Question

In the given figure (I hv attached it separately) ABC is a right angled triangle. AngleB=90 degree, AB = 28 cm and BC = 21 cm. With AC as diameter, a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

Answer 1

Is 428.75 the right answer????

Answer 2

(42.87500 times pi) + 294 = 428.695785
~ 428.70 is what I got.

Answer 3

thanks for the confirmation.
can u pls give detailed calculation/method as your answer matches the one I calculated??

Answer 4

Shaded Region Area = Area of ABC + Area of Semicircle w/ Diameter AC - Area of quartercircle w/ radius BC

Area of ABC = (1/2)bh b = 21 cm, h = 28 cm
= (1/2)(21)(28)
= (1/2)(588)
= 294 cm^2
Area of Semicircle = (1/2)pi r^2: r = (1/2)sqrt(AB^2 + BC^2)
= (1/2)sqrt((28)^2 + (21)^2)
= (1/2)sqrt(784 + 441)
= (1/2)sqrt(1225)
= (1/2)35 cm

= (1/2)pi (35/2)^2
= (1/2)pi (1225/4)
= 1225pi / 8 cm^2


Area of Quartercircle = (1/4)pi r^2: r = 21
= (1/4)pi (21)^2
= (1/4)pi (441)
= 441pi / 4 cm^2

Shaded Region Area = Area of ABC + Area of Semicircle w/ Diameter AC - Area of quartercircle w/ radius BC
= 294 cm^2 + (1225pi / 8) cm^2 - (441pi / 4) cm^2 (441pi / 4) = (882pi / 8)
= 294 + pi (1225 - 882) / 8
= 294 + 343pi / 8
= 294 + 42.875 pi cm^2

At that point, I just used Google and it calculated it out to be about 428.695785

Answer 5

Thanks, I arrived at the answer the same way but the answer key at the back of the book gave another answer....
Medal for u...☺

Answer 6

Same as access, but with a picture