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

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

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

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

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

Now, The LCM of 2,4 and 5 is 20. So let's make the denominator of all fractions, 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 the denominator of both fractions 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

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

$\Rightarrow$ Perimeter of Triangle > Perimeter of Rectangle.

Posted by

Pankaj Sanodiya

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