A right pyramid 6 m height has a square base of which the diagonal is $$\sqrt{1152} m$$. Volume of the pyramid is

It is given that base of pyramid is square and diagonal is given as $$\sqrt{1152} m$$

we know diagonal of sqaure is $$\sqrt{2}$$ a , where a is the side of sqaure

so, $$\sqrt{1152} m$$ = $$\sqrt{2} a$$

a = 24 m

Area of base = $$\text{side}^2$$ = $$24^2$$ = 576

Height of pyramid = 6 m

Volume of right pyramid = $$\frac{1}{3} {\times \text{Base Area} \times \text{Height}}$$ = $$\frac{1}{3} \times 576 \times 6$$ = $$1152m^{3}$$