I can help you solve these questions. Let's break down each one:

1. Tentukan jenis segitiga berikut berdasarkan panjang sisi-sisinya:

   • Segitiga PQR dengan panjang PQ = 5 cm, QR = 3 cm, dan PR = 4 cm.

   • To determine the type of triangle based on its sides, we can apply the Pythagorean theorem for a right triangle. Let's check if it holds true:

     $PQ^2 = QR^2 + PR^2 \quad \text{(if it forms a right triangle)}$ $5^2 = 3^2 + 4^2 \quad \Rightarrow \quad 25 = 9 + 16 \quad \Rightarrow \quad 25 = 25$ Since the Pythagorean theorem holds, this is a right triangle (segitiga siku-siku).

2. Segitiga ABC dengan panjang AB = 10 cm, BC = 6 cm, dan AC = 5 cm.

   • Again, applying the Pythagorean theorem to check if it's a right triangle: $AB^2 = BC^2 + AC^2$ $10^2 = 6^2 + 5^2 \quad \Rightarrow \quad 100 = 36 + 25 \quad \Rightarrow \quad 100 \neq 61$ Since this does not satisfy the Pythagorean theorem, this is not a right triangle. We can check if it’s an obtuse triangle (segitiga tumpul) or acute triangle (segitiga lancip) by comparing the square of the longest side (AB) with the sum of squares of the other two sides: $AB^2 > BC^2 + AC^2$ Since 100 > 61, it is an obtuse triangle.

3. Panjang alas segitiga sama kaki 14 cm, garis tinggi dari titik puncak ke sisi alas dan luas segitiga tersebut jika salah satu kakinya 25 cm.

   • In this isosceles triangle, we know the base is 14 cm, and each leg (sisi kaki) is 25 cm. To find the area, we first calculate the height (garis tinggi) using the Pythagorean theorem. The height splits the base into two equal parts of 7 cm each. $h^2 + 7^2 = 25^2$ $h^2 + 49 = 625 \quad \Rightarrow \quad h^2 = 576 \quad \Rightarrow \quad h = 24 \, \text{cm}$ Now, we can find the area of the triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 24 = 168 \, \text{cm}^2$

4. Pada segitiga ABC diketahui AB = 10 cm, BC = 24 cm, dan AC = 26 cm.

   • This is a typical Pythagorean triple (10, 24, 26), indicating that it is a right triangle (segitiga siku-siku). We can show that: $AC^2 = AB^2 + BC^2 \quad \Rightarrow \quad 26^2 = 10^2 + 24^2 \quad \Rightarrow \quad 676 = 100 + 576 \quad \Rightarrow \quad 676 = 676$ Therefore, $\triangle ABC$ is a right triangle.

5. Di titik manakah $\triangle ABC$ siku-siku?

   • In this right triangle, the right angle (siku-siku) is located at point B because the two sides forming the right angle are $AB$ and $BC$.

Would you like a detailed explanation for any of these steps?

Here are five related questions to consider:

1. How do you determine the type of a triangle based on its side lengths?
2. What is the general formula for the area of a triangle?
3. How do you use the Pythagorean theorem to check if a triangle is right-angled?
4. What other properties can classify triangles besides side lengths and angles?
5. Can you find the perimeter of a triangle given its side lengths?

Tip: When working with triangles, always remember to check if the triangle is valid by ensuring that the sum of the lengths of any two sides is greater than the length of the remaining side (triangle inequality theorem).