A joker-s cap is in the form of a cone of radius 7 cm and height 24 cm- Find the area of the cardboard required to the cap-

Question

Toppr Admin · Accepted Answer

Area of the cardboard required - curved surface area of the cap -or cone - - -x3C0-rlNow-l-x221A-r2-h2-x221A-72-242-25 cmCurved surface area - 227-xD7-7-xD7-25-550 cm2Therefore- area of the cardboard required - 550 cm2

A joker's cap is in the form of a cone of radius $7 c m$ and height $24 c m$ . Find the area of the cardboard required to make the cap.

Area of the cardboard required = curved surface area of the cap (or cone ) = $π r l$
Now,
$l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = 25$ $c m$
Curved surface area = $\frac{22}{7} \times 7 \times 25 = 550$ $c m^{2}$
Therefore, area of the cardboard required = $550$ $c m^{2}$

A joker's cap is in the form of a cone of radius 7cm and height 24cm. Find the area of the cardboard required to make the cap.

Area of the cardboard required = curved surface area of the cap (or cone ) = πrlNow,l=√r2+h2=√72+242=25 cmCurved surface area = 227×7×25=550 cm2Therefore, area of the cardboard required = 550 cm2

A joker's cap is in the form of a cone of radius $7 c m$ and height $24 c m$ . Find the area of the cardboard required to make the cap.

Area of the cardboard required = curved surface area of the cap (or cone ) = $π r l$
Now,
$l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = 25$ $c m$
Curved surface area = $\frac{22}{7} \times 7 \times 25 = 550$ $c m^{2}$
Therefore, area of the cardboard required = $550$ $c m^{2}$