【2024年最高峰の図形問題】難しいけど解けると楽しい最難関中学の図形の難問【中学受験の算数】

Manabi Square's book "A Book That Anyone Can Solve in Just One Day for Kaisei and Nada Junior High School Math Entrance Exams" → https://amzn.asia/d/4kUl4x2 [Difficulty: ★★★☆☆] This is a 2024 Todaiji Gakuen Junior High School entrance exam question. ▼Important Solution Points (1) Step 1: Check the problem's assumptions. There is a pentagon ABCDE, and angles C, D, and E are right angles. BC is 10.5cm, DE is 33cm, and the area of ​​triangle BCF is 27cm², and the area of ​​triangle AFD is 432cm². (2) Step 2: Consider the area of ​​triangle BCD. The height is 10.5cm relative to the base CD. The area can be calculated by multiplying the base by the height and dividing by 2, so CD x 10.5 ÷ 2. (3) Step 3: Next, consider the area of ​​triangle ACD. The height relative to the base CD is 33cm. Similarly, the area is CD × 33 ÷ 2. (4) Step 4: Consider the ratio of the areas of triangles BCD and ACD. BCD is CD × 5.25, and ACD is CD × 16.5, so the area ratio is 5.25:16.5. This can be easily converted to a ratio of 2:7. (5) Step 5: If the area of ​​triangle BCD is assumed to be 70, the area of ​​triangle ACD is 70 × 7 ÷ 2, or 220. Using this ratio, find the difference between the areas of triangles BCD and ACD. The difference is 220 - 70 = 150. (6) Step 6: The area of ​​triangle BCD is 27cm² plus the area of ​​triangle CDF, and the area of ​​triangle ACD is 432cm² plus the area of ​​triangle CDF. This difference is 432 - 27 = 405cm². (7) Step 7: The difference between the area of ​​triangle BCD (70) and the area of ​​triangle ACD (220), 150, is actually 405 cm². Therefore, 405 cm² relative to 70 is 27 cm² relative to 1. (8) Step 8: The area of ​​triangle BCD is 27 cm², which is 7 times 70. Therefore, the area of ​​70 is 27 × 7 = 189 cm². Similarly, the area of ​​triangle ACD is 7 times 220, so the area of ​​220 is 189 + 405 = 594 cm². (9) Step 9: Subtracting the area of ​​triangle BCF (27 cm²) from triangle BCD gives the area of ​​triangle CDF (162 cm²). Also, the length of CD can be found from the area of ​​triangle BCD (189 cm²), which is 36 cm. (10) Step 10: Find the area of ​​triangle ABF. Since triangles BCF and ADF have equal heights, the ratio of their base lengths is also the ratio of their areas. The ratio of their areas, 27:162, is 1:6, so the ratio of the bases BF:FD is also 1:6. (11) Step 11: The area of ​​triangle ABF is ADF's area of ​​432 cm² divided by 6. Therefore, the area of ​​triangle ABF is 72 cm². (12) Step 12: Find the area of ​​triangle ABC. This is the sum of the area of ​​triangle ABF, 72 cm², and the area of ​​triangle BCF, 27 cm², or 99 cm². AG is 18 cm. (13) Step 13: Quadrilateral GCDE is a rectangle, and GE has the same length as CD, 36 cm. Therefore, AE is 36 cm - 18 cm = 18 cm. (14) Step 14: Finally, find the area of ​​triangle ADE. Since the base AE is 18cm and the height DE is 33cm, the area is 18 × 33 ÷ 2 = 297cm². Following these steps, the area of ​​triangle ADE is 297cm². (This description was generated by AI.) ▼ManaviSquare (Manavi Square) related pages are available here. ・Website https://manavigate.co.jp/ ・Membership    / @manavisquare   ・ManaviSquare (Online Tutoring Platform) https://mnsq.jp/ ・Twitter   / manavisquare   ・Yuta Sugato's Twitter   / mrkeiosfc16no1   ▼Please feel free to contact us! [email protected] #JuniorHighSchoolEntranceExams #Math #Geometry