Question

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

Answer 1

Step 1: Calculate perimeter of $∆ABE$

From the figure we get the sides as

$AB=\frac{5}{2}cm,BE=2\frac{3}{4}cm,AE=3\frac{3}{5}cm$

Perimeter of a triangle $=$ sum of all three sides of a triangle

$\therefore p=AB+BE+AE$

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

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

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

$=\frac{177}{20}=8\frac{17}{20}cm$

Step 2: Calculate perimeter of rectangle $BCDE$

From the figure we get,

$BE=2\frac{3}{4}cm,DE=\frac{7}{6}cm$

We know that the perimeter of a rectangle is given as,

Perimeter $=2($ length $+$ breadth $)$

$\therefore p=2\left(2\frac{3}{4}+\frac{7}{6}\right)$

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

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

$=\frac{47}{6}=7\frac{5}{6}cm$

Step 3: Compare the perimeters

Perimeter of $∆ABE=8\frac{17}{20}cm=\frac{177}{20}$

Perimeter of rectangle $BCDE=7\frac{5}{6}cm=\frac{47}{6}$

The LCM of their denominator is $60$.

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

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

Hence the equivalent fractions can be written and compared as,

$\frac{531}{60}>\frac{470}{60}$

$\therefore \frac{177}{20}>\frac{47}{6}$

Hence, perimeter of $∆ABE>$perimeter of rectangle $BCDE$

Therefore,

(i) Perimeter of $∆ABE$ is $8\frac{17}{20}cm$.

(ii) Perimeter of Rectangle $BCDE$ is $7\frac{5}{6}cm$.

$\therefore$Hence, perimeter of $∆ABE>$perimeter of rectangle $BCDE$