variables

Length = 6x

Width = 3x

Height = 6x - 3

Commence calculations

For surface areas:

L * W surfaces = \(36x^2\)

L * H surfaces = \(72x^2-36x\)

W * H surfaces = \(36x^2-18x\)

Set up equation

\(144x^2-54x = 100\)

\(144x^2-54x - 100 = 0\) divide by 2

\(72x^2-27x-50=0\)

72x^2-27x-50=0 = {x=(25/24), x=-((2/3))}

So \(x=25/24\)

now solve for height

Well it is a 30-60-90 triangle.

Tell me what you are confused on

This is what the cutting is? Im just gonna solve it like that...

Now we turn the cube on its cut face (i suck at drawing so its hard to comprehend). Imagine that the cut face (triangle) is on the table.

Commence calculations

Sorry this may be EXTREMELY difficult to comprehend because I suck at drawing lol

Hypotenuse = \(\sqrt{2}\)

Long Leg = \(\frac{\sqrt{6}}{3}\)

By pythageoran theorem:

Answer = \(\sqrt{(\sqrt{2})^2-(\frac{\sqrt{6}}{3})^2}\)

We get a radical of

\(\boxed{\frac{2\sqrt{3}}{3}}\)

that is high-iq

In triangle \(ABC\), \(AP = BQ = \frac{PQ}{2}\) and \(CR/RA = 3/2 \). Find the ratio of the area of triangle \(BQC\)  to the area of triangle \(CRQ\).

Since AP = BQ = PQ/2. We can conclude that QB is 1/4 of AB.

Let the area of ABC = x

So BQC = \(\frac{x}{4}\)

Area of triangle CAQ = \(x-\frac{x}{4}=\frac{3x}{4}\)

Since CR/RA = 3/2, that means that CRQ = \(\frac{3x}{4}*\frac{3}{5}=\frac{9x}{20}\)

SO

Area of BQC is \(\frac{x}{4}\)

Area of CRQ is \(\frac{9x}{20}\)

Can you find the ratio now? (Hint: It's simple algebra!)

ok boomer