Question

Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and the other sides are 15 cm each.

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{Parallel sides of a trapezium are 25 cm and 13 cm.} \\ \text{Its nonparallel sides are equal in length and each is equal to 15 cm.} \\ \text{A rough skech of the trapezium is given below:} \end{gathered}$$


$$\begin{gathered} \text{In above figure, we observe that both the right angle triangles AMD and BNC are similar triangles.} \\ \text{This is because both have two common sides as 15 cm and the altitude as x and a right angle.} \\ \text{Hence, the remaining side of both the triangles will be equal.} \\ \therefore \text{AM=BN} \\ \text{Also MN=13} \\ \text{Now, since AB=AM+MN+NB:} \\ \therefore \text{25=AM+13+BN} \\ \text{AM+BN=25-13=12 cm} \\ \text{Or, BN+BN=12 cm (Because AM=BN)} \\ \text{2 BN=12} \\ \text{BN=}\frac{12}{2}=6\text{cm} \\ \therefore \text{AM=BN=6 cm} \\ \text{Now, to find the value of x, we will use the Pythagorian theorem in the right angle triangle AMD whose sides are 15, 6 and x.} \\ {\text{(Hypotenus)}}^2=(B\text{ase}{)}^2+(\mathrm{A}\text{ltitude}{)}^2 \\ (15{)}^2=(6{)}^2+\left(\text{x}\right)2 \\ 225=36+{\text{x}}^2 \\ {\text{x}}^2\text{= 225}-36=189 \\ \therefore \text{x =}\sqrt{189}\text{=}\sqrt{9\times 21}\text{= 3}\sqrt{21}\text{cm} \\ \therefore \mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{parallel}\mathrm{sides=}3\sqrt{21}\mathrm{cm} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{trapezium=}\frac{1}{2}\times (\mathrm{Sum}\mathrm{of}\mathrm{parallel}\mathrm{sides)\times (}\mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{parallel}\mathrm{sides}) \\ =\frac{1}{2}\times (25+13)\times (3\sqrt{21}) \\ =57\sqrt{21}{\text{cm}}^2 \end{gathered}$$