What is its answer plz explain...
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Join AE. AE is median.
We know, the cente of circle lies on centroid of equilateral triangle.
First let's find AE
In ∆AEB,
AB² = BD + AD
AD² = 12-3
AD = 3cm
We know,
The centroid divides the median in 2:1
Radius of circle is OE,
r = 1/3 ×AD = 1cm
We know by mid POINT theorem,
DF = BC/2
DF = √3 cm
Area of shaded region = Area of circle - Area of ∆DEF
= πr² - √3/4 × DF²
= 3.14 - 3√3/4
=3.14-1.29
=1.85cm²
We know, the cente of circle lies on centroid of equilateral triangle.
First let's find AE
In ∆AEB,
AB² = BD + AD
AD² = 12-3
AD = 3cm
We know,
The centroid divides the median in 2:1
Radius of circle is OE,
r = 1/3 ×AD = 1cm
We know by mid POINT theorem,
DF = BC/2
DF = √3 cm
Area of shaded region = Area of circle - Area of ∆DEF
= πr² - √3/4 × DF²
= 3.14 - 3√3/4
=3.14-1.29
=1.85cm²
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