ABCD is an isosceles trapezium such that OA = 3 cm, OD = 4 cm and the perimeter of $△$ AOB, $△$ DOC and $△$ AOD is 12 cm, 15 cm and 16 cm respectively, then find the perimeter of the trapezium.

31 cm

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32 cm

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40 cm

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35 cm

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Solution

The correct option is A 31 cm
Given,
ABCD is an isosceles trapezium
OA = OB = 3 cm (Diagonals are equal for an isosceles trapezium)
OD = OC = 4 cm (Diagonals are equal for an isosceles trapezium)

$P e r i m e t e r o f △ A O B = 12 c m$
$O A + O B + A B = 12 c m$
$3 c m + 3 c m + A B = 12 c m$
$A B = 6 c m$ .. (i)

$P e r i m e t e r o f △ D O C = 15 c m$
$O C + O D + D C = 15 c m$
$4 c m + 4 c m + D C = 15 c m$
$D C = 7 c m$ ... (ii)

$P e r i m e t e r o f △ A O D = 12 c m$
$O A + O D + A D = 16 c m$
$3 c m + 4 c m + A B = 16 c m$
$A D = 9 c m$

AD = BC = 9 cm ... (iii)
(Non parallel sides of an isosceles trapezium are equal)

Perimeter of the trapezium ABCD
= AB + BC + CD + DA
= 6 cm + 9 cm + 7 cm + 9 cm
[From (i), (ii) and (iii)]
= 18 cm + 13 cm
= 31 cm

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