Maths problem, thanks?

Question

Answer 1

最佳解答

△ABC : △ADE = AC : AE

△ABC : 1 = (AC/AD) : (AE/AD)

△ABC = (1/cos60°) : cos60°

△ABC = 2 : 1/2

△ABC = 4 cm²

The answer is B.

Answer 2

∠B=∠E=90⁰

∠BAC=30⁰=∠EDA

∠BCA=60⁰=∠EAD

∴ △ABC ～ △DEA (AAA)

=> area of △ABC ：area of △DEA =AC² ：DA²

=> area of △ABC ：1 = AC² ：DA²

=> area of △ABC = (AC/DA)² . . . . . . . . . . . . . . . ①

　　　　　　　　　　　　　　　DA

From rt-angled △ADC, cos 60⁰ = ------

　　　　　　　　　　　　　　　AC

　　　　　　　　　　 　　½ = DA/AC

　　　　　　　　　

From ①： area of △ABC = (2/1)² = 4　　(ans : B )

• SC147

  Lv 6

  3 月 前舉報

  Formula:
  =======
  area of △ABC ：area of △XYZ = AB² ：XY² (corresponding sides of the 2△s)
  ** In this problem : AC is the hypotenuse of △ABC, while DA is the hypotenuse of △DEA
  ∴ area of △ABC ：area of △DEA = AC² ：DA²
  

Answer 3

△ABC : △ADE = AC : AE

△ABC : 1 = (AC/AD) : (AE/AD)

△ABC = (1/cos60°) : cos60°

△ABC = 2 : 1/2

△ABC = 4 cm²

The answer is B.

Answer 4

△ADE ～ △CAB

area△ADE : area△ABC = AD ² : AC ² = 1² : 2² = 1 : 4

4cm²

B. #