Ncert exercise 12.3_In a circular table …
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Gaurav Seth 5 years, 3 months ago
Radius of the circle = 32 cm
Draw a median AD of the triangle passing through the centre of the circle.
⇒ BD = AB/2
Since, AD is the median of the triangle
∴ AO = Radius of the circle = 2/3 AD
⇒ 2/3 AD = 32 cm
⇒ AD = 48 cm
In ΔADB,
<a href="https://3.bp.blogspot.com/-gsz4sf7No4w/VYwFPSkyvbI/AAAAAAAABA4/6QBibJ98sAw/s1600/ch12-class10-maths-area-related-to-circles-6.png" rel="nofollow" target="_blank">
</a>
By Pythagoras theorem,
AB2 = AD2 + BD2
⇒ AB2 = 482 + (AB/2)2
⇒ AB2 = 2304 + AB2/4
⇒ 3/4 (AB2) = 2304
⇒ AB2 = 3072
⇒ AB = 32√3 cm
Area of ΔADB = √3/4 × (32√3)2 cm2 = 768√3 cm2
Area of circle = π R2 = 22/7 × 32 × 32 = 22528/7 cm2
Area of the design = Area of circle - Area of ΔADB
= (22528/7 - 768√3) cm2
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