Find the area of figure ABCP. (Take π= 22/7)
On diameter AC a semi-circle is drawn.
AB = BC = 7 cm.
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AnsweIn right angled triangle ABC,
we have
AC
2
=AB
2
+BC
2
AC
2
=14
2
+14
2
AC=
2×14
2
=14
2
cm
Now required Area
=Area APCQA
=Area ACQA - Area ACPA
=Area ACQA - (Area ABCPA - Area of Δ ABC)
=
2
1
×π×(
2
14
2
)
2
−[
4
1
×π(14)
2
−
2
1
×14×14]
=
2
1
×
7
22
×7
2
×7
2
−
4
1
×
7
22
×14×14+
2
1
×14×14
=154−154+98
=98 cm
2
r:
Step-by-step explanation:
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