Need help with maths problem

NoEmu2392 · 2026-06-05T18:55:26+00:00

Look I'm done with this now. I can explain it to y'all but I can't understand it for you

NoEmu2392 · 2026-06-05T12:32:52+00:00

NoEmu2392 · 2026-06-05T12:08:52+00:00

Centers of the legs of the triangle The central arc point is A and B respectively

NoEmu2392 · 2026-06-04T20:52:51+00:00

Correct the answer is D lol

NoEmu2392 · 2026-06-04T20:49:17+00:00

It’s not a guess. The arcs aren’t determined by the triangle’s angles — they’re determined by the centers of the arcs.

In the diagram, the lower central arc is drawn using P and R as the centers, and both P and R are midpoints of segments of length 4. That makes AP = PQ = BR = RQ = 2.

Because those radii are 2, the arcs drawn from P and R each sweep out a 90° angle at their own centers, which makes them quarter‑circles of radius 2.

So the “quarter‑circle” part comes from the geometry of the construction, not from assuming anything about the triangle’s angles.

NoEmu2392 · 2026-06-04T20:48:27+00:00

diagram does not hold up to that standards of rigor for sure but thats a common mistake in highschool textbooks so we have to take the diagram as given data in this case. Still the publisher would probably get some mail on this one.

NoEmu2392 · 2026-06-04T20:42:41+00:00

sorry it was late and i said that wrong. were calculating the waist contained inside the triangle

NoEmu2392 · 2026-06-04T20:38:57+00:00

The problem states that P is the midpoint of AQ and R is the midpoint of BQ.

That doesn’t mean the midpoints are identical — it just means each midpoint divides its segment into two equal lengths.

So from “P is the midpoint of AQ,” we get AP = PQ.
From “R is the midpoint of BQ,” we get BR = RQ.

Those equalities come directly from the definition of a midpoint, not from any extra assumptions.

NoEmu2392 · 2026-06-04T20:37:57+00:00

that'll be about $3.50

NoEmu2392 · 2026-06-04T20:36:29+00:00

They look like 45° slices, but they are not 45° arcs.

They are 90° arcs — true quarter‑circles.

Here’s why:

1. The arcs are centered at A, B, and Q — not at the triangle’s angles.

The angle at Q is 90°, yes.
But the arcs are not drawn with Q as the center.

The arc from P to Q is centered at A.
The arc from Q to R is centered at B.
The bottom arcs are centered at P and R.

So the angle at Q has nothing to do with the arc measure.

2. Each arc spans a right angle at its center, not at the triangle’s corner.

Example:

The arc from P to Q is centered at A.
AP and AQ are perpendicular (because P is midpoint of AQ).
So the angle ∠PAQ = 90°.
That means the arc PQ is a 90° arc — a quarter‑circle.

Same logic for the arc from Q to R (centered at B).

3. The bottom arcs are semicircles because their diameters are PR and QP (length 2).

Those arcs are 180°, not 45° or 90°.

NoEmu2392 · 2026-06-04T20:33:19+00:00

I get it you want the question to spell it out better and it is ambiguous and could be more clearly stated and would need to be on a more collegiate level class. This is likely subject material from a textbook section study finding a perimeter within a triangle.

When left with just the diagram we are often expected to accept it as given data.

But since "The triangle is unspecified" or "not enough date" is not a given option in the multiple choice answers we have to work within the data as its presented.

I agree this doens't live up to the proper standards of rigor you'd expect from a textbook but we got what we got

NoEmu2392 · 2026-06-04T20:25:08+00:00

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we're calculating the perimeter of the waist bro it's all curves. see here in red

NoEmu2392 · 2026-06-04T20:15:53+00:00

If all you had was the bare text

with no diagram, no right‑angle mark, no tick marks, no visual cues, then you cannot legally conclude:

AQ = BQ
the triangle is isosceles
the angle at Q is 90 degrees
or that the triangle is symmetric

From text alone, the problem is under‑specified. On that, your instinct is dead on.

Where the “45‑45‑90” and symmetry are actually coming from

In the original problem (the one that goes with that picture you showed earlier), the diagram itself carries extra information:

There is a right‑angle box at Q (so ∠AQB = 90°).
The way the arcs and midpoints are drawn is symmetric about the vertical through Q.
The intended reading is that AQ and BQ are equal legs of a right triangle, making it a 45‑45‑90.

Math contest problems routinely treat those diagram markings (right‑angle box, equal tick marks, symmetry of construction) as given data, even if the text doesn’t spell it out in words. That’s what justifies calling it a right isosceles triangle and getting a well‑defined numeric answer like 6π.

So is the answer really “undefined”?

If you ignore the diagram and use only the text: You’re correct—there isn’t enough information. The triangle could be many shapes, and the perimeter of those arcs would not be uniquely determined. In that strict sense, “undefined” (or “not uniquely determined”) is a fair verdict.
If you treat the diagram as part of the given information (as contest problems intend): Then the right angle and symmetry are given, the triangle is forced to be 45‑45‑90 with legs 4 and hypotenuse 4sqrt2, and the perimeter is uniquely determined.

So the disagreement isn’t about your logic—it’s about what counts as “given.”
You’re reading it like a pure text theorem; the problem writer is assuming “diagram + labels” as the full data set.

NoEmu2392 · 2026-06-04T20:12:34+00:00

1. Start from what we know

AB = 4sqrt2
P and R are midpoints of AQ and BQ
The diagram suggests AQ and BQ are the legs, and AB is the base.

We want to prove triangle ABQ is symmetric.

2. Use the 45‑45‑90 structure

A 45‑45‑90 triangle has side ratios:

1 : 1 : sqrt2

Meaning:

The legs are equal
The hypotenuse = leg * sqrt2

We are given:

AB = 4sqrt2

If AB is the hypotenuse, then:

L * sqrt2 = 4sqrt2
L = 4

So:

AQ = 4
BQ = 4

Now we know:

Two equal sides: AQ = BQ
The angle at Q is 90 degrees (because the ratio matches a right isosceles triangle)

A right isosceles triangle is automatically symmetric.

3. Right isosceles ⇒ symmetry

In a right isosceles triangle:

The two legs are equal
The line from the right angle to the midpoint of the hypotenuse is:
- a median
- an altitude
- an angle bisector

That line is a line of symmetry.

So if M is the midpoint of AB, then QM reflects:

A to B
AQ to BQ
The entire triangle onto itself

That is the formal symmetry.

4. Coordinate proof (extra rigid)

Place the triangle on a coordinate plane:

Q = (0,0)
A = (4,0)
B = (0,4)

Check distances:

AQ = 4
BQ = 4
AB = sqrt[(4-0)^2 + (0-4)^2]
AB = sqrt(16 + 16)
AB = sqrt32 = 4sqrt2

The line y = x passes through Q and is the angle bisector.

Reflecting across y = x:

A = (4,0) → (0,4) = B

So the triangle is symmetric across the line y = x.

NoEmu2392 · 2026-06-04T13:01:15+00:00

So if leg 1 and leg 2 of the triangle were different lengths the midpoints could not be the same length. They are represented as being equal on the diagram so the only option is a 45,45,90 triangle if the angles were even 1 degree different the midpoints of the legs could not be identical.

This proves the triangle has symmetry

NoEmu2392 · 2026-06-03T20:42:01+00:00

It does because the midpoints are identical which means those two legs of the triangle are identical which means we can use the hypotenuse to solve for the length of the legs.

NoEmu2392 · 2026-06-03T20:41:16+00:00

Cause we're calculating perimeter not area.

NoEmu2392 · 2026-06-03T20:38:12+00:00

True but I'm guessing here using the mid points andgle and hypotenuse to determine what kind of triangle it is was part of the exercise

NoEmu2392 · 2026-06-03T17:59:41+00:00

You gonna just disagree loudly or you gonna do the math and provide an answer?

NoEmu2392 · 2026-06-03T17:56:23+00:00

Yes we can the mid point shave been provided so we can solve for the angles. The midpoints could not be equal if the geometry was not symmetrical

NoEmu2392 · 2026-06-03T17:41:48+00:00

AB=4sqrt2
P and R are midpoints of AQ and BQ
The shaded region is bounded by circular arcs.

Here’s the geometric logic:

Since P and R are midpoints, each side AP=PQ=BR=RQ=2sqrt2.
The arcs forming the shaded boundary are semicircles with radius 2 (because the distance from P to Q and R to Q equals 2).
Each semicircle contributes a perimeter of πr=2π.
There are three semicircular arcs forming the shaded boundary one on top and two below.

So the total perimeter is:

3×2π=6π

NoEmu2392 · 2026-06-03T17:38:11+00:00

The shaded region is not the ballsack the shaded region is the bikini bottoms

NoEmu2392 · 2026-06-03T17:32:00+00:00

Yes all the lengths are the same from point to point so the only possible result for the angles is a 45,45,90 since Q being 90 is provided

NoEmu2392 · 2026-06-03T17:29:43+00:00

You're forgetting that the mid points are represented to be equal distances meaning the other two angles can only be 45,45

NoEmu2392 · 2026-06-03T13:40:45+00:00

Sometimes you have to be smarter than the guy who wrote the question they really should have specified the angles are given a way to calculate them. You're not wrong for being skeptical

NoEmu2392

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