Question

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

Answer 1

From the fig,

$AB=\frac{5}{2}\text{cm}$

$AE=3\frac{3}{5}=\frac{18}{5}\text{cm}$

$BE=2\frac{3}{4}=\frac{11}{4}\text{cm}$

$ED=\frac{7}{6}\text{cm}$

(i) We know that,

Perimeter of the triangle = Sum of all sides

Then,

Perimeter of triangle ABE = AB + BE + EA

$=\left(\frac{5}{2}\right)+\left(\frac{11}{4}\right)+\left(\frac{18}{5}\right)$

The LCM of 2, 4, 5 = 20

Now, let us change each of the given fraction into an equivalent fraction having 20 as the denominator.

$=\{[\left(\frac{5}{2}\right)\times \left(\frac{10}{10}\right)]+[\left(\frac{11}{4}\right)\times \left(\frac{5}{5}\right)]+[\left(\frac{18}{5}\right)\times \left(\frac{4}{4}\right)]\}$

$\left(\frac{50}{20}\right)+\left(\frac{55}{20}\right)+\left(\frac{72}{20}\right)$

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

$=\frac{177}{20}$

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

(ii) Now, we have to find the perimeter of the rectangle,

We know that,

Perimeter of the rectangle $=2\times$ (length + breadth)

Then,

Perimeter of rectangle BCDE $=2\times$ (BE + ED)

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

The LCM of 4, 6 = 12

Now, let us change each of the given fraction into an equivalent fraction having 16 as the denominator

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

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

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

$=\frac{47}{6}$

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

Finally, we have find which one is having greater perimeter.

Perimeter of triangle ABE $=\left(\frac{177}{20}\right)$

Perimeter of rectangle BCDE $=\left(\frac{47}{6}\right)$

The two perimeters are in the form of unlike fraction.

Changing perimeters into like fractions we have,

$\left(\frac{177}{20}\right)=\left(\frac{177}{20}\right)\times \left(\frac{3}{3}\right)=\frac{531}{60}$

$\left(\frac{47}{6}\right)=\left(\frac{47}{6}\right)\times \left(\frac{10}{10}\right)=\frac{470}{60}$

clearly,$\left(\frac{531}{60}\right)>\left(\frac{470}{60}\right)$

Hence,$\left(\frac{177}{20}\right)>\left(\frac{47}{6}\right)$

$\therefore$ Perimeter of Triangle ABE > Perimeter of Rectangle (BCDE)