Question

In a trapezium ABCD where AB is parallel to CD, E is the mid-point of BC, prove that ΔAED = 1/2 trapezium ABCD.

Answer 1

Answer:

The proof is explained below

Step-by-step explanation:

Given a trapezium ABCD where AB is parallel to CD, E is the mid-point of BC, we have to prove that ΔAED = $\frac{1}{2}$trapezium ABCD

Let h be the height of trapezium ABEF

In trapezium ABFE because E is the mid point of side BC

$ar(AFE) = ar(ABEF)-ar(AFE)$

$ar(AFE) = ar(ABEF)-\frac{1}{2}\times AB\times h$

In trapezium FECD,

$ar(FED) = ar(FECD)-ar(CDE)$

$ar(FED) = ar(FECD)-\frac{1}{2}\times CD\times h$

Adding above two equations, we get

$ar(AFE)+ar(FED)=ar(ABEF)-\frac{1}{2}\times AB\times h+ar(FECD)-\frac{1}{2}\times CD\times h$

⇒ $ar(AED)=ar(ABCD)-\frac{1}{2}[\frac{1}{2}\times 2h \times (AB+CD)]$

⇒$ar(AED)=ar(ABCD)-\frac{1}{2}ar(ABCD)$

⇒$ar(AED)=\frac{1}{2}ar(ABCD)$

Answer 2

Step-by-step explanation:

The proof is explained below

Step-by-step explanation:

Given a trapezium ABCD where AB is parallel to CD, E is the mid-point of BC, we have to prove that ΔAED = \frac{1}{2}

2

1

trapezium ABCD

Let h be the height of trapezium ABEF

In trapezium ABFE because E is the mid point of side BC

ar(AFE) = ar(ABEF)-ar(AFE)ar(AFE)=ar(ABEF)−ar(AFE)

ar(AFE) = ar(ABEF)-\frac{1}{2}\times AB\times har(AFE)=ar(ABEF)−

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×AB×h

In trapezium FECD,

ar(FED) = ar(FECD)-ar(CDE)ar(FED)=ar(FECD)−ar(CDE)

ar(FED) = ar(FECD)-\frac{1}{2}\times CD\times har(FED)=ar(FECD)−

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×CD×h

Adding above two equations, we get

ar(AFE)+ar(FED)=ar(ABEF)-\frac{1}{2}\times AB\times h+ar(FECD)-\frac{1}{2}\times CD\times har(AFE)+ar(FED)=ar(ABEF)−

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×AB×h+ar(FECD)−

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×CD×h

⇒ ar(AED)=ar(ABCD)-\frac{1}{2}[\frac{1}{2}\times 2h \times (AB+CD)]ar(AED)=ar(ABCD)−

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1

[

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×2h×(AB+CD)]

⇒ar(AED)=ar(ABCD)-\frac{1}{2}ar(ABCD)ar(AED)=ar(ABCD)−

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ar(ABCD)

⇒ar(AED)=\frac{1}{2}ar(ABCD)ar(AED)=

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ar(ABCD)