Question

ABCD is a trapezoid which AB║CD, CD = 3 cm and AB is long two times of CD, Perpendicular between AB and CD = 3.2 cm, E is the center point of BC. Find the area of ADF.

Answer 1

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YOUR ANSWER:

First let us prove that △DCE ≅ △BEF,

∠DEC= ∠BEF (vertically opposite angles )

CE=BE ( is the midpoint of line BC )

∠ EDC =∠ EFB ( as CD is parallel to AB hence DF is parallel to CD )

Therefore , by AAS

△DCE ≅ △BEF

Therefore , by CPCT

DC=BF

Now, let us consider △ADF

draw a perpendicular from DS to side AB

we know that the perpendicular between the lines ab and CD is equal to 3.2 cm

AF is equal to AB+BF=6+3=9 cm

( AB is equal to 6 because AB is twice the size of CD and BF is equal to 3 because as we proved above BF is equal to CD)

Therefore the area of the triangle ADF is,

$1 \div 2(b)(h)$

$= 1 \div 2(9)(3.2)$

$= 9 \times 1.6$

$= 14.4 \: cm {}^{2}$

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Answer 2

ANSWER:----------

DCE ≅ △BEF

Therefore , by CPCT

DC=BF

Now, let us consider △ADF

draw a perpendicular from DS to side AB

we know that the perpendicular between the lines ab and CD is equal to 3.2 cm

AF is equal to AB+BF=6+3=9 cm

( AB is equal to 6 because AB is twice the size of CD and BF is equal to 3 because as we proved above BF is equal to CD)

Therefore the area of the triangle ADF is,

1÷2(b)(h)1 \div 2(b)(h)1÷2(b)(h)

=1÷2(9)(3.2)= 1 \div 2(9)(3.2)=1÷2(9)(3.2)

=9×1.6= 9 \times 1.6=9×1.6

=14.4cm2= 14.4 \: cm {}^{2}=14.4cm

2

hope it helps:--------

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