Question

Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as
(i) the sum of the areas of two triangles and one rectangle.
(ii) the difference of the area of a rectangle and the sum of the areas of two triangles.

Answer 1

$$\begin{gathered} \text{Given:} \\ \text{Length of the parallel sides of a trapezium are 10 cm and 15 cm.} \\ \text{The distance between them is 6 cm.} \\ \mathrm{Let}\mathrm{us}\mathrm{e}\text{xtend the smaller side and then draw perpendiculars from the ends of both sides.} \end{gathered}$$


$$\begin{gathered} \text{(i)} \\ \text{Area of trapezium ABCD =(Area of rectangle EFCD)+(Area of triangle AED+Area of triangle BFC)} \\ \text{=(10×6)+[(}\frac{1}{2}\text{×AE×ED)+(}\frac{1}{2}\text{×BF×FC)]} \\ \text{=60+[}(\frac{1}{2}\text{×AE×6)+(}\frac{1}{2}\text{×BF×6)]} \\ \text{=60+[}3\text{AE+3 BF]} \\ \text{=60+3×(AE+BF)} \\ \text{Here, AE+EF+FB = 15cm} \\ A\text{nd EF = 10 cm} \\ \therefore \text{AE+10+BF=15} \\ \text{Or, AE+BF=15-10=5 cm} \\ \text{Putting this value in the above formula:} \\ \mathrm{A}{\text{rea of the trapezium=60+3×(5)=60+15=75 cm}}^2 \end{gathered}$$

$$\begin{gathered} \text{(ii)} \\ \text{In this case, the figure will look as follows:} \end{gathered}$$


$$\begin{gathered} \text{Area of trapezium ABCD=(Area of rectangle ABGH)-[(Area of triangle AHD)+(Area of triangle BGC)]} \\ =(15\times 6)-\left[\right(\frac{1}{2}\times \text{DH}\times 6)+(\frac{1}{2}\times \text{GC}\times 6\left)\right] \\ =90-[3\text{×DH}+3\times \text{GC}] \\ =90-3[\text{DH+GC]} \\ \text{Here, HD+DC+CG=15 cm} \\ \text{DC=10 cm} \\ \text{HD+10+CG=15} \\ \text{HD+GC=15-10=5 cm} \\ \text{Putting this value in the above equation:} \\ \mathrm{A}{\text{rea of the trapezium=90-3(5)=90-15=75 cm}}^2 \end{gathered}$$