## 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 = 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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