question_answer 5)

Find the perimeters of (i) \[\Delta ABE\](ii) the rectangle BCDE in this figure. Whose perimeter is greater?

Answer:

                (i) Perimeter of \[\Delta ABE\] \[=AB+BE+EA\] \[=\frac{5}{2}\text{cm}+2\frac{3}{4}\text{cm + 3}\frac{3}{5}\text{ cm}\] \[=\frac{5}{2}\text{cm + }\frac{11}{4}\text{cm +}\frac{18}{5}\text{cm}\] \[=\left( \frac{5}{2}+\frac{11}{4}+\frac{18}{5} \right)\text{cm}\] \[=\frac{50+55+72}{20}\text{cm}\]                 \[=\frac{177}{20}\text{ cm = 8 }\frac{17}{20}\text{ cm}\]               \[\left| \begin{align}   & \text{ 20}\overset{8}{\overline{\left){\text{ 177}}\right.}} \\  & \underline{\frac{-160}{17}} \\ \end{align} \right.\]        (ii) Perimeter of the rectangle BCDE \[=BC+CD+DE+EB\] \[=\frac{7}{6}\text{cm + 2}\frac{3}{4}\text{ cm +}\frac{7}{6}\text{ cm + 2}\frac{3}{4}\text{cm}\] \[=\left( \frac{7}{6}+\frac{11}{4}+\frac{7}{6}+\frac{11}{4} \right)\text{cm =}\frac{14+33+14+33}{12}\text{cm}\] \[\text{=}\frac{94}{12}\text{cm}\] \[=\frac{94\div 2}{12\div 2}\text{ cm =}\frac{47}{6}\text{cm = 7}\frac{5}{6}\]      \[\left| \begin{align}   & \text{ 6}\overset{7}{\overline{\left){\text{ 47}}\right.}} \\  & \underline{\frac{-42}{5}} \\ \end{align} \right.\] So, the perimeter of \[\Delta ABE\] is greater than the perimeter of the rectangle BCDE.