Question

In the given figure the semicircle centered at O has a diameter 6cm. The chord BC is parallel to
and BC = AD. Then the area of trapezium in cm’ is
AD
a) 4
b) 472
d) 8V2​

Answer 1

the area of trapezium = 8√2 cm²  where the semicircle centered at O has a diameter 6cm and BC ║ AD & BC = AD/3

Step-by-step explanation:

OA = OD = 6/2 = 3 cm

BC = AD/3  = 6/3  = 2 cm

in ΔOBC

OB = OC = radius = 3 cm

BC = 2 cm

=> OM ⊥BC will be at mid point of BC

=> BM = CM = 2/2 = 1 cm

=> OM² = OB² - BM²

=> OM² = 3² - 1²

=> OM² = 8

=> OM = 2√2

Area of trapezium = (1/2)(AD + BC) * OM

=  (1/2)( 6  + 2 ) 2√2

= 8√2

the area of trapezium = 8√2 cm²

Leran more about trapezium

The circle is inscribed in the isosceles trapezium ABCD in which AB is parallel DC

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ABCD is a trapezium with AB||CD

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Answer 2

Answer:

hi the answer is true from Think NTSE