Question

The area of trapezium ABCD, where AB = 45 cm, BC = 20 cm, CD = 20 cm and DA = 15 cm and AB || CD, is

• A. $24\frac{2}{5}c{m}^2$
• B. $390c{m}^2$
• C. $280c{m}^2$
• D. $58\frac{4}{5}c{m}^2$

Answer 1

The correct option is B $390c{m}^2$

We have, ABCD is a trapezium in which AB = 45 cm, BC = 20 cm, CD = 20 cm, AD = 15 cm.


By construction, CE || AD and CF ⊥ AB.
Thus, AECD is a parallelogram.
In ΔBCE, BC = 20 cm, CE = 15 cm, BE = 25 cm
$\therefore$ Semi-perimeter, (s) = $\frac{BC+CE+BE}{2}$
$=\frac{20+15+25}{2}$
$=30cm$
Area of ΔBCE = $\sqrt{s(s-BC)(s-CE)(s-BE)}$
$\Rightarrow \frac{1}{2}\times BE\times CF=\sqrt{30\times 10\times 15\times 5}$
$\Rightarrow \frac{1}{2}\times 25\times CF=\sqrt{10\times 3\times 10\times 5\times 3\times 5}$
$\Rightarrow \frac{25}{2}\times CF=10\times 3\times 5$
$\Rightarrow CF=\frac{50\times 3\times 2}{25}$
$\Rightarrow CF=12cm$
∴ Area of trapezium ABCD = $\frac{1}{2}\times (AB+CD)\times CF$
= $\frac{1}{2}\times (45+20)\times 12$
$=6\times 65=390c{m}^2$
Hence, the correct answer is option (b).