Fraction Midpoint F41D62

1. **Problem 1: Find the fraction exactly halfway between $\frac{1}{3}$ and $\frac{3}{7}$.\n\n2. To find the fraction halfway between two fractions, we calculate their average using the formula: $$\text{Midpoint} = \frac{\frac{1}{3} + \frac{3}{7}}{2}.$$\n\n3. Find a common denominator to add $\frac{1}{3}$ and $\frac{3}{7}$. The least common denominator (LCD) of 3 and 7 is 21.\n\n4. Convert each fraction: $$\frac{1}{3} = \frac{7}{21}, \quad \frac{3}{7} = \frac{9}{21}.$$\n\n5. Add the fractions: $$\frac{7}{21} + \frac{9}{21} = \frac{7+9}{21} = \frac{16}{21}.$$\n\n6. Now divide by 2 to find the midpoint: $$\frac{\frac{16}{21}}{2} = \frac{16}{21} \times \frac{1}{2} = \frac{16}{42}.$$\n\n7. Simplify $\frac{16}{42}$ by dividing numerator and denominator by their greatest common divisor 2: $$\frac{\cancel{16}^{8}}{\cancel{42}^{21}} = \frac{8}{21}.$$\n\n**Answer for Problem 1:** The fraction halfway between $\frac{1}{3}$ and $\frac{3}{7}$ is $\frac{8}{21}$.\n\n---\n\n1. **Problem 2: Which number is closest to $\frac{11}{12} + \frac{7}{8}$?**\n\n2. Find a common denominator for $\frac{11}{12}$ and $\frac{7}{8}$. The LCD of 12 and 8 is 24.\n\n3. Convert each fraction: $$\frac{11}{12} = \frac{22}{24}, \quad \frac{7}{8} = \frac{21}{24}.$$\n\n4. Add the fractions: $$\frac{22}{24} + \frac{21}{24} = \frac{43}{24}.$$\n\n5. Convert $\frac{43}{24}$ to a mixed number: $$43 \div 24 = 1 \text{ remainder } 19,$$ so $$\frac{43}{24} = 1 \frac{19}{24} \approx 1.79.$$\n\n6. Compare to the given options: 1, 2, 4, 18, 20. The closest number to 1.79 is 2.\n\n**Answer for Problem 2:** The number closest to $\frac{11}{12} + \frac{7}{8}$ is 2.\n\n---\n\n1. **Problem 3: Find the area of the kite with diagonals and side lengths given.**\n\n2. The kite has two pairs of equal sides: 10 cm and 17 cm, and one diagonal is given as 16 cm (horizontal).\n\n3. The area of a kite is given by: $$\text{Area} = \frac{d_1 \times d_2}{2},$$ where $d_1$ and $d_2$ are the lengths of the diagonals.\n\n4. We know $d_1 = 16$ cm. We need to find $d_2$.\n\n5. Using the properties of kites, the diagonals are perpendicular, and the diagonal $d_2$ is split into two segments by $d_1$. Let the half-lengths of $d_2$ be $x$ and $y$.\n\n6. Using the Pythagorean theorem on the two right triangles formed:\n- For side 10 cm: $$10^2 = x^2 + 8^2 \Rightarrow 100 = x^2 + 64 \Rightarrow x^2 = 36 \Rightarrow x = 6.$$\n- For side 17 cm: $$17^2 = y^2 + 8^2 \Rightarrow 289 = y^2 + 64 \Rightarrow y^2 = 225 \Rightarrow y = 15.$$\n\n7. Therefore, $$d_2 = x + y = 6 + 15 = 21 \text{ cm}.$$\n\n8. Calculate the area: $$\text{Area} = \frac{16 \times 21}{2} = \frac{336}{2} = 168 \text{ cm}^2.$$\n\n**Answer for Problem 3:** The area of the kite is $168$ cm$^2$.