Question

Find the magnitude of the shape obtained by rotating a right triangle around its hypotenuse, assuming that the base and height of the triangle are 7 m and 24 m, respectively.​

Answer 1

Answer:

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

$hypotenuse \: ac \: = \sqrt{ {3}^{2} + {4}^{2} } \\ = \sqrt{25} = 5 \: cm$

Area of ΔABC=

$\frac{1}{2} \times ab \times ac \\ = \frac{1}{2} \times 4 \times 3 \\ = \frac{1}{2 } \times 5 \times db \\ \\ = 6 \\ so \: db = \frac{12}{5} = 2.4cm$

The volume of double cone = Volume of cone 1 + Volume of cone 2

=31πr2h1+31πr2h2

=31πr2[h1+h2]=31πr2[DA+DC]

=31×3.14×2.42×5

=30.14  cm3

The surface area of double cone = Surface area of cone 1 + Surface area of cone 2

=πrl1+πrl2

=πr[4+3]=3.14×2.4×7

=52.75 cm2

Answer 2

Answer:

tqsm ☺️☺️ sis for your help ❤