In the figure, C lies on AE, ∆ABC and ∆CDE are equilateral triangles, F and G are
the midpoints of BC and DE respectively. If the area of ∆ABC is 24 cm2
, the area of ∆CDE is 60 cm2, and the area of ∆AFG is Q cm2, find the value of Q.
HKMO 2003-2004 area of ∆
2013-06-02 7:18 pm
回答 (4)
2013-06-03 6:28 am
✔ 最佳答案
As F, G are the mid pts of BC, DE respectively, soangle FAC = angle GCE = 30°
therefore, FA // GC, (corr angles are equal)
so,
Area of △AFG
= Area of △AFC ... ... ... (same base, same height)
= (1/2) * Area of △ABC ... (as F is the mid pt of AC)
= (1/2) * 24 cm2
= 12 cm2
Therefore, Q = 12
2013-06-05 1:03 am
伍家榮，你唔會唔識掛！
2013-06-02 10:24 pm
Sol
設 A(0，0)，C(a，0)，E(a+b，0)，a>0，b>0
B(a/2，√3a/2)，D(a+b/2，√3b/2)
F(3a/4，√3a/4)，G(a+3b/4，3√b/2)
｜１ ０ ０｜
｜１ ３ａ／４ √３ａ／４｜
｜１ ａ＋３ｂ／４ √３ｂ／４｜
｜ ３ａ／４ √３ａ／４｜
＝｜ａ＋３ｂ／４ √３ｂ／４｜
=3√3ab/16－√3a^2/4－3√3ab/16
=－√3a^2/4
=－24
△AFG=(1/2)*24=12
設 A(0，0)，C(a，0)，E(a+b，0)，a>0，b>0
B(a/2，√3a/2)，D(a+b/2，√3b/2)
F(3a/4，√3a/4)，G(a+3b/4，3√b/2)
｜１ ０ ０｜
｜１ ３ａ／４ √３ａ／４｜
｜１ ａ＋３ｂ／４ √３ｂ／４｜
｜ ３ａ／４ √３ａ／４｜
＝｜ａ＋３ｂ／４ √３ｂ／４｜
=3√3ab/16－√3a^2/4－3√3ab/16
=－√3a^2/4
=－24
△AFG=(1/2)*24=12
2013-06-02 7:25 pm
Where is the figure?
收錄日期: 2021-04-13 19:30:10
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130602000051KK00090