The base of a right pyramid is an equilateral triangle of side 4cm each. Each slant edge is 5cm long. The volume of pyramid is
For this question,
answer provided by sscadda.com is $\dfrac{20}{3}\sqrt{3}$ (see comments section at bottom)
(a) But this may be wrong because volume is $\dfrac{1}{3}×\text{base}×\text{height}$ and in the solution, slant edge is used in place of height. Am I correct here?
(b) So, for correct solution, we may need to first find height and then use the formula volume is $\dfrac{1}{3}×\text{base}×\text{height}$ , am I correct?
(c) Is the the volume is $\dfrac{4\sqrt{59}}{3}$ as provided in careerbless.com? If this approach is right, is there any shortcut formula for the same?
(d) If none of my comments is right, please guide how to handle this kind of a question.
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$\begingroup$By Heron's Formula, the area of the base is $$ \Delta=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{6(6-4)(6-4)(6-4)}=4\sqrt3 $$ so the radius of the circumcircle is $$ R=\frac{abc}{4\Delta}=\frac{4\cdot4\cdot4}{4\cdot4\sqrt3}=\frac4{\sqrt3} $$The Pythagorean Theorem says that the altitude of the pyramid is $$ h=\sqrt{5^2-R^2}=\sqrt{25-\frac{16}3}=\sqrt{\frac{59}3} $$ The volume is therefore $$ V=\frac13\Delta h=\frac43\sqrt{59} $$
$\endgroup$ $\begingroup$@Kiran An equilateral triangle's area is $\;\frac{\sqrt3}4s^2\;$ , with $\;s=$ the triangle's side, and then he center of the triangle is of length $\;\frac s{\sqrt3}\;$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:
$$\sqrt{T^2-\frac{s^2}3}\;\;,\;\;\;\;T=\text{ the slanted's pyramid's side's length}$$
Now input $\;s=4\;,\;\;T=5\;$$ in the above
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