SOLUTION: 8 multi step pythagorean theorem problems - Studypool - Free Printable

It looks like you've uploaded an image containing several geometry problems involving triangles, with labeled side lengths and heights (perpendiculars from the apex to the base). The goal appears to be calculating areas of these triangles using the formula:

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
$$

Let’s solve each problem one by one.

---

Problem 9)

- Base = 4
- Height = 3
- Sides: 7 and 4 (not needed for area)

$$
\text{Area} = \frac{1}{2} \times 4 \times 3 = 6
$$

✔ Answer: 6 square units

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Problem 10)

- Base = 6.4
- Height = ? (not given directly)
- Two equal sides = 9 → this is an isosceles triangle

We need to find the height using the Pythagorean theorem.

The height splits the base into two equal parts:
$$
\frac{6.4}{2} = 3.2
$$

Now, in one of the right triangles formed:
- Hypotenuse = 9
- One leg = 3.2
- Find the other leg (height):

$$
h^2 + 3.2^2 = 9^2 \\
h^2 + 10.24 = 81 \\
h^2 = 70.76 \\
h = \sqrt{70.76} \approx 8.41
$$

Now compute area:
$$
\text{Area} = \frac{1}{2} \times 6.4 \times 8.41 \approx 26.91
$$

✔ Answer: ≈ 26.91 square units

---

Problem 11)

- Base = 11
- Height = ?
- Sides: 29 and 26

We can use the Pythagorean theorem again, but we must first determine if the height splits the base into segments.

Let the foot of the height divide the base into two parts: $ x $ and $ 11 - x $

Then, from the two right triangles:
$$
x^2 + h^2 = 29^2 = 841 \\
(11 - x)^2 + h^2 = 26^2 = 676
$$

Subtract the equations:
$$
[x^2 + h^2] - [(11 - x)^2 + h^2] = 841 - 676 \\
x^2 - (121 - 22x + x^2) = 165 \\
x^2 - 121 + 22x - x^2 = 165 \\
22x = 286 \\
x = 13
$$

Wait — but $ x = 13 $, and base is only 11 → impossible.

So contradiction! That means our assumption about which side is opposite is wrong?

Wait — perhaps the height is not drawn from the vertex between the 29 and 26 sides? But it's shown as a perpendicular from the top vertex to the base (labeled 11), so the base is 11.

But the sides adjacent to the base are 29 and 26.

So let’s try again:

Let the height split the base into $ x $ and $ 11 - x $

Then:
$$
x^2 + h^2 = 29^2 = 841 \quad \text{(1)}\\
(11 - x)^2 + h^2 = 26^2 = 676 \quad \text{(2)}
$$

Subtract (2) from (1):
$$
[x^2 + h^2] - [(11 - x)^2 + h^2] = 841 - 676 = 165 \\
x^2 - (121 - 22x + x^2) = 165 \\
x^2 - 121 + 22x - x^2 = 165 \\
22x = 286 \\
x = 13
$$

But $ x = 13 > 11 $ → impossible!

This suggests that such a triangle cannot exist, or we’ve misread the diagram.

Wait — could the base be not the side of length 11?

Looking at the diagram: the base is labeled 11, and the height is drawn from the top vertex to the base. So yes, base = 11.

But then the two other sides are 29 and 26.

But the sum of any two sides must be greater than the third.

Check triangle inequality:
- 29 + 26 = 55 > 11 ✔
- 29 + 11 = 40 > 26 ✔
- 26 + 11 = 37 > 29 ✔

So triangle exists.

But earlier calculation gave $ x = 13 $, which is longer than the base — contradiction.

Wait — maybe the height falls outside the triangle? That would happen if the triangle is obtuse.

But the height is shown inside the triangle (since the perpendicular is drawn to the base, and the red square is on the base), so it should fall within.

But mathematically, the foot of the perpendicular may lie outside.

Wait — no, if the height is drawn from the vertex to the base, and the red square is on the base, then it's assumed the foot lies on the segment.

But our result $ x = 13 $ implies the foot is beyond the base.

So perhaps the height is not to the side of length 11?

Wait — the base is labeled 11, and the height is drawn from the opposite vertex to that base.

So unless the triangle is obtuse, the foot should lie within.

But let's recheck.

Let me denote:
- Let $ AB = 11 $, $ AC = 29 $, $ BC = 26 $
- Draw height from $ C $ to $ AB $, meeting at point $ D $, with $ AD = x $, $ DB = 11 - x $
- Then:
$$
CD^2 = AC^2 - AD^2 = 29^2 - x^2 = 841 - x^2 \\
CD^2 = BC^2 - BD^2 = 26^2 - (11 - x)^2 = 676 - (121 - 22x + x^2) = 676 - 121 + 22x - x^2 = 555 + 22x - x^2
$$

Set equal:
$$
841 - x^2 = 555 + 22x - x^2 \\
841 = 555 + 22x \\
286 = 22x \\
x = 13
$$

Again $ x = 13 $, so $ AD = 13 $, but $ AB = 11 $, so $ D $ lies outside the segment $ AB $, beyond $ B $

That means the triangle is obtuse at $ B $, and the height from $ C $ falls outside.

But the diagram shows the height inside the triangle, so either the diagram is misleading or we have a labeling issue.

But the problem says the base is 11, and the height is drawn to it.

In such cases, we still use the formula, but the height is measured from the vertex to the line of the base.

So even if the foot is outside, the area is still:
$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
$$

From above:
$$
CD^2 = 841 - x^2 = 841 - 169 = 672 \\
CD = \sqrt{672} \approx 25.92
$$

Then:
$$
\text{Area} = \frac{1}{2} \times 11 \times \sqrt{672} = \frac{11}{2} \times \sqrt{672}
$$

Simplify $ \sqrt{672} $:
$$
672 = 16 \times 42 \Rightarrow \sqrt{672} = 4\sqrt{42} \approx 4 \times 6.48 = 25.92
$$

So:
$$
\text{Area} \approx 5.5 \times 25.92 \approx 142.56
$$

Alternatively, use Heron’s formula to verify.

Heron’s Formula (for Problem 11)

Sides: $ a = 11 $, $ b = 26 $, $ c = 29 $

Semi-perimeter:
$$
s = \frac{11 + 26 + 29}{2} = \frac{66}{2} = 33
$$

Area:
$$
A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{33(33-11)(33-26)(33-29)} = \sqrt{33 \times 22 \times 7 \times 4}
$$

Calculate:
- $ 33 \times 22 = 726 $
- $ 7 \times 4 = 28 $
- $ 726 \times 28 = ? $

Break it down:
$$
726 \times 28 = 726 \times (30 - 2) = 726 \times 30 - 726 \times 2 = 21780 - 1452 = 20328
$$

So:
$$
A = \sqrt{20328} \approx 142.58
$$

✔ Matches previous calculation.

So area ≈ 142.58 square units

But since the height was calculated via Pythagoras, and we used the correct method, we can accept it.

However, note: the height does not land on the base segment, but on its extension.

But the area is still valid.

✔ Answer: ≈ 142.58 square units

---

Problem 12)

- Base = 9
- Height = ?
- Sides: 16 and 17

Let the height split the base into $ x $ and $ 9 - x $

Then:
$$
x^2 + h^2 = 16^2 = 256 \quad \text{(1)}\\
(9 - x)^2 + h^2 = 17^2 = 289 \quad \text{(2)}
$$

Subtract (1) from (2):
$$
[(9 - x)^2 + h^2] - [x^2 + h^2] = 289 - 256 = 33 \\
(81 - 18x + x^2) - x^2 = 33 \\
81 - 18x = 33 \\
-18x = -48 \\
x = \frac{48}{18} = \frac{8}{3} \approx 2.666...
$$

Now plug back into (1):
$$
h^2 = 256 - x^2 = 256 - \left(\frac{8}{3}\right)^2 = 256 - \frac{64}{9} = \frac{2304 - 64}{9} = \frac{2240}{9} \\
h = \sqrt{\frac{2240}{9}} = \frac{\sqrt{2240}}{3}
$$

Now area:
$$
\text{Area} = \frac{1}{2} \times 9 \times h = \frac{9}{2} \times \frac{\sqrt{2240}}{3} = \frac{3}{2} \times \sqrt{2240}
$$

Simplify $ \sqrt{2240} $:
$$
2240 = 64 \times 35 \Rightarrow \sqrt{2240} = 8\sqrt{35} \approx 8 \times 5.916 = 47.33
$$

So:
$$
\text{Area} \approx \frac{3}{2} \times 47.33 = 1.5 \times 47.33 \approx 70.995
$$

Or use Heron’s formula to verify.

Heron’s Formula (Problem 12)

Sides: 9, 16, 17

Semi-perimeter:
$$
s = \frac{9 + 16 + 17}{2} = \frac{42}{2} = 21
$$

Area:
$$
A = \sqrt{21(21-9)(21-16)(21-17)} = \sqrt{21 \times 12 \times 5 \times 4}
$$

Compute:
- $ 21 \times 12 = 252 $
- $ 5 \times 4 = 20 $
- $ 252 \times 20 = 5040 $

$$
A = \sqrt{5040} \approx 71.0
$$

Yes, matches.

✔ Answer: ≈ 71.0 square units

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Problems 13–16: Blurry Images

Unfortunately, the images for problems 13 to 16 are blurry and unreadable. I cannot extract the numbers or labels.

Please upload a clearer version of those problems, or describe them (e.g., "triangle with base 10, height 5", etc.).

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Summary of Answers:

| Problem | Area (Approximate) |
|--------|---------------------|
| 9 | 6 |
| 10 | 26.91 |
| 11 | 142.58 |
| 12 | 71.0 |

> ✔ For problems 13–16: Please clarify or provide clearer images.

Let me know if you want exact values (in radicals) instead of decimals!