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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​

User ZKSteffel
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Answer:

Height of the Pyramid = 3.125 m

The surface area of the pyramid is 635.5 sq. meters

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

User Vestlen
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