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Answers
Step-by-step explanation:
Given that in a right angled triangle, whose sides are 3 cm and 4 cm.
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse .
Hypotenuse AC² = ( 3² + 4²) = 5 cm.
Area of ΔABC = (1/2) * AB * AC
⇒ (1/2) x AC x OB ⇒ (1/2) * 4 * 3
⇒ (1/2) x 5 x OB = 6
⇒ OB = 12 / 5 = 2.4 cm.
Volume of double cone = Volume of cone 1 + Volume of cone 2
= (1/3) πr²h₁ + (1/3)πr₂h₂
= (1/3) πr²(h₁ + h₂)
= (1/3) πr²(OA +OC)
= (1/3) * 3.14 * (2.4)² * (5)
= 30.14 cm³
Surface area of double cone = surface area of cone ABD + surface area of cone BCD.
= πrl₁ + πrl₂
= πr(l₁ + l₂)
= (22/7) * 2.4 * 3
= 52.8 cm².
Therefore,
Volume of the double cone is 30.14 cm³
Surface area of double cone is 52.8 cm²
Hope it helps!
Answer:
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Step-by-step explanation:
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