Square Geometry Problem: Comparing Diagonal Sums vs Side Lengths in a 4-Unit Square

Q: Why can't I just add the side lengths to get the diagonal?

Because a diagonal cuts diagonally across the square! You need the Pythagorean theorem because the diagonal forms the hypotenuse of a right triangle with two sides of the square.

Q: What does √32 actually equal?

$ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \approx 5.66 $. You can leave it as 4√2 for exact answers or use 5.66 for decimal approximations.

Q: How do I know which is bigger without a calculator?

Since $ \sqrt{2} \approx 1.41 $, each diagonal is $ 4 \times 1.41 = 5.64 $. Two diagonals: 2 × 5.64 = 11.28. Three sides: 3 × 4 = 12. So 11.28

Q: Why are we comparing with only 3 sides instead of 4?

The question specifically asks about 3 sides of the square, not all 4 sides. This makes the comparison more interesting since 3 × 4 = 12 is closer to 2 × 5.66 = 11.31.

Q: Could the diagonals ever be longer than 3 sides?

Never! In any square, $ 2 \times \text{diagonal} = 2s\sqrt{2} \approx 2.83s $ while $ 3 \times \text{side} = 3s $. Since 2.83

Pythagorean Theorem with Diagonal Calculations

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Is the sum of diagonals greater than the sum of 3 sides of the square?

00:03 In a square all sides are equal

00:07 We'll use the Pythagorean theorem in triangle BCD

00:11 We'll substitute appropriate values and solve for BD

00:29 This is the length of diagonal BD

00:32 In a square the diagonals are equal

00:36 Let's calculate the sum of diagonals

00:40 We'll substitute appropriate values and solve for the sum

00:46 Let's calculate the sum of 3 sides of the square

00:54 We'll substitute appropriate values in the equation

01:01 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the square below:

Is the sum of the two diagonals greater than the sum of the 3 sides of the square?

Step-by-step solution

Let's look at triangle BCD, let's calculate the diagonal by the Pythagorean theorem:

$DC^2+BC^2=BD^2$

As we are given one side, we know that the other sides are equal to 4, so we will replace accordingly in the formula:

$4^2+4^2=BD^2$

$16+16=BD^2$

$32=BD^2$

We extract the root: $BD=AC=\sqrt{32}$

Now we calculate the sum of the diagonals:

$2\times\sqrt{32}=11.31$

Now we calculate the sum of the 3 sides of the square:

$4\times3=12$

And we reveal that the sum of the two diagonals is less than the sum of the 3 sides of the square.

$11.31 < 12$

Final Answer

Key Points to Remember

Essential concepts to master this topic

Rule: Square diagonal equals side squared plus side squared
Technique: Calculate $\sqrt{4^2 + 4^2} = \sqrt{32} = 5.66$ per diagonal
Check: Compare totals: 2(5.66) = 11.31 vs 3(4) = 12 ✓

Common Mistakes

Avoid these frequent errors

Adding diagonal lengths incorrectly
Don't just add 4 + 4 = 8 for each diagonal = wrong total of 16! This ignores the Pythagorean theorem completely. Always use $\sqrt{a^2 + b^2}$ to find diagonal length first.

Practice Quiz

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Look at the triangle in the diagram. How long is side AB?

FAQ

Everything you need to know about this question

Why can't I just add the side lengths to get the diagonal?

Because a diagonal cuts diagonally across the square! You need the Pythagorean theorem because the diagonal forms the hypotenuse of a right triangle with two sides of the square.

What does √32 actually equal?

$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \approx 5.66$ . You can leave it as 4√2 for exact answers or use 5.66 for decimal approximations.

How do I know which is bigger without a calculator?

Since $\sqrt{2} \approx 1.41$ , each diagonal is $4 \times 1.41 = 5.64$ . Two diagonals: 2 × 5.64 = 11.28. Three sides: 3 × 4 = 12. So 11.28 < 12!

Why are we comparing with only 3 sides instead of 4?

The question specifically asks about 3 sides of the square, not all 4 sides. This makes the comparison more interesting since 3 × 4 = 12 is closer to 2 × 5.66 = 11.31.

Could the diagonals ever be longer than 3 sides?

Never! In any square, $2 \times \text{diagonal} = 2s\sqrt{2} \approx 2.83s$ while $3 \times \text{side} = 3s$ . Since 2.83 < 3, the diagonals are always shorter.

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Square Geometry Problem: Comparing Diagonal Sums vs Side Lengths in a 4-Unit Square

Pythagorean Theorem with Diagonal Calculations

❤️ Continue Your Math Journey!

Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why can't I just add the side lengths to get the diagonal?

What does √32 actually equal?

How do I know which is bigger without a calculator?

Why are we comparing with only 3 sides instead of 4?

Could the diagonals ever be longer than 3 sides?

🌟 Unlock Your Math Potential

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