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Posted on 2/14/25 at 5:27 pm to DVinBR
This post was edited on 2/16/25 at 3:44 pm
Posted on 2/14/25 at 5:28 pm to anc
quote:
disagree. Something is amiss on how our education system is functioning.
How dumb are you lol
Posted on 2/14/25 at 5:34 pm to anc
I came up with 300, 20x15=300
Posted on 2/14/25 at 5:38 pm to anc
My wife is 10X smarter than your dumbass. We made really smart kids.
Posted on 2/14/25 at 5:39 pm to anc
quote:
Could your wife or girlfriend solve this?
I can google it so sure.
I'm great at math, but never in my life have I had to solve that in the real world with out Google (or ever) since I was taught it so I dont remember.
I produced great off-spring. He's in the 96% for math.
Posted on 2/14/25 at 5:53 pm to anc
From ChatGTP:
To determine the area of the square, let's analyze the given right triangle.
Given:
The base of the large right triangle = 5
The height of the large right triangle = 20
A square is inscribed within the triangle.
Step 1: Find the slope (similar triangles approach)
The large right triangle has a base-to-height ratio of:
\frac{20}{5} = 4
Since the square is inside the triangle, let the side of the square be s. The small right triangle adjacent to the square is similar to the large triangle.
Step 2: Solve for s
Since the base-to-height ratio remains the same, the remaining portion of the base (outside the square) is also .
Using the total base equation:
s + \frac{s}{4} = 5
Multiply everything by 4 to eliminate the fraction:
4s + s = 20
5s = 20
s = 4
Step 3: Find the Area
Since the side length of the square is 4, the area of the square is:
4^2 = 16
Final Answer:
The area of the square is 16 square units.
To determine the area of the square, let's analyze the given right triangle.
Given:
The base of the large right triangle = 5
The height of the large right triangle = 20
A square is inscribed within the triangle.
Step 1: Find the slope (similar triangles approach)
The large right triangle has a base-to-height ratio of:
\frac{20}{5} = 4
Since the square is inside the triangle, let the side of the square be s. The small right triangle adjacent to the square is similar to the large triangle.
Step 2: Solve for s
Since the base-to-height ratio remains the same, the remaining portion of the base (outside the square) is also .
Using the total base equation:
s + \frac{s}{4} = 5
Multiply everything by 4 to eliminate the fraction:
4s + s = 20
5s = 20
s = 4
Step 3: Find the Area
Since the side length of the square is 4, the area of the square is:
4^2 = 16
Final Answer:
The area of the square is 16 square units.
Posted on 2/14/25 at 5:57 pm to anc
One thing I don’t want my wife to be good at is measurements.
Posted on 2/14/25 at 7:25 pm to Korkstand
quote:
Now this one *does* require assumptions. We know that the green area is square, but we can't be sure that the large "square" is square.
It can't be square.
30^2 + 40^2 = 900 + 1600 = 2500.
v2500 = 50.
So the side of the blue shape is equal to the hypotenuse of the triangle in the lower left.
Posted on 2/14/25 at 8:23 pm to Bestbank Tiger
What
Posted on 2/14/25 at 8:40 pm to anc
Considering there is no unit of measurement given, nobody can solve this.
100 acres? 100 meters? Miles?
100 acres? 100 meters? Miles?
This post was edited on 2/14/25 at 8:45 pm
Posted on 2/14/25 at 8:42 pm to anc
She was a landscape architect, so yeah. And probably in her head.
Posted on 2/14/25 at 8:43 pm to collegefootballisbroken
quote:
name 20 starting QBs in the NFL
My cousin and I tried that to see who could name 20 starting quarterbacks first. We are going to try WRs next.
Posted on 2/14/25 at 8:48 pm to anc
It's just a ratio.
20/x=x/5
x=10
Area=x*x=100
20/x=x/5
x=10
Area=x*x=100
Posted on 2/14/25 at 8:49 pm to anc
49???? Lol
Posted on 2/14/25 at 9:09 pm to Korkstand
quote:
What
I couldn't quote the photo.
A squared plus b squared equals c squared.
A right triangle with sides of 30 and 40 has a hypotenuse of 50.
But the side of the blue area is also 50. And the hypotenuse of the triangle is at an angle. So it looks like a square or rectangle but it can't be, since the opposite sides aren't parallel.
Posted on 2/14/25 at 9:15 pm to Bestbank Tiger
Not sure what the hell you are talking about. The green area as well as the whole bundle can certainly both be squares.
This post was edited on 2/14/25 at 9:16 pm
Posted on 2/14/25 at 9:20 pm to poochie
There’s enough information.
Posted on 2/14/25 at 9:28 pm to 91TIGER
I made it harder than I needed to.
Posted on 2/14/25 at 9:31 pm to Korkstand
This one is not solvable. You can easily imagine a slightly larger square that wouldn't violate anything you've been told in the problem.
ETA: I am talking about the 2nd problem with the green square in the square.
ETA: I am talking about the 2nd problem with the green square in the square.
This post was edited on 2/15/25 at 7:01 pm
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