abc is a tight angled triangle in which /_A=90° AB=21cm ans AC=28cm. Semi-circle on AB,BC,and AC as diameters . find the area of shaded region.
Attachments:
Answers
Answered by
2
Answer:
Step-by-step explanation:
In right ΔABC, by Pythagoras theorem:
AC2 = AB2+BC2
AC2 = 282 + 212
AC2 =35
Area of ΔABC = ½*BC*AB
= ½*21*28
=294
Semicircle's area = ½* 22/7*35/2*35/2
= 481.25 Quadrant's area
= ¼*22/7*21*21
=346.5 Area of the shaded region
= Semicircle's area +Area of ΔABC - Quadrant's area
= 481.25 + 346.5 - 294 = 428.75 cm²
Answered by
1
Answer:
Step-by-step explanation:
In right ΔABC, by Pythagoras theorem:
AC2 = AB2+BC2
AC2 = 282 + 212
AC2 =35
Area of ΔABC = ½*BC*AB
= ½*21*28
=294
Semicircle's area = ½* 22/7*35/2*35/2
= 481.25 Quadrant's area
= ¼*22/7*21*21
=346.5 Area of the shaded region
= Semicircle's area +Area of ΔABC - Quadrant's area
= 481.25 + 346.5 - 294 = 428.75 cm²
Similar questions