GRADE 8 GEOMETRY PRACTICE 1 • Teacha! - Free Printable

The image shows four problems involving the Pythagorean theorem. I will solve each one step by step.

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Problem 1:
We are given a right triangle with:
- One leg = 8
- Hypotenuse = 20
- Find the other leg (let’s call it x)

Using the Pythagorean theorem:
a² + b² = c²
→ 8² + x² = 20²
→ 64 + x² = 400
→ x² = 400 - 64 = 336
→ x = √336

Simplify √336:
336 = 16 × 21 → √336 = √(16×21) = 4√21 ≈ 18.33

But let’s check if we made a mistake — maybe they want exact form or decimal? Since the example in the image used whole numbers, perhaps this is meant to be simplified.

Wait — actually, 336 is not a perfect square. Let me double-check:

8² = 64
20² = 400
400 - 64 = 336 → correct.

So answer is √336 or 4√21. But maybe we should leave it as simplified radical? Or perhaps the problem expects a decimal? The example showed 5, which is exact. Let’s keep it as √336 for now and simplify later if needed.

Actually, let’s see if 336 can be simplified further:
336 ÷ 16 = 21 → yes, so 4√21 is simplest radical form.

But wait — maybe I misread? Let me look again at the diagram.

In Problem 1: vertical side is 8, hypotenuse is 20, base is unknown. Yes.

Alternatively, maybe it's designed to have an integer answer? Let me check 8-15-17 or something... no, 8-? -20.

Wait — 20² - 8² = 400 - 64 = 336 → not a perfect square. So answer is √336 or 4√21.

But let’s proceed to others and come back.

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Problem 2:
Given:
- One leg = 7
- Hypotenuse = 25
- Find other leg (x)

a² + b² = c²
→ x² + 7² = 25²
→ x² + 49 = 625
→ x² = 625 - 49 = 576
→ x = √576 = 24

Perfect! Integer answer.

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Problem 3:
Given:
- One leg = 12
- Hypotenuse = 14
- Find other leg (x)

x² + 12² = 14²
→ x² + 144 = 196
→ x² = 196 - 144 = 52
→ x = √52 = √(4×13) = 2√13 ≈ 7.21

Again, not a whole number. But mathematically correct.

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Problem 4:
Given:
- One leg = 14
- Hypotenuse = 15
- Find other leg (x)

x² + 14² = 15²
→ x² + 196 = 225
→ x² = 225 - 196 = 29
→ x = √29 ≈ 5.39

Also not a whole number.

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Wait — let me re-express all answers clearly:

Problem 1: √(20² - 8²) = √(400 - 64) = √336 = 4√21
Problem 2: √(25² - 7²) = √(625 - 49) = √576 = 24
Problem 3: √(14² - 12²) = √(196 - 144) = √52 = 2√13
Problem 4: √(15² - 14²) = √(225 - 196) = √29

These are all correct mathematically. Perhaps the worksheet expects simplified radicals or decimals? But since the example used exact values (like 5), I’ll present exact forms.

But let me check Problem 1 again — maybe I misidentified the sides?

Looking back at the image description: In Problem 1, the vertical side is labeled 8, hypotenuse is 20, and the horizontal side is unknown. That’s correct.

Similarly for others.

Perhaps the student is expected to use calculator and round? But the instruction says “solve accurately”, and the example didn’t round.

Alternatively, maybe there’s a typo in my reading? Let me think differently.

Wait — in Problem 4, the diagram shows: vertical side is ?, horizontal is 14, hypotenuse is 15. So yes, same as above.

All calculations are correct.

So final answers:

1. √336 or 4√21
2. 24
3. √52 or 2√13
4. √29

But to match the style of the example (which gave 5, a whole number), perhaps only Problem 2 has a nice answer. Others are irrational.

Maybe I should write them as simplified radicals.

Let me confirm simplifications:

- √336 = √(16*21) = 4√21
- √52 = √(4*13) = 2√13
- √29 is already simplified

Yes.

Now, since the user said "solve accurately", I’ll provide exact answers.

But let me see if the problems might have been intended to have integer answers — perhaps I misread the diagrams.

For example, in Problem 1: if it were 6-8-10 scaled, but 8 and 20 — 20/8=2.5, not matching 6-8-10.

Or 12-16-20? But here it’s 8 and 20.

No, I think my calculations are correct.

Perhaps the worksheet allows radicals.

I’ll go with exact simplified radicals where applicable.

Final Answers:

1. \( 4\sqrt{21} \)
2. 24
3. \( 2\sqrt{13} \)
4. \( \sqrt{29} \)

But let me write them neatly.

Actually, for consistency, since Problem 2 is integer, others are radicals.

Perhaps the student is to leave as square roots.

I think that's fine.

So, summarizing:

Problem 1: The missing side is \( \sqrt{20^2 - 8^2} = \sqrt{400 - 64} = \sqrt{336} = 4\sqrt{21} \)

Problem 2: \( \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \)

Problem 3: \( \sqrt{14^2 - 12^2} = \sqrt{196 - 144} = \sqrt{52} = 2\sqrt{13} \)

Problem 4: \( \sqrt{15^2 - 14^2} = \sqrt{225 - 196} = \sqrt{29} \)

All verified.

Now, for the final answer, since the question is to solve the problems, I'll list them in order.

But the user didn't specify how to present multiple answers. Looking at the image, each problem has a line for answer, so probably four separate answers.

In the response, I should provide all four.

Since the instruction says "Final Answer" section, and it's for the entire task, I'll list them.

Perhaps as:

1. \( 4\sqrt{21} \)
2. 24
3. \( 2\sqrt{13} \)
4. \( \sqrt{29} \)

To make it clear.

I think that's accurate.

Let me box them as per requirement.

But in the Final Answer section, I need to put only the final correct answer — since there are four, I'll present them clearly.

Perhaps write:

The lengths of the unknown sides are:
Problem 1: \( 4\sqrt{21} \)
Problem 2: 24
Problem 3: \( 2\sqrt{13} \)
Problem 4: \( \sqrt{29} \)

Yes.

I think that's it.