Question

The perimeter of an isosceles trapezium (having non-parallel sides equal) is 7.07 cm. If the parallel sides of the trapezium measure 37/20cm and 58/25cm, find the measure of the equal non-parallel sides.

Answer 1

ABCD is a isosceles trapezium.

AB is parallel to DC .

$AB = \frac{58}{25} = 2.32 \:cm \\and \: DC = \frac{37}{20} = 1.85 \:cm$

Let AD = AC = x .

/* According to the problem given */

$Perimeter \: of \: ABCD = 7.07 \:cm$

$\implies AB + BC + CD + DA = 7.07 \:cm$

$\implies 2.32 + x + 1.85 + x = 7.07$

$\implies 4.17 + 2x = 7.07$

$\implies 2x = 7.07 - 4.17$

$\implies x = \frac{2.9}{2}$

$\implies x = 1.45 \:cm$

Therefore.,

$\red { Measure \:of \:the \:equal \:and }$

$\red { non - parallel \: sides } \green {=1.45\:cm }$

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Answer 2

Answer:

Measure of the equal non-parallel sides is 29/20 cm

Step-by-step explanation:

Assume a isosceles trapezium ABCD having, AB parallel to DC and AD = BC.

AB = 58/25 = 2.32 cm , CD = 37/20 = 1.85 cm

Perimeter of ABCD is 7.07 cm

AB + BC + CD + AA = 7.07cm

2.32 + BC + 1.85 + BC = 7.07

2BC + 4.17 = 7.07

2BC = 7.07 - 4.17

2BC = 2.9

BC = 2.9/2

BC = 29/20 cm and AD = 29/20 cm