In rectangle ABCD, C is trisected by CE and CF where Elies on AB and Fon AD. IF BE = 6 cm and AF = 2 cm, which of the
following integers is nearest to the area of the rectangle ABCD in sq.cm.:
190
170
O 150
0 130
Answers
Given : In rectangle ABCD, ∠C is trisected by CE and CF where E lies on AB and F on AD. If BE = 6 cm and AF = 2 cm
To find : area of the rectangle ABCD in sq.cm. nearest to integers
Solution:
∠C = 90° trisected by CE & CF
∠BCE = ∠ECF - ∠FCD = 90°/3 = 30°
in ΔCBE
tan30° = BE/BC
=> 1/√3 = 6/BC
=> BC = 6√3
AD = BC ( opposite sides of rectangle)
=> AD = 6√3
DF = AD - AF
=> DF = 6√3 - 2
in ΔCDF
Tan30° = DF/DC
=> 1/√3 = (6√3 - 2)/DC
=> DC = 18 - 2√3
Area of rectangle = BC * CD
= 6√3 (18 - 2√3)
= 108√3 - 36
√3 = 1.73
= 150.84
area of the rectangle ABCD = 150 cm²
150 is the correct answer
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