Question

Fearghal Hurley

請問這題 謝謝

10. 已知P為正方形ABCD內部的一點,若
AP=7.BP=5.CP=1,試求正方形
ABCD 的面積是
x
B
D
x
C

Answer 1

✨ คำตอบที่ดีที่สุด ✨

qn

cosθ = (x²+5²-1²)/(2·5·x)
cosφ = (x²+5²-7²)/(2·5·x) = sinθ

sin²θ + cos²θ = 1
[ (x²-24)/(10x) ]² + [ (x²+24)/(10x) ]² = 1
(x²-24)²/(100x²) + (x²+24)²/(100x²) = 1
(x²-24)² + (x²+24²) = 100x²
2[(x²)² + 24²] = 100x²
(x²)² + 484 = 100x²
(x²)² - 100x² + 484 = 0
x² = [ 100 ± √(10000 - 1936) ]/2
= 50 ± √(8064)/2
= 50 ± √2016
= 50 ± 12√14

但由於 7 < AC
⇒ 7² < 2x²
⇒ x² > 24.5
而其中 50-12√14 = 50 - √2016 < 50 - 44 = 6
所以“±”取“+”，故所求 = x² = 50+12√14

Fearghal Hurley

不好意思再問一個問題 cosφ = sinθ是如何判斷的

qn

θ + φ = ∠ABC = 90°