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Montano1993 [528]
2 years ago
13

Find the Total surface area of a cone if it's slant height is 21m and diameter of it's base is 24m.

Mathematics
2 answers:
kherson [118]2 years ago
8 0
<h3><u>Answer :- </u></h3>

  • The total surfAce area of cone is <u>1244.57m².</u>

<h3><u>Step-by-step</u><u> </u><u>explanation</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>To </u><u>find </u><u>:</u><u>-</u><u> </u>

  • The total surface area of cone..

<h3><u>Solution :- </u></h3>

Given that ,

  • The slant height of the cone = 21m.
  • Diameter of it's base = 24m.

<h3><u>♦</u><u> </u><u>Radius is </u></h3>

<u>=</u>> Diameter / 2

=> 24 / 2

=> 12m

<h3>As we know that , </h3>

<u>Total surface area of cone = πr ( r + L ) .</u>

<h3><u>Where</u><u> </u><u>we </u><u>know</u><u>,</u></h3>

  • π = 22/7
  • r = Radius ( radii )..
  • L = Slant height.

<h3>According to the question :- </h3>

The total surface of cone is,

<u>=> Total surface area = πr { r + L } ..</u>

=  \frac{22}{7}  \times 12 \times (12 + 21)

=  \frac{22}{7}  \times 12 \times 33

= (22 \times 12 \times 33) \div 7

= (726 \times 12) \div 7

= 8712 \div 7

= 1244.57 {m}^{2}

• Therefore , The total surface area of cone is <u>1244.57m².</u>

Hope this helps you :)

tigry1 [53]2 years ago
8 0

1244.57 cm²

Step-by-step explanation:

<u>Given:</u><u> </u>

  • Slant height (l) is 21m
  • Diameter (d) is 24m

Hence, radius will be :

➝ diameter/2

➝ 24/2

➝ 12m

\:

<u>To </u><u>Find</u><u>:</u><u> </u>

  • Total Surface Area (TSA) of the cone.

<u>Solution</u><u>:</u><u> </u>

As, we know:

\star \quad{ \underline{ \green{ \boxed{TSA_{(cone)} = \pi r( l+r ) }}}}  \quad\star \quad

<u>Here,</u><u> </u>

  • π = 22/7
  • r = 12m
  • l = 21m

\rightarrow \:  \frac{22}{7}  \times 12 \: (21 + 12)

\rightarrow \:  \frac{22}{7}  \times 12 \: (33)

\rightarrow \:  \frac{8712}{7}  {cm}^{2}

Therefore, Total Surface Area of Cone is <u>8712/7</u> <u>cm²</u> or <u>1244.57cm²</u>.

<h2>_____________________</h2><h3><em><u>Additional</u></em><em><u> Information</u></em><em><u>:</u></em><em><u> </u></em></h3>

\footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}

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