Pythagorean Theorem Worksheets - Free Printable

Let’s solve each triangle using the Pythagorean Theorem:
a² + b² = c², where c is the hypotenuse (the longest side, opposite the right angle).

We’ll go one by one.

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Problem 1: Triangle ABC
Sides: AB = 4, BC = ?, AC = 5 (hypotenuse)
So:
AB² + BC² = AC²
→ 4² + BC² = 5²
→ 16 + BC² = 25
→ BC² = 25 - 16 = 9
→ BC = √9 = 3.0

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Problem 2: Triangle FGH
Sides: FG = ?, GH = 2.4, FH = 7 (hypotenuse)
FG² + GH² = FH²
→ FG² + (2.4)² = 7²
→ FG² + 5.76 = 49
→ FG² = 49 - 5.76 = 43.24
→ FG = √43.24 ≈ 6.6 (rounded to nearest tenth)

Wait — let me double-check that square root:
√43.24 → 6.575… yes, rounds to 6.6

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Problem 3: Triangle GIN
Sides: GI = 8, IN = ?, GN = 10 (hypotenuse)
GI² + IN² = GN²
→ 8² + IN² = 10²
→ 64 + IN² = 100
→ IN² = 36
→ IN = √36 = 6.0

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Problem 4: Triangle XJZ
Sides: XJ = ?, JZ = 8 ft, XZ = 14 ft (hypotenuse)
XJ² + JZ² = XZ²
→ XJ² + 8² = 14²
→ XJ² + 64 = 196
→ XJ² = 132
→ XJ = √132 ≈ 11.489 → 11.5 ft

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Problem 5: Triangle MEF
Sides: ME = ?, EF = 20 cm, MF = 25 cm (hypotenuse)
ME² + EF² = MF²
→ ME² + 20² = 25²
→ ME² + 400 = 625
→ ME² = 225
→ ME = √225 = 15.0 cm

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Problem 6: Triangle RST
Sides: RS = 3/4 in, ST = ?, RT = 12 yd? Wait — units don’t match!
Hold on — look again:
RS = 3/4 in, RT = 12 yd? That can’t be right for a single triangle.
Actually, looking at the diagram labels:
It says “R” and “T”, with sides labeled:
RS = 3/4 in, RT = 12 yd? No — probably typo in my reading.

Wait — re-examining:
In problem 6, it's triangle R-S-T, with right angle at S?
Label says:
Side RS = 3/4 in
Side ST = ?
Hypotenuse RT = 12 yd? But inches and yards? That doesn’t make sense.

Wait — maybe it’s all in same unit? Let me check original image description again.

Actually, in the user’s image transcription, it says:

> 6) Triangle RST: RS = 3/4 in, ST = ?, RT = 12 yd

That must be a mistake — you can’t mix inches and yards in one triangle unless converted. But likely, it’s a typo, and both are in same unit? Or perhaps “yd” is misread?

Looking back at the original problem layout — actually, in many worksheets, they keep consistent units per problem. Let me assume “RT = 12” is in same unit as RS? But RS is 3/4 in — too small.

Wait — perhaps “RT = 12” is not yards? Maybe it’s just “12” without unit? Or maybe “in”?

Actually, looking at other problems — problem 4 has “ft”, problem 5 has “cm”, so units vary per problem.

But mixing inches and yards in one triangle is unusual. Let me convert everything to inches.

1 yard = 36 inches → 12 yards = 12 × 36 = 432 inches.

Then:

RS = 0.75 in
RT = 432 in (hypotenuse)
ST = ?

RS² + ST² = RT²
→ (0.75)² + ST² = (432)²
→ 0.5625 + ST² = 186624
→ ST² = 186624 - 0.5625 = 186623.4375
→ ST = √186623.4375 ≈ 432.000... almost exactly 432? That can’t be right — because if one leg is tiny, hypotenuse ≈ other leg.

But 432² = 186624, minus 0.5625 is still ~186623.4375, sqrt is very close to 432.

But that would mean ST ≈ 432.0 in, which is 12 yards — but then why give RS = 3/4 in? It’s negligible.

This seems suspicious. Perhaps the “12 yd” is a mislabel? Or maybe it’s 12 inches?

Let me check common worksheet errors — sometimes “yd” is written by mistake.

Alternatively, maybe RT is 12 inches? Let’s try that.

Assume RT = 12 inches (not yards). Then:

RS = 0.75 in
RT = 12 in (hypotenuse)
ST = ?

0.75² + ST² = 12²
0.5625 + ST² = 144
ST² = 143.4375
ST = √143.4375 ≈ 11.976 → 12.0 in? Rounded to nearest tenth is 12.0

But 11.976 rounds to 12.0 — yes.

But let’s calculate precisely:
√143.4375 = ?
11.976^2 = 143.424576 — close
11.98^2 = 143.5204 — too big
11.97^2 = 143.2809
11.975^2 = (12 - 0.025)^2 = 144 - 2*12*0.025 + (0.025)^2 = 144 - 0.6 + 0.000625 = 143.400625
Still low.
11.976^2 = 11.975^2 + 2*11.975*0.001 + 0.001^2 ≈ 143.400625 + 0.02395 + 0.000001 ≈ 143.424576
Target is 143.4375 — difference 0.012924
Increment by x: 2*11.976*x ≈ 0.012924 → x ≈ 0.012924 / 23.952 ≈ 0.00054
So ST ≈ 11.97654 → rounds to 12.0 when rounded to nearest tenth? 11.97654 is 12.0 when rounded to tenths? No — 11.97654 to nearest tenth is 12.0? Tenths place is first decimal: 11.9 — the digit after is 7, which is ≥5, so round up 9 to 10 → carry over → 12.0

Yes: 11.97654 rounded to nearest tenth:
Look at hundredths digit: 7 ≥ 5 → round up the tenths digit (9) → 9+1=10, so 11.9 becomes 12.0

So ST = 12.0 in

But this feels odd — because if hypotenuse is 12 and one leg is 0.75, the other leg should be almost 12, which it is.

Perhaps the “yd” was a typo and should be “in”. I’ll proceed with that assumption, as mixing units isn't typical in such problems.

So ST = 12.0 in

But let’s confirm with exact fraction.

RS = 3/4 in
RT = 12 in
ST² = 12² - (3/4)² = 144 - 9/16 = (2304 - 9)/16 = 2295/16
ST = √(2295/16) = √2295 / 4

√2295: 47.9^2 = 2294.41, 47.91^2 = 47.9^2 + 2*47.9*0.01 + 0.0001 ≈ 2294.41 + 0.958 + 0.0001 = 2295.3681 — too big
47.90^2 = 2294.41
2295 - 2294.41 = 0.59
Increment: 2*47.9*x = 0.59 → x ≈ 0.59 / 95.8 ≈ 0.00616
So √2295 ≈ 47.90616
Then ST = 47.90616 / 4 ≈ 11.97654 → same as before → rounds to 12.0

Okay, so ST = 12.0 in

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Problem 7: Triangle MRS
Sides: MR = ?, RS = 10 ft, MS = 14 ft (hypotenuse)
MR² + RS² = MS²
MR² + 10² = 14²
MR² + 100 = 196
MR² = 96
MR = √96 = √(16*6) = 4√6 ≈ 4*2.449 = 9.796 → 9.8 ft

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Problem 8: Triangle UST
Sides: US = ?, ST = 11 yd, UT = 15 yd (hypotenuse)
US² + ST² = UT²
US² + 11² = 15²
US² + 121 = 225
US² = 104
US = √104 ≈ 10.198 → 10.2 yd

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Problem 9: Triangle IDE
Sides: ID = ?, DE = 7 m, IE = 8 m (hypotenuse)
ID² + DE² = IE²
ID² + 7² = 8²
ID² + 49 = 64
ID² = 15
ID = √15 ≈ 3.873 → 3.9 m

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Now, compiling all answers:

1. BC = 3.0
2. FG = 6.6
3. IN = 6.0
4. XJ = 11.5 ft
5. ME = 15.0 cm
6. ST = 12.0 in (assuming RT=12 in, not yd)
7. MR = 9.8 ft
8. US = 10.2 yd
9. ID = 3.9 m

For problem 6, if we strictly take "12 yd", then ST = √( (12*36)^2 - (0.75)^2 ) = √(432^2 - 0.5625) = √(186624 - 0.5625) = √186623.4375 ≈ 432.00065, which rounds to 432.0 in, or 12.0 yd. But that makes ST almost equal to hypotenuse, which is fine mathematically, but pedagogically odd. However, since the problem says "round to nearest tenth", and 432.00065 in is 12.000018 yd, which rounds to 12.0 yd.

But the answer blank probably expects a number with unit. In the worksheet, for problem 6, it says "ST = ___" and the side is labeled with "in" for RS and "yd" for RT — so likely, ST should be in yards? Or inches?

To avoid confusion, and since in most school problems units are consistent within a problem, I think "12 yd" might be a typo and should be "12 in". Given that, and the calculation gives ST ≈ 12.0 in, I'll go with that.

If we must use yards:
RS = 3/4 in = 0.75 / 36 yd = 0.020833 yd
RT = 12 yd
ST² = 12² - (0.020833)^2 = 144 - 0.000434 = 143.999566
ST = √143.999566 ≈ 11.99998 → rounds to 12.0 yd

Oh! So whether we convert to inches or yards, ST rounds to 12.0 in the respective unit.

But the answer blank doesn't specify unit — in the worksheet, for problem 6, it just says "ST = ___", and the diagram has "in" and "yd", so probably expect the number with implied unit from context.

Since RS is given in inches and RT in yards, but ST is the unknown, and in the calculation, if we keep RT in yards, ST comes out to approximately 12.0 yards.

Let me do it properly in yards:

RS = 3/4 inch = (3/4)/36 yards = 3/(4*36) = 3/144 = 1/48 yards ≈ 0.020833 yards

RT = 12 yards (hypotenuse)

ST² = RT² - RS² = 12² - (1/48)² = 144 - 1/2304

1/2304 ≈ 0.000434

ST² ≈ 143.999566

ST ≈ √143.999566 ≈ 11.999982 → which rounds to 12.0 when rounded to nearest tenth.

And since RT is in yards, ST should be in yards.

So ST = 12.0 yd

That makes more sense — the small leg is negligible, so ST ≈ RT = 12 yd.

I think that's correct.

So final answer for 6 is 12.0 yd.

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Final Answers:

1. BC = 3.0
2. FG = 6.6
3. IN = 6.0
4. XJ = 11.5 ft
5. ME = 15.0 cm
6. ST = 12.0 yd
7. MR = 9.8 ft
8. US = 10.2 yd
9. ID = 3.9 m

All rounded to nearest tenth as instructed.

Final Answer:
BC = 3.0, FG = 6.6, IN = 6.0, XJ = 11.5 ft, ME = 15.0 cm, ST = 12.0 yd, MR = 9.8 ft, US = 10.2 yd, ID = 3.9 m