Math worksheet for solving similar right triangles, featuring eight problems with labeled sides and a missing side to find.

Final Answer:
1) $ x = 49\sqrt{2} $
2) $ x = 81\sqrt{2} $
3) $ x = 15 $
4) $ x = \sqrt{55} $
5) $ x = \sqrt{1156 - 1296} $ → Wait, that’s negative — correction: In problem 5, the side labeled 40 is the hypotenuse (longest side), and 36 is a leg, so $ x = \sqrt{40^2 - 36^2} = \sqrt{1600 - 1296} = \sqrt{304} = 4\sqrt{19} $
6) $ x = \sqrt{4^2 - 12^2} $ → impossible; but 12 > 4, so 12 must be hypotenuse. So $ x = \sqrt{12^2 - 4^2} = \sqrt{144 - 16} = \sqrt{128} = 8\sqrt{2} $
7) $ x = \sqrt{73^2 - 64^2} = \sqrt{5329 - 4096} = \sqrt{1233} = \sqrt{9 \cdot 137} = 3\sqrt{137} $
8) $ x = \sqrt{41^2 - 25^2} = \sqrt{1681 - 625} = \sqrt{1056} = \sqrt{16 \cdot 66} = 4\sqrt{66} $

But wait — let’s double-check each based on triangle labeling:

In all diagrams, the right angle is at the bottom-left corner (square mark), the vertical leg is labeled x, the horizontal leg is given, and the hypotenuse is the slanted side (sometimes labeled, sometimes not). The dashed line with “x” is the vertical leg.

So for each:
1) Isosceles right triangle: legs = 49, 49 → hypotenuse = 49√2, but here x is a leg, and both legs are 49? Wait — diagram shows vertical leg = x, horizontal leg = 49, hypotenuse = 49. That’s impossible (hypotenuse must be longest). So likely: right angle at bottom-left, vertical leg = x, horizontal leg = 49, hypotenuse = 49? No — can’t be.

Actually, looking carefully: In problem 1, the two sides shown are both 49, and the right angle is between them → so it's an isosceles right triangle with legs = 49, and x is the hypotenuse. Yes! The “x” is along the hypotenuse (dashed line across the triangle). So x = √(49² + 49²) = 49√2.

Same for #2: legs = 81 and x? No — right angle at bottom-left, vertical leg = x, horizontal leg = 81, hypotenuse = 90? Wait, label says “90” near hypotenuse, and x on vertical leg. So:
Legs: x and 81, hypotenuse = 90 → x = √(90² − 81²) = √(8100 − 6561) = √1539 = √(81·19) = 9√19.

Ah! I misread earlier. Let’s correctly interpret each:

Standard convention in these diagrams:
- Right angle at bottom-left (□).
- Horizontal leg (bottom) is given number.
- Vertical leg (left side) is labeled x (dashed line with “x”).
- Slanted side (hypotenuse) is the other given number (or sometimes missing).

So:
1) Horizontal = 49, hypotenuse = 49 → impossible. But diagram shows both legs labeled 49, and x on hypotenuse → yes, x is hypotenuse. So legs = 49, 49 → x = 49√2.

2) Horizontal = 81, hypotenuse = 90, x = vertical leg → x = √(90² − 81²) = √(8100 − 6561) = √1539 = √(81×19) = 9√19.

3) Horizontal = 16, hypotenuse = 25, x = vertical leg → x = √(25² − 16²) = √(625 − 256) = √369 = √(9×41) = 3√41. Wait, but 369 = 9×41, yes.

But standard Pythagorean triple: 7-24-25, 15-20-25 — 16-?–25 is not integer. √(625−256)=√369≈19.21, okay.

4) Horizontal = 9, hypotenuse = 64 → x = √(64² − 9²) = √(4096 − 81) = √4015. 4015 = 5×803 = 5×11×73 → no squares → x = √4015.

But that seems messy. Maybe in #4, x is the hypotenuse? No, dashed line is vertical leg.

Wait — perhaps in some problems, the given number is the vertical leg, and x is horizontal? No, the dashed line with “x” is always the side opposite the right angle? No, dashed line goes from right angle to hypotenuse — it's the altitude? No, in all, the dashed line is one leg, marked x, and the right angle is at its corner.

Let me instead use the most consistent interpretation used in Math-Aids worksheets (since logo is there):
These are *right triangles* where two sides are given, and you solve for the third using Pythagorean theorem:
a² + b² = c², where c = hypotenuse (longest side).

For each:
1) Legs: 49 and 49 → hypotenuse x = √(49²+49²) = 49√2
2) Leg = 81, hypotenuse = 90 → other leg x = √(90²−81²) = √1539 = 9√19
3) Leg = 16, hypotenuse = 25 → x = √(625−256) = √369 = 3√41
4) Leg = 9, hypotenuse = 64 → x = √(4096−81) = √4015 (simplify? 4015 = 5×11×73 → no perfect square) → leave as √4015
5) Leg = 36, hypotenuse = 40 → x = √(1600−1296) = √304 = √(16×19) = 4√19
6) Leg = 4, hypotenuse = 12 → x = √(144−16) = √128 = 8√2
7) Leg = 64, hypotenuse = 73 → x = √(5329−4096) = √1233 = √(9×137) = 3√137
8) Leg = 25, hypotenuse = 41 → x = √(1681−625) = √1056 = √(16×66) = 4√66

These are all simplified radical forms.

So final answers:
1) $ 49\sqrt{2} $
2) $ 9\sqrt{19} $
3) $ 3\sqrt{41} $
4) $ \sqrt{4015} $
5) $ 4\sqrt{19} $
6) $ 8\sqrt{2} $
7) $ 3\sqrt{137} $
8) $ 4\sqrt{66} $

But the problem says “Leave your answer in the simplest radical form”, and all above are simplified.

However, double-check #4: 64² = 4096, 9² = 81, difference 4015. 4015 ÷ 5 = 803, 803 ÷ 11 = 73, so 4015 = 5×11×73 — no square factors. Correct.

Thus:

Final Answer:
1) $ 49\sqrt{2} $
2) $ 9\sqrt{19} $
3) $ 3\sqrt{41} $
4) $ \sqrt{4015} $
5) $ 4\sqrt{19} $
6) $ 8\sqrt{2} $
7) $ 3\sqrt{137} $
8) $ 4\sqrt{66} $