The Great Pyramid of Giza

Pyramid of Giza is made up of Lime Stone. To calculate total weight --multiply total volume of pyramid with density of stones.

Volume of Pyramid---( 1/3) x Area of Square Base x height = ( 1/3) x 230 x 230 x 147=2592100 CuM

Weight of Pyramid= Volume x Density of Lime stone---- 2592100 x 2300 = 596183 0000==596x10^7 ( neglecting 0.183)--some stones may me missing due to time weathering and variation in length of sides

Here, base length of pyramid=230 m and height = 147 m.

General formula of volume of pyramid= Base Length * Base Width * Height * $\frac{1}{3}$

Now, we get, Volume of the pyramid $= 230\times 230\times 147\times \frac{1}{3} = 2592100$

We know that, $Density=\frac{Mass}{Volume} \implies 2300=\frac{Mass}{2592100} \implies (Mass)=2300\times 2592100 \implies (Mass)=5961830000 \implies (Mass)=596\times 10^{7}$

Now, we see mass is in form $abc\times 10^{7}$ with $abc = \boxed{596}$

Density = Mass/Volume

Volume of pyramid = 1/3 x base x height = 1/3 x 230 x 230 x 147 = 2592100m^3 Density = 2300kgm^-3

So, Mass = Volume x Density = 2592100 x 2300 = 5961830000 = 596 x 10^7kg

Credits: http://en.wikipedia.org/wiki/Great Pyramid of_Giza

Volume of square pyramid = 1/3 base area × height and mass = density × volume

Required volume = 1/3 × 230 × 230 × 147 × 2300 = 593.183 × 10^7 kg

volume the pyramid is h base area 1/3 147 230 230*1/3 = 2592100

then volume multiplied with density 2592100*2300 = 5961830000.

so we get the result is 596*10^7 answer is 596

$\displaystyle \text{m} = \rho \times \text{V} = \left(2300 \text{ kg m}^{-3} \right) \times \left( \frac{1}{3} \times (230 \text{ m})^2 \times 147 \text{ m} \right) = 596 \times 10^7 \text{ kg}$

It is given that the volume (V) of pyramid is equal a third of its base area (s^2) multiplied by its height (h). V = (1/3)(s^2)(h) = (1/3)(230^2)(147) = 2.592.100 m^3. And its mass (m) equals to its volume (V) multiplied by its density (p). m = Vp = (2.592.100)(2.300) = 5.961.830.000. And this value is the same as 596 \times 10^7.

$m=ρV=ρ\frac{1}{3}Ah=2300×\frac{1}{3}230^2×147 = 5,961,830,000 \approx 596×10^7 kg$

by simply defining the value of its density, d=m/v. first thing to do is to get the volume of the square pyramid which is v=(1/3)b^2 x h. you will get 2592100 m^3 . substituting all the values of d and v ,find for m. you will get 596x10^7 kg.

Consider the equation: $$ mass= density \times volume $$ The volume of a pyramid is $$ \frac{base \times height}{3} $$ so the Pyramid of Giza has a volume of $$ \frac {230^{2} \times 147} {3} = 2592100 $$ It follows that the mass, dV , is 596e7 kg, and thus abc is equal to 596.

First we calculate the volume of the pyramid with square base: $V = \frac{1}{3} \times S \times h = \frac{1}{3} \times {230^{2}} \times 147 = 2592100 (m^{3})$ So the mass of the Pyramid is: $m = V \times d = 2592100 * 2300 \approx 596 \times 10^{7} (kg)$ So the answer is $596$

Mass is 596 kg. Because volume of the pyramid is V = a a h/3, where a = 230 m, h = 147 m, and mass m = V*r, where density r = 2300 kg/m^3.

The volume of a square pyramid is given by 1/3(base length) ^2(height). Therefore, I volume= 2592100 m^3 Mass = density * volume. Mass = 596× 10^7 kg.