Jun Heng's answer to keith lim's Secondary 4 E Maths Singapore question.
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7a)
Angle BCE = Angle ACF (same angle)
Angle AFC = Angle BEC (corresponding angle AF//BE)
Therefore ∆BCE is similar to ∆ACF
7b)
AC = 20+30 = 50cm
BE/AF = BC/AC
x/25 = 30/50
x = 15
7c)
Angle BAE = Angle CAD (same angle)
Angle AEB = Angle ADC (corresponding angle BE//CD)
Therefore ∆BAE is similar to ∆CAD
CD/BE = AC/AB
y/15 = 50/20
y = 37.5
Angle BCE = Angle ACF (same angle)
Angle AFC = Angle BEC (corresponding angle AF//BE)
Therefore ∆BCE is similar to ∆ACF
7b)
AC = 20+30 = 50cm
BE/AF = BC/AC
x/25 = 30/50
x = 15
7c)
Angle BAE = Angle CAD (same angle)
Angle AEB = Angle ADC (corresponding angle BE//CD)
Therefore ∆BAE is similar to ∆CAD
CD/BE = AC/AB
y/15 = 50/20
y = 37.5
Date Posted:
1 year ago
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