In A trapezium ABCD, AB= 7cm, CD = 16cm, BE || AD is drawn. P,Q,R,S are mid point

Question

I got of area of BEC = 144 after this i calculated height and
I got area of traingle ABD =126.

In question options given are (a) 208 (b) 56 (c) 28 (d) 112 Is my answer right or wrong?

Answer 1

Note that :

Area ($\bigtriangleup$ EBC) = $2$ Area ($\bigtriangleup$ EPC) = $4$ Area ($\bigtriangleup$ EQC) = $8$ Area ($\bigtriangleup$ EQR) = $16$ Area ($\bigtriangleup$ ESR) = $16 \cdot9$ = $144$ cm$^{2}$

Now, let the height of the trapezium be $x$ cm. Calculating the area of trapezium ABCD by two methods, we have:

$7x+144 = \frac {1}{2}\cdot (7+16)\cdot x$

which gives $x= 32$$ cm.

Now, ABDE is a parallelogram with a side of $7$ cm and height $32$ cm.

Hence, the area of $\bigtriangleup$ ABD $ =\frac {1}{2}\cdot7\cdot32 = 112$ cm$^{2}.$