5. Find the perimeters of (i)\bigtriangleup\! ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

         

Answers (1)

The perimeter of \bigtriangleup\! ABE^{} = AB + BE + AE 

                                        =\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}

                                        =\frac{5}{2}+\frac{4\times2+3}{4}+\frac{3\times5+3}{5}

                                        =\frac{5}{2}+\frac{11}{4}+\frac{18}{5}

Noe, The LCM of 2,4 and 5 is 20. So let's make the denominator of all fraction, 20.

So,

The perimeter of \bigtriangleup\! ABE^{} :

    =\frac{5}{2}\times\frac{10}{10}+\frac{11}{4}\times\frac{5}{5}+\frac{18}{5}\times\frac{4}{4}

   =\frac{50}{20}+\frac{55}{20}+\frac{72}{20}

    =\frac{50+55+72}{20}

    =\frac{177}{20}cm

 

Now,

The perimeter of rectangle BCDE = 2 x ( BE + ED )

=2\times\left (2\frac{3}{4}+\frac{7}{6} \right )

=2\times\left (\frac{4\times2+3}{4}+\frac{7}{6} \right )

=2\times\left (\frac{11}{4}+\frac{7}{6} \right )

The LCM of 4 and 6 is 12. So let's make the denominator of both fractions equal to 12.

=2\times\left (\frac{11}{4}\times\frac{3}{3}+\frac{7}{6}\times\frac{2}{2} \right )

=2\times\left (\frac{33}{12}+\frac{14}{12} \right )

=2\times\left (\frac{33+14}{12} \right )

=2\times\left (\frac{47}{12} \right )

=\frac{47}{6}cm

Hence The perimeter of the Triangle is 177/20 \:cm and the perimeter of Rectangle is  47/6 \: cm.

Now, we have 

\frac{177}{20} \:\: and\:\:\frac{47}{6}

LCM of 20 and 6 is 60, so let's make denominator of both fraction equal to 60.

So,

\frac{177}{20} =\frac{177}{20}\times\frac{3}{3}=\frac{531}{60}

And

\frac{47}{6}=\frac{47}{6}\times\frac{10}{10}=\frac{470}{60}

Now, Since 531 > 470 

\Rightarrow \frac{177}{20}>\frac{47}{6}.

\Rightarrow Area of Triangle > Area of Rectangle. 

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