Gripe Trafikkork skråningen converse of pythagoras sy Cosmic Natur - Free Printable

It looks like you've uploaded an image of a triangle with side lengths labeled:

- One side is 24 units,
- Another side is 22 units,
- The base is 26 units.

Since no specific question was stated, I’ll assume you're asking for a common problem related to this triangle — such as finding the area. Let's solve for the area of the triangle using Heron’s formula, which is ideal when all three sides are known.

---

Step 1: Use Heron’s Formula

Heron’s formula states that the area $ A $ of a triangle with side lengths $ a $, $ b $, and $ c $ is:

$$
A = \sqrt{s(s - a)(s - b)(s - c)}
$$

where $ s $ is the semi-perimeter:

$$
s = \frac{a + b + c}{2}
$$

Let’s assign:
- $ a = 24 $
- $ b = 22 $
- $ c = 26 $

---

Step 2: Compute the semi-perimeter $ s $

$$
s = \frac{24 + 22 + 26}{2} = \frac{72}{2} = 36
$$

---

Step 3: Plug into Heron’s formula

$$
A = \sqrt{36(36 - 24)(36 - 22)(36 - 26)} = \sqrt{36 \times 12 \times 14 \times 10}
$$

Now compute the product inside the square root:

$$
36 \times 12 = 432 \\
14 \times 10 = 140 \\
432 \times 140 = ?
$$

Break it down:

$$
432 \times 140 = 432 \times (100 + 40) = 43200 + 17280 = 60480
$$

So,

$$
A = \sqrt{60480}
$$

Now simplify $ \sqrt{60480} $:

Let’s factor it:

$$
60480 = 16 \times 3780 \\
= 16 \times 9 \times 420 \\
= 16 \times 9 \times 4 \times 105 \\
= 16 \times 9 \times 4 \times 3 \times 35 \\
= 16 \times 9 \times 4 \times 3 \times 5 \times 7
$$

Group perfect squares:

$$
= (4^2) \times (3^2) \times (2^2) \times 3 \times 5 \times 7 \\
= (4 \times 3 \times 2)^2 \times (3 \times 5 \times 7) \\
= (24)^2 \times 105
$$

Wait — actually:

$$
\sqrt{60480} = \sqrt{16 \times 9 \times 4 \times 105} = \sqrt{16} \times \sqrt{9} \times \sqrt{4} \times \sqrt{105} = 4 \times 3 \times 2 \times \sqrt{105} = 24\sqrt{105}
$$

But let’s check numerically:

$$
\sqrt{60480} \approx \sqrt{60516} \approx 246 \quad (\text{since } 246^2 = 60516)
$$

Try $ 245^2 = 60025 $, $ 246^2 = 60516 $, so:

$$
\sqrt{60480} \approx 245.93
$$

So the area is approximately:

$$
A \approx 245.93 \text{ square units}
$$

But let’s do exact calculation:

$$
A = \sqrt{36 \times 12 \times 14 \times 10}
$$

We can simplify step by step:

$$
= \sqrt{(36 \times 12) \times (14 \times 10)} = \sqrt{432 \times 140}
$$

Alternatively, factor each number:

- $ 36 = 6^2 $
- $ 12 = 4 \times 3 = 2^2 \times 3 $
- $ 14 = 2 \times 7 $
- $ 10 = 2 \times 5 $

So total:

$$
36 \times 12 \times 14 \times 10 = (6^2) \times (2^2 \times 3) \times (2 \times 7) \times (2 \times 5) = 2^{2+1+1} \times 3^{1+1} \times 5 \times 7 \times 6^2
$$

Wait — better to write all prime factors:

- $ 36 = 2^2 \times 3^2 $
- $ 12 = 2^2 \times 3 $
- $ 14 = 2 \times 7 $
- $ 10 = 2 \times 5 $

Total:

$$
2^{2+2+1+1} \times 3^{2+1} \times 5 \times 7 = 2^6 \times 3^3 \times 5 \times 7
$$

Now take square root:

$$
\sqrt{2^6 \times 3^3 \times 5 \times 7} = 2^3 \times 3^{1} \times \sqrt{3 \times 5 \times 7} = 8 \times 3 \times \sqrt{105} = 24\sqrt{105}
$$

So the exact area is:

$$
\boxed{24\sqrt{105}} \text{ square units}
$$

Approximately:

$$
\sqrt{105} \approx 10.24695 \\
24 \times 10.24695 \approx 245.93 \text{ square units}
$$

---

✔ Final Answer:

$$
\boxed{24\sqrt{105}} \text{ or approximately } \boxed{245.93} \text{ square units}
$$

Let me know if you wanted something else (like angles, perimeter, height, etc.)!