Pythagorean Theorem worksheet with six problems requiring calculation of missing side lengths using the theorem.

Let's solve each problem step by step using the Pythagorean Theorem, which states:

> For a right triangle:
> $$
> a^2 + b^2 = c^2
> $$
> where $ c $ is the hypotenuse (the side opposite the right angle), and $ a $ and $ b $ are the other two sides.

We’ll apply this to each shape, identifying right triangles within them and solving for the missing side. Round answers to the nearest tenth.

---

1) Parallelogram with height and base

Given:
- Base = 3 m
- Height = 8 m (this is perpendicular to the base)
- We need to find the slant side $ y $

This forms a right triangle with:
- One leg = 3 m (base)
- Other leg = 8 m (height)
- Hypotenuse = $ y $

$$
y^2 = 3^2 + 8^2 = 9 + 64 = 73
$$
$$
y = \sqrt{73} \approx 8.544
$$
Rounded to nearest tenth:
$ y = 8.5 $ m

✔ Answer: $ y = 8.5 $

---

2) Triangle with altitude

Given:
- Base = 10 yd
- Altitude = 11 yd (perpendicular to base)
- We need to find $ x $, one of the equal sides (assuming it's an isosceles triangle)

The altitude splits the base into two equal parts:
Each half = $ \frac{10}{2} = 5 $ yd

Now we have a right triangle with:
- Leg 1 = 5 yd
- Leg 2 = 11 yd
- Hypotenuse = $ x $

$$
x^2 = 5^2 + 11^2 = 25 + 121 = 146
$$
$$
x = \sqrt{146} \approx 12.083
$$
Rounded to nearest tenth:
$ x = 12.1 $ yd

✔ Answer: $ x = 12.1 $

---

3) Kite-shaped figure

Given:
- Diagonal = 13 ft (horizontal)
- Vertical segment = 13 ft (from center to top)
- Need to find $ b $, the length from center to bottom vertex?

Wait — actually, looking at the diagram: it appears the vertical diagonal is split into two segments: one labeled 13 ft (top part), and we're to find $ b $, the full vertical diagonal? But that doesn’t make sense.

Actually, let’s interpret carefully:

It looks like a kite with diagonals intersecting at right angles. One diagonal is horizontal: total length 13 ft → so half is 6.5 ft.

The vertical diagonal has a segment labeled 13 ft from the center to the top point, and we’re to find $ b $, which is the full vertical diagonal? But wait — no, in the diagram, it shows a right triangle formed by:
- One leg = 6.5 ft (half of 13 ft horizontal)
- Other leg = 13 ft (vertical from center to top)
- Hypotenuse = $ b $? No — wait, label says “b” is the side of the kite.

So likely, $ b $ is the length of one of the kite’s edges.

In the right triangle:
- One leg = 6.5 ft (half of horizontal diagonal)
- Other leg = 13 ft (vertical from center to top)
- Hypotenuse = $ b $

So:
$$
b^2 = 6.5^2 + 13^2 = 42.25 + 169 = 211.25
$$
$$
b = \sqrt{211.25} \approx 14.534
$$
Rounded to nearest tenth:
$ b = 14.5 $ ft

✔ Answer: $ b = 14.5 $

---

4) Rectangle with diagonal

Given:
- Width = 7 in
- Length = 6 in
- Diagonal = $ n $

Use Pythagorean Theorem:
$$
n^2 = 7^2 + 6^2 = 49 + 36 = 85
$$
$$
n = \sqrt{85} \approx 9.219
$$
Rounded to nearest tenth:
$ n = 9.2 $ in

✔ Answer: $ n = 9.2 $

---

5) Trapezoid with height and legs

Given:
- Top base = 3 yd
- Bottom base = 11 yd
- Height = 9 yd
- Need to find $ p $, the slant side (non-parallel side)

The difference in bases is $ 11 - 3 = 8 $ yd. This extra length is distributed on both sides.

Assuming it's symmetric, each overhang is $ \frac{8}{2} = 4 $ yd.

So, the slant side $ p $ forms a right triangle with:
- One leg = 4 yd (horizontal)
- Other leg = 9 yd (vertical)
- Hypotenuse = $ p $

$$
p^2 = 4^2 + 9^2 = 16 + 81 = 97
$$
$$
p = \sqrt{97} \approx 9.849
$$
Rounded to nearest tenth:
$ p = 9.8 $ yd

✔ Answer: $ p = 9.8 $

---

6) Rectangle with diagonal

Given:
- One side = 13 ft
- Other side = 12 ft
- Diagonal = $ h $

$$
h^2 = 13^2 + 12^2 = 169 + 144 = 313
$$
$$
h = \sqrt{313} \approx 17.692
$$
Rounded to nearest tenth:
$ h = 17.7 $ ft

✔ Answer: $ h = 17.7 $

---

✔ Final Answers:

1) $ y = 8.5 $
2) $ x = 12.1 $
3) $ b = 14.5 $
4) $ n = 9.2 $
5) $ p = 9.8 $
6) $ h = 17.7 $

All rounded to the nearest tenth.

Let me know if you'd like a visual explanation or help with any specific step!