Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

Consider the centre of circle to be O, ends of chord as A, B and their mid-point as M.
We have, radius, r = 5√2 cm and chord length, l = 10 cm.
Now you can observe, ∆OAM & ∆OBM are isosceles & right angled at M.
Here, angle AOM = angle BOM = 45°, then area of ∆ AOB = 25 cm²
and the area of sector AOB = ¼(circle area) = (12.5)π cm², since angle AOB = 90°.
Then area of segments are 50(π-2)/4 and 50(3π+2)/4.

we have r=√50cm cord=10cm we know of the segment make angle Ф at centre we have sinФ/2=5/√50=45 degree ⇒ Ф=90 Area of segment=(Фπ/180-sinФ)r²/2 =(90π/180-sin90)50/2 =(.5π-1)25 =14.26cm²