Question

a pyramid has a rectangular base of side 16m by14m. Given that the volume of the pyramid is 700m cube, find its height and surface area​

Answer 1

Answer:

Height of the Pyramid = 3.125 m

The surface area of the pyramid is 635.5 sq. meters

Step-by-step explanation:

Here, the length  and width of the rectangle =  16 m and 14 m

Let the height of the pyramid  = h m

Volume of the pyramid = 700 cubic meters

Volume of Rectangular Pyramid  =  Length  x Width x Height

or, 700   =  16 x  14 x h

or, [tex]h = \frac{700}{14 \times 16} = 3.125[/t)ex]

or, the Height of the Pyramid = 3.125 m

Now, the Surface area of Rectangular Pyramid   =  2( LW + WH + HL)

Here, Surface Area  =  2 ( (16  x 14 ) + (14 x 3.125) + (3.125 x 16))

= 2( 224 + 43.75 + 50)  = 2 x  317.75

or, Surface Area =  635.5 sq. meters

Hence, the surface area of the pyramid is 635.5 sq. meters