Who is the centre of the semicircle of a semicircular arc acb is a straight line find the shaded region?
Answers
Answered by
0
Answer:
Step-by-step explanation:
In right triangle AED AD2=AE2+DE2
= =92+122 = 81+144 = 225
Therefore,AD2=225
⇒AD=15
Area of the shaded region = Area of rectangle – Area of triangle AED + Area of semicircle
=AB×BC–12×AE×DE+12π(BC2)2
=20×15–12×9×12+12×3.14(152)2
=300–54+88.31
= 334.31 cm2
Hence, the area of shaded region is 334.31 cm2
Similar questions