Can you please provide me the explanation video for this above question. Where can I get important and additional questions for this chapter.

Ncert class 10 chapter 12

Page No. 235

Exercise 12.3

Question No. 6

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Radius of circle = 32

Centroid O divides median AD into 2:1, therefore A0:0D = 2:1

= AO/OD = 2/1 ⇒ 32/OD = 2/1 ⇒ OD = 16

Therefore, AD = 32 + 16 = 48 cm

In AABD,

AB² = AD² + BD² = (48)² + (AB/2)²

⇒ 3/4 AB² = (48)² ⇒ AB = (48 × 2)/√3 = (96√3)/3 = 32√3

Area of equilateral triangle ABC

= √3/4 (32√3)² = 768√3 cm²

Area of circle

= πr² = π(32)² = 22/7 × 32 × 32 = 22528/7 cm²

Area of design = Area of circle – Area of equilateral triangle ABC

= (22528/7 – 768√3) cm²