Q.87 If ABCD is an isosceles trapezium inscribed in a semi-circle with diameter AD and AB = CD = 2 cm and the radius
of the semi – circle is 4 cm, what is the length of BC?
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Answers
Given : ABCD is an isosceles trapezium inscribed in a semi-circle with diameter AD and AB = CD = 2 cm . radius of the semi – circle is 4 cm
To Find : what is the length of BC
Solution:
AD is Diameter
Hence ∠ABD = 90°
AB² + BD² = AD²
AB = 2 , AD = 2 * 4 = 8 cm
=> 2² + BD² = 8²
=> BD² = 60
=> BD = 2√15 cm
Draw BE ⊥ AD
Area of triangle ABD = (1/2) * AB * BD
Area of triangle ABD = (1/2) * AD * BE
(1/2) * AB * BD = (1/2) * AD * BE
=> AB * BD = AD * BE
=> 2 * 2√15 = 8 * BE
=> BE = √15 / 2
BE = OM = √15 / 2 as AD || BC ( O center of circle)
Let say M is mid point of BC
BM² = OB² - OM²
=> BM² = 4² - 15./4
=> BM² = 49/4
=> BM = 7/2
BC = 2 * BM = 2 * 7/2 = 7 cm
length of BC = 7 cm
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