AB=DC=20cm. AE=3cm. Area of AEFD= area of EBCF. Prove that DF=17 cm.
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Given : AB=DC=20cm. AE=3cm. Area of AEFD= area of EBCF.
To find : Prove that DF=17 cm
Solution:
AB=DC=20 cm
AE = 3 cm
EB = AB - AE = 20 - 3 = 17 cm
Let say
DF = x cm => CF = DC - DF = 20 - x cm
Height = h cm
Area of trapezium AEFD
= (1/2) ( AE + DF) * height
= (1/2) ( 3 + x) * h
Area of trapezium EBCF
= (1/2) ( EB + CF) * height
= (1/2) ( 17 + 20 - x ) * h
= (1/2) ( 37 - x ) * h
Area of AEFD= area of EBCF
=> (1/2) ( 3 + x) * h = (1/2) ( 37 - x ) * h
=> 3 + x = 37 - x
=> 2x = 34
=> x = 17
DF = 17 cm
QED
Hence proved
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