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A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.

Mensuration

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Answer

The toy is shown in the figure below:

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm. Mensuration, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Height of the cylindrical part (H) = 13 cm.

Radius = 5 cm.

Radius of cone (r) = 5 cm

Height of cone (h) = 12 cm.

Slant height of cone, l = r2+h2\sqrt{r^2 + h^2}

Putting values we get,

l=52+122=25+144=169=13 cm.l = \sqrt{5^2 + 12^2} \\[1em] = \sqrt{25 + 144} \\[1em] = \sqrt{169} \\[1em] = 13 \text{ cm}.

Surface area of toy(S) = Curved surface area of cylinder + Curved surface area of hemisphere + Curved surface area of cone.

S=2πrH+2πr2+πrl=πr(2H+2r+l)=227×5×(2×13+2×5+13)=1107×(26+10+13)=1107×49=110×7=770 cm2.S = 2πrH + 2πr^2 + πrl \\[1em] = πr(2H + 2r + l) \\[1em] = \dfrac{22}{7} \times 5 \times (2 \times 13 + 2 \times 5 + 13) \\[1em] = \dfrac{110}{7} \times (26 + 10 + 13) \\[1em] = \dfrac{110}{7} \times 49 \\[1em] = 110 \times 7 \\[1em] = 770 \text{ cm}^2.

Hence, the surface area of the toy is 770 cm2.

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