Question

Two parallelograms $\text{A}BCD$ and $\text{E}FCD$ that have the same base $\text{C}D$ and lie on the same parallels $\text{A}F$ and $\text{C}D$, then

• A. Area($\text{A}BCD$) = Area($\text{A}FCD$)
• B. Area($\text{A}BCD$) = Area($\text{E}FCD$)
• C. Area($\text{E}BDC$) = Area($\text{E}FCD$)
• D. None of these

Answer 1

The correct option is B Area($\text{A}BCD$) = Area($\text{E}FCD$)

ABCD is a parallelogram, then $AD\parallel BC$

& AB is transversal , therefore

$\angle DAB=\angle CBF\to \left(1\right)$ (corresponding angles)

Also $AD=BC\to \left(2\right)$ (opposite sides of $\parallel gm$)

Similarly in $\parallel gm$ CDEF,

$ED\parallel CF$ & EF is transversal , therefore

$\angle DEA=\angle CFE\to \left(3\right)$

(corresponding angles)

In $△AED$ & $△BFC$

$\angle DEA=\angle CFE$ [from(3)]

$AD=BC$ [from(2)]

$\angle DAB=\angle CBF$ [from (1)]

$\therefore △AED\cong △BFC$ (By ASA congruency)

$ar△AED=ar△BFC$ (areas of congruent part)

Now, $arABCD=ar△ADE+arEBCD$

$=ar△BFC+arEBCD$

$\therefore arABCD=arEFCD$