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The lateral surface area of a square based pyramid having equal edges is $144√3$ cm^2. Find the area of base.

My Attempt: let,

length of base =$a$ slant height=$l$

Then,

$$L.S.A. of pyramid=144√3$$ $$2al=144√3$$ $$al=72√3$$.

Then, what should I do to go further.?

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  • $\begingroup$ Please give your questions informative titles. "LSA of pyramid" is neither informative nor a question. $\endgroup$ – symplectomorphic Dec 25 '16 at 5:35
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Write $T$ for the area of one of the 4 lateral triangles. Since every triangle is equilater we know that $$T=a\cdot a\sqrt 3/2\cdot 1/2=a^2\sqrt3 /4$$

We know that 4 triangles form the lateral area, thus: $$LSA=4T=a^2\sqrt 3$$

We need $a^2$ of course: $$a^2=LSA/\sqrt 3=144$$

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  • $\begingroup$ Hello, How is every triangle equilateral? $\endgroup$ – pi-π Dec 25 '16 at 11:06
  • $\begingroup$ Because the pyramid has equal edges. $\endgroup$ – Maffred Dec 26 '16 at 0:24

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