Some Geometry Questions
Some friends have requested for some help for EOY questions in the HCI paper. Here are the solutions.
Q15)
Q15)
a)
angle EAB = 65 (base of isoc triangle)
angle AEF = angle EAB = 65 (alt. angles)
cos 65 = FE/EA
6 cos 65 = FE = roughly 2.54 (3sf)
therefore FE = roughly 2.54 (3sf)
b)
angle EAB = 65 (base of isoc triangle)
angle AEF = angle EAB = 65 (alt. angles)
cos 65 = FE/EA
6 cos 65 = FE = roughly 2.54 (3sf)
therefore FE = roughly 2.54 (3sf)
b)
AF = ^2} "\sqrt {AE^2 - FE^2} = \sqrt {6^2 - (6 cos 65)^2}")
AF = roughly 5.44 (3sf)
sin angle D = sin 55 = 5.44/AD
1/sin55 = AD/5.44
5.44/sin 55 = AD
therefore AD = roughly 6.64 (3sf)
sin angle D = sin 55 = 5.44/AD
1/sin55 = AD/5.44
5.44/sin 55 = AD
therefore AD = roughly 6.64 (3sf)
Q16)
The question:
ABC is a right-angled triangle with angle ABC - 90 degrees. Given that AD = 10cm, DB = 6cm, BE = 8cm and EC = 10cm, find the shaded area.
See the solution below. We'll make use of menelaus theorem
ABC is a right-angled triangle with angle ABC - 90 degrees. Given that AD = 10cm, DB = 6cm, BE = 8cm and EC = 10cm, find the shaded area.
See the solution below. We'll make use of menelaus theorem
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