A right angle triangle with sides 3 cm and 4 cm forming a right angle is revolved about its hypotenuse . Find the volume of the solid so obtaibed .
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I'M ASKING THIS QUESTION FOR THE SECOND TIME PLEASE ANSWER CORRECTLY.
Answers
Step-by-step explanation:
In order to find the volume and surface area, we need to find BD or radius of the double cone
From the figure it can be seen that BD ⊥ AC
In ΔABC right-angled at B using Pythagoras theorem,
AC² = AB² + BC²
AC = (3 cm)² + (4 cm)²
= 9 cm² + 16 cm²
= 25 cm²
= 5 cm
Consider ΔABC and ΔBDC
∠ABC = ∠CDB = 90° (BD ⊥ AC)
∠BCA = ∠BCD (common)
By AA criterion of similarity ΔABC ∼ ΔBDC
Therefore,
AB/BD = AC/BC (Corresponding sides of similar triangles are in proportion)
BD = (AB × BC)/AC
= (3 cm × 4 cm)/5 cm
= 12/5 cm
= 2.4 cm
We know that, Volume of the cone = 1/3πr²h
Volume of double cone = Volume of Cone ABB’ + Volume of Cone BCB’
= 1/3 × π(BD)² × AD + 1/3π (BD)² × DC
= 1/3 × π(BD)² [AD + DC]
= 1/3 × π(BD)² × AC
= 1/3 × 3.14 × 2.4 cm × 2.4 cm × 5 cm
= 30.144 cm³
We know that, CSA of frustum of a cone = πrl
CSA of double Cone = CSA of cone ABB’ + CSA of cone BCB’
= π × BD × AB + π × BD × BC
= π × BD [ AB + BC]
= 3.14 × 2.4 cm × [3 cm + 4 cm]
= 3.14 × 2.4 cm × 7 cm
= 52.752 cm²
= 52.75 cm²