RRB Junior Engineer Stage 1 Model Papers 7

Question : 26 of 100

Marks: +1, -0

The base of a right pyramid is an equilateral triangle of side 4 cm each. Each slant edge is 5 cm long. The volume of the pyramid is

Solution:

Height of base

=√3/2a

where a = 4

=√3/2×4=2√3

Ao=2/3×√3/2×4=4/√3cm

Slant edge BD = 5cm
Vertical height

Do^2=(5)^2-(4/√3)^2=25-16/3

={75-16}/3=√59/√3=Do

Volume of Pyramid=

1/3

ar of base × height

=1/3×√3/4×4×4×√59/√3={4√59}/√3

Go to Question:

Prev Question

Next Question