Question

Find the perimeter of

(i) an isosceles trapezium $ABCD$ (in which $A B ∥ D C$ and $A D = B C$ ).

(ii) square $EFGH$

Which of the two has the greater perimeter?

Accepted Answer

From the figure, we obtain, $D E = \frac{5}{6} cm, E H = 2 \frac{3}{4} cm, H C = \frac{5}{6} cm$ and $B C = 1 \frac{3}{5} cm$ .

$C D = D E + E H + H C = (\frac{5}{6} + 2 \frac{3}{4} + \frac{5}{6}) cm$

Also, we have, $A D = B C = 1 \frac{3}{5} cm$ .

We note that $F E$ and $G H$ , when produced will meet the points $A$ and $B$ , respectively.

Given, $E F G H$ is a square.

So, $F E ∥ G H \Rightarrow F A ∥ G B \Rightarrow E A ∥ H B$ [A square is a special parallelogram]

In $A E H B$ , $A B ∥ E H$ and $E A ∥ H B$ . So, $A E H B$ is a parallelogram.

Hence, $A B = E H = 2 \frac{3}{4} cm$ . [The opposite sides of a parallelogram are equal]

(i) So, the perimeter of the isosceles trapezium $A B C D = A B + B C + C D + D A$

$= [2 \frac{3}{4} + 1 \frac{3}{5} + (\frac{5}{6} + 2 \frac{3}{4} + \frac{5}{6}) + 1 \frac{3}{5}] cm$

$= [\frac{11}{4} + \frac{8}{5} + \frac{5}{6} + \frac{11}{4} + \frac{5}{6} + \frac{8}{5}] cm$

$= [\frac{11 \times 15}{4 \times 15} + \frac{8 \times 12}{5 \times 12} + \frac{5 \times 10}{6 \times 10} + \frac{11 \times 15}{4 \times 15} + \frac{5 \times 10}{6 \times 10} + \frac{8 \times 12}{5 \times 12}] cm$

$= [\frac{165}{60} + \frac{96}{60} + \frac{50}{60} + \frac{165}{60} + \frac{50}{60} + \frac{96}{60}] cm$

$= [\frac{165 + 96 + 50 + 165 + 50 + 96}{60}] cm$

$= \frac{622}{60} cm$

$= \frac{622 \div 2}{60 \div 2} cm$

$= \frac{311}{30} cm$

$= 10 \frac{11}{30} cm$ .

(ii) The perimeter of the square $= 4 \times$ side $= 4 \times 2 \frac{3}{4} cm = 4 \times \frac{11}{4} cm = 11 cm$ .

We can write $10 \frac{11}{30} cm$ as $10 cm + \frac{11}{30} cm$ .

$\frac{11}{30} cm$ is less than $1$ .

Hence, $10 \frac{11}{30} cm$ is less than $11 cm$ .

Therefore, the square has greater perimeter.

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