ABC is a right angled triangle right angled at B. Semicircles is drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter. Please send me real answer
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Area of shaded region = area of semicircle AB + area of semi-circle AC - (area of semicircle on side BC) + area of ΔABC
Area of semicircle AB = πr
2
=
7
22
×
2
3
×
2
3
=7.07 sq. units
Area of semicircle AC = πr
2
=
7
22
×2×2=12.57 sq. units
Area of semicircle BC = πr
2
=π×(
2
BC
)
2
Now, BC=
4
2
+3
2
=
25
=5
∴ Area of semi-circle BC =
7
22
×(
2
5
)
2
=19.64 sq. units.
Area ΔABC=
2
1
×AB×AC=
2
1
×3×4=6 sq. units
Therefore, area of shaded region= 7.07+12.57−19.64+6
= 6 sq. units
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