Question

The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{The parallel sides of a trapezium are 25 cm and 13 cm.} \\ \text{Its nonparallel sides are equal in length and each is equal to 10 cm.} \\ \text{A rough sketch for the given trapezium is given below:} \end{gathered}$$


$$\begin{gathered} \text{In above figure, we observe that both the right angle trangles AMD and BNC are congruent triangles.} \\ \mathrm{AD}=\mathrm{BC}=10\mathrm{cm} \\ \mathrm{D}\hspace{0.17em}=\mathrm{CN}=x\mathrm{cm} \\ \angle \mathrm{DMA}=\angle \mathrm{CNB}=90° \\ \text{Hence, the third side of both the triangles will also be equal.} \\ \therefore \text{AM=BN} \\ \text{Also, MN=13} \\ \mathrm{S}\text{ince AB = AM+MN+NB:} \\ \therefore \text{25=AM+13+BN} \\ \text{AM+BN=25-13=12 cm} \\ \text{Or, BN+BN=12 cm (Because}AM=BN) \\ \text{2 BN=12} \\ \text{BN=}\frac{12}{2}\text{=6 cm} \\ \therefore \text{AM = BN = 6 cm.} \\ \text{Now, to find the value of}x\text{, we will use the Pythagoras theorem in the right angle triangle AMD, whose sides are 10, 6 and x.} \\ {\text{(Hypotenuse)}}^2{\text{=(Base)}}^2{\text{+(Altitude)}}^2 \\ {\text{(10)}}^2{\text{=(6)}}^2{\text{+(x)}}^2 \\ {\text{100=36+x}}^2 \\ {\text{x}}^2\text{=100-36=64} \\ \text{x=}\sqrt{64}\text{=8 cm} \\ \therefore \mathrm{D}\text{istance between the parallel sides = 8 cm} \\ \therefore \text{Area of trapezium =}\frac{1}{2}\text{×(Sum of parallel sides)×(Distance between parallel sides)} \\ \text{=}\frac{1}{2}\text{×(25+13)×(8)} \\ {\text{=152 cm}}^2 \end{gathered}$$