The problem asks us to find the value of $x$ in 10 different diagrams. All lengths are in cm. Several diagrams involve right triangles, which implies we might need to use the Pythagorean theorem. We need to apply appropriate geometric principles to determine $x$ in each case.

1. Problem Description

The problem asks us to find the value of $x$ in 10 different diagrams. All lengths are in cm. Several diagrams involve right triangles, which implies we might need to use the Pythagorean theorem. We need to apply appropriate geometric principles to determine $x$ in each case.

2. Solution Steps

Let's solve each problem.

1. Applying the Pythagorean theorem: $x^2 = 6^2 + 8^2$. So, $x^2 = 36 + 64 = 100$. Then, $x = \sqrt{100} = 10$.

2. Applying the Pythagorean theorem: $9^2 = x^2 + 8^2$. So, $81 = x^2 + 64$. Then, $x^2 = 81 - 64 = 17$. Therefore, $x = \sqrt{17}$.

3. Applying the Pythagorean theorem: $x^2 = x^2 + 6^2$. This implies that the two sides labelled as 'x' are adjacent and equal to each other. Therefore, this is an isosceles right triangle, so $x = 6$.

4. The diagram is not clear, so it is not possible to provide a numerical answer for $x$. It can't be determined without more information. Assume this is a right triangle. Then $x^2 + x^2 = x^2$, which would only be true when $x = 0$, which doesn't make sense for a side length. Assume the sides have lengths x, x, and side length a. Applying the law of cosines: $x^2 = x^2 + x^2 - 2(x)(x) cos{\theta}$. $x^2 = 2x^2 - 2x^2 cos{\theta}$. $0 = x^2 - 2x^2 cos{\theta}$. We still need the angle or one side to solve this. Assume the triangle is isosceles so $x=x$ for the two sides. Assume angle of 90 degrees between the two sides, thus $x^2 + x^2 = side3^2$. $2x^2 = x^2$. Therefore $x=0$, which does not make sense. Can't solve this without more information.

5. The figure contains two right triangles, one with sides 3 and 6 and a hypotenuse of 7, and another one with sides 3 and x and a hypotenuse of x. From the diagram, it looks like we have two parallel lines, the one with length 6, and the one with length x. From similar triangles we can say $\frac{3}{3+3} = \frac{7}{7+x}$. Solving this equation: $\frac{3}{6} = \frac{7}{7+x}$. Therefore $1/2 = \frac{7}{7+x}$. Thus $7+x=14$ and $x=7$.

6. The diagram has an altitude splitting the triangle into two congruent right triangles. The triangle is isosceles. From the right triangles, $5^2 + 3^2 = x^2$. $25+9=x^2$. $x^2 = 34$. So, $x=\sqrt{34}$.

7. There is a right triangle with legs of length 4, and an unknown length, and hypotenuse $x+8$. The side with the length 2 is part of the hypotenuse $x$. So $4^2 + unknown^2 = x^2$, $16 + unknown^2 = x^2$. Also using triangle similarity we have $4/6 = x/8$. So $4*8 = 6x$. So $x = 32/6 = 16/3$.

8. Applying the Pythagorean theorem: $12^2 + x^2 = (x+8)^2$. $144 + x^2 = x^2 + 16x + 64$. $144 = 16x + 64$. $80 = 16x$. Then, $x = \frac{80}{16} = 5$.

9. Applying the Pythagorean theorem: $8^2 + (x-2)^2 = x^2$. $64 + x^2 - 4x + 4 = x^2$. $68 - 4x = 0$. So, $4x = 68$. Then, $x = \frac{68}{4} = 17$.

1

0. Applying the Pythagorean theorem: $(x+3)^2 + [4(x+2)]^2 = 25^2$. $x^2 + 6x + 9 + 16(x^2 + 4x + 4) = 625$. $x^2 + 6x + 9 + 16x^2 + 64x + 64 = 625$. $17x^2 + 70x + 73 = 625$. $17x^2 + 70x - 552 = 0$. This is a quadratic equation. Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $x = \frac{-70 \pm \sqrt{70^2 - 4(17)(-552)}}{2(17)}$. $x = \frac{-70 \pm \sqrt{4900 + 37536}}{34}$. $x = \frac{-70 \pm \sqrt{42436}}{34}$. $x = \frac{-70 \pm 206.0}{34}$. Since $x$ must be positive, $x = \frac{-70 + 206}{34} = \frac{136}{34} = 4$.

3. Final Answer

1. $x = 10$

2. $x = \sqrt{17}$

3. $x = 6$

4. Cannot be solved.

5. $x = 7$

6. $x = \sqrt{34}$

7. $x = 16/3$

8. $x = 5$

9. $x = 17$

1

0. $x = 4$