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

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

        

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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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