Question : A pyramid has an equilateral triangle as its base, of which each side is 8 cm. Its slant edge is 24 cm. The whole surface area of the pyramid (in cm2) is:
Option 1: $(16 \sqrt{3}+24 \sqrt{35})$
Option 2: $(12 \sqrt{3}+24 \sqrt{35})$
Option 3: $(24\sqrt{3}+36\sqrt{35})$
Option 4: $(16\sqrt{3}+48\sqrt{35})$
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Correct Answer: $(16\sqrt{3}+48\sqrt{35})$
Solution :
Total Surface area of a pyramid = Base area + 3 × The area of each triangular face
The area of a equilateral triangle with sides 8 cm = $\frac{\sqrt{3}}{4}×(\text{side})^2=\frac{\sqrt{3}}{4}×8^2=16\sqrt{3}\ \text{cm}^2$
Let the height of the pyramid be $h$.
$h^2=\text{Slant height}^2-(\frac{1}{2}×\text{base})^2$
$⇒h^2=24^2-(\frac{1}{2}×8)^2$
$h^2=576-16$
$\therefore h=\sqrt{560}=4\sqrt{35}\ \text{cm}$
The area of each triangular face = $\frac{1}{2}×$ base × height = $\frac{1}{2}×8×4\sqrt{35}=16\sqrt{35}\ \text{cm}^2$
Total surface area of a pyramid = $16\sqrt{3}+3×16\sqrt{35}=(16\sqrt{3}+48\sqrt{35})\ \text{cm}^2$
Hence, the correct answer is $(16\sqrt{3}+48\sqrt{35})\ \text{cm}^2$.
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